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er 1 year Annually $1100 Semiannually $1102.50 Quarterly $1103.81 Monthly $1104.71 Daily $1105.16 Table 6.5 The Compound Interest Formula Compound interest can be calculated using the formula A(t) = P⎛ ⎝1 + r n nt ⎞ ⎠ (6.2) where • A(t) is the account value, • t is measured in years, • P is the starting amount of the account, often called the principal, or more generally present value, • • r is the annual percentage rate (APR) expressed as a decimal, and n is the number of compounding periods in one year. Example 6.8 Calculating Compound Interest If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years? Solution Because we are starting with $3,000, P = 3000. Our interest rate is 3%, so r = 0.03. Because we are compounding quarterly, we are compounding 4 times per year, so n = 4. We want to know the value of the account in 10 years, so we are looking for A(10), the value when t = 10. A(t) nt = P⎛ ⎝1 + r n ⎞ ⎠ A(10) = 3000 4⋅10 ⎛ ⎝1 + 0.03 4 ⎞ ⎠ Use the compound interest formula. Substitute using given values. ≈ $4045.05 Round to two decimal places. 658 Chapter 6 Exponential and Logarithmic Functions The account will be worth about $4,045.05 in 10 years. An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What 6.8 will the investment be worth in 30 years? Example 6.9 Using the Compound Interest Formula to Solve for the Principal A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now? Solution The nominal interest rate is 6%, so r = 0.06. Interest is compounded twice a year, so k = 2. We want to find the initial investment, P, needed so that the value of the account will be worth $40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for P. A(t) nt = P⎛ ⎝1 + r n ⎞ ⎠ Use the compound interest formula. 2(18) ⎞ ⎠ = P⎛ ⎝1 + 0.06 2 = P(1.03)36 40,000 40,000 40,000 (1.03)36 = P Substitute using given values A, r, n, and t. Simplify. Isolate P. P ≈ $13, 801 Divide and round to the nearest dollar. Lily will need to invest $13,801 to have $40,000 in 18 years. Refer to Example 6.9. To the nearest dollar, how much would Lily need to invest if the account is 6.9 compounded quarterly? Evaluating Functions with Base e As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 6.6 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue. Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 6.6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 659 Frequency A(t) = ⎛ ⎝1 + 1 n n ⎞ ⎠ Annually Semiannually Quarterly Monthly Daily Hourly 1 2 4 ⎛ ⎝1 + 1 1 ⎞ ⎠ ⎛ ⎝1 + 1 2 ⎞ ⎠ ⎛ ⎝1 + 1 4 ⎞ ⎠ 12 ⎛ ⎝1 + 1 12 ⎞ ⎠ 365 ⎛ ⎝1 + 1 365 ⎞ ⎠ 8766 ⎛ ⎝1 + 1 8766 ⎞ ⎠ Value $2 $2.25 $2.441406 $2.613035 $2.714567 $2.718127 Once per minute 525960 ⎛ ⎝1 + 1 525960 ⎞ ⎠ $2.718279 Once per second ⎛ ⎝1 + 1 31557600 ⎞ ⎠ 31557600 $2.718282 Table 6.6 These values appear to be approaching a limit as n increases without bound. In fact, as n gets larger and larger, the expression ⎛ ⎝1 + 1 n n ⎞ ⎠ approaches a number used so frequently in mathematics that it has its own name: the letter e. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below. The Number e The letter e represents the irrational number n ⎛ ⎝1 + 1 n ⎞ ⎠ , as n increases without bound The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, e ≈ 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties. 660 Chapter 6 Exponential and Logarithmic Functions Example 6.10 Using a Calculator to Find Powers of e Calculate e3.14. Round to five decimal places. Solution On a calculator, press the button labeled [e x ⎦. Type 3.14 and then close parenthesis, ]. The window shows ⎡ ⎦. Press [ENTER]. Rounding to 5 decimal places, e3.14 ≈ 23.10387. Caution: Many scientific calculators ⎡ ⎣)⎤ have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of e. ⎣e ^ ( ⎤ 6.10 Use a calculator to find e−0.5. Round to five decimal places. Investigating Continuous Growth So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics. The Continuous Growth/Decay Formula For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula A(t) = aert (6.3) where • a is the initial value, • r is the continuous growth rate per unit time, • and t is the elapsed time. If r > 0 , then the formula represents continuous growth. If r < 0 , then the formula represents continuous decay. For business applications, the continuous growth formula is called the continuous compounding formula and takes the form A(t) = Pert where • P is the principal or the initial invested, • r is the growth or interest rate per unit time, • and t is the period or term of the investment. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 661 Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function. 1. Use the information in the problem to determine a , the initial value of the function. 2. Use the information in the problem to determine the growth rate r. a. b. If the problem refers to continuous growth, then r > 0. If the problem refers to continuous decay, then r < 0. 3. Use the information in the problem to determine the time t. 4. Substitute the given information into the continuous growth formula and solve for A(t). Example 6.11 Calculating Continuous Growth A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year? Solution Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year: A(t) = Pert = 1000(e)0.1 ≈ 1105.17 Use the continuous compounding formula. r, and t. Substitute known values for P, Use a calculator to approximate. The account is worth $1,105.17 after one year. A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be 6.11 the value of the investment in 30 years? Example 6.12 Calculating Continuous Decay Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days? Solution Since the substance is decaying, radon-222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days: is negative. So, r = − 0.173. The initial amount of the rate, 17.3% , A(t) = aert = 100e−0.173(3) ≈ 59.5115 Use the continuous growth formula. Substitute known values for a, r, and t. Use a calculator to approximate. 662 Chapter 6 Exponential and Logarithmic Functions So 59.5115 mg of radon-222 will remain. 6.12 Using the data in Example 6.12, how much radon-222 will remain after one year? Access these online resources for additional instruction and practice with exponential functions. • Exponential Growth Function (http://openstaxcollege.org/l/expgrowth) • Compound Interest (http://openstaxcollege.org/l/compoundint) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 663 6.1 EXERCISES Verbal 1. Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function. Given a formula for an exponential function, 2. is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain. 3. The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.”[3] Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest. Algebraic For the following exercises, identify whether the statement represents an exponential function. Explain. The average annual population increase of a pack of 4. wolves is 25. 5. A population of bacteria decreases by a factor of 1 8 every 24 hours. The value of a coin collection has increased by 3.25% 6. annually over the last 20 years. For each training session, a personal trainer charges his 7. clients $5 less than the previous training session. The height of a projectile at time t is represented by the 8. function h(t) = − 4.9t 2 + 18t + 40. the population of a forest of For the following exercises, consider this scenario: For each year t, trees is t . In a represented by the function A(t) = 115(1.025) neighboring forest, the population of the same type of tree is represented by the function B(t) = 82(1.029) . (Round answers to the ne
arest whole number.) t Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after 100 years? By how many? Discuss the above results from the previous four 13. exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model? For the the following exercises, determine whether equation represents exponential growth, exponential decay, or neither. Explain. 14. y = 300(1 − t)5 15. y = 220(1.06) x 16. y = 16.5(1.025) 1 x 17. y = 11, 701(0.97) t For the following exercises, find the formula for an exponential function that passes through the two points given. 18. (0, 6) and (3, 750) 19. (0, 2000) and (2, 20) 20. ⎛ ⎝−1, 3 2 ⎞ ⎠ and (3, 24) 21. (−2, 6) and (3, 1) 22. (3, 1) and (5, 4) For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. 9. Which forest’s population is growing at a faster rate? Which forest had a greater number of trees initially? 10. By how many? Assuming the population growth models continue to 11. represent the growth of the forests, which forest will have a greater number of trees after 20 years? By how many? 12. 23. 24. x 1 2 3 4 f(x) 70 40 10 -20 3. Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina. 664 25. 26. 27. x 1 2 3 4 h(x) 70 49 34.3 24.01 x 1 2 3 4 m(x) 80 61 42.9 25.61 x 1 2 3 4 f(x) 10 20 40 80 x 1 g(x) -3.25 2 2 3 4 7.25 12.5 Chapter 6 Exponential and Logarithmic Functions 34. Use the formula found in the previous exercise to calculate the initial deposit of an account that is worth $14, 472.74 after compounded monthly for 5 years. (Round to the nearest dollar.) earning 5.5% interest How much more would the account in the previous two 35. exercises be worth if it were earning interest for 5 more years? Use properties of rational exponents to solve the 36. compound interest formula for the interest rate, r. Use the formula found in the previous exercise to 37. calculate the interest that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years. rate for an account Use the formula found in the previous exercise to 38. calculate the interest that was compounded monthly, had an initial deposit of $5,500, and was worth $38,455 after 30 years. rate for an account For the the following exercises, determine whether equation represents continuous growth, continuous decay, or neither. Explain. 39. y = 3742(e)0.75t 40. 3.25 t y = 150(e) 41. y = 2.25(e)−2t For the following exercises, use the compound interest formula, A(t) = P⎛ nt ⎞ ⎠ . ⎝1 + r n After a certain number of years, 28. investment account the value of an represented by the equation is deposit 42. initial Suppose an investment account is opened with an of $12, 000 earning 7.2% interest compounded continuously. How much will the account be worth after 30 years? ⎛ ⎝1 + 0.04 10, 250 12 ⎞ ⎠ 120 . What is the value of the account? What was the initial deposit made to the account in the 29. previous exercise? How much less would the account from Exercise 42 be 43. worth after 30 years if it were compounded monthly instead? Numeric How many years had the account from the previous 30. exercise been accumulating interest? For the following exercises, evaluate each function. Round answers to four decimal places, if necessary. 31. An account is opened with an initial deposit of $6,500 and earns 3.6% interest compounded semi-annually. What will the account be worth in 20 years? 44. f (x) = 2(5) x , for f (−3) 45. f (x) = − 42x + 3, for f (−1) 32. How much more would the account in the previous exercise have been worth if the interest were compounding weekly? Solve the compound interest formula for the principal, 33. P . 46. f (x) = e x , for f (3) 47. 48. f (x) = − 2e x − 1, for f (−1) f (x) = 2.7(4)−x + 1 + 1.5, for f (−2) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 665 49. 50. f (x) = 1.2e2x − 0.3, for f (3) f (x) = − 3 2 (3)−x + 3 2 , for f (2) Technology For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve. 51. (0, 3) and (3, 375) 52. (3, 222.62) and (10, 77.456) 53. (20, 29.495) and (150, 730.89) 54. (5, 2.909) and (13, 0.005) 55. (11,310.035) and (25,356.3652) Extensions The annual percentage yield (APY) of an investment 56. account is a representation of the actual interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly can be found with the formula APY = 12 ⎛ ⎝1 + r 12 ⎞ ⎠ − 1. 57. Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function I(n) for the APY of any account that compounds n times per year. 58. Recall that an exponential function is any equation written in the form f (x) = a ⋅ b x such that a and b are positive numbers and b ≠ 1. Any positive number b can be written as b = en for some value of n . Use this fact to rewrite the formula for an exponential function that uses the number e as a base. In an exponential decay function, 59. the base of the exponent is a value between 0 and 1. Thus, for some number b > 1, the exponential decay function can be x ⎞ ⎠ ⎛ ⎝ . Use this formula, along with 1 written as f (x) = a ⋅ b the fact that b = en to show that an exponential decay function takes the form f (x) = a(e)−nx for some positive number n . , The formula for the amount A in an investment 60. account with a nominal interest rate r at any time t is given by A(t) = a(e) rt , where a is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time t can be calculated with the formula I(t) = ert − 1. Real-World Applications The fox population in a certain region has an annual 61. growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020? A scientist begins with 100 milligrams of a radioactive 62. substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain after 54 hours? In the year 1985, a house was valued at $110,000. By 63. the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010? A car was valued at $38,000 in the year 2007. By 2013, 64. the value had depreciated to $11,000 If the car’s value continues to drop by the same percentage, what will it be worth by 2017? Jamal wants to save $54,000 for a down payment on a 65. home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years? Kyoko has $10,000 that she wants to invest. Her bank 66. has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (Hint: solve the compound interest formula for the interest rate.) Alyssa opened a retirement account with 7.25% APR 67. in the year 2000. Her initial deposit was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously? An investment account with an annual interest rate of 68. 7% was opened with an initial deposit of $4,000 Compare the values of the account after 9 years when the interest is compounded and continuously. quarterly, monthly, annually, 666 Chapter 6 Exponential and Logarithmic Functions 6.2 \| Graphs of Exponential Functions Learning Objectives 6.2.1 Graph exponential functions. 6.2.2 Graph exponential functions using transformations. As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events. Graphing Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form f (x) = b x whose base is greater than one. We’ll use the function f (x) = 2 . Observe how the output values in Table 6.7 change as the input increases by 1. x x −3 −2 −1 f(x Table 6.7 Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for any exponential function with the form f (x) = ab x b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. , Notice from the table that • the output values are positive for all values of x; • as x increases, the output values increase without bound; and • as x decreases, the output values grow smaller, approaching zero. Figure 6.8 shows the exponential growth function f (x) = 2 x . This content is available for fr
ee at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 667 Figure 6.8 Notice that the graph gets close to the x-axis, but never touches it. The domain of f (x) = 2 To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f (x) = b x x is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0. whose base is between zero and one. We’ll use the function g(x) = change as the input increases by 1. x -3 -2 -1 g(x) = ⎛ ⎝ x ⎞ ⎠ 1 2 Table 6. . Observe how the output values in Table 6. Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio 1 2 . Notice from the table that • the output values are positive for all values of x; • as x increases, the output values grow smaller, approaching zero; and • as x decreases, the output values grow without bound. Figure 6.9 shows the exponential decay function, g(x 668 Chapter 6 Exponential and Logarithmic Functions Figure 6.9 The domain of g(x) = x ⎞ ⎠ ⎛ ⎝ 1 2 is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0. Characteristics of the Graph of the Parent Function f(x) = bx An exponential function with the form f (x) = b x b > 0, , b ≠ 1, has these characteristics: • one-to-one function • horizontal asymptote: y = 0 • domain: ( – ∞, ∞) • range: (0, ∞) • x-intercept: none • y-intercept: (0, 1) • increasing if b > 1 • decreasing if b < 1 Figure 6.10 compares the graphs of exponential growth and decay functions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 669 Figure 6.10 Given an exponential function of the form f(x) = b x, graph the function. 1. Create a table of points. 2. Plot at least 3 point from the table, including the y-intercept (0, 1). 3. Draw a smooth curve through the points. 4. State the domain, (−∞, ∞), the range, (0, ∞), and the horizontal asymptote, y = 0. Example 6.13 Sketching the Graph of an Exponential Function of the Form f(x) = bx Sketch a graph of f (x) = 0.25 x . State the domain, range, and asymptote. Solution Before graphing, identify the behavior and create a table of points for the graph. • Since b = 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. • Create a table of points as in Table 6.9. x −3 −2 −1 f(x) = 0.25 x 64 16 4 0 1 1 2 3 0.25 0.0625 0.015625 Table 6.9 • Plot the y-intercept, (0, 1), along with two other points. We can use (−1, 4) and (1, 0.25). 670 Chapter 6 Exponential and Logarithmic Functions Draw a smooth curve connecting the points as in Figure 6.11. Figure 6.11 The domain is (−∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. 6.13 Sketch the graph of f (x) = 4 x . State the domain, range, and asymptote. Graphing Transformations of Exponential Functions Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function f (x) = b x without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied. Graphing a Vertical Shift The first transformation occurs when we add a constant d to the parent function f (x) = b x d units in the same direction as the sign. For example, if we begin by graphing a parent function, f (x) = 2 can then graph two vertical shifts alongside it, using d = 3 : the upward shift, g(x) = 2 h(x) = 2 , giving us a vertical shift , we + 3 and the downward shift, − 3. Both vertical shifts are shown in Figure 6.12. x x x This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 671 Figure 6.12 Observe the results of shifting f (x) = 2 x vertically: • The domain, (−∞, ∞) remains unchanged. • When the function is shifted up 3 units to g(x) = 2 x + 3 : ◦ The y-intercept shifts up 3 units to (0, 4). ◦ The asymptote shifts up 3 units to y = 3. ◦ The range becomes (3, ∞). • When the function is shifted down 3 units to h(x) = 2 x − 3 : ◦ The y-intercept shifts down 3 units to (0, − 2). ◦ The asymptote also shifts down 3 units to y = − 3. ◦ The range becomes (−3, ∞). Graphing a Horizontal Shift The next transformation occurs when we add a constant c to the input of the parent function f (x) = b x , giving us a horizontal shift c units in the opposite direction of the sign. For example, if we begin by graphing the parent function x + 3, and the shift f (x) = 2 , we can then graph two horizontal shifts alongside it, using c = 3 : the shift left, g(x) = 2 x right, h(x) = 2 x − 3. Both horizontal shifts are shown in Figure 6.13. 672 Chapter 6 Exponential and Logarithmic Functions Figure 6.13 Observe the results of shifting f (x) = 2 x horizontally: • The domain, (−∞, ∞), remains unchanged. • The asymptote, y = 0, remains unchanged. • The y-intercept shifts such that: ◦ When the function is shifted left 3 units to g(x) = 2 x + 3, , so the initial value of the function is 8. 2 x + 3 = (8)2 x the y-intercept becomes (0, 8). This is because ◦ When the function is shifted right 3 units to h(x) = 2 x − 3, the y-intercept becomes ⎛ ⎝0, 1 8 ⎞ ⎠. Again, see that 2 x − 3 = x ⎞ ⎠2 ⎛ ⎝ 1 8 , so the initial value of the function is 1 8 . Shifts of the Parent Function f(x) = bx For any constants c and d, the function f (x) = b x + c + d shifts the parent function f (x) = b x • vertically d units, in the same direction of the sign of d. • horizontally c units, in the opposite direction of the sign of c. ⎝0, bc • The y-intercept becomes ⎛ + d⎞ ⎠. • The horizontal asymptote becomes y = d. • The range becomes (d, ∞). • The domain, (−∞, ∞), remains unchanged. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 673 Given an exponential function with the form f(x) = b x + c + d, graph the translation. 1. Draw the horizontal asymptote y = d. 2. Identify the shift as (−c, d). Shift the graph of f (x) = b x left c units if c is positive, and right c units if c is negative. 3. Shift the graph of f (x) = b x up d units if d is positive, and down d units if d is negative. 4. State the domain, (−∞, ∞), the range, (d, ∞), and the horizontal asymptote y = d. Example 6.14 Graphing a Shift of an Exponential Function Graph f (x) = 2 x + 1 − 3. State the domain, range, and asymptote. Solution We have an exponential equation of the form f (x) = b x + c Draw the horizontal asymptote y = d , so draw y = −3. Identify the shift as (−c, d), Shift the graph of f (x) = b x left 1 units and down 3 units. so the shift is (−1, −3). + d, with b = 2, c = 1, and d = −3. Figure 6.14 The domain is (−∞, ∞); the range is (−3, ∞); the horizontal asymptote is y = −3. 674 Chapter 6 Exponential and Logarithmic Functions 6.14 Graph f (x) = 2 x − 1 + 3. State domain, range, and asymptote. Given an equation of the form f(x) = b x + c solution. + d for x, use a graphing calculator to approximate the • Press [Y=]. Enter the given exponential equation in the line headed “Y1=”. • Enter the given value for f (x) in the line headed “Y2=”. • Press [WINDOW]. Adjust the y-axis so that it includes the value entered for “Y2=”. • Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of f (x). • To find the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function. Example 6.15 Approximating the Solution of an Exponential Equation Solve 42 = 1.2(5) x + 2.8 graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 1.2(5) 3 for x and –5 to 55 for y. Press [GRAPH]. The graphs should intersect somewhere near x = 2. + 2.8 next to Y1=. Then enter 42 next to Y2=. For a window, use the values –3 to x For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth, x ≈ 2.166. 6.15 Solve 4 = 7.85(1.15) x − 2.27 graphically. Round to the nearest thousandth. Graphing a Stretch or Compression While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function f (x) = b x by a constant \|a\| > 0. For example, if we begin by graphing x as shown on the left in the parent function f (x) = 2 Figure 6.15, and the compression, using a = 1 3 , we can then graph the stretch, using a = 3, x as shown on the right in Figure 6.15. to get h(x) = 1 3 to get g(x) = 3(2) (2) x , This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 675 Figure 6.15 (a) g(x) = 3(2) x stretches the graph of f (x) = 2 x vertically by a factor of 3. (b) h(x) = 1 3 (2) x compresses the graph of f (x) = 2 x vertically by a factor of 1 3 . Stretches and Compressions of the Parent Function f(x) = bx For any factor a > 0, the function f (x) = a(b) x • • is stretched vertically by a factor of a if \|a\| > 1. is compressed vertically by a factor of a if \|a\| < 1. • has a y-intercept of (0, a). • has a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function. Example 6.16 Graphing the Stretch of an Exponential Function ⎛ Sketch a graph of f (x) = 4 ⎝ x ⎞ ⎠ 1 2 . State the doma
in, range, and asymptote. Solution Before graphing, identify the behavior and key points on the graph. • Since b = 1 2 is between zero and one, the left tail of the graph will increase without bound as x decreases, and the right tail will approach the x-axis as x increases. • Since a = 4, the graph of f (x) = x ⎞ ⎠ ⎛ ⎝ 1 2 will be stretched by a factor of 4. 676 Chapter 6 Exponential and Logarithmic Functions • Create a table of points as shown in Table 6.10. x −3 −2 −1 f(x Table 6.10 32 16 8 0 4 1 2 2 1 3 0.5 • Plot the y-intercept, (0, 4), along with two other points. We can use (−1, 8) and (1, 2). Draw a smooth curve connecting the points, as shown in Figure 6.16. Figure 6.16 The domain is (−∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. 6.16 Sketch the graph of f (x) = 1 2 x (4) . State the domain, range, and asymptote. Graphing Reflections In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function f (x) = b x by −1, we get a reflection about the x-axis. When we multiply the input by −1, we get a reflection about the y-axis. For example, if we begin by graphing the parent function f (x) = 2 , we can then graph the two reflections alongside it. The reflection about the x-axis, g(x) = −2 is shown on the left side of Figure 6.17, and the reflection about the y-axis h(x) = 2−x is shown on the right side of Figure 6.17. x x , , This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 677 Figure 6.17 (a) g(x) = − 2 f (x) = 2 x about the y-axis. x reflects the graph of f (x) = 2 x about the x-axis. (b) g(x) = 2−x reflects the graph of Reflections of the Parent Function f(x) = bx The function f (x) = − b x • reflects the parent function f (x) = b x about the x-axis. • has a y-intercept of (0, − 1). • has a range of (−∞, 0) • has a horizontal asymptote at y = 0 and domain of (−∞, ∞), which are unchanged from the parent function. The function f (x) = b−x • reflects the parent function f (x) = b x about the y-axis. • has a y-intercept of (0, 1), a horizontal asymptote at y = 0, a range of (0, ∞), and a domain of (−∞, ∞), which are unchanged from the parent function. Example 6.17 Writing and Graphing the Reflection of an Exponential Function 678 Chapter 6 Exponential and Logarithmic Functions Find and graph the equation for a function, g(x), that reflects f (x) = x ⎞ ⎠ ⎛ ⎝ 1 4 about the x-axis. State its domain, range, and asymptote. Solution Since we want to reflect the parent function f (x) = x ⎞ ⎠ ⎛ ⎝ 1 4 about the x-axis, we multiply f (x) by − 1 to get, g(x . Next we create a table of points as in Table 6.11. x −3 −2 −1 0 1 2 3 g(x Table 6.11 −64 −16 −4 −1 −0.25 −0.0625 −0.0156 Plot the y-intercept, (0, −1), along with two other points. We can use (−1, −4) and (1, −0.25). Draw a smooth curve connecting the points: Figure 6.18 The domain is (−∞, ∞); the range is (−∞, 0); the horizontal asymptote is y = 0. 6.17 Find and graph the equation for a function, g(x), that reflects f (x) = 1.25 x about the y-axis. State its domain, range, and asymptote. Summarizing Translations of the Exponential Function Now that we have worked with each type of translation for the exponential function, we can summarize them in Table 6.12 to arrive at the general equation for translating exponential functions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 679 Translations of the Parent Function f(x) = b x Translation Form Shift • Horizontally c units to the left • Vertically d units up f (x) = b x + c + d Stretch and Compress • Stretch if \|a\| > 1 • Compression if 0 < \|a\| < 1 f (x) = ab x Reflect about the x-axis f (x) = − b x Reflect about the y-axis f (x) = b−x = x ⎛ ⎝ 1 b ⎞ ⎠ General equation for all translations f (x) = ab x + c + d Table 6.12 Translations of Exponential Functions A translation of an exponential function has the form f (x) = ab x + c + d (6.4) Where the parent function, y = b x , b > 1, is • • shifted horizontally c units to the left. stretched vertically by a factor of \|a\| if \|a\| > 0. • compressed vertically by a factor of \|a\| if 0 < \|a\| < 1. • shifted vertically d units. • reflected about the x-axis when a < 0. Note the order of the shifts, transformations, and reflections follow the order of operations. Example 6.18 Writing a Function from a Description Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range. • f (x) = e x is vertically stretched by a factor of 2 , reflected across the y-axis, and then shifted up 4 units. 680 Chapter 6 Exponential and Logarithmic Functions Solution We want to find an equation of the general form f (x) = ab x + c a, b, c, and d. • We are given the parent function f (x) = e x , so b = e. + d. We use the description provided to find • The function is stretched by a factor of 2 , so a = 2. • The function is reflected about the y-axis. We replace x with − x to get: e−x • The graph is shifted vertically 4 units, so d = 4. . Substituting in the general form we get, f (x) = ab x + c + d = 2e−x + 0 + 4 = 2e−x + 4 The domain is (−∞, ∞); the range is (4, ∞); the horizontal asymptote is y = 4. Write the equation for function described below. Give the horizontal asymptote, the domain, and the 6.18 range. • f (x) = e x is compressed vertically by a factor of 1 3 , reflected across the x-axis and then shifted down 2 units. Access this online resource for additional instruction and practice with graphing exponential functions. • Graph Exponential Functions (http://openstaxcollege.org/l/graphexpfunc) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 681 6.2 EXERCISES Verbal What 69. exponential function play in telling us about behavior of the graph? role does the horizontal asymptote of an the end What is the advantage of knowing how to recognize function the graph of a parent 70. transformations of algebraically? Algebraic 71. The graph of f (x) = 3 x is reflected about the y-axis and stretched vertically by a factor of 4. What is the equation of the new function, g(x) ? State its y-intercept, domain, and range. 72. The graph of f (x) = −x ⎞ ⎠ ⎛ ⎝ 1 2 is reflected about the y- axis and compressed vertically by a factor of 1 5 . What is the equation of the new function, g(x) ? State its y- intercept, domain, and range. 73. The graph of f (x) = 10 x is reflected about the x-axis and shifted upward 7 units. What is the equation of the new function, g(x) ? State its y-intercept, domain, and range. 74. The graph of f (x) = (1.68) x is shifted right 3 units, stretched vertically by a factor of 2, reflected about the xaxis, and then shifted downward 3 units. What is the equation of the new function, g(x) ? State its y-intercept (to the nearest thousandth), domain, and range. 75 is shifted left The graph of f (x) = − 1 2 2 units, stretched vertically by a factor of 4, reflected about the x-axis, and then shifted downward 4 units. What is the equation of the new function, g(x) ? State its yintercept, domain, and range. Graphical For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept. 76. f (x 77. g(x) = − 2(0.25) x 78. h(x) = 6(1.75)−x For the following exercises, graph each set of functions on the same axes. 79. 80. f (x(x) = 3(2) f (x) = 1 4 x (3) , g(x) = 2(3) x x , and h(x) = 3(4) , and h(x) = 4(3) x x For the following exercises, match each function with one of the graphs in Figure 6.19. Figure 6.19 81. 82. 83. 84. 85. 86. f (x) = 2(0.69) f (x) = 2(1.28) f (x) = 2(0.81) f (x) = 4(1.28) f (x) = 2(1.59) f (x) = 4(0.69) x x x x x x For the following exercises, use the graphs shown in Figure 6.20. All have the form f (x) = ab x . 682 Chapter 6 Exponential and Logarithmic Functions 99. f (x) = 3(4)−x + 2 the following exercises, start with the graph of . Then write a function that results from the For f (x) = 4 given transformation. x 100. Shift f (x) 4 units upward 101. Shift f (x) 3 units downward 102. Shift f (x) 2 units left 103. Shift f (x) 5 units right 104. Reflect f (x) about the x-axis 105. Reflect f (x) about the y-axis For the following exercises, each graph is a transformation of y = 2 the transformation. describing . Write equation an x 106. Figure 6.20 87. Which graph has the largest value for b ? 88. Which graph has the smallest value for b ? 89. Which graph has the largest value for a ? 90. Which graph has the smallest value for a ? For the following exercises, graph the function and its reflection about the x-axis on the same axes. 91. f (x) = 1 2 x (4) 92. f (x) = 3(0.75) x − 1 93. f (x) = − 4(2) x + 2 For the following exercises, graph the transformation of . Give the horizontal asymptote, the domain, f (x) = 2 and the range. x 107. 94. f (x) = 2−x 95. h(x) = 2 x + 3 96. f (x) = 2 x − 2 For the following exercises, describe the end behavior of the graphs of the functions. 97. f (x) = − 5(4) x − 1 98. f (x This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 683 108. 110. For the following exercises, find an exponential equation for the graph. 109. Numeric the following exercises, evaluate the exponential For functions for the indicated value of x. 111. g(x) = 1 3 (7) x − 2 for g(6). 112. f (x) = 4(2) x − 1 − 2 for f (5). 113. x h(x for h( − 7). Technology For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth. f (x) = ab x + d. 114. 684 Chapter 6 Exponential and Logarithmic Functions −50 = − −x ⎞ ⎠ ⎛ ⎝ 1 2 115. x 116 = 1 4 ⎞ ⎠ ⎛ ⎝ 1 8 116. 12 = 2(3) x + 1 117 118. −30 = − 4(2) x + 2 + 2 Extensions 119. Explore and discuss the graphs of F(x) = (b) x and x ⎛ ⎝ 1 b ⎞ ⎠ a conjecture . Then make G(x) =
relationship between the graphs of the functions b x and ⎛ ⎝ for any real number b > 0. about the ⎞ ⎠ x 1 b 120. Prove the conjecture made in the previous exercise. 121. Explore and discuss the graphs of f (x) = 4 x , g(x) = 4 x − 2, and h(x) = x ⎛ ⎝ 1 16 ⎞ ⎠4 . Then make a conjecture about the relationship between the graphs of the functions b x and ⎛ ⎞ ⎠b x for any real number n and real ⎝ 1 bn number b > 0. 122. Prove the conjecture made in the previous exercise. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 685 6.3 \| Logarithmic Functions Learning Objectives In this section, you will: 6.3.1 Convert from logarithmic to exponential form. 6.3.2 Convert from exponential to logarithmic form. 6.3.3 Evaluate logarithms. 6.3.4 Use common logarithms. 6.3.5 Use natural logarithms. Figure 6.21 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce) In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes[4]. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[5] like those shown in Figure 6.21. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[6] whereas the Japanese earthquake registered a 9.0.[7] The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is 108 − 4 = 104 = 10,000 times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends. Converting from Logarithmic to Exponential Form In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in x magnitude. The equation that represents this problem is 10 = 500, where x represents the difference in magnitudes on the Richter Scale. How would we solve for x ? We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve 10 so it is clear that x must be some value between 2 and 3, since y = 10 x is increasing. We can examine a graph, as in Figure 6.22, to better estimate the solution. = 500. We know that 102 = 100 and 103 = 1000, x 4. http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/#summary. Accessed 3/4/2013. 5. http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#summary. Accessed 3/4/2013. 6. http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/. Accessed 3/4/2013. 7. http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#details. Accessed 3/4/2013. 686 Chapter 6 Exponential and Logarithmic Functions Figure 6.22 Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure 6.22 passes the horizontal line test. The exponential function y = b x is one-to-one, so its inverse, x = b y is also a function. As is the case with all inverse functions, we simply interchange x and y and solve for y to find the inverse function. To represent y as a function of x, we use a logarithmic function of the form y = logb (x). The base b logarithm of a number is the exponent by which we must raise b to get that number. We read a logarithmic expression as, “The logarithm with base b of x is equal to y, ” or, simplified, “log base b of x is y. ” We can also say, “ b raised to the power of y is x, ” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since 25 = 32, we can write log2 32 = 5. We read this as “log base 2 of 32 is 5.” We can express the relationship between logarithmic form and its corresponding exponential form as follows: Note that the base b is always positive. logb (x) = y ⇔ b y = x, b > 0, b ≠ 1 Because logarithm is a function, it is most correctly written as logb(x), using parentheses to denote function evaluation, just as we would with f (x). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as logb x. Note that many calculators require parentheses around the x. We can illustrate the notation of logarithms as follows: Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means y = logb (x) and y = b x are inverse functions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 687 Definition of the Logarithmic Function A logarithm base b of a positive number x satisfies the following definition. For x > 0, b > 0, b ≠ 1, where, y = logb (x) is equivalent to b y = x (6.5) • we read logb (x) as, “the logarithm with base b of x ” or the “log base b of x. " • the logarithm y is the exponent to which b must be raised to get x. Also, since the logarithmic and exponential functions switch the x and y values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore, • • the domain of the logarithm function with base b is (0, ∞). the range of the logarithm function with base b is ( − ∞, ∞). Can we take the logarithm of a negative number? No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number. Given an equation in logarithmic form logb (x) = y, convert it to exponential form. 1. Examine the equation y = logb x and identify b, y, andx. 2. Rewrite logb x = y as b y = x. Example 6.19 Converting from Logarithmic Form to Exponential Form Write the following logarithmic equations in exponential form. a. b. log6 ⎛ ⎝ 6⎞ ⎠ = 1 2 log3 (9) = 2 Solution First, identify the values of b, y, andx. Then, write the equation in the form b y = x. a. log6 ⎛ ⎝ 6⎞ ⎠ = 1 2 Here, b = 6, y = 1 2 , and x = 6. Therefore, the equation log6 ⎛ ⎝ 6⎞ ⎠ = 1 2 is equivalent to 6 1 2 = 6. b. log3 (9) = 2 Here, b = 3, y = 2, and x = 9. Therefore, the equation log3 (9) = 2 is equivalent to 32 = 9. 688 Chapter 6 Exponential and Logarithmic Functions 6.19 Write the following logarithmic equations in exponential form. a. b. log10 (1,000,000) = 6 log5 (25) = 2 Converting from Exponential to Logarithmic Form To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base b, exponent x, and output y. Then we write x = logb (y). Example 6.20 Converting from Exponential Form to Logarithmic Form Write the following exponential equations in logarithmic form. a. b. c. 23 = 8 52 = 25 10−4 = 1 10,000 Solution First, identify the values of b, y, andx. Then, write the equation in the form x = logb (y). a. b. 23 = 8 Here, b = 2, 52 = 25 Here, b = 5, x = 3, and y = 8. Therefore, the equation 23 = 8 is equivalent to log2(8) = 3. x = 2, and y = 25. Therefore, the equation 52 = 25 is equivalent to log5(25) = 2. c. 10−4 = 1 10,000 Here, b = 10, x = − 4, and y = 1 10,000 . Therefore, the equation 10−4 = 1 10,000 is equivalent to log10( 1 10,000 ) = − 4. 6.20 Write the following exponential equations in logarithmic form. a. 32 = 9 b. 53 = 125 c. 2−1 = 1 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 689 Evaluating Logarithms Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider log2 8. We ask, “To what exponent must 2 be raised in order to get 8?” Because we already know 23 = 8, it follows that log2 8 = 3. Now consider solving log7 49 and log3 27 mentally. • We ask, “To what exponent must 7 be raised in order to get 49?” We know 72 = 49. Therefore, log7 49 = 2 • We ask, “To what exponent must 3 be raised in order to get 27?” We know 33 = 27. Therefore, log3 27 = 3 Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate log 2 3 mentally. 4 9 • We ask, “To what exponent must 2 3 be raised in order to get 4 9 ? ” We know 22 = 4 and 32 = 9, so ⎛ ⎝ 2 ⎞ ⎠ 2 3 = 4 9 . Therefore, log 2 3 ⎞ ⎠ = 2. ⎛ ⎝ 4 9 Given a logarithm of the form y = logb (x), evaluate it mentally. 1. Rewrite the argument x as a power of b : b y 2. Use previous knowledge of powers of b identify y by asking, “To what exponent should b be raised in = x. order to get x ? ” Example 6.21 Solving Logarithms Mentally Solve y = log4 (64) without using a calculator. Solution First we rewrite the logarithm in exponential form: 4 in order to get 64?” y = 64. Next, we ask, “To what exponent must 4 be raised We know Therefore, 43 = 64 log(64) = 3 6.21 Solve y = log121 (11) without using a calculator. Example 6.22 690 Chapter 6 Exponential and Logarithmic Functions Evaluating the Logarithm of a Reciprocal Evaluate y = log3 ⎛ ⎝ 1 27 ⎞ ⎠ without using a calculator. Solution First we rewrite the logarithm in exponential form: 3 y = 1 27 . Next, we ask, “To what exponent must 3 be raised in order to get 1 27 ? ” We know 33 = 27, but what must we do to get the reciprocal, 1 27 ? Recall from working with exponents that b−a = 1 ba. We use this information to write 3−3 = 1 33 = 1 27 Therefore, log3 ⎛ ⎝ 1 27 ⎞ ⎠ = − 3. 6.22 Evaluate y = log2 ⎛ ⎝
1 32 ⎞ ⎠ without using a calculator. Using Common Logarithms Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression log(x) means log10 (x). We call a base-10 logarithm a common logarithm. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms. Definition of the Common Logarithm A common logarithm is a logarithm with base 10. We write log10 (x) simply as log(x). The common logarithm of a positive number x satisfies the following definition. For x > 0, y = log(x) is equivalent to 10 y = x (6.6) We read log(x) as, “the logarithm with base 10 of x ” or “log base 10 of x. ” The logarithm y is the exponent to which 10 must be raised to get x. Given a common logarithm of the form y = log(x), evaluate it mentally. 1. Rewrite the argument x as a power of 10 : 10 y = x. 2. Use previous knowledge of powers of 10 to identify y by asking, “To what exponent must 10 be raised in order to get x ? ” This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 691 Example 6.23 Finding the Value of a Common Logarithm Mentally Evaluate y = log(1000) without using a calculator. Solution First we rewrite the logarithm in exponential form: 10 raised in order to get 1000?” We know y = 1000. Next, we ask, “To what exponent must 10 be Therefore, log(1000) = 3. 103 = 1000 6.23 Evaluate y = log(1,000,000). Given a common logarithm with the form y = log(x), evaluate it using a calculator. 1. Press [LOG]. 2. Enter the value given for x, followed by [ ) ]. 3. Press [ENTER]. Example 6.24 Finding the Value of a Common Logarithm Using a Calculator Evaluate y = log(321) to four decimal places using a calculator. Solution • Press [LOG]. • Enter 321, followed by [ ) ]. • Press [ENTER]. Rounding to four decimal places, log(321) ≈ 2.5065. Analysis Note that 102 = 100 and that 103 = 1000. Since 321 is between 100 and 1000, we know that log(321) must be between log(100) and log(1000). This gives us the following: 100 < 321 < 1000 2 < 2.5065 < 3 6.24 Evaluate y = log(123) to four decimal places using a calculator. 692 Chapter 6 Exponential and Logarithmic Functions Example 6.25 Rewriting and Solving a Real-World Exponential Model The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation 10 = 500 represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? x Solution We begin by rewriting the exponential equation in logarithmic form. x 10 = 500 log(500) = x Use the definition of he common log. Next we evaluate the logarithm using a calculator: • Press [LOG]. • Enter 500, followed by [ ) ]. • Press [ENTER]. • To the nearest thousandth, log(500) ≈ 2.699. The difference in magnitudes was about 2.699. The amount of energy released from one earthquake was 8,500 times greater than the amount of energy = 8500 represents this situation, where x is the difference in 6.25 released from another. The equation 10 magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? x Using Natural Logarithms The most frequently used base for logarithms is e. Base e logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base e logarithm, loge (x), has its own notation, ln(x). Most values of ln(x) can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, ln1 = 0. For other natural logarithms, we can use the ln key that can be found on most scientific calculators. We can also find the natural logarithm of any power of e using the inverse property of logarithms. Definition of the Natural Logarithm A natural logarithm is a logarithm with base e. We write loge (x) simply as ln(x). The natural logarithm of a positive number x satisfies the following definition. For x > 0, y = ln(x) is equivalent to e y = x (6.7) We read ln(x) as, “the logarithm with base e of x ” or “the natural logarithm of x. ” The logarithm y is the exponent to which e must be raised to get x. Since the functions y = e and y = ln(x) are inverse functions, ln(e x ) = x for all x and e = x for x > 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 693 Given a natural logarithm with the form y = ln(x), evaluate it using a calculator. 1. Press [LN]. 2. Enter the value given for x, followed by [ ) ]. 3. Press [ENTER]. Example 6.26 Evaluating a Natural Logarithm Using a Calculator Evaluate y = ln(500) to four decimal places using a calculator. Solution • Press [LN]. • Enter 500, followed by [ ) ]. • Press [ENTER]. Rounding to four decimal places, ln(500) ≈ 6.2146 6.26 Evaluate ln(−500). Access this online resource for additional instruction and practice with logarithms. • Introduction to Logarithms (http://openstaxcollege.org/l/intrologarithms) 694 Chapter 6 Exponential and Logarithmic Functions 6.3 EXERCISES Verbal What is a base b logarithm? Discuss the meaning by = x 123. interpreting each part of the equivalent equations b y and logb x = y for b > 0, b ≠ 1. How is 124. related to the exponential function g(x) = b x result of composing these two functions? the logarithmic function f (x) = logb x ? What is the 125. How can the logarithmic equation logb x = y be solved for x using the properties of exponents? Discuss the meaning of the common logarithm. What and how 126. is its relationship to a logarithm with base b, does the notation differ? Discuss the meaning of the natural logarithm. What is 127. its relationship to a logarithm with base b, and how does the notation differ? Algebraic For the following exercises, rewrite each equation in exponential form. 128. log4(q) = m 129. loga(b) = c 130. log16 (y) = x 131. log x (64) = y 132. log y (x) = −11 133. log15 (a) = b 134. log y (137) = x 135. log13 (142) = a 136. log(v) = t 137. ln(w) = n 140. m−7 = n 141. x 19 = y 142. − 10 13 = y x 143. n4 = 103 144 145. y x = 39 100 146. a 10 = b 147. ek = h For the following exercises, solve for x by converting the logarithmic equation to exponential form. 148. log3(x) = 2 149. log2(x) = − 3 150. log5(x) = 2 151. log3 (x) = 3 152. log2(x) = 6 153. log9(x) = 1 2 154. log18(x) = 2 155. log6 (x) = − 3 156. log(x) = 3 157. ln(x) = 2 For the following exercises, use the definition of common and natural logarithms to simplify. For the following exercises, rewrite each equation in logarithmic form. 138. x 4 = y 139. cd = k 158. log(1008) 159. log(32) 10 160. 2log(.0001) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 695 161. eln(1.06) 162. ⎝e−5.03⎞ ⎛ ln ⎠ 163. eln(10.125) + 4 Numeric ⎛ ln ⎝ ⎞ ⎠ 4 5 179. log( 2) 180. ln( 2) Extensions the For logarithmic expression without using a calculator. following exercises, evaluate the base b 164. log3 ⎛ ⎝ 1 27 ⎞ ⎠ 165. 166. log6( 6) log2 ⎞ ⎠ + 4 ⎛ ⎝ 1 8 167. 6log8(4) the Is x = 0 in function domain 181. f (x) = log(x) ? If so, what is the value of the function when x = 0 ? Verify the result. the of the Is f (x) = 0 in function 182. f (x) = log(x) ? If so, for what value of x ? Verify the result. range the of Is there a number x such that lnx = 2 ? If so, what is 183. that number? Verify the result. the following exercises, evaluate the common For logarithmic expression without using a calculator. 184. Is the following true: log3(27) ⎞ 1 ⎠ 64 log4 ⎛ ⎝ = −1 ? Verify the 168. log(10, 000) 169. log(0.001) 170. log(1) + 7 171. 2log(100−3) the For logarithmic expression without using a calculator. following exercises, evaluate the natural 172. 1 3) ln(e 173. ln(1) 174. ln(e−0.225) − 3 175. 2 5) 25ln(e Technology For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth. 176. log(0.04) 177. ln(15) 178. result. 185. Is the following true: ⎝e1.725⎞ ⎛ ln ⎠ ln(1) = 1.725 ? Verify the result. Real-World Applications The exposure index EI for a 35 millimeter camera is 186. a measurement of the amount of light that hits the film. It is determined by the equation EI = log2 ⎛ ⎜ ⎝ f 2 t ⎞ ⎟ , where f is ⎠ the “f-stop” setting on the camera, and t is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be? Refer to the previous exercise. Suppose the light and the 187. meter on a camera indicates an EI of − 2, desired exposure time is 16 seconds. What should the f-stop setting be? 188. a The intensity levels I of two earthquakes measured on formula compared by the log = M1 − M2 where M is the magnitude given by seismograph can be I1 I2 the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same devastating region earthquake, this time with a magnitude of 9.0.[8] How many another, more experienced yet 8. http://earthquake.usgs.gov/earthquakes/world/historical.php. Accessed 3/4/2014. 696 Chapter 6 Exponential and Logarithmic Functions times greater was the intensity of the 2011 earthquake? Round to the nearest whole number. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 697 6.4 \| Graphs of Logarithmic Functions Learning Objectives In this section, you will: 6.4.1 Identify the domain of a logarithmic function. 6.4.2 Graph logarithmic functions. In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function
is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect. To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year t can be found with the equation A = 2500e0.05t But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 6.23 shows this point on the logarithmic graph. . Figure 6.23 In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions. Finding the Domain of a Logarithmic Function Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined. Recall that the exponential function is defined as y = b x for any real number x and constant b > 0, b ≠ 1, where • The domain of y is (−∞, ∞). 698 Chapter 6 Exponential and Logarithmic Functions • The range of y is (0, ∞). In the last section we learned that the logarithmic function y = logb (x) is the inverse of the exponential function y = b x So, as inverse functions: . • The domain of y = logb (x) is the range of y = b x • The range of y = logb (x) is the domain of y = b x : (0, ∞). : (−∞, ∞). Transformations of the parent function y = logb (x) behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of shape. In Graphs of Exponential Functions we saw that certain transformations can change the range of y = b x . Similarly, applying transformations to the parent function y = logb (x) can change the domain. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. That is, the argument of the logarithmic function must be greater than zero. For example, consider f (x) = log4 (2x − 3). This function is defined for any values of x such that the argument, in this case 2x − 3, is greater than zero. To find the domain, we set up an inequality and solve for x : 2x − 3 > 0 2x > 3 x > 1.5 Show the argument greater than zero. Add 3. Divide by 2. In interval notation, the domain of f (x) = log4 (2x − 3) is (1.5, ∞). Given a logarithmic function, identify the domain. 1. Set up an inequality showing the argument greater than zero. 2. Solve for x. 3. Write the domain in interval notation. Example 6.27 Identifying the Domain of a Logarithmic Shift What is the domain of f (x) = log2(x + 3) ? Solution The logarithmic function is defined only when the input is positive, so this function is defined when x + 3 > 0. Solving this inequality The input must be positive. Subtract 3. The domain of f (x) = log2(x + 3) is (−3, ∞). 6.27 What is the domain of f (x) = log5(x − 2) + 1 ? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 699 Example 6.28 Identifying the Domain of a Logarithmic Shift and Reflection What is the domain of f (x) = log(5 − 2x) ? Solution The logarithmic function is defined only when the input is positive, so this function is defined when 5 – 2x > 0. Solving this inequality, The input must be positive. 5 − 2x > 0 − 2x > − 5 Subtract 5. x < 5 2 Divide by − 2 and switch the inequality. The domain of f (x) = log(5 − 2x) is ⎛ ⎝ – ∞, 5 2 ⎞ ⎠. 6.28 What is the domain of f (x) = log(x − 5) + 2 ? Graphing Logarithmic Functions Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function y = logb (x) along with all its transformations: shifts, stretches, compressions, and reflections. We begin with the parent function y = logb (x). Because every logarithmic function of this form is the inverse of an exponential function with the form y = b x their graphs will be reflections of each other across the line y = x. To x and its equivalent illustrate this, we can observe the relationship between the input and output values of y = 2 x = log2(y) in Table 6.13. , x −3 −2 − log2 (y) = x −3 −2 − Table 6.13 Using the inputs and outputs from Table 6.13, we can build another table to observe the relationship between points on the x and g(x) = log2(x). See Table 6.14. graphs of the inverse functions f (x) = 2 700 Chapter 6 Exponential and Logarithmic Functions f(x) = 2 x ⎛ ⎝−3, 1 8 ⎞ ⎠ ⎛ ⎝−2, 1 4 ⎞ ⎠ ⎛ ⎝−1, 1 2 ⎞ ⎠ (0, 1) (1, 2) (2, 4) (3, 8) g(x) = log2 (x1, 0) (2, 1) (4, 2) (8, 3) Table 6.14 As we’d expect, the x- and y-coordinates are reversed for the inverse functions. Figure 6.24 shows the graph of f and g. Figure 6.24 Notice that the graphs of f (x) = 2 g(x) = log2 (x) are reflections about the line y = x. x and Observe the following from the graph: • f (x) = 2 x has a y-intercept at (0, 1) and g(x) = log2 (x) has an x- intercept at (1, 0). • The domain of f (x) = 2 x , (−∞, ∞), is the same as the range of g(x) = log2 (x). • The range of f (x) = 2 x , (0, ∞), is the same as the domain of g(x) = log2 (x). Characteristics of the Graph of the Parent Function, f(x) = logb(x) For any real number x and constant b > 0, b ≠ 1, we can see the following characteristics in the graph of f (x) = logb (x) : • one-to-one function • vertical asymptote: x = 0 • domain: (0, ∞) • range: (−∞, ∞) • x-intercept: (1, 0) and key point (b, 1) • y-intercept: none This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 701 • increasing if b > 1 • decreasing if 0 < b < 1 See Figure 6.25. Figure 6.25 Figure 6.26 shows how changing the base b in f (x) = logb (x) can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function ln(x) has base e ≈ 2.718.) Figure 6.26 The graphs of three logarithmic functions with different bases, all greater than 1. Given a logarithmic function with the form f(x) = logb (x), graph the function. 1. Draw and label the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Plot the key point (b, 1). 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (−∞,∞), and the vertical asymptote, x = 0. 702 Chapter 6 Exponential and Logarithmic Functions Example 6.29 Graphing a Logarithmic Function with the Form f(x) = logb(x). Graph f (x) = log5 (x). State the domain, range, and asymptote. Solution Before graphing, identify the behavior and key points for the graph. • Since b = 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote x = 0, and the right tail will increase slowly without bound. • The x-intercept is (1, 0). • The key point (5, 1) is on the graph. • We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure 6.27). Figure 6.27 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. 6.29 Graph f (x) = log 1 5 (x). State the domain, range, and asymptote. Graphing Transformations of Logarithmic Functions As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function y = logb (x) without loss of shape. Graphing a Horizontal Shift of f(x) = logb(x) When a constant c is added to the input of the parent function f (x) = logb(x), the result is a horizontal shift c units in the opposite direction of the sign on c. To visualize horizontal shifts, we can observe the general graph of the parent function f (x) = logb (x) and for c > 0 alongside the shift left, g(x) = logb (x + c), and the shift right, h(x) = logb (x − c). See Figure 6.28. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 703 Figure 6.28 Horizontal Shifts of the Parent Function y = logb(x) For any constant c, the function f (x) = logb (x + c) • • shifts the parent function y = logb (x) left c units if c > 0. shifts the parent function y = logb (x) right c units if c < 0. • has the vertical asymptote x = − c. • has domain (−c, ∞). • has range (−∞, ∞). Given a logarithmic function with the form f(x) = logb (x + c), graph the translation. 1. Identify the horizontal shift: a. b. If c > 0, shift the graph of f (x) = logb (x) left c units. If c < 0, shift the graph of f (x) = logb (x) right c units. 2. Draw the vertical asymptote x = − c. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting c from the x coordinate. 4. Label the three points. 5. The Domain is (−c, ∞), the range is (−∞, ∞), and the vertical asymptote is x = − c. 704 Chapter 6 Exponential and Logarithmic Functions Example 6.30 Graphing a Horizontal Shift of the Parent Function y = logb(x) Sketch the horizontal shift f (x) = log3(x − 2) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x − 2), we notice x + (−2) = x – 2. Thus c = − 2, so c < 0. This means we will shift the function f (x) = log3(x) right 2 units. The vertical asymptote is x = − ( − 2) or x = 2. Consider the three key points from the pare
nt function, ⎛ ⎝ ⎞ ⎠, , −1 1 3 (1, 0), and (3, 1). The new coordinates are found by adding 2 to the x coordinates. Label the points ⎛ ⎝ ⎞ ⎠, , −1 7 3 (3, 0), and (5, 1). The domain is (2, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 2. Figure 6.29 6.30 Sketch a graph of f (x) = log3(x + 4) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Graphing a Vertical Shift of y = logb(x) When a constant d is added to the parent function f (x) = logb (x), the result is a vertical shift d units in the direction of the sign on d. To visualize vertical shifts, we can observe the general graph of the parent function f (x) = logb (x) alongside the shift up, g(x) = logb (x) + d and the shift down, h(x) = logb (x) − d. See Figure 6.30. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 705 Figure 6.30 Vertical Shifts of the Parent Function y = logb(x) For any constant d, the function f (x) = logb (x) + d • • shifts the parent function y = logb (x) up d units if d > 0. shifts the parent function y = logb (x) down d units if d < 0. • has the vertical asymptote x = 0. • has domain (0, ∞). • has range (−∞, ∞). 706 Chapter 6 Exponential and Logarithmic Functions Given a logarithmic function with the form f(x) = logb (x) + d, graph the translation. 1. Identify the vertical shift: ◦ ◦ If d > 0, shift the graph of f (x) = logb (x) up d units. If d < 0, shift the graph of f (x) = logb (x) down d units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d to the y coordinate. 4. Label the three points. 5. The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x = 0. Example 6.31 Graphing a Vertical Shift of the Parent Function y = logb(x) Sketch a graph of f (x) = log3(x) − 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x) − 2, we will notice d = – 2. Thus d < 0. This means we will shift the function f (x) = log3(x) down 2 units. The vertical asymptote is x = 0. Consider the three key points from the parent function, ⎛ ⎝ ⎞ ⎠, , −1 1 3 (1, 0), and (3, 1). The new coordinates are found by subtracting 2 from the y coordinates. Label the points ⎛ ⎝ ⎞ ⎠, , −3 1 3 (1, −2), and (3, −1). The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 707 Figure 6.31 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. 6.31 Sketch a graph of f (x) = log2(x) + 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Graphing Stretches and Compressions of y = logb(x) When the parent function f (x) = logb (x) is multiplied by a constant a > 0, of the original graph. To visualize stretches and compressions, we set a > 1 and observe the general graph of the parent function f (x) = logb (x) alongside the vertical stretch, g(x) = alogb (x) and the vertical compression, h(x) = 1 alogb (x). See Figure 6.32. the result is a vertical stretch or compression 708 Chapter 6 Exponential and Logarithmic Functions Figure 6.32 Vertical Stretches and Compressions of the Parent Function y = logb(x) For any constant a > 1, the function f (x) = alogb (x) • stretches the parent function y = logb (x) vertically by a factor of a if a > 1. • compresses the parent function y = logb (x) vertically by a factor of a if 0 < a < 1. • has the vertical asymptote x = 0. • has the x-intercept (1, 0). • has domain (0, ∞). • has range (−∞, ∞). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 709 Given a logarithmic function with the form f(x) = alogb (x), a > 0, graph the translation. 1. Identify the vertical stretch or compressions: ◦ ◦ If \|a\| > 1, If \|a\| < 1, the graph of f (x) = logb (x) is stretched by a factor of a units. the graph of f (x) = logb (x) is compressed by a factor of a units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the y coordinates by a. 4. Label the three points. 5. The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. Example 6.32 Graphing a Stretch or Compression of the Parent Function y = logb(x) Sketch a graph of f (x) = 2log4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = 2log4(x), we will notice a = 2. This means we will stretch the function f (x) = log4(x) by a factor of 2. The vertical asymptote is x = 0. Consider the three key points from the parent function, ⎛ ⎝ ⎞ , −1 ⎠, 1 4 (1, 0), and (4, 1). The new coordinates are found by multiplying the y coordinates by 2. Label the points ⎛ ⎝ ⎞ , −2 ⎠, 1 4 (1, 0) , and (4, 2). The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. See Figure 6.33. 710 Chapter 6 Exponential and Logarithmic Functions Figure 6.33 The domain is (0, ∞), the range is (−∞, ∞), and the vertical asymptote is x = 0. 6.32 Sketch a graph of f (x) = 1 2 log4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Example 6.33 Combining a Shift and a Stretch Sketch a graph of f (x) = 5log(x + 2). State the domain, range, and asymptote. Solution Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in Figure 6.34. The vertical asymptote will be shifted to x = −2. The x-intercept will be (−1,0). The domain will be (−2, ∞). Two points will help give the shape of the graph: (−1, 0) and (8, 5). We chose x = 8 as the x-coordinate of one point to graph because when x = 8, x + 2 = 10, the base of the common logarithm. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 711 Figure 6.34 The domain is (−2, ∞), the range is (−∞, ∞), and the vertical asymptote is x = − 2. 6.33 Sketch a graph of the function f (x) = 3log(x − 2) + 1. State the domain, range, and asymptote. Graphing Reflections of f(x) = logb(x) When the parent function f (x) = logb (x) is multiplied by −1, the result is a reflection about the y-axis. To visualize reflections, we restrict b > 1, and observe the is multiplied by −1, general graph of the parent function f (x) = logb (x) alongside the reflection about the x-axis, g(x) = −logb (x) and the reflection about the y-axis, h(x) = logb (−x). the result is a reflection about the x-axis. When the input 712 Chapter 6 Exponential and Logarithmic Functions Figure 6.35 Reflections of the Parent Function y = logb(x) The function f (x) = −logb (x) • reflects the parent function y = logb (x) about the x-axis. range, (−∞, ∞), and vertical asymptote, x = 0, which are unchanged from the • has domain, (0, ∞), parent function. The function f (x) = logb (−x) • reflects the parent function y = logb (x) about the y-axis. • has domain (−∞, 0). • has range, (−∞, ∞), and vertical asymptote, x = 0, which are unchanged from the parent function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 713 Given a logarithmic function with the parent function f(x) = logb (x), graph a translation. If f(x) = − logb(x) If f(x) = logb( − x) 1. Draw the vertical asymptote, x = 0. 1. Draw the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 2. Plot the x-intercept, (1, 0). 3. Reflect the graph of the parent function f (x) = logb (x) about the x-axis. 3. Reflect the graph of the parent function f (x) = logb (x) about the y-axis. 4. Draw a smooth curve through the points. 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (−∞, ∞), and the vertical asymptote x = 0. 5. State the domain, (−∞, 0), the range, (−∞, ∞), and the vertical asymptote x = 0. Example 6.34 Graphing a Reflection of a Logarithmic Function Sketch a graph of f (x) = log( − x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Before graphing f (x) = log( − x), identify the behavior and key points for the graph. • Since b = 10 is greater than one, we know that the parent function is increasing. Since the input value f is a reflection of the parent graph about the y-axis. Thus, f (x) = log( − x) will is multiplied by −1, be decreasing as x moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote x = 0. • The x-intercept is (−1, 0). • We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points. 714 Chapter 6 Exponential and Logarithmic Functions Figure 6.36 The domain is (−∞, 0), the range is (−∞, ∞), and the vertical asymptote is x = 0. 6.34 Graph f (x) = − log( − x). State the domain, range, and asymptote. Given a logarithmic equation, use a graphing calculator to approximate solutions. 1. Press [Y=]. Enter the given logarithm equation or equations as Y1= and, if needed, Y2=. 2. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection. 3. To find the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x, for the point(s) of intersection. Example 6.35 Approximating the Solution of a Logarithmic Equation Solve 4l
n(x) + 1 = − 2ln(x − 1) graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 4ln(x) + 1 next to Y1=. Then enter − 2ln(x − 1) next to Y2=. For a window, use the values 0 to 5 for x and –10 to 10 for y. Press [GRAPH]. The graphs should intersect somewhere a little to right of x = 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 715 For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for Guess?) So, to the nearest thousandth, x ≈ 1.339. 6.35 Solve 5log(x + 2) = 4 − log(x) graphically. Round to the nearest thousandth. Summarizing Translations of the Logarithmic Function Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table 6.15 to arrive at the general equation for translating exponential functions. Translations of the Parent Function y = logb (x) Translation Form Shift • Horizontally c units to the left • Vertically d units up y = logb (x + c) + d Stretch and Compress • Stretch if \|a\| > 1 • Compression if \|a\| < 1 y = alogb (x) Reflect about the x-axis y = − logb (x) Reflect about the y-axis y = logb (−x) General equation for all translations y = alogb(x + c) + d Table 6.15 Translations of Logarithmic Functions All translations of the parent logarithmic function, y = logb (x), have the form f (x) = alogb (x + c) + d (6.8) where the parent function, y = logb (x), b > 1, is • • • shifted vertically up d units. shifted horizontally to the left c units. stretched vertically by a factor of \|a\| if \|a\| > 0. 716 Chapter 6 Exponential and Logarithmic Functions • compressed vertically by a factor of \|a\| if 0 < \|a\| < 1. • reflected about the x-axis when a < 0. For f (x) = log(−x), the graph of the parent function is reflected about the y-axis. Example 6.36 Finding the Vertical Asymptote of a Logarithm Graph What is the vertical asymptote of f (x) = −2log3(x + 4) + 5 ? Solution The vertical asymptote is at x = − 4. Analysis The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to x = −4. 6.36 What is the vertical asymptote of f (x) = 3 + ln(x − 1) ? Example 6.37 Finding the Equation from a Graph Find a possible equation for the common logarithmic function graphed in Figure 6.37. Figure 6.37 Solution This graph has a vertical asymptote at x = –2 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 717 It appears the graph passes through the points (–1, 1) and (2, –1). Substituting (–1, 1), f (x) = − alog(x + 2) + k 1 = − alog(−1 + 2) + k Substitute (−1, 1). 1 = − alog(1) + k 1 = k Arithmetic. log(1) = 0. Next, substituting in (2, –1) , −1 = − alog(2 + 2) + 1 Plug in (2, −1). −2 = − alog(4) a = 2 Solve for a. Arithmetic. log(4) This gives us the equation f (x) = – 2 log(4) log(x + 2) + 1. Analysis We can verify this answer by comparing the function values in Table 6.15 with the points on the graph in Figure 6.37. x −1 f(x) x 1 4 0 0 5 f(x) −1.5850 −1.8074 Table 6.15 1 2 3 −0.58496 −1 −1.3219 6 −2 7 8 −2.1699 −2.3219 718 Chapter 6 Exponential and Logarithmic Functions 6.37 Give the equation of the natural logarithm graphed in Figure 6.38. Figure 6.38 Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph? Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure 6.38. The graph approaches x = −3 (or thereabouts) more and more closely, so x = −3 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, { x \| x > −3}. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as x → − 3+ , f (x) → − ∞ and as x → ∞, f (x) → ∞. Access these online resources for additional instruction and practice with graphing logarithms. • Graph an Exponential Function and Logarithmic Function (http://openstaxcollege.org/l/ graphexplog) • Match Graphs with Exponential and Logarithmic Functions (http://openstaxcollege.org/l/ matchexplog) • Find the Domain of Logarithmic Functions (http://openstaxcollege.org/l/domainlog) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 719 6.4 EXERCISES Verbal 189. The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each? What type(s) of translation(s), if any, affect the range 190. of a logarithmic function? 206. h(x) = − log(3x − 4) + 3 207. g(x) = ln(2x + 6) − 5 208. f (x) = log3 (15 − 5x) + 6 For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE. What 191. domain of a logarithmic function? type(s) of translation(s), if any, affect the Consider the 192. f (x) = logb (x). Why can’t x be zero? general logarithmic 209. h(x) = log4 (x − 1) + 1 function 210. f (x) = log(5x + 10) + 3 Does the graph of a general logarithmic function have 193. a horizontal asymptote? Explain. Algebraic For the following exercises, state the domain and range of the function. 211. g(x) = ln(−x) − 2 212. f (x) = log2 (x + 2) − 5 213. h(x) = 3ln(x) − 9 Graphical For the following exercises, match each function in Figure 6.39 with the letter corresponding to its graph. 194. f (x) = log3 (x + 4) 195. ⎛ h(x) = ln ⎝ − x⎞ ⎠ 1 2 196. g(x) = log5 (2x + 9) − 2 197. h(x) = ln(4x + 17) − 5 198. f (x) = log2 (12 − 3x) − 3 For the following exercises, state the domain and the vertical asymptote of the function. 199. f (x) = logb(x − 5) 200. g(x) = ln(3 − x) 201. f (x) = log(3x + 1) 202. f (x) = 3log( − x) + 2 203. g(x) = − ln(3x + 9) − 7 Figure 6.39 214. d(x) = log(x) For the following exercises, state the domain, vertical asymptote, and end behavior of the function. 215. f (x) = ln(x) 204. f (x) = ln(2 − x) 205. ⎛ ⎝x − 3 f (x) = log 7 ⎞ ⎠ 216. g(x) = log2 (x) 217. h(x) = log5 (x) 218. j(x) = log25 (x) 720 Chapter 6 Exponential and Logarithmic Functions For the following exercises, match each function in Figure 6.40 with the letter corresponding to its graph. 229. f (x) = log2(x + 2) 230. f (x) = 2log(x) 231. f (x) = ln( − x) 232. g(x) = log(4x + 16) + 4 233. g(x) = log(6 − 3x) + 1 234. h(x) = − 1 2 ln(x + 1) − 3 For the following exercises, write a logarithmic equation corresponding to the graph shown. 235. Use y = log2(x) as the parent function. Figure 6.40 219. f (x) = log 1 3 (x) 220. g(x) = log2 (x) 221. h(x) = log 3 4 (x) For the following exercises, sketch the graphs of each pair of functions on the same axis. 222. f (x) = log(x) and g(x) = 10 x 223. f (x) = log(x) and g(x) = log 1 2 (x) 224. f (x) = log4(x) and g(x) = ln(x) 225. f (x) = e x and g(x) = ln(x) For the following exercises, match each function in Figure 6.41 with the letter corresponding to its graph. Graph of three logarithmic functions. Figure 6.41 226. f (x) = log4 (−x + 2) 227. g(x) = − log4 (x + 2) 228. h(x) = log4 (x + 2) For the following exercises, sketch the graph of the indicated function. This content is available for free at https://cnx.org/content/col11758/1.5 236. Use f (x) = log3(x) as the parent function. 237. Use f (x) = log4(x) as the parent function. Chapter 6 Exponential and Logarithmic Functions 721 Explore and discuss the graphs of f (x) = log 1 2 (x) and g(x) = − log2 (x). Make a conjecture based on the result. 246. Prove the conjecture made in the previous exercise. What 247. ⎛ f (x) = ln ⎝ is x + 2 x − 4 the domain of ⎞ ⎠? Discuss the result. the function Use properties of exponents to find the x-intercepts of 248. ⎞ ⎛ ⎝x2 + 4x + 4 ⎠ algebraically. Show the function f (x) = log the steps for solving, and then verify the result by graphing the function. 238. Use f (x) = log5(x) as the parent function. Technology For the following exercises, use a graphing calculator to find approximate solutions to each equation. 239. log(x − 1) + 2 = ln(x − 1) + 2 240. log(2x − 3) + 2 = − log(2x − 3) + 5 241. ln(x − 2) = − ln(x + 1) 242. 243. 2ln(5x + 1) = 1 2 ln(−5x) + 1 1 3 log(1 − x) = log(x + 1) + 1 3 Extensions Let b be any positive real number such that b ≠ 1. 244. What must logb 1 be equal to? Verify the result. 245. 722 Chapter 6 Exponential and Logarithmic Functions 6.5 \| Logarithmic Properties Learning Objectives In this section, you will: 6.5.1 Use the product rule for logarithms. 6.5.2 Use the quotient rule for logarithms. 6.5.3 Use the power rule for logarithms. 6.5.4 Expand logarithmic expressions. 6.5.5 Condense logarithmic expressions. 6.5.6 Use the change-of-base formula for logarithms. Figure 6.42 The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan) In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances: • Battery acid: 0.8 • Stomach acid: 2.7 • Orange juice: 3.3 • Pure water: 7 (at 25° C) • Human blood: 7.35 • Fresh coconut: 7.8 • Sodium hydroxide (lye): 14 To determine whether a so
lution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where a is the concentration of hydrogen ion in the solution pH = − log([H+ ]) ⎞ ⎛ 1 = log ⎠ ⎝ [H+ ] This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 723 The equivalence of − log⎛ ⎡ ⎝ ⎣H+ ⎤ ⎛ ⎜ 1 ⎞ ⎠ and log ⎦ ⎣H+ ⎤ ⎡ ⎝ ⎦ ⎞ ⎟ is one of the logarithm properties we will examine in this section. ⎠ Using the Product Rule for Logarithms Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove. logb 1 = 0 logb b = 1 For example, log5 1 = 0 since 50 = 1. And log5 5 = 1 since 51 = 5. Next, we have the inverse property. ) = x logb(b x logb x b = x, x > 0 For example, to evaluate log(100), we can rewrite the logarithm as log10 logb (b x ) = x to get log10 ⎛ ⎝102⎞ ⎠ = 2. ⎝102⎞ ⎛ ⎠, and then apply the inverse property To evaluate eln(7) eloge 7 = 7. , we can rewrite the logarithm as eloge 7 , and then apply the inverse property b logb x = x to get Finally, we have the one-to-one property. logb M = logb N if and only if M = N We can use the one-to-one property to solve the equation log3 (3x) = log3 (2x + 5) for x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x : 3x = 2x + 5 Set the arguments equal. x = 5 Subtract 2x. But what about the equation log3 (3x) + log3 (2x + 5) = 2 ? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation. = xa + b Recall that we use the product rule of exponents to combine the product of exponents by adding: xa xb . We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below. Given any real number x and positive real numbers M, N, and b, where b ≠ 1, we will show logb (MN)=logb (M) + logb (N). Let m = logb M and n = logb N. In exponential form, these equations are bm = M and bn = N. It follows that logb (MN) = logb (bm bn ⎝bm + n⎞ ⎛ ) Substitute for M and N. Apply the product rule for exponents. ⎠ = logb = m + n = logb (M) + logb (N) Substitute for m and n. Apply the inverse property of logs. Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider logb(wxyz). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors: 724 Chapter 6 Exponential and Logarithmic Functions logb(wxyz) = logb w + logb x + logb y + logb z The Product Rule for Logarithms The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms. logb(MN) = logb (M) + logb (N) for b > 0 (6.9) Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms. 1. Factor the argument completely, expressing each whole number factor as a product of primes. 2. Write the equivalent expression by summing the logarithms of each factor. Example 6.38 Using the Product Rule for Logarithms Expand log3 (30x(3x + 4)). Solution We begin by factoring the argument completely, expressing 30 as a product of primes. log3 (30x(3x + 4)) = log3 ⎛ ⎝2 ⋅ 3 ⋅ 5 ⋅ x ⋅ (3x + 4)⎞ ⎠ Next we write the equivalent equation by summing the logarithms of each factor. log3 (30x(3x + 4)) = log3 (2) + log3 (3) + log3 (5) + log3 (x) + log3 (3x + 4) 6.38 Expand logb(8k). Using the Quotient Rule for Logarithms For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: x . The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule. a b = xa − b Given any real number x and positive real numbers M, N, and b, where b ≠ 1, we will show logb ⎛ ⎝ M N ⎞ ⎠=logb (M) − logb (N). Let m = logb M and n = logb N. In exponential form, these equations are bm = M and bn = N. It follows that logb ⎛ ⎝ M N bm bn ⎞ ⎠ = logb Substitute for M and N. ⎛ ⎞ ⎝ ⎠ = logb (bm − n = m − n = logb (M) − logb (N) Substitute for m and n. ) Apply the quotient rule for exponents. Apply the inverse property of logs. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 725 ⎛ For example, to expand log ⎝ 2x2 + 6x 3x + 9 ⎞ ⎠, we must first express the quotient in lowest terms. Factoring and canceling we get, ⎛ log ⎝ 2x2 + 6x 3x + 9 ⎞ 2x(x + 3) ⎛ ⎠ = log ⎝ 3(x + 3) ⎛ 2x = log ⎝ 3 ⎞ ⎠ ⎞ ⎠ Factor the numerator and denominator. Cancel the common factors. Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule. 2x 3 ⎛ log ⎝ ⎞ ⎠ = log(2x) − log(3) = log(2) + log(x) − log(3) The Quotient Rule for Logarithms The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms. logb ⎛ ⎝ M N ⎞ ⎠ = logb M − logb N (6.10) Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms. 1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms. 2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator. 3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely. Example 6.39 Using the Quotient Rule for Logarithms Expand log2 ⎛ ⎝ 15x(x − 1) (3x + 4)(2 − x) ⎞ ⎠. Solution First we note that the quotient is factored and in lowest terms, so we apply the quotient rule. log2 ⎛ ⎝ 15x(x − 1) (3x + 4)(2 − x) ⎞ ⎠ = log2 ⎛ ⎝15x(x − 1)⎞ ⎠ − log2 ((3x + 4)(2 − x)) Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5. log2(15x(x − 1)) − log2((3x + 4)(2 − x)) = [log2(3) + log2(5) + log2(x) + log2(x − 1)] − [log2(3x + 4) + log2(2 − x)] = log2(3) + log2(5) + log2(x) + log2(x − 1) − log2(3x + 4) − log2(2 − x) Analysis 726 Chapter 6 Exponential and Logarithmic Functions There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x = − 4 and x = 2. Also, since the argument of a logarithm must be positive, 3 we note as we observe the expanded logarithm, that x > 0, x > 1, x > − 4 3 and x < 2. Combining these , conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises. 6.39 Expand log3 ⎛ ⎝ 7x2 + 21x 7x(x − 1)(x − 2) ⎞ ⎠. Using the Power Rule for Logarithms We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as x2 ? One method is as follows: logb ⎝x2⎞ ⎛ ⎠ = logb (x ⋅ x) = logb x + logb x = 2logb x Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example, 100 = 102 3 = 3 1 2 1 e = e−1 The Power Rule for Logarithms The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base. logb (M n ) = nlogb M (6.11) Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm. 1. Express the argument as a power, if needed. 2. Write the equivalent expression by multiplying the exponent times the logarithm of the base. Example 6.40 Expanding a Logarithm with Powers Expand log2 x5. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 727 The argument is already written as a power, so we identify the exponent, 5, and the base, x, equivalent expression by multiplying the exponent times the logarithm of the base. and rewrite the log2 ⎝x5⎞ ⎛ ⎠ = 5log2 x 6.40 Expand lnx2. Example 6.41 Rewriting an Expression as a Power before Using the Power Rule Expand log3 (25) using the power rule for logs. Solution Expressing the argument as a power, we get log3 (25) = log3 ⎛ ⎝52⎞ ⎠. Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log3 ⎛ ⎝52⎞ ⎠ = 2log3 (5) 6.41 ⎛ Expand ln ⎝ ⎞ ⎠. 1 x2 Example 6.42 Using the Power Rule in Reverse Rewrite 4ln(x) using the power rule for logs to a single logarithm with a leading coefficient of 1. Solution Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression 4ln(x), we identify the factor, 4, as the exponent and the argument, x, as the base, and rewrite the product as a logarithm of a power: 4ln(x) = ln(x4). 6.42 Rewrite 2log3 4 using the power rule for logs to a single logarithm with a leading coefficient of 1. 728 Chapter 6 Exponential and Logarithmic
Functions Expanding Logarithmic Expressions Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example: logb ⎛ ⎝ 6x y ⎞ ⎠ = logb (6x) − logb y = logb 6 + logb x − logb y We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power: logb ⎛ ⎝ A C ⎞ ⎠ = logb ⎝AC −1⎞ ⎛ ⎠ ⎝C −1⎞ ⎛ = logb (A) + logb ⎠ = logb A + ( − 1)logb C = logb A − logb C We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm. Example 6.43 Expanding Logarithms Using Product, Quotient, and Power Rules ⎛ Rewrite ln ⎜ ⎝ x4 y 7 ⎞ ⎟ as a sum or difference of logs. ⎠ Solution First, because we have a quotient of two expressions, we can use the quotient rule: ⎛ ⎜ ln ⎝ x4 y 7 ⎞ ⎛ ⎝x4 y⎞ ⎟ = ln ⎠ ⎠ − ln(7) Then seeing the product in the first term, we use the product rule: ⎛ ⎝x4 y⎞ ln ⎛ ⎝x4⎞ ⎠ − ln(7) = ln ⎠ + ln(y) − ln(7) Finally, we use the power rule on the first term: ⎝x4⎞ ⎛ ln ⎠ + ln(y) − ln(7) = 4ln(x) + ln(y) − ln(7) 6.43 ⎛ Expand log ⎜ ⎝ x2 y3 z4 ⎞ ⎟. ⎠ Example 6.44 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 729 Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression Expand log( x). Solution 6.44 ⎛ 3 ⎝ x2 Expand ln ⎞ ⎠. ⎛ ⎝x2 + y2⎞ Can we expand ln ⎠? log( x) = logx ⎞ ⎠ ⎛ ⎝ 1 2 = 1 2 logx No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Example 6.45 Expanding Complex Logarithmic Expressions Expand log6 ⎛ 64x3 (4x + 1) ⎝ (2x − 1) ⎞ ⎠. Solution We can expand by applying the Product and Quotient Rules. log6 ⎛ ⎜ 64x3(4x + 1) (2x − 1) ⎝ ⎞ ⎟ = log6 64 + log6 x3 + log6(4x + 1) − log6(2x − 1) Apply the Quotient Rule. ⎠ = log6 26 + log6 x3 + log6(4x + 1) − log6(2x − 1) Simplify by writing 64 as 26. = 6log6 2 + 3log6 x + log6(4x + 1) − log6(2x − 1) Apply the Power Rule. 6.45 ⎛ ⎜ (x − 1)(2x + 1)2 Expand ln (x2 − 9) ⎝ ⎞ ⎟. ⎠ Condensing Logarithmic Expressions We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing. 730 Chapter 6 Exponential and Logarithmic Functions Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm. 1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. 2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product. 3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient. Example 6.46 Using the Product and Quotient Rules to Combine Logarithms Write log3 (5) + log3 (8) − log3 (2) as a single logarithm. Solution Using the product and quotient rules log3 (5) + log3 (8) = log3 (5 ⋅ 8) = log3 (40) This reduces our original expression to Then, using the quotient rule log3(40) − log3(2) log3 (40) − log3 (2) = log3 ⎛ ⎝ 40 2 ⎞ ⎠ = log3 (20) 6.46 Condense log3 − log4 + log5 − log6. Example 6.47 Condensing Complex Logarithmic Expressions Condense log2 ⎝x2⎞ ⎛ ⎠ + 1 2 log2 (x − 1) − 3log2 ⎛ ⎝(x + 3)2⎞ ⎠. Solution We apply the power rule first: log2 ⎛ ⎝x2⎞ ⎠ + 1 2 log2 (x − 1) − 3log2 ⎛ ⎝(x + 3)2⎞ ⎠ = log2 ⎛ ⎝x2⎞ ⎠ + log2 Next we apply the product rule to the sum: ⎛ ⎝ x − 1⎞ ⎠ − log2 ⎝(x + 3)6⎞ ⎛ ⎠ log2 ⎝x2⎞ ⎛ ⎠ + log2 ⎛ ⎝ x − 1⎞ ⎠ − log2 ⎝(x + 3)6⎞ ⎛ ⎠ = log2 ⎛ ⎝x2 x − 1 ⎞ ⎠ − log2 ⎝(x + 3)6⎞ ⎛ ⎠ Finally, we apply the quotient rule to the difference: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 731 log2 ⎞ ⎛ ⎝x2 x − 1 ⎠ − log2 ⎝(x + 3)6⎞ ⎛ ⎠ = log2 x2 x − 1 (x + 3)6 Example 6.48 Rewriting as a Single Logarithm Rewrite 2logx − 4log(x + 5) + 1 xlog(3x + 5) as a single logarithm. Solution We apply the power rule first: 2logx − 4log(x + 5) + 1 ⎛ ⎝x2⎞ xlog(3x + 5) = log ⎛ ⎝(x + 5)4⎞ ⎠ − log ⎛ ⎝(3x + 5) ⎠ + log Next we apply the product rule to the sum: ⎛ ⎝x2⎞ log ⎛ ⎝(x + 5)4⎞ ⎠ − log ⎛ ⎝(3x + 5) ⎠ + log x−1⎞ ⎛ ⎝x2⎞ ⎠ = log ⎛ ⎝(x + 5)4 (3x + 5) ⎠ − log x−1⎞ ⎠ x−1⎞ ⎠ Finally, we apply the quotient rule to the difference: ⎛ ⎝x2⎞ log ⎛ ⎝(x + 5)4 (3x + 5) ⎠ − log x−1⎞ ⎛ ⎜ ⎜ ⎠ = log ⎜ ⎝ (x + 5)4 ⎛ x2 ⎝(3x + 5) x−1⎞ ⎠ ⎞ ⎟ ⎟ ⎟ ⎠ 6.47 Rewrite log(5) + 0.5log(x) − log(7x − 1) + 3log(x − 1) as a single logarithm. 6.48 Condense 4⎛ ⎝3log(x) + log(x + 5) − log(2x + 3)⎞ ⎠. Example 6.49 Applying of the Laws of Logs Recall that, in chemistry, pH = − log[H+ ]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH? Solution Suppose C is the original concentration of hydrogen ions, and P is the original pH of the liquid. Then P = – log(C). If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is 732 Chapter 6 Exponential and Logarithmic Functions Using the product rule of logs pH = − log(2C) pH = − log(2C) = − ⎛ ⎝log(2) + log(C)⎞ ⎠ = − log(2) − log(C) Since P = – log(C), the new pH is When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301. pH = P − log(2) ≈ P − 0.301 6.49 How does the pH change when the concentration of positive hydrogen ions is decreased by half? Using the Change-of-Base Formula for Logarithms Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or e, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs. To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms. Given any positive real numbers M, b, and n, where n ≠ 1 and b ≠ 1, we show Let y = logb M. By taking the log base n of both sides of the equation, we arrive at an exponential form, namely b y It follows that = M. logb M= logn M logn b logn(b y ) = logn M Apply the one-to-one property. ylogn b = logn M Apply the power rule for logarithms. y = logb M = logn M logn b logn M logn b Isolate y. Substitute for y. For example, to evaluate log5 36 using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log. log5 36 = log(36) log(5) Apply the change of base formula using base 10. ≈ 2.2266 Use a calculator to evaluate to 4 decimal places. The Change-of-Base Formula The change-of-base formula can be used to evaluate a logarithm with any base. For any positive real numbers M, b, and n, where n ≠ 1 and b ≠ 1, logb M= logn M logn b . (6.12) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 733 It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs. and logb M = lnM lnb logb M = logM logb Given a logarithm with the form logb M, use the change-of-base formula to rewrite it as a quotient of logs with any positive base n, where n ≠ 1. 1. Determine the new base n, ln(x), has base e. remembering that the common log, log(x), has base 10, and the natural log, 2. Rewrite the log as a quotient using the change-of-base formula ◦ The numerator of the quotient will be a logarithm with base n and argument M. ◦ The denominator of the quotient will be a logarithm with base n and argument b. Example 6.50 Changing Logarithmic Expressions to Expressions Involving Only Natural Logs Change log5 3 to a quotient of natural logarithms. Solution Because we will be expressing log5 3 as a quotient of natural logarithms, the new base, n = e. We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5. logb M = lnM lnb log5 3 = ln3 ln5 6.50 Change log0.5 8 to a quotient of natural logarithms. Can we change common logarithms to natural logarithms? Yes. Remember that log9 means log10 9. So, log9 = ln9 ln10 . Example 6.51 734 Chapter 6 Exponential and Logarithmic Functions Using the Change-of-Base Formula with a Calculator Evaluate log2(10) using the change-of-base formula with a calculator. Solution According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e. log2 10 = ln10 ln2 Apply the change of base formula using base e. ≈ 3.3219 Use a calculator to evaluate to 4 decimal places. 6.51 Evaluate log5(100) using the change-of-base formula. Access these online resources for additional instruction and practice with laws of logarithms. • The Properties of Logarithms (http://openstaxcollege.org/l/proplog) • Expand Logarithmic Expressions (http://openstaxcollege.org/l/expandlog) • Evaluate a Natural Logarithmic Expression (http://openstaxcollege.org/l/evaluatelog) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 735 6.5 EXERCISES Verbal How does the power rule for logarithms help when 249. solving logarithms with the form logb ( xn ) ? What does the change-of-base formula do? Why is it 250. useful when using a calculator? A
lgebraic ⎛ a−2 ln ⎝ b−4 c5 ⎞ ⎠ 265. ⎝ x3 y−4⎞ ⎛ log ⎠ 266. ⎛ ⎝y ln y 1 − y ⎞ ⎠ For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. 267. 3 ⎞ ⎛ ⎝x2 y3 x2 y5 log ⎠ 251. logb ⎛ ⎝7x ⋅ 2y⎞ ⎠ 252. ln(3ab ⋅ 5c) 253. logb ⎛ ⎝ 13 17 ⎞ ⎠ 254. x ⎛ z w ⎝ ⎞ ⎠ log4 255. ⎛ 1 ln ⎝ k 4 ⎞ ⎠ 256. log2 ⎛ ⎝y x⎞ ⎠ For the following exercises, condense to a single logarithm if possible. 257. ln(7) + ln(x) + ln(y) 258. log3(2) + log3(a) + log3(11) + log3(b) 259. logb(28) − logb(7) 260. ln(a) − ln(d) − ln(c) 261. −logb ⎞ ⎠ ⎛ ⎝ 1 7 262. ln(8) 1 3 the following exercises, use the properties of For logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs. ⎛ ⎜ log ⎝ x15 y13 z19 ⎞ ⎟ ⎠ 263. 264. For the following exercises, condense each expression to a single logarithm using the properties of logarithms. 268. ⎝2x4⎞ ⎛ log ⎝3x5⎞ ⎛ ⎠ + log ⎠ 269. ln(6x9) − ln(3x2) 270. 2log(x) + 3log(x + 1) 271. 272. log(x) − 1 2 log(y) + 3log(z) 4log7 (c) + log7 (a) 3 + log7 (b) 3 For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base. 273. log7 (15) to base e 274. log14 (55.875) to base 10 the following exercises, suppose log5 (6) = a and For log5 (11) = b. Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of a and b. Show the steps for solving. 275. log11 (5) 276. log6 (55) 277. log11 ⎛ ⎝ 6 11 ⎞ ⎠ Numeric For the following exercises, use properties of logarithms to evaluate without using a calculator. 278. log3 ⎛ ⎝ 1 9 ⎞ ⎠ − 3log3 (3) Chapter 6 Exponential and Logarithmic Functions 736 279. 280. 6log8 (2) + log8 (64) 3log8 (4) 2log9 (3) − 4log9 (3) + log9 ⎛ ⎝ 1 729 ⎞ ⎠ the following exercises, use the change-of-base For formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places. 281. log3 (22) 282. log8 (65) 283. log6 (5.38) 284. log4 ⎛ ⎝ 15 2 ⎞ ⎠ 285. (4.7) log 1 2 Extensions 286. values Use the product rule for logarithms to find all x that log12 (2x + 6) + log12 (x + 2) = 2. such Show the steps for solving. 287. values Use the quotient rule for logarithms to find all x such that log6 (x + 2) − log6 (x − 3) = 1. Show the steps for solving. Can the power property of logarithms be derived from 288. the power property of exponents using the equation b x = m ? If not, explain why. If so, show the derivation. 289. Prove that logb (n) = 1 logn (b) for any positive integers b > 1 and n > 1. 290. Does log81 (2401) = log3 (7) ? Verify the claim algebraically. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 737 6.6 \| Exponential and Logarithmic Equations Learning Objectives In this section, you will: 6.6.1 Use like bases to solve exponential equations. 6.6.2 Use logarithms to solve exponential equations. 6.6.3 Use the definition of a logarithm to solve logarithmic equations. 6.6.4 Use the one-to-one property of logarithms to solve logarithmic equations. 6.6.5 Solve applied problems involving exponential and logarithmic equations. Figure 6.43 Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr) In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions. Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions. Using Like Bases to Solve Exponential Equations The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where b > 0, b ≠ 1, bS = bT if and only if S = T. In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-toone to set the exponents equal to one another, and solve for the unknown. For example, consider the equation 34x − 7 = 32x 3 the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x : . To solve for x, we use the division property of exponents to rewrite 738 Chapter 6 Exponential and Logarithmic Functions 34x − 7 = 32x 3 34x − 7 = 32x 31 Rewrite 3 as 31. 34x − 7 = 32x − 1 Use the division property of exponents. 4x − 7 = 2x − 1 Apply the one-to-one property of exponents. 2x x Subtract 2x and add 7 to both sides. Divide by 3. = 6 = 3 Using the One-to-One Property of Exponential Functions to Solve Exponential Equations For any algebraic expressions S and T, and any positive real number b ≠ 1, bS = bT if and only if S = T (6.13) Given an exponential equation with the form bS = bT, where S and T are algebraic expressions with an unknown, solve for the unknown. 1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT . 2. Use the one-to-one property to set the exponents equal. 3. Solve the resulting equation, S = T, for the unknown. Example 6.52 Solving an Exponential Equation with a Common Base Solve 2 x − 1 = 22x − 4. Solution 2 x − 1 = 22x − 4 x − 1 = 2x − 4 x = 3 The common base is 2. By the one-to-one property the exponents must be equal. Solve for x. 6.52 Solve 52x = 53x + 2. Rewriting Equations So All Powers Have the Same Base Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property. For example, consider the equation 256 = 4 apply the rules of exponents, along with the one-to-one property, to solve for x : x − 5. We can rewrite both sides of this equation as a power of 2. Then we This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 739 x − 5 x − 5 256 = 4 ⎝22⎞ ⎛ 28 = ⎠ 28 = 22x − 10 8 = 2x − 10 18 = 2x x = 9 Rewrite each side as a power with base 2. Use the one-to-one property of exponents. Apply the one-to-one property of exponents. Add 10 to both sides. Divide by 2. Given an exponential equation with unlike bases, use the one-to-one property to solve it. 1. Rewrite each side in the equation as a power with a common base. 2. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT . 3. Use the one-to-one property to set the exponents equal. 4. Solve the resulting equation, S = T, for the unknown. Example 6.53 Solving Equations by Rewriting Them to Have a Common Base Solve 8 x + 2 = 16 x + 1. Solution = x + 1 x + 1 x + 2 = 16 x + 2 ⎛ ⎝24⎞ ⎠ 8 ⎝23⎞ ⎛ ⎠ 23x + 6 = 24x + 4 To take a power of a power, multiply exponents . 3x + 6 = 4x + 4 Use the one-to-one property to set the exponents equal. x = 2 Write 8 and 16 as powers of 2. Solve for x. 6.53 Solve 52x = 253x + 2. Example 6.54 Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base Solve 25x = 2. Solution 740 Chapter 6 Exponential and Logarithmic Functions 1 2 25x = 2 5x = 1 2 x = 1 10 Write the square root of 2 as a power of 2. Use the one-to-one property. Solve for x. 6.54 Solve 5 x = 5. Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process? No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 741 Example 6.55 Solving an Equation with Positive and Negative Powers Solve 3 x + 1 = −2. Solution This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive. Analysis Figure 6.44 shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution. Figure 6.44 6.55 Solve 2 x = −100. Solving Exponential Equations Using Logarithms Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since log(a) = log(b) is equivalent to a = b, we may apply logarithms with the same base on both sides of an exponential equation. Given an exponential equation in which a common base cannot be found, solve for the unknown. 1. Apply the logarithm of both sides of the equation. ◦ ◦ If one of the terms in the equation has base 10, use the common logarithm. If none of the terms in the equation has base 10, use the natural logarithm. 2. Use the rules of logarithms to solve for the unknown. 742 Chapter 6 Exponential and Logarithmic Functions Example 6.56 Solving an Equation Containing Powers of Different Bases Solve 5 x + 2 = 4 x . Solution = ln4 x ln5 (x + 2)ln5 = xln4 xln5 + 2ln5 = xln4 xln5 − xln4 = − 2ln5 x(ln5 − ln4) = − 2ln5 ⎞ ⎛ ⎞ ⎛ 5 1 xln ⎠ = ln ⎝ ⎠ ⎝ 4 25 ⎛ ⎞ 1 ln ⎝ ⎠ 25 ⎞ ⎛ 5 ln ⎠ ⎝ 4 x = There is no easy way to get the powers to have the same base . Take ln of both sides. Use laws of logs. Use the distributive law. Get terms containing x on one side, terms without x on t
he other. On the left hand side, factor out an x. Use the laws of logs. Divide by the coefficient o x. 6.56 Solve 2 x = 3 x + 1. Is there any way to solve 2 x = 3 x? Yes. The solution is x = 0. Equations Containing e One common type of exponential equations are those with base e. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it. Given an equation of the form y = Aekt , solve for t. 1. Divide both sides of the equation by A. 2. Apply the natural logarithm of both sides of the equation. 3. Divide both sides of the equation by k. Example 6.57 Solve an Equation of the Form y = Aekt Solve 100 = 20e2t . Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 743 100 = 20e2t 5 = e2t ln5 = 2t = ln5 t 2 Divide by the coefficient of he power . Take ln of both sides. Use the fact that ln(x) and e x are inverse functions. Divide by the coefficient o t. Analysis Using laws of logs, we can also write this answer in the form t = ln 5. If we want a decimal approximation of the answer, we use a calculator. 6.57 Solve 3e0.5t = 11. Does every equation of the form y = Aekt have a solution? No. There is a solution when k ≠ 0, and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is 2 = −3et. Example 6.58 Solving an Equation That Can Be Simplified to the Form y = Aekt Solve 4e2x + 5 = 12. Solution 4e2x 4e2x e2x + 5 = 12 = 7 = 7 4 ⎛ 2x = ln ⎝ Combine like terms. Divide by the coefficient of he power . ⎞ 7 ⎠ Take ln of both sides. 4 ⎛ ln ⎝ ⎞ ⎠ Solve for x. 7 4 x = 1 2 6.58 Solve 3 + e2t = 7e2t . Extraneous Solutions Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output. 744 Chapter 6 Exponential and Logarithmic Functions Example 6.59 Solving Exponential Functions in Quadratic Form Solve e2x − e x = 56. Solution e2x − e x + 7)(e x e x e2x (e x − e x = 56 − 56 = 0 − 8 or e x = − 7 or e = 8 = ln8 Get one side of the equation equal to zero. Factor by the FOIL method. If a product is zero, then one factor must be zero. Isolate the exponentials. Reject the equation in which the power equals a negative number. Solve the equation in which the power equals a positive number . Analysis When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x = −7 because a positive number never equals a negative number. The solution x = ln(−7) is not a real number, and in the real number system this solution is rejected as an extraneous solution. 6.59 Solve e2x = e x + 2. Does every logarithmic equation have a solution? No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions. Using the Definition of a Logarithm to Solve Logarithmic Equations We have already seen that every logarithmic equation logb (x) = y is equivalent to the exponential equation b y = x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log2 (2) + log2 (3x − 5) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x : log2(2) + log2(3x − 5) = 3 log2(2(3x − 5)) = 3 log2(6x − 10) = 3 23 = 6x − 10 8 = 6x − 10 18 = 6x x = 3 Apply the product rule of logarithms. Distribute. Apply the definition of a lo arithm. Calculate 23. Add 10 to both sides. Divide by 6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 745 Using the Definition of a Logarithm to Solve Logarithmic Equations For any algebraic expression S and real numbers b and c, where b > 0, b ≠ 1, logb(S) = c if and only if bc = S (6.14) Example 6.60 Using Algebra to Solve a Logarithmic Equation Solve 2lnx + 3 = 7. Solution 2lnx + 3 = 7 2lnx = 4 Subtract 3. lnx = 2 Divide by 2. x = e2 Rewrite in exponential form. 6.60 Solve 6 + lnx = 10. Example 6.61 Using Algebra Before and After Using the Definition of the Natural Logarithm 2ln(6x) = 7 ln(6x) = 7 2 6x = Divide by 2. Use the definition of ln. Divide by 6. Solve 2ln(6x) = 7. Solution 6.61 Solve 2ln(x + 1) = 10. Example 6.62 Using a Graph to Understand the Solution to a Logarithmic Equation 746 Chapter 6 Exponential and Logarithmic Functions Solve lnx = 3. Solution lnx = 3 x = e3 Use the definition of he natural logarithm. Figure 6.45 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words e3 ≈ 20. A calculator gives a better approximation: e3 ≈ 20.0855. Figure 6.45 The graphs of y = lnx and y = 3 cross at the point (e3, 3), which is approximately (20.0855, 3). Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2 6.62 2 decimal places. x = 1000 to Using the One-to-One Property of Logarithms to Solve Logarithmic Equations As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b ≠ 1, For example, logb S = logb T if and only if S = T. If log2(x − 1) = log2(8), then x − 1 = 8. So, if x − 1 = 8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log2 (9 − 1) = log2 (8) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation log(3x − 2) − log(2) = log(x + 4). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x : This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 747 log(3x − 2) − log(2) = log(x + 4) 3x − 2 ⎛ log ⎝ 2 3x − 2 = x + 4 3x − 2 = 2x + 8 x = 10 2 ⎞ ⎠ = log(x + 4) Apply the quotient rule of logarithms. Apply the one to one property of a logarithm. Multiply both sides of the equation by 2. Subtract 2x and add 2. To check the result, substitute x = 10 into log(3x − 2) − log(2) = log(x + 4). log(3(10) − 2) − log(2) = log((10) + 4) log(28) − log(2) = log(14) ⎛ log ⎝ ⎞ ⎠ = log(14) 28 2 The solution checks. Using the One-to-One Property of Logarithms to Solve Logarithmic Equations For any algebraic expressions S and T and any positive real number b, where b ≠ 1, logb S = logb T if and only if S = T (6.15) Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. Given an equation containing logarithms, solve it using the one-to-one property. 1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form logb S = logb T. 2. Use the one-to-one property to set the arguments equal. 3. Solve the resulting equation, S = T, for the unknown. Example 6.63 Solving an Equation Using the One-to-One Property of Logarithms Solve ln(x2) = ln(2x + 3). Solution ln(x2) = ln(2x + 3) x2 = 2x + 3 x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 x − 3 = 0 or x + 1 = 0 x = 3 or x = − 1 Use the one-to-one property of the logarithm. Get zero on one side before factoring. Factor using FOIL. If a product is zero, one of the factors must be zero. Solve for x. Analysis There are two solutions: x = 3 or x = −1. The solution x = −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive. 748 Chapter 6 Exponential and Logarithmic Functions 6.63 Solve ln(x2) = ln1. Solving Applied Problems Using Exponential and Logarithmic Equations In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm. One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 6.16 lists the half-life for several of the more common radioactive substances. Substance Use Half-life gallium-67 nuclear medicine 80 hours cobalt-60 manufacturing 5.3 years technetium-99m nuclear medicine 6 hours americium-241 construction 432 years carbon-14 archeological dating 5,715 years uranium-235 atomic power 703,800,000 years Table 6.16 We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount re
maining after a specified time. We can use the formula for radioactive decay: ln(0.5) T t ln(0.5) t T A(t) = A0 e A(t) = A0 e A(t) = A0 (eln(0.5) ) A(t) = A0 where • A0 is the amount initially present • T is the half-life of the substance • • t is the time period over which the substance is studied y is the amount of the substance present after time t This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 749 Example 6.64 Using the Formula for Radioactive Decay to Find the Quantity of a Substance How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay? Solution y = 1000e 900 = 1000e t ln(0.5) 703,800,000 ln(0.5) 703,800,000 t ln(0.5) 703,800,000 t 0.9 = e ⎛ ⎜e ln(0.9) = ln ⎜ ⎝ ln(0.5) 703,800,000 t⎞ ⎟ ⎟ ⎠ After 10% decays, 900 grams are left. Divide by 1000. Take ln of both sides. ln(0.9) = ln(0.5) 703,800,000 t t = 703,800,000× ln(0.9) ln(0.5) years t ≈ 106,979,777 years ln(e M ) = M Solve for t. Analysis Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams. 6.64 How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed? Access these online resources for additional instruction and practice with exponential and logarithmic equations. • Solving Logarithmic Equations (http://openstaxcollege.org/l/solvelogeq) • Solving Exponential Equations with Logarithms (http://openstaxcollege.org/l/ solveexplog) 750 Chapter 6 Exponential and Logarithmic Functions 6.6 EXERCISES Verbal 291. How can an exponential equation be solved? When does an extraneous solution occur? How can an 292. extraneous solution be recognized? 311. e2x − e x − 132 = 0 312. 7e8x + 8 − 5 = −95 313. 10e8x + 3 + 2 = 8 When can the one-to-one property of logarithms be 293. used to solve an equation? When can it not be used? 314. 4e3x + 3 − 7 = 53 Algebraic 315. 8e−5x − 2 − 4 = −90 For the following exercises, use like bases to solve the exponential equation. 316. 32x + 1 = 7 x − 2 294. 4−3v − 2 = 4−v 295. 64 ⋅ 43x = 16 296. 32x + 1 ⋅ 3 x = 243 297. 2−3n ⋅ 1 4 n + 2 = 2 298. 625 ⋅ 53x + 3 = 125 299. 363b 362b = 2162 − b 300. 3n ⎛ ⎝ 1 64 ⎞ ⎠ ⋅ 8 = 26 For the following exercises, use logarithms to solve. 301. x − 10 = 1 9 302. 2e6x = 13 303. er + 10 − 10 = −42 304. 2 ⋅ 109a = 29 305. p + 7 −8 ⋅ 10 − 7 = −24 306. 7e3n − 5 + 5 = −89 307. e−3k + 6 = 44 308. −5e9x − 8 − 8 = −62 309. −6e9x + 8 + 2 = −74 310. x + 1 = 52x − 1 2 This content is available for free at https://cnx.org/content/col11758/1.5 317. e2x − e x − 6 = 0 318. 3e3 − 3x + 6 = −31 For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation. 319. 320. ⎛ log ⎝ 1 100 ⎞ ⎠ = −2 log324 (18) = 1 2 the following exercises, use the definition of a For logarithm to solve the equation. 321. 5log7 n = 10 322. −8log9 x = 16 323. 4 + log2 (9k) = 2 324. 2log(8n + 4) + 6 = 10 325. 10 − 4ln(9 − 8x) = 6 For the following exercises, use the one-to-one property of logarithms to solve. 326. ln(10 − 3x) = ln(−4x) 327. log13 (5n − 2) = log13 (8 − 5n) 328. log(x + 3) − log(x) = log(74) 329. ⎝x2 − 6x⎞ ⎛ ln(−3x) = ln ⎠ 330. log4 (6 − m) = log4 3m Chapter 6 Exponential and Logarithmic Functions 751 331. ln(x − 2) − ln(x) = ln(54) 332. log9 ⎝2n2 − 14n⎞ ⎛ ⎠ = log9 ⎝−45 + n2⎞ ⎛ ⎠ 333. ⎞ ⎛ ⎝x2 − 10 ⎠ + ln(9) = ln(10) ln For the following exercises, solve each equation for x. 334. log(x + 12) = log(x) + log(12) 335. ln(x) + ln(x − 3) = ln(7x) 336. log2(7x + 6) = 3 337. ⎛ ⎝2 − 4x2⎞ ln(7) + ln ⎠ = ln(14) 338. log8 (x + 6) − log8 (x) = log8 (58) 339. ln(3) − ln(3 − 3x) = ln(4) 340. log3 (3x) − log3 (6) = log3 (77) Graphical For the following exercises, solve the equation for x, there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. if 341. log9 (x) − 5 = −4 342. log3 (x) + 3 = 2 343. ln(3x) = 2 344. ln(x − 5) = 1 345. log(4) + log(−5x) = 2 346. −7 + log3 (4 − x) = −6 347. ln(4x − 10) − 6 = − 5 348. log(4 − 2x) = log(−4x) 353. 3 log2 (10) − log(x − 9) = log(44) 354. ln(x) − ln(x + 3) = ln(6) For the following exercises, solve for the indicated value, and graph the situation showing the solution point. 355. An account with an initial deposit of $6,500 earns 7.25% annual interest, compounded continuously. How much will the account be worth after 20 years? 356. I I0 ⎛ D is defined by the equation D = 10log ⎝ The formula for measuring sound intensity in decibels ⎞ ⎠, where I is the intensity of the sound in watts per square meter and I0 = 10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.3 ⋅ 102 watts per square meter? The population of a small town is modeled by the 357. equation P = 1650e0.5t where t is measured in years. In approximately how many years will the town’s population reach 20,000? Technology For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x to 3 decimal places. 358. 1000(1.03) t = 5000 using the common log. 359. e5x = 17 using the natural log 360. 3(1.04)3t = 8 using the common log 361. 34x − 5 = 38 using the common log 362. 50e−0.12t = 10 using the natural log For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth. 349. log11 ⎝−2x2 − 7x⎞ ⎛ ⎠ = log11 (x − 2) 350. ln(2x + 9) = ln(−5x) 351. log9 (3 − x) = log9 (4x − 8) 352. ⎞ ⎛ ⎝x2 + 13 ⎠ = log(7x + 3) log 363. 7e3x − 5 + 7.9 = 47 364. ln(3) + ln(4.4x + 6.8) = 2 365. log(−0.7x − 9) = 1 + 5log(5) 366. 752 Chapter 6 Exponential and Logarithmic Functions Atmospheric pressure P in pounds per square inch is represented by the formula P = 14.7e−0.21x , where x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile) 367. The magnitude M of an earthquake is represented by ⎛ log ⎝ the equation M = 2 3 ⎞ where E is the amount of ⎠ earthquake E E0 energy released by the and E0 = 104.4 is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing 1.4 ⋅ 1013 joules of energy? in joules Extensions 368. Use the definition of a logarithm along with the one- to-one property of logarithms to prove that b logb x = x. 369. the formula for continually compounding Recall interest, y = Aekt . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm. kt the interest compound Recall ⎝1 + r k 370. A = a⎛ with properties of logarithms to solve the formula for time t. . Use the definition of a logarithm along formula ⎞ ⎠ ⎞ ⎝T0 − Ts 371. Newton’s Law of Cooling states that the temperature T of an object at any time t can be described by the ⎠e−kt equation T = Ts + ⎛ the temperature of the surrounding environment, T0 is the initial temperature of the object, and k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t such that t is equal to a single logarithm. where Ts is , This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 753 6.7 \| Exponential and Logarithmic Models Learning Objectives In this section, you will: 6.7.1 Model exponential growth and decay. 6.7.2 Use Newton’s Law of Cooling. 6.7.3 Use logistic-growth models. 6.7.4 Choose an appropriate model for data. 6.7.5 Express an exponential model in base e . Figure 6.46 A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus (credit: Georgia Tech Research Institute) We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling. Modeling Exponential Growth and Decay In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function: y = A0 ekt is equal to the value at time zero, e is Euler’s constant, and k is a positive constant that determines the rate where A0 (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model. On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form y = A0 ekt where A0 is the starting value, and e is Euler’s constant. Now k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes. 754 Chapter 6 Exponential and Logarithmic Functions In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a
distinctive shape, as we can see in Figure 6.47 and Figure 6.48. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis. Figure 6.47 A graph showing exponential growth. The equation is y = 2e3x . Figure 6.48 A graph showing exponential decay. The equation is y = 3e−2x . Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 × 1013. So, we could describe this number as having order of magnitude 1013. Characteristics of the Exponential Function, y = A0ekt An exponential function with the form y = A0 ekt has the following characteristics: • one-to-one function • horizontal asymptote: y = 0 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 755 • domain: ( – ∞, ∞) • range: (0, ∞) • x intercept: none • y-intercept: ⎛ ⎝0, A0 ⎞ ⎠ • increasing if k > 0 (see Figure 6.49) • decreasing if k < 0 (see Figure 6.49) Figure 6.49 An exponential function models exponential growth when k > 0 and exponential decay when k < 0. Example 6.65 Graphing Exponential Growth A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time. Solution When an amount grows at a fixed percent per unit time, the growth is exponential. To find A0 A0 doubles from 10 to 20. The formula is derived as follows we use the fact that is the amount at time zero, so A0 = 10. To find k, use the fact that after one hour (t = 1) the population 20 = 10ek ⋅ 1 2 = ek ln2 = k Divide by 10 Take the natural logarithm so k = ln(2). Thus the equation we want to graph is y = 10e(ln2)t Figure 6.50. = 10(eln2) t = 10 · 2 t . The graph is shown in 756 Chapter 6 Exponential and Logarithmic Functions Figure 6.50 The graph of y = 10e(ln2)t Analysis The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude 104. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 107, so we could say that the population has increased by three orders of magnitude in ten hours. Half-Life We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is halflife, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay. To find the half-life of a function describing exponential decay, solve the following equation: We find that the half-life depends only on the constant k and not on the starting quantity A0. The formula is derived as follows A0 = Ao ekt 1 2 1 2 A0 = Ao ekt = ekt 1 2 ⎛ ⎞ 1 ln ⎠ = kt ⎝ 2 −ln(2) = kt ln(2) k = t − Divide by A0. Take the natural log. Apply laws of logarithms. Divide by k. Since t, the time, is positive, k must, as expected, be negative. This gives us the half-life formula t = − ln(2) k (6.16) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 757 Given the half-life, find the decay rate. 1. Write A = Ao ekt . 2. Replace A by 1 2 A0 and replace t by the given half-life. 3. Solve to find k. Express k as an exact value (do not round). Note: It is also possible to find the decay rate using k = − ln(2) t . Example 6.66 Finding the Function that Describes Radioactive Decay The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t. Solution This formula is derived as follows. () The continuous growth formula. A = A0 ekt 0.5A0 = A0 ek ⋅ 5730 Substitute the half-life for t and 0.5A0 for f (t). 0.5 = e5730k ln(0.5) = 5730k ln(0.5) 5730 ⎛ ⎝ Divide by A0. Take the natural log of both sides. Divide by the coefficient o k. k = ⎞ ⎠t ln(0.5) 5730 Substitute for r in the continuous growth formula. A = A0 e The function that describes this continuous decay is f (t) = A0 e ln(0.5) 5730 ≈ − 1.2097 is negative, as expected in the case of exponential decay. ⎛ ⎝ ln(0.5) 5730 ⎞ ⎠t . We observe that the coefficient of t, The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 6.65 remaining as a function of time, measured in years. Radiocarbon Dating The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years. Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries. 758 Chapter 6 Exponential and Logarithmic Functions As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated. Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t years is ⎛ ⎝ ln(0.5) 5730 ⎞ ⎠t A ≈ A0 e where • A is the amount of carbon-14 remaining • A0 is the amount of carbon-14 when the plant or animal began decaying. This formula is derived as follows: A = A0 ekt The continuous growth formula. 0.5A0 = A0 ek ⋅ 5730 Substitute the half-life for t and 0.5A0 for f (t). 0.5 = e5730k ln(0.5) = 5730k ln(0.5) 5730 ⎛ ⎝ Divide by A0. Take the natural log of both sides. Divide by the coefficient o k. k = ⎞ ⎠t ln(0.5) 5730 Substitute for r in the continuous growth formula. A = A0 e To find the age of an object, we solve this equation for t : ⎞ ⎠ ⎛ A ln ⎝ A0 −0.000121 t = (6.17) Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation A ≈ A0 e−0.000121t we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is r = A to A0 . We solve this equation for t, ≈ e−0.000121t get t = ln(r) −0.000121 Given the percentage of carbon-14 in an object, determine its age. 1. Express the given percentage of carbon-14 as an equivalent decimal, k. 2. Substitute for k in the equation t = ln(r) −0.000121 and solve for the age, t. Example 6.67 Finding the Age of a Bone A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone? Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 759 We substitute 20% = 0.20 for k in the equation and solve for t : t = ln(r) −0.000121 ln(0.20) −0.000121 = ≈ 13301 Use the general form of the equation. Substitute for r. Round to the nearest year. The bone fragment is about 13,301 years old. Analysis The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as 13,301 years ± 1% or 13,301 years ± 133 years. Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or 6.66 less than 230 years until only 1 milligram remains? Calculating Doubling Time For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time. Given the basic exponential growth equation A = A0 ekt quantity has doubled, that is, by solving 2A0 = A0 ekt The formula is derived as follows: , doubling time can be found by solving for when the original . 2A0 = A0 ekt 2 = ekt ln2 = kt t = ln2 k Divide by A0. Take the natural logarithm. Divide by the coefficient o t. Thus the doubling time is Example 6.68 t = ln2 k (6.18) Finding a Function That Describes Exponential Growth According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior. Solution The formula is derived as follows: 760 Chapter 6 Exponential and Logarithmic Functions t = ln2 k 2 = ln2 k k = ln2 2 t ln2 2 A = A0 e The doubling time formula. Use a doubling time of two years. Multiply by k and divide by 2. Substitute k into the continuous growth formula. The function is A = A0 e
t ln2 2 . 6.67 Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account. Using Newton’s Law of Cooling Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature This formula is derived as follows: T(t) = aekt + Ts ) T(t) = Abct + Ts T(t) = Aeln(bct T(t) = Aectlnb T(t) = Aekt + Ts + Ts + Ts Laws of logarithms. Laws of logarithms. Rename the constant c ln b, calling it k. Newton’s Law of Cooling The temperature of an object, T, in surrounding air with temperature Ts will behave according to the formula T(t) = Aekt + Ts (6.19) where • t is time • A is the difference between the initial temperature of the object and the surroundings • k is a constant, the continuous rate of cooling of the object This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 761 Given a set of conditions, apply Newton’s Law of Cooling. 1. Set Ts equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature). 2. Substitute the given values into the continuous growth formula T(t) = Aek t + Ts to find the parameters A and k. 3. Substitute in the desired time to find the temperature or the desired temperature to find the time. Example 6.69 Using Newton’s Law of Cooling A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, and is placed into a 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F. If we must wait until the cheesecake has cooled to 70°F before we eat it, how long will we have to wait? Solution Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation We know the initial temperature was 165, so T(0) = 165. T(t) = Aekt + 35 165 = Aek0 + 35 Substitute (0, 165). A = 130 Solve for A. We were given another data point, T(10) = 150, which we can use to solve for k. = e10k Substitute (10, 150). 150 = 130ek10 + 35 115 = 130ek10 115 130 ⎞ 115 ⎠ = 10k 130 ⎛ 115 ln ⎝ 130 10 This gives us the equation for the cooling of the cheesecake: T(t) = 130e – 0.0123t k = Divide by 130. ⎛ ln ⎝ Subtract 35. ⎞ ⎠ = − 0.0123 Divide by the coefficient o k. + 35. Take the natural log of both sides. Now we can solve for the time it will take for the temperature to cool to 70 degrees. 70 = 130e−0.0123t 35 = 130e−0.0123t 35 130 ln( 35 130 ) = − 0.0123t = e−0.0123t + 35 Substitute in 70 for T(t). Subtract 35. Divide by 130. Take the natural log of both sides t = ln( 35 130) −0.0123 ≈ 106.68 Divide by the coefficient o t. It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°F. 762 Chapter 6 Exponential and Logarithmic Functions 6.68 A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees? Using Logistic Growth Models Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-halfbillion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value. The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants a, b, and c, the logistic growth of a population over time x is represented by the model f (x) = c 1 + ae−bx The graph in Figure 6.51 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases. Figure 6.51 Logistic Growth The logistic growth model is f (x) = c 1 + ae−bx where • • • c 1 + a is the initial value c is the carrying capacity, or limiting value b is a constant determined by the rate of growth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 763 Example 6.70 Using the Logistic-Growth Model An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population. For example, at time t = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed. Solution We substitute the given data into the logistic growth model f (x) = c 1 + ae−bx Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is c = 1000. To find a, we use the formula that the number of cases at time t = 0 is c from which 1 + a = 1, it follows that a = 999. This model predicts that, after ten days, the number of people who have had the flu is 1 + 999e−0.6030x ≈ 293.8. Because the actual number must be a whole number (a person has either f (x) = 1000 had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, c = 1000. Analysis Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph in Figure 6.52 gives a good picture of how this model fits the data. 764 Chapter 6 Exponential and Logarithmic Functions Figure 6.52 The graph of f (x) = 1000 1 + 999e−0.6030x 6.69 Using the model in Example 6.70, estimate the number of cases of flu on day 15. Choosing an Appropriate Model for Data Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015. Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered. In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down. A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection. After using the graph to help us choose a type of function to use as a model,
we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 765 Example 6.71 Choosing a Mathematical Model Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 6.17? Find the model, and use a graph to check your choice.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394 x y 1 0 Table 6.17 Solution First, plot the data on a graph as in Figure 6.53. For the purpose of graphing, round the data to two significant digits. Figure 6.53 Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try y = aln(bx). Plugging in the first point, (1,0), gives 0 = alnb. We reject the case that a = 0 (if it were, all outputs would be 0), so we know ln(b) = 0. Thus b = 1 and y = aln(x). Next we can use the point (9,4.394) to solve for a : y = aln(x) 4.394 = aln(9) a = 4.394 ln(9) 766 Chapter 6 Exponential and Logarithmic Functions Because a = 4.394 ln(9) ≈ 2, an appropriate model for the data is y = 2ln(x). To check the accuracy of the model, we graph the function together with the given points as in Figure 6.54. Figure 6.54 The graph of y = 2lnx. We can conclude that the model is a good fit to the data. ⎛ ⎝x2⎞ Compare Figure 6.54 to the graph of y = ln ⎠ shown in Figure 6.55. ⎝x2⎞ ⎛ Figure 6.55 The graph of y = ln ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 767 ⎝x2⎞ ⎛ The graphs appear to be identical when x > 0. A quick check confirms this conclusion: y = ln x > 0. ⎠ = 2ln(x) for ⎝x2⎞ ⎛ the graph of y = ln However, if x < 0, ⎠ includes a “extra” branch, as shown in Figure 6.56. This occurs because, while y = 2ln(x) cannot have negative values in the domain (as such values would force the argument ⎛ ⎝x2⎞ to be negative), the function y = ln ⎠ can have negative domain values. Figure 6.56 6.70 Does a linear, exponential, or logarithmic model best fit the data in Table 6.18? Find the model.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034 Table 6.18 Expressing an Exponential Model in Base e While powers and logarithms of any base can be used in modeling, the two most common bases are 10 and e. In science and mathematics, the base e is often preferred. We can use laws of exponents and laws of logarithms to change any base to base e. Given a model with the form y = ab x, change it to the form y = A0 ekx. ⎝b x⎞ . 1. Rewrite y = ab x as y = ae ln⎛ ⎠ 2. Use the power rule of logarithms to rewrite y as y = ae xln(b) = aeln(b)x . 3. Note that a = A0 and k = ln(b) in the equation y = A0 ekx . Example 6.72 768 Chapter 6 Exponential and Logarithmic Functions Changing to base e Change the function y = 2.5(3.1) x so that this same function is written in the form y = A0 ekx . Solution The formula is derived as follows x y = 2.5(3.1) ln⎛ ⎝3.1 x⎞ ⎠ = 2.5e = 2.5e xln3.1 = 2.5e(ln3.1) x Insert exponential and its inverse. Laws of logs. Commutative law of multiplication 6.71 Change the function y = 3(0.5) x to one having e as the base. Access these online resources for additional instruction and practice with exponential and logarithmic models. • Logarithm Application – pH (http://openstaxcollege.org/l/logph) • Exponential Model – Age Using Half-Life (http://openstaxcollege.org/l/expmodelhalf) • Newton’s Law of Cooling (http://openstaxcollege.org/l/newtoncooling) • Exponential Growth Given Doubling Time (http://openstaxcollege.org/l/expgrowthdbl) • Exponential Growth – Find Initial Amount Given Doubling Time (http://openstaxcollege.org/l/initialdouble) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 769 6.7 EXERCISES Verbal 372. With what kind of exponential model would half-life be associated? What role does half-life play in these models? What is carbon dating? Why does it work? Give an 373. example in which carbon dating would be useful. 374. With what kind of exponential model would doubling time be associated? What role does doubling time play in these models? Define Newton’s Law of Cooling. Then name at least 375. three real-world situations where Newton’s Law of Cooling would be applied. What is an order of magnitude? Why are orders of 376. magnitude useful? Give an example to explain. Numeric The temperature of an object in degrees Fahrenheit is equation represented by + 72. To the nearest degree, what is 377. after t minutes T(t) = 68e−0.0174t the temperature of the object after one and a half hours? the x f(x) –2 0.694 –1 0.833 0 1 2 3 4 5 1 1.2 1.44 1.728 2.074 2.488 For the following exercises, use the logistic growth model f (x) = 150 1 + 8e−2x. 384. Rewrite f (x) = 1.68(0.65) x as an exponential equation with base e to five significant digits. 378. Find and interpret f (0). Round to the nearest tenth. Technology 379. Find and interpret f (4). Round to the nearest tenth. 380. Find the carrying capacity. 381. Graph the model. For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that linear, exponential, or logarithmic. is 382. Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data. 385. 383. 770 Chapter 6 Exponential and Logarithmic Functions x f(x) x f(x.079 5.296 6.159 6.828 7.375 7.838 8.238 8.592 .4 2.88 3.456 4.147 4.977 5.972 7.166 8.6 10.32 10 8.908 10 12.383 386. 387. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 771 x f(x) 4 5 6 7 8 9 9.429 9.972 10.415 10.79 11.115 11.401 10 11.657 11 11.889 12 12.101 13 12.295 388. x f(x) 1.25 5.75 2.25 8.75 3.56 12.68 4.2 14.6 5.65 18.95 6.75 22.25 7.25 23.75 8.6 27.8 9.25 29.75 10.5 33.5 For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t years is modeled by the equation P(t) = 1000 1 + 9e−0.6t. 389. Graph the function. 390. What is the initial population of fish? To the nearest tenth, what is the doubling time for the 391. fish population? To the nearest whole number, what will the fish 392. population be after 2 years? To the nearest tenth, how long will it take for the 393. population to reach 900 ? What is the carrying capacity for the fish population? 394. Justify your answer using the graph of P. Extensions 395. A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many 772 Chapter 6 Exponential and Logarithmic Functions half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay? 396. The formula for an increasing population is given by P(t) = P0 ert where P0 is the initial population and r > 0. Derive a general formula for the time t it takes for the population to increase by a factor of M. 397. Recall the formula for calculating the magnitude of an The half-life of Radium-226 is 1590 years. What is the to four annual decay rate? Express the decimal result significant digits and the percentage to two significant digits. The half-life of Erbium-165 is 10.4 hours. What is 407. the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. earthquake, M = 2 3 S S0 this equation algebraically for the seismic moment S. ⎞ ⎠. Show each step for solving ⎛ log ⎝ 398. y = What is the y-intercept of the logistic growth model 1 + ae−rx ? Show the steps for calculation. What c does this point tell us about the population? 399. Prove that b x = e xln(b) for positive b ≠ 1. Real-World Applications A wooden artifact from an archeological dig contains 408. 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.) A research student is working with a culture of 409. bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours? For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. 400. To the nearest hour, what is the half-life of the drug? Write an exponential model representing the amount 401. of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram. Using the model found in the previous exercise, find the result. Round to the nearest 402. f (10) and interpret hundredth. For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the 403. Iodine-125 to decay? Write an exponential model representing the amount 404. of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram. A scientist begins with 250 grams of a radioactive 405. substance. After 250 minutes, the sample has decayed to 32 grams. R
ounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance? 406. This content is available for free at https://cnx.org/content/col11758/1.5 To the nearest whole number, what was the initial 410. population in the culture? an Rounding to six significant digits, write 411. exponential equation representing this situation. To the nearest minute, how long did it take the population to double? For the following exercises, use this scenario: A pot of temperature of 100° boiling soup with an internal Fahrenheit was taken off the stove to cool in a 69° F room. After fifteen minutes, the internal temperature of the soup was 95° F. Use Newton’s Law of Cooling to write a formula that 412. models this situation. To the nearest minute, how long will it take the soup 413. to cool to 80° F? To the nearest degree, what will the temperature be 414. after 2 and a half hours? For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165°F and is allowed to cool in a 75°F room. After half an hour, the internal temperature of the turkey is 145°F. 415. Write a formula that models this situation. To the nearest degree, what will the temperature be 416. after 50 minutes? Chapter 6 Exponential and Logarithmic Functions 773 To the nearest minute, how long will it take the turkey 417. to cool to 110° F? D. f (t) = 4.75 1 + 13e−0.83925t For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth. 418. 419. 420. sounds on a logarithmic scale: Whisper: 10−10 Plot each set of approximate values of intensity of W m2, Vacuum: 10−4 W m2, Jet: 102 W m2 421. S S0 ⎛ log ⎝ earthquake, M = 2 3 Recall the formula for calculating the magnitude of an ⎞ ⎠. One magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth. earthquake has For the following exercises, use this scenario: The equation N(t) = models the number of people in a 500 1 + 49e−0.7t town who have heard a rumor after t days. 422. How many people started the rumor? To the nearest whole number, how many people will 423. have heard the rumor after 3 days? 424. As t increases without bound, what value does N(t) approach? Interpret your answer. For the following exercise, choose the correct answer choice. A doctor and injects a patient with 13 milligrams of 425. radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation? A. B. C. f (t) = 13(0.0805) t f (t) = 13e0.9195t f (t) = 13e( − 0.0839t) 774 Chapter 6 Exponential and Logarithmic Functions 6.8 \| Fitting Exponential Models to Data Learning Objectives In this section, you will: 6.8.1 Build an exponential model from data. 6.8.2 Build a logarithmic model from data. 6.8.3 Build a logistic model from data. In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from realworld observations. With regression analysis, we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events. Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation. We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model realworld phenomena. Building an Exponential Model from Data As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay. Recall that exponential functions have the form y = ab x or y = A0 ekx form most commonly used on graphing utilities, y = ab x learned about the exponential function y = ab x (assume a > 0) : . When performing regression analysis, we use the . Take a moment to reflect on the characteristics we’ve already • b must be greater than zero and not equal to one. • The initial value of the model is y = a. ◦ ◦ If b > 1, slowly at first, but then increase more and more rapidly, without bound. the function models exponential growth. As x increases, the outputs of the model increase If 0 < b < 1, the function models exponential decay. As x increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero. As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable r, or r 2. (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of r 2 instead of r, but the closer either value is to 1, the better the regression equation approximates the data. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 775 Exponential Regression Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, y = ab x Note that: • b must be non-negative. • when b > 1, we have an exponential growth model. • when 0 < b < 1, we have an exponential decay model. Given a set of data, perform exponential regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow an exponential pattern. 3. Find the equation that models the data. a. Select “ExpReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = ab x . 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 6.73 Using Exponential Regression to Fit a Model to Data In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table 6.19 shows results from the study [9]. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol. 9. Source: Indiana University Center for Studies of Law in Action, 2007 776 Chapter 6 Exponential and Logarithmic Functions BAC Relative Risk of Crashing 0 1 0.01 0.03 0.05 0.07 0.09 1.03 1.06 1.38 2.09 3.54 BAC 0.11 0.13 0.15 0.17 0.19 0.21 Relative Risk of Crashing 6.41 12.6 22.1 39.05 65.32 99.78 Table 6.19 a. Let x represent the BAC level, and let y represent the corresponding relative risk. Use exponential regression to fit a model to these data. b. After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth. Solution a. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 6.57: Figure 6.57 Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model, Converting from scientific notation, we have: y = 0.58304829(2.20720213E10) y = 0.58304829(22,072,021,300) x x This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 777 Notice that r 2 ≈ 0.97 which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 6.58: Figure 6.58 b. Use the model to estimate the risk associated with a BAC of 0.16. Su
bstitute 0.16 for x in the model and solve for y. y = 0.58304829(22,072,021,300) x Use the regression model found in part (a). = 0.58304829(22,072,021,300)0.16 Substitute 0.16 for x. ≈ 26.35 Round to the nearest hundredth. If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober. 6.72 Table 6.20 shows a recent graduate’s credit card balance each month after graduation. Month 1 2 3 4 5 6 7 8 620.00 761.88 899.80 1039.93 1270.63 1589.04 1851.31 2154.92 Debt ($) Table 6.20 a. Use exponential regression to fit a model to these data. b. If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating? 778 Chapter 6 Exponential and Logarithmic Functions Is it reasonable to assume that an exponential regression model will represent a situation indefinitely? No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation). Building a Logarithmic Model from Data Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best. Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, y = a + bln(x). For this function • All input values, x, must be greater than zero. • The point (1, a) is on the graph of the model. • • If b > 0, If b < 0, the model is increasing. Growth increases rapidly at first and then steadily slows over time. the model is decreasing. Decay occurs rapidly at first and then steadily slows over time. Logarithmic Regression Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form, y = a + bln(x) Note that • all input values, x, must be non-negative. • when b > 0, the model is increasing. • when b < 0, the model is decreasing. Given a set of data, perform logarithmic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logarithmic pattern. 3. Find the equation that models the data. a. Select “LnReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = a + bln(x). 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 779 Example 6.74 Using Logarithmic Regression to Fit a Model to Data Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century. Table 6.21 shows the average life expectancies, in years, of Americans from 1900–2010[10]. Year 1900 1910 1920 1930 1940 1950 Life Expectancy(Years) 47.3 50.0 54.1 59.7 62.9 68.2 Year 1960 1970 1980 1990 2000 2010 Life Expectancy(Years) 69.7 70.8 73.7 75.4 76.8 78.7 Table 6.21 a. Let x represent time in decades starting with x = 1 for the year 1900, x = 2 for the year 1910, and so on. Let y represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data. b. Use the model to predict the average American life expectancy for the year 2030. Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern as shown in Figure 6.59: Figure 6.59 10. Source: Center for Disease Control and Prevention, 2013 780 Chapter 6 Exponential and Logarithmic Functions Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model, y = 42.52722583 + 13.85752327ln(x) Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 6.60: Figure 6.60 b. To predict the life expectancy of an American in the year 2030, substitute x = 14 for the in the model and solve for y : y = 42.52722583 + 13.85752327ln(x) Use the regression model found in part (a). = 42.52722583 + 13.85752327ln(14) Substitute 14 for x. ≈ 79.1 Round to the nearest tenth. If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 781 6.73 Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 6.22 shows the number of games sold, in thousands, from the years 2000–2010. Year 2000 2001 2002 2003 2004 2005 Number Sold (thousands) 142 149 154 155 159 161 Year 2006 2007 2008 2009 2010 Number Sold (thousands) 163 164 164 166 167 - - Table 6.22 a. Let x represent time in years starting with x = 1 for the year 2000. Let y represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data. b. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand. Building a Logistic Model from Data Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients. It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities: y = c 1 + ae−bx Recall that: c 1 + a • is the initial value of the model. • when b > 0, the model increases rapidly at first until it reaches its point of maximum growth rate, ⎛ ⎝ ln(a) b , ⎞ ⎠. At c 2 that point, growth steadily slows and the function becomes asymptotic to the upper bound y = c. • c is the limiting value, sometimes called the carrying capacity, of the model. Logistic Regression Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form y = c 1 + ae−bx Note that • The initial value of the model is c 1 + a. 782 Chapter 6 Exponential and Logarithmic Functions • Output values for the model grow closer and closer to y = c as time increases. Given a set of data, perform logistic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logistic pattern. 3. Find the equation that models the data. a. Select “Logistic” from the STAT then CALC menu. b. Use the values returned for a, b, and c to record the model, y = c 1 + ae−bx. 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 6.75 Using Logistic Regression to Fit a Model to Data Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 6.23 shows the percentage of Americans with cellular service between the years 1995 and 2012 [11]. 11. Source: The World Bank, 2013 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 783 Year Americans with Cellular Service (%) Year Americans with Cellular Service (%) 1995 12.69 1996 16.35 1997 20.29 1998 25.08 1999 30.81 2000 38.75 2001 45.00 2002 49.16 2003 55.15 Table 6.23 2004 62.852 2005 68.63 2006 76.64 2007 82.47 2008 85.68 2009 89.14 2010 91.86 2011 95.28 2012 98.17 a. Let x represent time in years starting with x = 0 for the year 1995. Let y represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data. b. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent. c. Discuss the value returned for the upper limit, c. What does thi
s tell you about the model? What would the limiting value be if the model were exact? Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 6.61: 784 Chapter 6 Exponential and Logarithmic Functions Figure 6.61 Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model, y = 105.7379526 1 + 6.88328979e−0.2595440013x Next, graph the model in the same window as shown in Figure 6.62 the scatterplot to verify it is a good fit: Figure 6.62 b. To approximate the percentage of Americans with cellular service in the year 2013, substitute x = 18 for the in the model and solve for y : This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 785 y = = 105.7379526 1 + 6.88328979e−0.2595440013x 105.7379526 1 + 6.88328979e−0.2595440013(18) Use the regression model found in part (a). Substitute 18 for x. ≈ 99.3 Round to the nearest tenth According to the model, about 98.8% of Americans had cellular service in 2013. c. The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be c = 100 and the model’s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service! Table 6.24 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 6.74 2012. Year Seal Population (Thousands) Year Seal Population (Thousands) 1997 3.493 2005 19.590 1998 5.282 2006 21.955 1999 6.357 2007 22.862 2000 9.201 2008 23.869 2001 11.224 2009 24.243 2002 12.964 2010 24.344 2003 16.226 2011 24.919 2004 18.137 2012 25.108 Table 6.24 a. Let x represent time in years starting with x = 0 for the year 1997. Let y represent the number of seals in thousands. Use logistic regression to fit a model to these data. b. Use the model to predict the seal population for the year 2020. c. To the nearest whole number, what is the limiting value of this model? Access this online resource for additional instruction and practice with exponential function models. • Exponential Regression on a Calculator (http://openstaxcollege.org/l/pregresscalc) 786 Chapter 6 Exponential and Logarithmic Functions this website (http://openstaxcollege.org/l/PreCalcLPC04) Visit Learningpod. for additional practice questions from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 787 6.8 EXERCISES Verbal What situations are best modeled by a logistic 426. equation? Give an example, and state a case for why the example is a good fit. What is a carrying capacity? What kind of model has 427. a carrying capacity built into its formula? Why does this make sense? What is regression analysis? Describe the process of 428. performing regression analysis on a graphing utility. What might a scatterplot of data points look like if it 429. were best described by a logarithmic model? What does the y-intercept on the graph of a logistic 430. equation correspond to for a population modeled by that equation? Graphical For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 6.63 through Figure 6.67. Answer using the letter beneath the matching graph. Figure 6.64 Figure 6.63 Figure 6.65 788 Chapter 6 Exponential and Logarithmic Functions x as Rewrite the exponential model A(t) = 1550(1.085) an equivalent model with base e. Express the exponent to four significant digits. A logarithmic model is given by the equation 438. h(p) = 67.682 − 5.792ln(p). To the nearest hundredth, for what value of p does h(p) = 62 ? 439. P(t) = A logistic model 90 1 + 5e−0.42t. To the nearest hundredth, for what is given by the equation Figure 6.66 Figure 6.67 431. y = 10.209e−0.294x 432. y = 5.598 − 1.912ln(x) 433. y = 2.104(1.479) x 434. y = 4.607 + 2.733ln(x) 435. y = 14.005 1 + 2.79e−0.812x Numeric To the nearest whole number, what is the initial value 436. of a population modeled by the logistic equation P(t) = 1 + 6.995e−0.68t ? What is the carrying capacity? 175 437. This content is available for free at https://cnx.org/content/col11758/1.5 value of t does P(t) = 45 ? What is the y-intercept on the graph of the logistic 440. model given in the previous exercise? Technology the following exercises, use this scenario: The For population P of a koi pond over x months is modeled by 68 1 + 16e−0.28x. the function P(x) = Graph the population model to show the population 441. over a span of 3 years. 442. What was the initial population of koi? How many koi will the pond have after one and a half 443. years? How many months will it take before there are 20 koi 444. in the pond? Use the intersect feature to approximate the number 445. of months it will take before the population of the pond reaches half its carrying capacity. For the following exercises, use this scenario: The population P of an endangered species habitat for wolves 558 1 + 54.8e−0.462x, is modeled by the function P(x) = where x is given in years. Graph the population model to show the population 446. over a span of 10 years. What was the initial population of wolves transported 447. to the habitat? How many wolves will 448. years? the habitat have after 3 How many years will it take before there are 100 449. wolves in the habitat? 450. Chapter 6 Exponential and Logarithmic Functions 789 Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity. For the following exercises, refer to Table 6.25. x 1 2 3 4 5 6 f(x) 1125 1495 2310 3294 4650 6361 Table 6.25 Use a graphing calculator to create a scatter diagram 451. of the data. Use the regression feature to find an exponential 452. function that best fits the data in the table. Write the exponential function as an exponential 453. equation with base e. x 1 2 3 4 5 6 f(x) 555 383 307 210 158 122 Table 6.26 Use a graphing calculator to create a scatter diagram 456. of the data. Use the regression feature to find an exponential 457. function that best fits the data in the table. Write the exponential function as an exponential 458. equation with base e. Graph the exponential equation on the scatter 459. diagram. Use the intersect feature to find the value of x for 460. which f (x) = 250. Graph the exponential equation on the scatter 454. diagram. For the following exercises, refer to Table 6.27. Use the intersect feature to find the value of x for 455. which f (x) = 4000. For the following exercises, refer to Table 6.26. 790 Chapter 6 Exponential and Logarithmic Functions x 1 2 3 4 5 6 f(x) 5.1 6.3 7.3 7.7 8.1 8.6 Table 6.27 Use a graphing calculator to create a scatter diagram 461. of the data. Use the LOGarithm option of the REGression feature 462. to find a logarithmic function of the form y = a + bln(x) that best fits the data in the table. Use the logarithmic function to find the value of the 463. function when x = 10. Graph the logarithmic equation on the scatter 464. diagram(x) 7.5 6 5.2 4.3 3.9 3.4 3.1 2.9 Table 6.28 Use a graphing calculator to create a scatter diagram 466. of the data. 467. Use the LOGarithm option of the REGression feature to find a logarithmic function of the form y = a + bln(x) that best fits the data in the table. Use the intersect feature to find the value of x for 465. which f (x) = 7. Use the logarithmic function to find the value of the 468. function when x = 10. For the following exercises, refer to Table 6.28. Graph the logarithmic equation on the scatter 469. diagram. Use the intersect feature to find the value of x for 470. which f (x) = 8. For the following exercises, refer to Table 6.29. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 791 (x) 8.7 12.3 15.4 18.5 20.7 22.5 23.3 24 24.6 10 24.8 Table 6.29 x f (x) 0 2 4 5 7 8 12 28.6 52.8 70.3 99.9 112.5 10 125.8 11 127.9 15 135.1 17 135.9 Table 6.30 Use a graphing calculator to create a scatter diagram 471. of the data. Use a graphing calculator to create a scatter diagram 476. of the data. 472. logistic growth model of the form y = Use the LOGISTIC regression option to find a that best c 1 + ae−bx 477. logistic growth model of the form y = Use the LOGISTIC regression option to find a that best c 1 + ae−bx fits the data in the table. fits the data in the table. 473. Graph the logistic equation on the scatter diagram. 478. Graph the logistic equation on the scatter diagram. To the nearest whole number, what is the predicted 474. carrying capacity of the model? To the nearest whole number, what is the predicted 479. carrying capacity of the model? Use the intersect feature to find the value of x for 475. which the model reaches half its carrying capacity. Use the intersect feature to find the value of x for 480. which the model reaches half its carrying capacity. For the following exercises, refer to Table 6.30. Extensions 481. a population is given by P(t) = Recall that the general form of a logistic equation for such that the c 1 + ae−bt, initial population at time t = 0 is P(0) = P0. Show algebraically that c − P(t) P(t) = c − P0 P0 e−bt . 792 Chapter 6 Exponential and Logarithmic Functions 482. Use a graphing utility to find an exponential regression formula f (x) and a logarithmic regression formula g(x) for the points (1.5, 1.5) and (8.5, 8.5). Round all numbers to 6 decimal places. Graph the points and both formulas along with the line y = x on the same the relationship of the axis. Make a conjecture about regression formulas. Verify the conjecture
made in the previous exercise. 483. Round all numbers to six decimal places when necessary. 484. Find the inverse function f −1 (x) for the logistic 1 + ae−bx. Show all steps. c function f (x) = Use the result from the previous exercise to graph the 485. logistic model P(t) = 20 1 + 4e−0.5t on the same axis. What are the intercepts and asymptotes of each function? along with its inverse This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 793 CHAPTER 6 REVIEW KEY TERMS annual percentage rate (APR) the yearly interest rate earned by an investment account, also called nominal rate carrying capacity in a logistic model, the limiting value of the output change-of-base formula a formula for converting a logarithm with any base to a quotient of logarithms with any other base. common logarithm the exponent to which 10 must be raised to get x; log10 (x) is written simply as log(x). compound interest interest earned on the total balance, not just the principal doubling time the time it takes for a quantity to double exponential growth a model that grows by a rate proportional to the amount present extraneous solution a solution introduced while solving an equation that does not satisfy the conditions of the original equation half-life the length of time it takes for a substance to exponentially decay to half of its original quantity logarithm the exponent to which b must be raised to get x; written y = logb (x) logistic growth model a function of the form f (x) = c 1 + ae−bx where c 1 + a is the initial value, c is the carrying capacity, or limiting value, and b is a constant determined by the rate of growth natural logarithm the exponent to which the number e must be raised to get x; loge (x) is written as ln(x). Newton’s Law of Cooling the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature nominal rate the yearly interest rate earned by an investment account, also called annual percentage rate order of magnitude the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal power rule for logarithms a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base product rule for logarithms a rule of logarithms that states that the log of a product is equal to a sum of logarithms quotient rule for logarithms a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms KEY EQUATIONS 794 Chapter 6 Exponential and Logarithmic Functions definition of the exponential function f (x) = b x , where b > 0, b ≠ 1 definition of exponential growth f (x) = ab x , where a > 0, b > 0, b ≠ 1 compound interest formula nt ⎝1 + r n , where ⎞ A(t) = P⎛ ⎠ A(t) is the account value at time t t is the number of years P is the initial investment, often called the principal r is the annual percentage rate (APR), or nominal rate n is the number of compounding periods in one year continuous growth formula , where A(t) = aert t is the number of unit time periods of growth a is the starting amount (in the continuous compounding formula a is replaced with P, the principal) e is the mathematical constant, e ≈ 2.718282 General Form for the Translation of the Parent Function f (x) = b x f (x) = ab x + c + d Definition of the logarithmic function For x > 0, b > 0, b ≠ 1, y = logb (x) if and only if b y = x. Definition of the common logarithm For x > 0, y = log(x) if and only if 10 y = x. Definition of the natural logarithm For x > 0, y = ln(x) if and only if e y = x. General Form for the Translation of the Parent Logarithmic Function f (x) = logb (x) f (x) = alogb (x + c) + d The Product Rule for Logarithms logb(MN) = logb (M) + logb (N) The Quotient Rule for Logarithms logb ⎛ ⎝ M N ⎞ ⎠ = logb M − logb N The Power Rule for Logarithms logb (M n ) = nlogb M The Change-of-Base Formula logb M= logn M logn b n > 0, n ≠ 1, b ≠ 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 795 One-to-one property for exponential functions Definition of a logarithm For any algebraic expressions S and T and any positive real number b, where bS if and only if S = T. = bT For any algebraic expression S and positive real numbers b and c, where b ≠ 1, logb(S) = c if and only if bc = S. One-to-one property for logarithmic functions For any algebraic expressions S and T and any positive real number b, where b ≠ 1, logb S = logb T if and only if S = T. Half-life formula If A = A0 ekt , k < 0, the half-life is t = − ln(2) k . Carbon-14 dating ⎞ ⎠ ⎛ A ln ⎝ A0 −0.000121 . t = A0 A is the amount of carbon-14 when the plant or animal died t is the amount of carbon-14 remaining today is the age of the fossil in years Doubling time formula If A = A0 ekt , k > 0, the doubling time is t = ln2 k Newton’s Law of Cooling T(t) = Aekt the continuous rate of cooling. + Ts, where Ts is the ambient temperature, A = T(0) − Ts, and k is KEY CONCEPTS 6.1 Exponential Functions • An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent. See Example 6.1. • A function is evaluated by solving at a specific value. See Example 6.2 and Example 6.3. • An exponential model can be found when the growth rate and initial value are known. See Example 6.4. • An exponential model can be found when the two data points from the model are known. See Example 6.5. • An exponential model can be found using two data points from the graph of the model. See Example 6.6. • An exponential model can be found using two data points from the graph and a calculator. See Example 6.7. • The value of an account at any time t can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example 6.8. • The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example 6.9. 796 Chapter 6 Exponential and Logarithmic Functions • The number e is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is e ≈ 2.718282. • Scientific and graphing calculators have the key [e x ] or ⎡ ⎣exp(x)⎤ ⎦ for calculating powers of e. See Example 6.10. • Continuous growth or decay models are exponential models that use e as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example 6.11 and Example 6.12. 6.2 Graphs of Exponential Functions • The graph of the function f (x) = b x has a y-intercept at (0, 1), domain (−∞, ∞), range (0, ∞), and horizontal asymptote y = 0. See Example 6.13. • • If b > 1, will increase without bound. the function is increasing. The left tail of the graph will approach the asymptote y = 0, and the right tail If 0 < b < 1, will approach the asymptote y = 0. the function is decreasing. The left tail of the graph will increase without bound, and the right tail • The equation f (x) = b x + d represents a vertical shift of the parent function f (x) = b x . • The equation f (x) = b x + c represents a horizontal shift of the parent function f (x) = b x . See Example 6.14. • Approximate solutions of the equation f (x) = b x + c + d can be found using a graphing calculator. See Example 6.15. • The equation f (x) = ab x , where a > 0, represents a vertical stretch if \|a\| > 1 or compression if 0 < \|a\| < 1 of the parent function f (x) = b x . See Example 6.16. • When the parent function f (x) = b x is multiplied by − 1, the result, f (x) = − b x , is a reflection about the x- axis. When the input is multiplied by − 1, 6.17. the result, f (x) = b−x , is a reflection about the y-axis. See Example • All translations of the exponential function can be summarized by the general equation f (x) = ab x + c + d. See Table 6.9. • Using the general equation f (x) = ab x + c + d, we can write the equation of a function given its description. See Example 6.18. 6.3 Logarithmic Functions • The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function. • Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm. See Example 6.19. • Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm See Example 6.20. • Logarithmic functions with base b can be evaluated mentally using previous knowledge of powers of b. See Example 6.21 and Example 6.22. • Common logarithms can be evaluated mentally using previous knowledge of powers of 10. See Example 6.23. • When common logarithms cannot be evaluated mentally, a calculator can be used. See Example 6.24. • Real-world exponential problems with base 10 can be rewritten as a common logarithm and then evaluated using a calculator. See Example 6.25. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 797 • Natural logarithms can be evaluated using a calculator Example 6.26. 6.4 Graphs of Logarithmic Functions • To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for x. See Example 6.27 and Example 6.28 • The graph of the parent function f (x) = logb (x) has an x-intercept at (1, 0), domain (0, ∞), range (−∞, ∞), vertical asymptote x = 0, and ◦ ◦ if b > 1, the function is increasing. if 0 < b < 1, the function is decreasing. See Example 6.29. • The equation f (x) = logb (x + c) shifts the parent function y = logb (x) horizontally ◦ ◦ left c units if c > 0. right c units if c < 0. See Example 6.30. • The equation f (x) = logb (x) + d shifts the p
arent function y = logb (x) vertically ◦ up d units if d > 0. ◦ down d units if d < 0. See Example 6.31. • For any constant a > 0, the equation f (x) = alogb (x) ◦ stretches the parent function y = logb (x) vertically by a factor of a if \|a\| > 1. ◦ compresses the parent function y = logb (x) vertically by a factor of a if \|a\| < 1. See Example 6.32 and Example 6.33. • When the parent function y = logb (x) is multiplied by − 1, the result is a reflection about the x-axis. When the input is multiplied by − 1, the result is a reflection about the y-axis. ◦ The equation f (x) = − logb (x) represents a reflection of the parent function about the x-axis. ◦ The equation f (x) = logb (−x) represents a reflection of the parent function about the y-axis. See Example 6.34. ◦ A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 6.35. • All translations of the logarithmic function can be summarized by the general equation f (x) = alogb (x + c) + d. See Table 6.15. • Given an equation with the general form f (x) = alogb (x + c) + d, we can identify the vertical asymptote x = − c for the transformation. See Example 6.36. • Using the general equation f (x) = alogb (x + c) + d, we can write the equation of a logarithmic function given its graph. See Example 6.37. 798 Chapter 6 Exponential and Logarithmic Functions 6.5 Logarithmic Properties • We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example 6.38. • We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example 6.39. • We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example 6.40, Example 6.41, and Example 6.42. • We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example 6.43, Example 6.44, and Example 6.45. • The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example 6.46, Example 6.47, Example 6.48, and Example 6.49. • We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See Example 6.50. • The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and e as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example 6.51. 6.6 Exponential and Logarithmic Equations • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown. • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example 6.52. • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example 6.53, Example 6.54, and Example 6.55. • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example 6.56. • We can solve exponential equations with base e, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example 6.57 and Example 6.58. • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example 6.59. • When given an equation of the form logb(S) = c, where S is an algebraic expression, we can use the definition of = S, and solve for the unknown. See a logarithm to rewrite the equation as the equivalent exponential equation bc Example 6.60 and Example 6.61. • We can also use graphing to solve equations with the form logb(S) = c. We graph both equations y = logb(S) and y = c on the same coordinate plane and identify the solution as the x-value of the intersecting point. See Example 6.62. • When given an equation of the form logb S = logb T, where S and T are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation S = T for the unknown. See Example 6.63. • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example 6.64. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 799 6.7 Exponential and Logarithmic Models • The basic exponential function is f (x) = ab x . If b > 1, we have exponential growth; if 0 < b < 1, we have exponential decay. • We can also write this formula in terms of continuous growth as A = A0 ekx , where A0 is the starting value. If A0 is positive, then we have exponential growth when k > 0 and exponential decay when k < 0. See Example 6.65. • In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See Example 6.66. • We can find the age, t, of an organic artifact by measuring the amount, k, of carbon-14 remaining in the artifact and using the formula t = ln(k) −0.000121 to solve for t. See Example 6.67. • Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See Example 6.68. • We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See Example 6.69. • We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See Example 6.70. • We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See Example 6.71. • Any exponential function with the form y = ab x can be rewritten as an equivalent exponential function with the form y = A0 ekx where k = lnb. See Example 6.72. 6.8 Fitting Exponential Models to Data • Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. • We use the command “ExpReg” on a graphing utility to fit function of the form y = ab x to a set of data points. See Example 6.73. • Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. • We use the command “LnReg” on a graphing utility to fit a function of the form y = a + bln(x) to a set of data points. See Example 6.74. • Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit. • We use the command “Logistic” on a graphing utility to fit a function of the form y = c 1 + ae−bx to a set of data points. See Example 6.75. CHAPTER 6 REVIEW EXERCISES Exponential Functions t 486. Determine whether the function y = 156(0.825) represents exponential growth, exponential decay, or neither. Explain 487. The population of a herd of deer is represented by the , where t is given in years. To function A(t) = 205(1.13) the nearest whole number, what will the herd population be after 6 years? t 800 Chapter 6 Exponential and Logarithmic Functions 488. Find an exponential equation that passes through the points (2, 2.25) and (5, 60.75). 489. Determine whether Table 6.31 could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. 2 3 4 0.9 0.27 0.081 x f(x) 1 3 Table 6.31 490. A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? 491. Hsu-Mei wants to save $5,000 for a down payment on a car. To the nearest dollar, how much will she need to invest in an account now with 7.5% APR, compounded daily, in order to reach her goal in 3 years? 492. Does equation y = 2.294e−0.654t represent continuous growth, continuous decay, or neither? Explain. the Logarithmic Functions 498. Rewrite log17 (4913) = x as exponential equation. an equivalent 499. Rewrite ln(s) = t as an equivalent exponential equation. 500. Rewrite a equation. − 2 5 = b as an equivalent logarithmic deposit 493. Suppose an investment account is opened with an of $10,500 earning 6.25% interest, initial compounded continuously. How much will the account be worth after 25 years? 501. Rewrite e−3.5 = h as an equivalent logarithmic equation. 502. Solve for x log64(x) = 1 3 to exponential form. Graphs of Exponential Functions the 494. Graph function f (x) = 3.5(2) domain and range and give the y-intercept. x . State the 503. Evaluate log5 ⎛ ⎝ 1 125 ⎞ ⎠ without using a calculator. 495. Graph the function f (x and its reflection about the y-axis on the same axes, and give the y-intercept. x is reflected about the y496. The graph of f (x) = 6.5 axis and stretched vertically by a factor of 7. What is the equation of the new function, g(x) ? State its y-intercept, domain, and range. 497. The graph below shows transformations of the graph . What is the equation for the transformation? of f (x) = 2 x 504. Evaluate log(0.000001) without using a calculator. 505. Evaluate log(4.005) using a calculator. Round to the nearest thousandth. ⎝e−0.
8648⎞ ⎛ 506. Evaluate ln ⎠ without using a calculator. ⎛ 3 507. Evaluate ln ⎝ 18 ⎞ using a calculator. Round to the ⎠ nearest thousandth. Graphs of Logarithmic Functions 508. Graph the function g(x) = log(7x + 21) − 4. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 801 509. Graph the function h(x) = 2ln(9 − 3x) + 1. 524. Solve State the domain, vertical asymptote, and end 510. behavior of the function g(x) = ln(4x + 20) − 17. Logarithmic Properties 511. Rewrite ln(7r ⋅ 11st) in expanded form. 512. Rewrite log8 (x) + log8 (5) + log8 (y) + log8 (13) in compact form. 513. Rewrite logm ⎛ ⎝ 67 83 ⎞ ⎠ in expanded form. 514. Rewrite ln(z) − ln(x) − ln(y) in compact form. ⎛ 515. Rewrite ln ⎝ ⎞ as a product. ⎠ 1 x5 516. Rewrite − log y ⎛ ⎝ 1 12 ⎞ ⎠ as a single logarithm. ⎛ r 2 s11 517. Use properties of logarithms to expand log ⎝ t 14 ⎞ ⎠. 518. Use ⎛ ⎝2b b + 1 ln b − 1 properties of logarithms to expand ⎞ ⎠. 519. Condense the expression 5ln(b) + ln(c) + ln(4 − a) 2 to a single logarithm. 520. Condense the expression 3log7 v + 6log7 w − log7 u 3 to a single logarithm. 521. Rewrite log3 (12.75) to base e. 522. Rewrite 512x − 17 = 125 as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth. Exponential and Logarithmic Equations 523. Solve 2163x ⋅ 216 side with a common base. x = 363x + 2 by rewriting each 125 −x − 3 = 53 by rewriting each side with ⎞ ⎠ 1 625 a common base. ⎛ ⎝ 525. Use logarithms to find the exact solution for 7 ⋅ 17−9x − 7 = 49. If there is no solution, write no solution. 526. Use logarithms to find the exact solution for 3e6n − 2 + 1 = − 60. If there is no solution, write no solution. 527. Find the exact solution for 5e3x no solution, write no solution. − 4 = 6 . If there is 528. Find the exact solution for 2e5x − 2 − 9 = − 56. If there is no solution, write no solution. 529. Find the exact solution for 52x − 3 = 7 is no solution, write no solution. x + 1. If there 530. Find the exact solution for e2x there is no solution, write no solution. − e x − 110 = 0. If 531. Use the definition of a logarithm to solve. − 5log7 (10n) = 5. 532. 47. Use the definition of a logarithm to find the exact solution for 9 + 6ln(a + 3) = 33. 533. Use the one-to-one property of logarithms to find an exact solution for log8 (7) + log8 (−4x) = log8 (5). If there is no solution, write no solution. 534. Use the one-to-one property of logarithms to find an ⎞ ⎛ ⎝5x2 − 5 ⎠ = ln(56). If there is exact solution for ln(5) + ln no solution, write no solution. 535. The formula for measuring sound intensity in ⎛ decibels D is defined by the equation D = 10log ⎝ ⎞ ⎠, where I is the intensity of the sound in watts per square I I0 meter and I0 = 10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a large orchestra with a sound intensity of 6.3 ⋅ 10−3 watts per square meter? 802 Chapter 6 Exponential and Logarithmic Functions 536. The population of a city is modeled by the equation P(t) = 256, 114e0.25t where t is measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million? 546. 537. Find the inverse function f −1 for the exponential function f (x. 538. Find the inverse function f −1 for the logarithmic function f (x) = 0.25 ⋅ log2 ⎞ ⎛ ⎝x3 + 1 ⎠. Exponential and Logarithmic Models For the following exercises, use this scenario: A doctor prescribes 300 milligrams of a therapeutic drug that decays by about 17% each hour. 539. To the nearest minute, what is the half-life of the drug? 540. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 24 hours. Round to the nearest hundredth of a gram. For the following exercises, use this scenario: A soup with an internal temperature of 350° Fahrenheit was taken off the stove to cool in a 71°F room. After fifteen minutes, the internal temperature of the soup was 175°F. 541. Use Newton’s Law of Cooling to write a formula that models this situation. 547. 542. How many minutes will it take the soup to cool to 85°F(x) 3.05 4.42 6.4 9.28 13.46 19.52 28.3 41.04 59.5 10 86.28 For the following exercises, use this scenario: The equation N(t) = models the number of people in 1200 1 + 199e−0.625t a school who have heard a rumor after t days. 543. How many people started the rumor? 544. To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity? 545. What is the carrying capacity? For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 803 x f(x) 0.5 18.05 model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. 551. 1 3 5 7 10 12 13 15 17 20 17 15.33 14.55 14.04 13.5 13.22 13.1 12.88 12.69 12.45 (x) 409.4 260.7 170.4 110.6 74 44.7 32.4 19.5 12.7 10 8.1 548. Find a formula for an exponential equation that goes through the points (−2, 100) and (0, 4). Then express the formula as an equivalent equation with base e. 552. by 250, 000 Fitting Exponential Models to Data 549. What modeled is the carrying capacity for a population equation logistic the P(t) = 1 + 499e−0.45t ? What is the initial population for the model? 550. The population of a culture of bacteria is modeled by the logistic equation P(t) = 14, 250 1 + 29e−0.62t, where t is in days. To the nearest tenth, how many days will it take the culture to reach 75% of its carrying capacity? For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic 804 Chapter 6 Exponential and Logarithmic Functions x f(x) 0.15 36.21 0.25 28.88 0.5 24.39 0.75 18.28 1 16.5 1.5 12.99 2 9.91 2.25 8.57 2.75 7.23 3 5.99 3.5 4.81 x 0 2 4 5 7 8 f(x) 9 22.6 44.2 62.1 96.9 113.4 10 133.4 11 137.6 15 148.4 17 149.3 553. CHAPTER 6 PRACTICE TEST 554. The population of a pod of bottlenose dolphins is modeled by the function A(t) = 8(1.17) , where t is given in years. To the nearest whole number, what will the pod population be after 3 years? t 558. Graph the function f (x) = 5(0.5)−x and its reflection across the y-axis on the same axes, and give the y-intercept. 555. Find an exponential equation that passes through the points (0, 4) and (2, 9). f (x) = x ⎞ ⎠ ⎛ ⎝ 1 2 . What is the equation for the transformation? 559. The graph shows transformations of the graph of 556. Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25% APR, compounding daily, in order to reach his goal in 4 years? 557. An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 805 571. 4ln(c) + ln(d) + Condense ln(a) 3 + the expression ln(b + 3) 3 to a single logarithm. 572. Rewrite 163x − 5 = 1000 as a logarithm. Then apply the change of base formula to solve for x using the natural log. Round to the nearest thousandth. x 573. Solve ⎛ ⎝ ⋅ 1 243 side with a common base. 1 81 ⎞ ⎠ −3x − 1 = ⎞ ⎠ ⎛ ⎝ 1 9 by rewriting each 574. Use logarithms to find the exact solution for − 9e10a − 8 − 5 = − 41 . If there is no solution, write no solution. 560. Rewrite log8.5 (614.125) = a as an equivalent exponential equation. 575. Find the exact solution for 10e4x + 2 + 5 = 56. If there is no solution, write no solution. Rewrite e 561. equation. 1 2 = m as an equivalent logarithmic 576. Find the exact solution for − 5e−4x − 1 − 4 = 64. If there is no solution, write no solution. 562. Solve for x by converting the logarithmic equation log 1 7 (x) = 2 to exponential form. 563. Evaluate log(10,000,000) without using a calculator. 564. Evaluate ln(0.716) using a calculator. Round to the nearest thousandth. 565. Graph the function g(x) = log(12 − 6x) + 3. State the domain, vertical asymptote, and end 566. behavior of the function f (x) = log5 (39 − 13x) + 7. 577. Find the exact solution for 2 is no solution, write no solution. x − 3 = 62x − 1. If there 578. Find the exact solution for e2x there is no solution, write no solution. − e x − 72 = 0. If 579. Use the definition of a logarithm to find the exact solution for 4log(2n) − 7 = − 11 580. Use the one-to-one property of logarithms to find ⎛ ⎞ ⎝4x2 − 10 an exact solution for log ⎠ + log(3) = log(51) If there is no solution, write no solution. 581. The formula for measuring sound intensity in 567. Rewrite log(17a ⋅ 2b) as a sum. 568. Rewrite logt (96) − logt (8) in compact form. 569. Rewrite log8 ⎛ ⎜a ⎝ 1 b ⎞ ⎟ as a product. ⎠ properties 570. Use ⎞ ⎛ 3 ⎝y3 z2 ⋅ x − 4 ln ⎠. of logarithm to expand I I0 ⎛ decibels D is defined by the equation D = 10log ⎝ ⎞ ⎠, where I is the intensity of the sound in watts per square meter and I0 = 10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a rock concert with a sound intensity of 4.7 ⋅ 10−1 watts per square meter? 582. A radiation safety officer is working with 112 grams of a radioactive substance. After 17 days, the sample has decayed to 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day
, what is the half-life of this substance? 806 Chapter 6 Exponential and Logarithmic Functions 587. The population of a lake of fish is modeled by the logistic equation P(t) = 16, 120 1 + 25e−0.75t, where t is time in years. To the nearest hundredth, how many years will it take the lake to reach 80% of its carrying capacity? For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. 588(x) 20 21.6 29.2 36.4 46.6 55.7 72.6 87.1 107.2 10 138.1 589. 583. Write the formula found in the previous exercise as an equivalent equation with base e. Express the exponent to five significant digits. 584. A bottle of soda with a temperature of 71° Fahrenheit was taken off a shelf and placed in a refrigerator with an internal temperature of 35° F. After ten minutes, the soda was 63° F. Use the internal Newton’s Law of Cooling to write a formula that models the this situation. To the nearest degree, what will temperature of the soda be after one hour? temperature of 585. The population of a wildlife habitat is modeled by 360 the equation P(t) = 1 + 6.2e−0.35t, where t is given in years. How many animals were originally transported to the habitat? How many years will it take before the habitat reaches half its capacity? 586. Enter the data from Table 6.32 into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic(x) 3 8.55 11.79 14.09 15.88 17.33 18.57 19.64 20.58 10 21.42 Table 6.32 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 807 x 3 4 5 6 7 8 9 f(x) 13.98 17.84 20.01 22.7 24.1 26.15 27.37 10 28.38 11 29.97 12 31.07 13 31.43 590. x 0 f(x) 2.2 0.5 2.9 1 3.9 1.5 4.8 2 3 4 5 6 7 8 6.4 9.3 12.3 15 16.2 17.3 17.9 808 Chapter 6 Exponential and Logarithmic Functions This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 809 7 \| THE UNIT CIRCLE: SINE AND COSINE FUNCTIONS Figure 7.1 The tide rises and falls at regular, predictable intervals. (credit: Andrea Schaffer, Flickr) Chapter Outline 7.1 Angles 7.2 Right Triangle Trigonometry 7.3 Unit Circle 7.4 The Other Trigonometric Functions Introduction Life is dense with phenomena that repeat in regular intervals. Each day, for example, the tides rise and fall in response to the gravitational pull of the moon. Similarly, the progression from day to night occurs as a result of Earth’s rotation, and the pattern of the seasons repeats in response to Earth’s revolution around the sun. Outside of nature, many stocks that mirror a company’s profits are influenced by changes in the economic business cycle. In mathematics, a function that repeats its values in regular intervals is known as a periodic function. The graphs of such functions show a general shape reflective of a pattern that keeps repeating. This means the graph of the function has the same output at exactly the same place in every cycle. And this translates to all the cycles of the function having exactly the same length. So, if we know all the details of one full cycle of a true periodic function, then we know the state of the function’s outputs at all times, future and past. In this chapter, we will investigate various examples of periodic functions. 810 Chapter 7 The Unit Circle: Sine and Cosine Functions 7.1 \| Angles In this section you will: Learning Objectives 7.1.1 Draw angles in standard position. 7.1.2 Convert between degrees and radians. 7.1.3 Find coterminal angles. 7.1.4 Find the length of a circular arc. 7.1.5 Use linear and angular speed to describe motion on a circular path. A golfer swings to hit a ball over a sand trap and onto the green. An airline pilot maneuvers a plane toward a narrow runway. A dress designer creates the latest fashion. What do they all have in common? They all work with angles, and so do all of us at one time or another. Sometimes we need to measure angles exactly with instruments. Other times we estimate them or judge them by eye. Either way, the proper angle can make the difference between success and failure in many undertakings. In this section, we will examine properties of angles. Drawing Angles in Standard Position Properly defining an angle first requires that we define a ray. A ray is a directed line segment. It consists of one point on a line and all points extending in one direction from that point. The first point is called the endpoint of the ray. We can refer to a specific ray by stating its endpoint and any other point on it. The ray in Figure 7.2 can be named as ray EF, or in symbol ⟶ form EF . Figure 7.2 An angle is the union of two rays having a common endpoint. The endpoint is called the vertex of the angle, and the two ⟶ . Angles can be named using a point ⟶ and EF rays are the sides of the angle. The angle in Figure 7.3 is formed from ED on each ray and the vertex, such as angle DEF, or in symbol form ∠ DEF. Figure 7.3 Greek letters are often used as variables for the measure of an angle. Table 7.1 is a list of Greek letters commonly used to represent angles, and a sample angle is shown in Figure 7.4. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 811 θ φ or ϕ α β γ theta phi alpha beta gamma Table 7.1 Figure 7.4 Angle theta, shown as ∠ θ Angle creation is a dynamic process. We start with two rays lying on top of one another. We leave one fixed in place, and rotate the other. The fixed ray is the initial side, and the rotated ray is the terminal side. In order to identify the different sides, we indicate the rotation with a small arrow close to the vertex as in Figure 7.5. Figure 7.5 As we discussed at the beginning of the section, there are many applications for angles, but in order to use them correctly, we must be able to measure them. The measure of an angle is the amount of rotation from the initial side to the terminal side. Probably the most familiar unit of angle measurement is the degree. One degree is 1 360 of a circular rotation, so a complete circular rotation contains 360 degrees. An angle measured in degrees should always include the unit “degrees” after the number, or include the degree symbol ° . For example, 90 degrees = 90°. To formalize our work, we will begin by drawing angles on an x-y coordinate plane. Angles can occur in any position on the coordinate plane, but for the purpose of comparison, the convention is to illustrate them in the same position whenever possible. An angle is in standard position if its vertex is located at the origin, and its initial side extends along the positive x-axis. See Figure 7.6. 812 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.6 If the angle is measured in a counterclockwise direction from the initial side to the terminal side, the angle is said to be a positive angle. If the angle is measured in a clockwise direction, the angle is said to be a negative angle. Drawing an angle in standard position always starts the same way—draw the initial side along the positive x-axis. To place the terminal side of the angle, we must calculate the fraction of a full rotation the angle represents. We do that by dividing the angle measure in degrees by 360°. For example, to draw a 90° angle, we calculate that 90° . So, the terminal side 360° = 1 4 will be one-fourth of the way around the circle, moving counterclockwise from the positive x-axis. To draw a 360° angle, we calculate that 360° = 1. So the terminal side will be 1 complete rotation around the circle, moving counterclockwise 360° from the positive x-axis. In this case, the initial side and the terminal side overlap. See Figure 7.7. Figure 7.7 Since we define an angle in standard position by its terminal side, we have a special type of angle whose terminal side lies on an axis, a quadrantal angle. This type of angle can have a measure of 0°, 90°, 180°, 270°, or 360°. See Figure 7.8. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 813 Figure 7.8 Quadrantal angles have a terminal side that lies along an axis. Examples are shown. Quadrantal Angles An angle is a quadrantal angle if its terminal side lies on an axis, including 0°, 90°, 180°, 270°, or 360°. Given an angle measure in degrees, draw the angle in standard position. 1. Express the angle measure as a fraction of 360°. 2. Reduce the fraction to simplest form. 3. Draw an angle that contains that same fraction of the circle, beginning on the positive x-axis and moving counterclockwise for positive angles and clockwise for negative angles. Example 7.1 Drawing an Angle in Standard Position Measured in Degrees a. Sketch an angle of 30° in standard position. b. Sketch an angle of −135° in standard position. Solution a. Divide the angle measure by 360°. To rewrite the fraction in a more familiar fraction, we can recognize that 30° 360° = 1 12 1 12 = 1 3 ⎞ ⎠ ⎛ ⎝ 1 4 One-twelfth equals one-third of a quarter, so by dividing a quarter rotation into thirds, we can sketch a line at 30°, as in Figure 7.9. 814 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.9 b. Divide the angle measure by 360°. In this case, we can recognize that −135° 360 Negative three-eighths is one and one-half times a quarter, so we place a line by moving clockwise one full quarter and one-half of another quarter, as in Figure 7.10. Figure 7.10 7.1 Show an angle of 240° on a circle in standard position. Converting Between Degrees and Radians Dividing a circle into 360 parts is a
n arbitrary choice, although it creates the familiar degree measurement. We may choose other ways to divide a circle. To find another unit, think of the process of drawing a circle. Imagine that you stop before the This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 815 circle is completed. The portion that you drew is referred to as an arc. An arc may be a portion of a full circle, a full circle, or more than a full circle, represented by more than one full rotation. The length of the arc around an entire circle is called the circumference of that circle. The circumference of a circle is C = 2πr. If we divide both sides of this equation by r, we create the ratio of the circumference, which is always 2π, to the radius, regardless of the length of the radius. So the circumference of any circle is 2π ≈ 6.28 times the length of the radius. That means that if we took a string as long as the radius and used it to measure consecutive lengths around the circumference, there would be room for six full string-lengths and a little more than a quarter of a seventh, as shown in Figure 7.11. Figure 7.11 This brings us to our new angle measure. One radian is the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle. A central angle is an angle formed at the center of a circle by two radii. Because the total circumference equals 2π times the radius, a full circular rotation is 2π radians. 2π radians = 360° π radians = 360° 1 radian = 180° = 180° 2 π ≈ 57.3° See Figure 7.12. Note that when an angle is described without a specific unit, it refers to radian measure. For example, an angle measure of 3 indicates 3 radians. In fact, radian measure is dimensionless, since it is the quotient of a length (circumference) divided by a length (radius) and the length units cancel. Figure 7.12 The angle t sweeps out a measure of one radian. Note that the length of the intercepted arc is the same as the length of the radius of the circle. 816 Chapter 7 The Unit Circle: Sine and Cosine Functions Relating Arc Lengths to Radius An arc length s is the length of the curve along the arc. Just as the full circumference of a circle always has a constant ratio to the radius, the arc length produced by any given angle also has a constant relation to the radius, regardless of the length of the radius. This ratio, called the radian measure, is the same regardless of the radius of the circle—it depends only on the angle. This property allows us to define a measure of any angle as the ratio of the arc length s to the radius r. See Figure 7.13. If s = r, then θ = r r = 1 radian. s = rθ s θ = r Figure 7.13 (a) In an angle of 1 radian, the arc length s equals the radius r. (b) An angle of 2 radians has an arc length s = 2r. (c) A full revolution is 2π, or about 6.28 radians. To elaborate on this idea, consider two circles, one with radius 2 and the other with radius 3. Recall the circumference of a circle is C = 2πr, where r is the radius. The smaller circle then has circumference 2π(2) = 4π and the larger has circumference 2π(3) = 6π. Now we draw a 45° angle on the two circles, as in Figure 7.14. Figure 7.14 A 45° angle contains one-eighth of the circumference of a circle, regardless of the radius. Notice what happens if we find the ratio of the arc length divided by the radius of the circle. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 817 Smaller circle: Larger circle the angle measures of both circles are the same, even though the arc length and radius differ. Since both ratios are 1 4 π, Radians One radian is the measure of the central angle of a circle such that the length of the arc between the initial side and the terminal side is equal to the radius of the circle. A full revolution (360°) equals 2π radians. A half revolution (180°) is equivalent to π radians. The radian measure of an angle is the ratio of the length of the arc subtended by the angle to the radius of the circle. In other words, if s is the length of an arc of a circle, and r is the radius of the circle, then the central angle containing that arc measures s r radians. In a circle of radius 1, the radian measure corresponds to the length of the arc. A measure of 1 radian looks to be about 60°. Is that correct? Yes. It is approximately 57.3°. Because 2π radians equals 360°, 1 radian equals 360° 2π ≈ 57.3°. Using Radians Because radian measure is the ratio of two lengths, it is a unitless measure. For example, in Figure 7.13, suppose the radius were 2 inches and the distance along the arc were also 2 inches. When we calculate the radian measure of the angle, the “inches” cancel, and we have a result without units. Therefore, it is not necessary to write the label “radians” after a radian measure, and if we see an angle that is not labeled with “degrees” or the degree symbol, we can assume that it is a radian measure. Considering the most basic case, the unit circle (a circle with radius 1), we know that 1 rotation equals 360 degrees, 360°. We can also track one rotation around a circle by finding the circumference, C = 2πr, and for the unit circle C = 2π. These two different ways to rotate around a circle give us a way to convert from degrees to radians. 1 rotation = 360° = 2π radians 1 rotation = 180° = π radians 2 1 4 rotation = 90° = radians π 2 Identifying Special Angles Measured in Radians In addition to knowing the measurements in degrees and radians of a quarter revolution, a half revolution, and a full revolution, there are other frequently encountered angles in one revolution of a circle with which we should be familiar. It is common to encounter multiples of 30, 45, 60, and 90 degrees. These values are shown in Figure 7.15. Memorizing these angles will be very useful as we study the properties associated with angles. 818 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.15 Commonly encountered angles measured in degrees Now, we can list the corresponding radian values for the common measures of a circle corresponding to those listed in Figure 7.15, which are shown in Figure 7.16. Be sure you can verify each of these measures. Figure 7.16 Commonly encountered angles measured in radians Example 7.2 Finding a Radian Measure Find the radian measure of one-third of a full rotation. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 819 For any circle, the arc length along such a rotation would be one-third of the circumference. We know that So, 1 rotation = 2πr (2πr) s = 1 3 = 2πr 3 The radian measure would be the arc length divided by the radius. radian measure = 2πr 3 r = 2πr 3r = 2π 3 7.2 Find the radian measure of three-fourths of a full rotation. Converting Between Radians and Degrees Because degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion where θ is the measure of the angle in degrees and θR is the measure of the angle in radians. This proportion shows that the measure of angle θ in degrees divided by 180 equals the measure of angle θ in radians divided by π. Or, phrased another way, degrees is to 180 as radians is to π. θ 180 θ R π = Degrees 180 = Radians π Converting between Radians and Degrees To convert between degrees and radians, use the proportion θR π = θ 180 Example 7.3 Converting Radians to Degrees Convert each radian measure to degrees. a. π 6 b. 3 820 Chapter 7 The Unit Circle: Sine and Cosine Functions Solution Because we are given radians and we want degrees, we should set up a proportion and solve it. a. We use the proportion, substituting the given information. θ 180 θ 180 θR π π 6 π = = θ = 180 6 θ = 30° b. We use the proportion, substituting the given information. θ 180 θ 180 = θ R π = 3 π 3(180) θ = π θ ≈ 172° 7.3 Convert − 3π 4 radians to degrees. Example 7.4 Converting Degrees to Radians Convert 15 degrees to radians. Solution In this example, we start with degrees and want radians, so we again set up a proportion, but we substitute the given information into a different part of the proportion 180 15 180 15π 180 π 12 Analysis Another way to think about this problem is by remembering that 30° = π 6 . Because 15° = 1 2 (30°), we can find that 1 2 ⎛ ⎝ π 6 ⎞ ⎠ is π 12 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 821 7.4 Convert 126° to radians. Finding Coterminal Angles Converting between degrees and radians can make working with angles easier in some applications. For other applications, we may need another type of conversion. Negative angles and angles greater than a full revolution are more awkward to work with than those in the range of 0° to 360°, or 0 to 2π. It would be convenient to replace those out-of-range angles with a corresponding angle within the range of a single revolution. It is possible for more than one angle to have the same terminal side. Look at Figure 7.17. The angle of 140° is a positive angle, measured counterclockwise. The angle of –220° is a negative angle, measured clockwise. But both angles have the same terminal side. If two angles in standard position have the same terminal side, they are coterminal angles. Every angle greater than 360° or less than 0° is coterminal with an angle between 0° and 360°, and it is often more convenient to find the coterminal angle within the range of 0° to 360° than to work with an angle that is outside that range. Figure 7.17 An angle of 140° and an angle of –220° are coterminal angles. Any angle has infinitely many coterminal angles because each time we add 360° to that angle—or subtract 360° from it—the resulting value has a terminal side in the same location. For example, 100° and 460° are coterminal for this reason, as is −260°. A
n angle’s reference angle is the measure of the smallest, positive, acute angle t formed by the terminal side of the angle t and the horizontal axis. Thus positive reference angles have terminal sides that lie in the first quadrant and can be used as models for angles in other quadrants. See Figure 7.18 for examples of reference angles for angles in different quadrants. 822 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.18 Coterminal and Reference Angles Coterminal angles are two angles in standard position that have the same terminal side. An angle’s reference angle is the size of the smallest acute angle, t′, formed by the terminal side of the angle t and the horizontal axis. Given an angle greater than 360°, find a coterminal angle between 0° and 360° 1. Subtract 360° from the given angle. 2. If the result is still greater than 360°, subtract 360° again till the result is between 0° and 360°. 3. The resulting angle is coterminal with the original angle. Example 7.5 Finding an Angle Coterminal with an Angle of Measure Greater Than 360° Find the least positive angle θ that is coterminal with an angle measuring 800°, where 0° ≤ θ < 360 ° . Solution An angle with measure 800° is coterminal with an angle with measure 800 − 360 = 440°, but 440° is still greater than 360°, so we subtract 360° again to find another coterminal angle: 440 − 360 = 80°. The angle θ = 80° is coterminal with 800°. To put it another way, 800° equals 80° plus two full rotations, as shown in Figure 7.19. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 823 Figure 7.19 7.5 Find an angle α that is coterminal with an angle measuring 870°, where 0° ≤ α < 360°. Given an angle with measure less than 0°, find a coterminal angle having a measure between 0° and 360°. 1. Add 360° to the given angle. 2. If the result is still less than 0°, add 360° again until the result is between 0° and 360°. 3. The resulting angle is coterminal with the original angle. Example 7.6 Finding an Angle Coterminal with an Angle Measuring Less Than 0° Show the angle with measure −45° on a circle and find a positive coterminal angle α such that 0° ≤ α < 360°. Solution Since 45° is half of 90°, we can start at the positive horizontal axis and measure clockwise half of a 90° angle. Because we can find coterminal angles by adding or subtracting a full rotation of 360°, we can find a positive coterminal angle here by adding 360°. We can then show the angle on a circle, as in Figure 7.20. −45° + 360° = 315° 824 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.20 7.6 Find an angle β that is coterminal with an angle measuring −300° such that 0° ≤ β < 360°. Finding Coterminal Angles Measured in Radians We can find coterminal angles measured in radians in much the same way as we have found them using degrees. In both cases, we find coterminal angles by adding or subtracting one or more full rotations. Given an angle greater than 2π, find a coterminal angle between 0 and 2π. 1. Subtract 2π from the given angle. 2. If the result is still greater than 2π, subtract 2π again until the result is between 0 and 2π. 3. The resulting angle is coterminal with the original angle. Example 7.7 Finding Coterminal Angles Using Radians Find an angle β that is coterminal with 19π 4 , where 0 ≤ β < 2π. Solution When working in degrees, we found coterminal angles by adding or subtracting 360 degrees, a full rotation. Likewise, in radians, we can find coterminal angles by adding or subtracting full rotations of 2π radians: 19π 4 − 2π = 19π 4 = 11π 4 − 8π 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 825 The angle 11π 4 is coterminal, but not less than 2π, so we subtract another rotation. 11π 4 − 2π = 11π 4 = 3π 4 − 8π 4 The angle 3π 4 is coterminal with 19π 4 , as shown in Figure 7.21. Figure 7.21 7.7 Find an angle of measure θ that is coterminal with an angle of measure − 17π 6 where 0 ≤ θ < 2π. Determining the Length of an Arc Recall that the radian measure θ of an angle was defined as the ratio of the arc length s of a circular arc to the radius r of the circle, θ = s r. From this relationship, we can find arc length along a circle, given an angle. Arc Length on a Circle In a circle of radius r, the length of an arc s subtended by an angle with measure θ in radians, shown in Figure 7.22, is s = rθ (7.1) 826 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.22 Given a circle of radius r, calculate the length s of the arc subtended by a given angle of measure θ. 1. If necessary, convert θ to radians. 2. Multiply the radius r θ : s = rθ. Example 7.8 Finding the Length of an Arc Assume the orbit of Mercury around the sun is a perfect circle. Mercury is approximately 36 million miles from the sun. a. In one Earth day, Mercury completes 0.0114 of its total revolution. How many miles does it travel in one day? b. Use your answer from part (a) to determine the radian measure for Mercury’s movement in one Earth day. Solution a. Let’s begin by finding the circumference of Mercury’s orbit. C = 2πr = 2π(36 million miles) ≈ 226 million miles Since Mercury completes 0.0114 of its total revolution in one Earth day, we can now find the distance traveled. b. Now, we convert to radians. (0.0114)226 million miles = 2.58 million miles radian = arclength radius = 2.58 million miles 36 million miles = 0.0717 7.8 Find the arc length along a circle of radius 10 units subtended by an angle of 215°. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 827 Finding the Area of a Sector of a Circle In addition to arc length, we can also use angles to find the area of a sector of a circle. A sector is a region of a circle bounded by two radii and the intercepted arc, like a slice of pizza or pie. Recall that the area of a circle with radius r can be found using the formula A = πr 2. If the two radii form an angle of θ, measured in radians, then θ 2π angle measure to the measure of a full rotation and is also, therefore, the ratio of the area of the sector to the area of the circle. Thus, the area of a sector is the fraction θ 2π multiplied by the entire area. (Always remember that this formula only is the ratio of the applies if θ is in radians.) Area of sector = = ⎞ ⎠πr 2 ⎛ θ ⎝ 2π θπr 2 2π θr 2 = 1 2 Area of a Sector The area of a sector of a circle with radius r subtended by an angle θ, measured in radians, is A = 1 2 θr 2 (7.2) See Figure 7.23. Figure 7.23 The area of the sector equals half the square of the radius times the central angle measured in radians. Given a circle of radius r, find the area of a sector defined by a given angle θ. 1. If necessary, convert θ to radians. 2. Multiply half the radian measure of θ by the square of the radius r : A = 1 2 θr 2. Example 7.9 Finding the Area of a Sector 828 Chapter 7 The Unit Circle: Sine and Cosine Functions An automatic lawn sprinkler sprays a distance of 20 feet while rotating 30 degrees, as shown in Figure 7.24. What is the area of the sector of grass the sprinkler waters? Figure 7.24 The sprinkler sprays 20 ft within an arc of 30°. Solution First, we need to convert the angle measure into radians. Because 30 degrees is one of our special angles, we already know the equivalent radian measure, but we can also convert: 30 degrees = 30 ⋅ π 180 radians = π 6 The area of the sector is then So the area is about 104.72 ft2. Area = 1 2 ≈ 104.72 π 6 ⎛ ⎝ ⎞ ⎠(20)2 7.9 In central pivot irrigation, a large irrigation pipe on wheels rotates around a center point. A farmer has a central pivot system with a radius of 400 meters. If water restrictions only allow her to water 150 thousand square meters a day, what angle should she set the system to cover? Write the answer in radian measure to two decimal places. Use Linear and Angular Speed to Describe Motion on a Circular Path In addition to finding the area of a sector, we can use angles to describe the speed of a moving object. An object traveling in a circular path has two types of speed. Linear speed is speed along a straight path and can be determined by the distance it moves along (its displacement) in a given time interval. For instance, if a wheel with radius 5 inches rotates once a second, a point on the edge of the wheel moves a distance equal to the circumference, or 10π inches, every second. So the linear speed of the point is 10π in./s. The equation for linear speed is as follows where v is linear speed, s is displacement, and t is time. v = s t Angular speed results from circular motion and can be determined by the angle through which a point rotates in a given time interval. In other words, angular speed is angular rotation per unit time. So, for instance, if a gear makes a full rotation every 4 seconds, we can calculate its angular speed as 360 degrees 4 seconds in radians per second, rotations per minute, or degrees per hour for example. The equation for angular speed is as follows, where ω (read as omega) is angular speed, θ is the angle traversed, and t is time. = 90 degrees per second. Angular speed can be given ω = θ t This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 829 Combining the definition of angular speed with the arc length equation, s = rθ, we can find a relationship between angular and linear speeds. The angular speed equation can be solved for θ, giving θ = ωt. Substituting this into the arc length equation gives: Substituting this into the linear speed equation gives: s = rθ = rωt v = s t rωt = t = rω Angular and Linear Speed As a point moves along a circle of radius r, its angular speed, ω, is the angular rotation θ per unit time, t. ω = θ t The linear speed, v, of the point can be found as the distance traveled, arc length s, per unit time, t. v = s t (
7.3) (7.4) When the angular speed is measured in radians per unit time, linear speed and angular speed are related by the equation This equation states that the angular speed in radians, ω, representing the amount of rotation occurring in a unit of time, can be multiplied by the radius r to calculate the total arc length traveled in a unit of time, which is the definition of linear speed. v = rω (7.5) Given the amount of angle rotation and the time elapsed, calculate the angular speed. 1. If necessary, convert the angle measure to radians. 2. Divide the angle in radians by the number of time units elapsed: ω = θ t . 3. The resulting speed will be in radians per time unit. Example 7.10 Finding Angular Speed A water wheel, shown in Figure 7.25, completes 1 rotation every 5 seconds. Find the angular speed in radians per second. 830 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.25 Solution The wheel completes 1 rotation, or passes through an angle of 2π radians in 5 seconds, so the angular speed would be ω = 2π 5 ≈ 1.257 radians per second. An old vinyl record is played on a turntable rotating clockwise at a rate of 45 rotations per minute. Find 7.10 the angular speed in radians per second. Given the radius of a circle, an angle of rotation, and a length of elapsed time, determine the linear speed. 1. Convert the total rotation to radians if necessary. 2. Divide the total rotation in radians by the elapsed time to find the angular speed: apply ω = θ t . 3. Multiply the angular speed by the length of the radius to find the linear speed, expressed in terms of the length unit used for the radius and the time unit used for the elapsed time: apply v = rω. Example 7.11 Finding a Linear Speed A bicycle has wheels 28 inches in diameter. A tachometer determines the wheels are rotating at 180 RPM (revolutions per minute). Find the speed the bicycle is traveling down the road. Solution Here, we have an angular speed and need to find the corresponding linear speed, since the linear speed of the outside of the tires is the speed at which the bicycle travels down the road. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 831 We begin by converting from rotations per minute to radians per minute. It can be helpful to utilize the units to make this conversion: 180 rotations minute ⋅ 2π radians rotation = 360πradians minute Using the formula from above along with the radius of the wheels, we can find the linear speed: ⎛ ⎝360πradians v = (14 inches) minute ⎞ ⎠ = 5040π inches minute Remember that radians are a unitless measure, so it is not necessary to include them. Finally, we may wish to convert this linear speed into a more familiar measurement, like miles per hour. 5040π inches minute ⋅ 1 feet 12 inches ⋅ 1 mile 5280 feet ⋅ 60 minutes 1 hour = 14.99 miles per hour (mph) A satellite is rotating around Earth at 0.25 radian per hour at an altitude of 242 km above Earth. If the 7.11 radius of Earth is 6378 kilometers, find the linear speed of the satellite in kilometers per hour. Access these online resources for additional instruction and practice with angles, arc length, and areas of sectors. • Angles in Standard Position (http://openstaxcollege.org/l/standardpos) • Angle of Rotation (http://openstaxcollege.org/l/angleofrotation) • Coterminal Angles (http://openstaxcollege.org/l/coterminal) • Determining Coterminal Angles (http://openstaxcollege.org/l/detcoterm) • Positive and Negative Coterminal Angles (http://openstaxcollege.org/l/posnegcoterm) • Radian Measure (http://openstaxcollege.org/l/radianmeas) • Coterminal Angles in Radians (http://openstaxcollege.org/l/cotermrad) • Arc Length and Area of a Sector (http://openstaxcollege.org/l/arclength) 832 Chapter 7 The Unit Circle: Sine and Cosine Functions 21. − 4π 3 For the following exercises, refer to Figure 7.26. Round to two decimal places. 7.1 EXERCISES Verbal Draw an angle in standard position. Label the vertex, 1. initial side, and terminal side. Explain why there are an infinite number of angles that 2. are coterminal to a certain angle. State what a positive or negative angle signifies, and 3. explain how to draw each. 4. How does radian measure of an angle compare to the degree measure? Include an explanation of 1 radian in your paragraph. 5. Explain the differences between linear speed and angular speed when describing motion along a circular path. Graphical For the following exercises, draw an angle in standard position with the given measure. Figure 7.26 6. 7. 8. 9. 30° 300° −80° 135° 10. −150° 11. 12. 13. 14. 15. 2π 3 7π 4 5π 6 π 2 − π 10 16. 415° 17. −120° 18. −315° 19. 20. 22π 3 − π 6 This content is available for free at https://cnx.org/content/col11758/1.5 22. Find the arc length. 23. Find the area of the sector. For the following exercises, refer to Figure 7.27. Round to two decimal places. Figure 7.27 24. Find the arc length. 25. Find the area of the sector. Algebraic For the following exercises, convert angles in radians to degrees. 26. Chapter 7 The Unit Circle: Sine and Cosine Functions 833 3π 4 27. 28. 29. 30. 31. 32. radians radians π 9 − 5π 4 radians radians π 3 − 7π 3 − 5π 12 radians radians 11π 6 radians For the following exercises, convert angles in degrees to radians. 33. 90° 34. 100° 35. −540° 36. −120° 37. 180° 38. −315° 39. 150° For the following exercises, use the given information to find the length of a circular arc. Round to two decimal places. Find the length of the arc of a circle of radius 12 inches 40. subtended by a central angle of π 4 . radians. Find the length of the arc of a circle of radius 5.02 41. miles subtended by the central angle of π 3 . Find the length of the arc of a circle of diameter 14 42. meters subtended by the central angle of 5π 6 . Find the length of the arc of a circle of radius 10 43. centimeters subtended by the central angle of 50°. Find the length of the arc of a circle of radius 5 inches 44. subtended by the central angle of 220°. Find the length of the arc of a circle of diameter 12 meters subtended by the central angle is 63°. For the following exercises, use the given information to find the area of the sector. Round to four decimal places. A sector of a circle has a central angle of 45° and a 46. radius 6 cm. A sector of a circle has a central angle of 30° and a 47. radius of 20 cm. A sector of a circle with diameter 10 feet and an angle radians. 48. of π 2 A sector of a circle with radius of 0.7 inches and an 49. angle of π radians. For the following exercises, find the angle between 0° and 360° that is coterminal to the given angle. 50. −40° 51. −110° 52. 700° 53. 1400° For the following exercises, find the angle between 0 and 2π in radians that is coterminal to the given angle. 54. 55. 56. 57. − π 9 10π 3 13π 6 44π 9 Real-World Applications A truck with 32-inch diameter wheels is traveling at 60 58. mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make? A bicycle with 24-inch diameter wheels is traveling at 59. 15 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make? A wheel of radius 8 inches is rotating 15°/s. What is the the angular speed in RPM, and the angular 60. linear speed v, speed in rad/s? 45. 61. 834 Chapter 7 The Unit Circle: Sine and Cosine Functions A wheel of radius 14 inches is rotating 0.5 rad/s. What is the linear speed v, the angular speed in RPM, and the angular speed in deg/s? 72. A bicycle has wheels 28 inches in diameter. A tachometer determines that the wheels are rotating at 180 RPM (revolutions per minute). Find the speed the bicycle is travelling down the road. A car travels 3 miles. Its tires make 2640 revolutions. 73. What is the radius of a tire in inches? 74. A wheel on a tractor has a 24-inch diameter. How many revolutions does the wheel make if the tractor travels 4 miles? A CD has diameter of 120 millimeters. When playing 62. audio, the angular speed varies to keep the linear speed constant where the disc is being read. When reading along the outer edge of the disc, the angular speed is about 200 RPM (revolutions per minute). Find the linear speed. When being burned in a writable CD-R drive, the 63. angular speed of a CD is often much faster than when playing audio, but the angular speed still varies to keep the linear speed constant where the disc is being written. When writing along the outer edge of the disc, the angular speed of one drive is about 4800 RPM (revolutions per minute). Find the linear speed if the CD has diameter of 120 millimeters. A person is standing on the equator of Earth (radius 64. 3960 miles). What are his linear and angular speeds? Find the distance along an arc on the surface of Earth . The radius of Earth is 3960 65. that subtends a central angle of 5 minutes ⎛ ⎝1 minute = 1 60 miles. ⎞ degree ⎠ 66. that subtends a central angle of 7 minutes ⎛ ⎝1 minute = 1 60 miles. ⎞ degree ⎠ Find the distance along an arc on the surface of Earth . The radius of Earth is 3960 67. Consider a clock with an hour hand and minute hand. What is the measure of the angle the minute hand traces in 20 minutes? Extensions 68. Two cities have the same longitude. The latitude of city A is 9.00 degrees north and the latitude of city B is 30.00 degree north. Assume the radius of the earth is 3960 miles. Find the distance between the two cities. 69. A city is located at 40 degrees north latitude. Assume the radius of the earth is 3960 miles and the earth rotates once every 24 hours. Find the linear speed of a person who resides in this city. 70. A city is located at 75 degrees north latitude. Assume the radius of the earth is 3960 miles and the earth rotates once every 24 hours. Find the linear speed of a person who resides in this city. 71. Find the linear speed of the moon if the average distance between the earth and moon is 239,000 miles, assuming the orbit of the moon is
circular and requires about 28 days. Express answer in miles per hour. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 835 7.2 \| Right Triangle Trigonometry Learning Objectives In this section you will: 7.2.1 Use right triangles to evaluate trigonometric functions. ⎛ 7.2.2 Find function values for 30° ⎝ π 6 ⎞ ⎛ ⎠, 45° ⎝ π 4 ⎛ ⎞ ⎠, and 60° ⎝ ⎞ ⎠. π 3 7.2.3 Use equal cofunctions of complementary angles. 7.2.4 Use the deﬁnitions of trigonometric functions of any angle. 7.2.5 Use right-triangle trigonometry to solve applied problems. Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains. Using Right Triangles to Evaluate Trigonometric Functions Figure 7.28 shows a right triangle with a vertical side of length y and a horizontal side has length x. Notice that the triangle is inscribed in a circle of radius 1. Such a circle, with a center at the origin and a radius of 1, is known as a unit circle. Figure 7.28 We can define the trigonometric functions in terms an angle t and the lengths of the sides of the triangle. The adjacent side is the side closest to the angle, x. (Adjacent means “next to.”) The opposite side is the side across from the angle, y. The hypotenuse is the side of the triangle opposite the right angle, 1. These sides are labeled in Figure 7.29. Figure 7.29 The sides of a right triangle in relation to angle t Given a right triangle with an acute angle of t, the first three trigonometric functions are listed. Sine sin t = Cosine cos t = opposite hypotenuse adjacent hypotenuse (7.6) (7.7) 836 Chapter 7 The Unit Circle: Sine and Cosine Functions Tangent tan t = opposite adjacent (7.8) A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.” For the triangle shown in Figure 7.28, we have the following. sin t = cos t = sec t = y 1 x 1 y x Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle. 1. Find the sine as the ratio of the opposite side to the hypotenuse. 2. Find the cosine as the ratio of the adjacent side to the hypotenuse. 3. Find the tangent as the ratio of the opposite side to the adjacent side. Example 7.12 Evaluating a Trigonometric Function of a Right Triangle Given the triangle shown in Figure 7.30, find the value of cos α. Figure 7.30 Solution The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17. cos(α) = adjacent hypotenuse = 15 17 7.12 Given the triangle shown in Figure 7.31, find the value of sin t. Figure 7.31 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 837 Reciprocal Functions In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle. Secant sec t = Cosecant csc t = Cotangent cot t = hypotenuse adjacent hypotenuse opposite adjacent opposite Take another look at these definitions. These functions are the reciprocals of the first three functions. sin t = cos t = tan t = 1 csc t 1 sec t 1 cot t csc t = sec t = cot t = 1 sin t 1 cos t 1 tan t (7.9) (7.10) (7.11) (7.12) When working with right triangles, keep in mind that the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 7.32. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa. Figure 7.32 The side adjacent to one angle is opposite the other angle. Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals. Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles. 1. 2. If needed, draw the right triangle and label the angle provided. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle. 3. Find the required function: ◦ sine as the ratio of the opposite side to the hypotenuse ◦ cosine as the ratio of the adjacent side to the hypotenuse ◦ ◦ tangent as the ratio of the opposite side to the adjacent side secant as the ratio of the hypotenuse to the adjacent side ◦ cosecant as the ratio of the hypotenuse to the opposite side ◦ cotangent as the ratio of the adjacent side to the opposite side Example 7.13 Evaluating Trigonometric Functions of Angles Not in Standard Position 838 Chapter 7 The Unit Circle: Sine and Cosine Functions Using the triangle shown in Figure 7.33, evaluate sin α, cos α, tan α, sec α, csc α, and cot α. Figure 7.33 Solution sin α = cos α = tan α = sec α = csc α = cot α = opposite α = 4 hypotenuse 5 adjacent to α = 3 hypotenuse 5 opposite α adjacent to α = 4 3 hypotenuse adjacent to α = 5 3 hypotenuse = 5 opposite α 4 adjacent to α opposite α = 3 4 Analysis Another approach would have been to find sine, cosine, and tangent first. Then find their reciprocals to determine the other functions. sec α = 1 cos α = 1 3 5 csc α = 1 csc α = 1 4 5 cot α = 1 tan .13 Using the triangle shown in Figure 7.34,evaluate sin t, cos t, tan t, sec t, csc t, and cot t. Figure 7.34 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 839 Finding Trigonometric Functions of Special Angles Using Side Lengths It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of 30°, 60°, and 45°. Remember, however, that when dealing with right triangles, we are limited to angles between 0° and 90°. Suppose we have a 30°, 60°, 90° triangle, which can also be described as a π 6 relation s, s 3, 2s. The sides of a 45°, 45°, 90° triangle, which can also be described as a π 4 triangle. The sides have lengths in the triangle, have lengths π 2 π 3 π 4 π 2 , , , , in the relation s, s, 2s. These relations are shown in Figure 7.35. Figure 7.35 Side lengths of special triangles We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles. Given trigonometric functions of a special angle, evaluate using side lengths. 1. Use the side lengths shown in Figure 7.35 for the special angle you wish to evaluate. 2. Use the ratio of side lengths appropriate to the function you wish to evaluate. Example 7.14 Evaluating Trigonometric Functions of Special Angles Using Side Lengths Find the exact value of the trigonometric functions of π 3 , using side lengths. Solution 840 Chapter 7 The Unit Circle: Sine and Cosine Functions 2s = 3 2 s 2s = 1 2 s = 3 ⎛ sin ⎝ ⎛ cos ⎝ ⎛ tan ⎝ ⎛ sec ⎝ ⎛ csc ⎝ cot ⎛ ⎝ ⎞ ⎠ = ⎞ ⎠ = ⎞ ⎠ = ⎞ ⎠ = ⎞ ⎠ = ⎞ ⎠ = = = 3s = 3s opp hyp adj hyp opp adj hyp = 2s s adj hyp opp = 2s 3s adj s opp = 3s = .14 Find the exact value of the trigonometric functions of π 4 , using side lengths. Using Equal Cofunction of Complements If we look more closely at the relationship between the sine and cosine of the special angles, we notice a pattern. In a right triangle with angles of π and π 3 6 is also the cosine of π 3 , we see that the sine of π 3 is also the cosine of π 6 , while the sine of π 6 , namely 3 2 namely 1 2 , , , . sin sin π 3 π 6 = cos = cos π 6 π 3 = 3s 2s = 3 2 s 2s = 1 2 = See Figure 7.36. Figure 7.36 The sine of π 3 equals the cosine of π 6 and vice versa. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 841 This result should not be surprising because, as we see from Figure 7.36, the side opposite the angle of π 3 is also the side adjacent to π 6 ⎛ , so sin ⎝ ⎛ ⎞ ⎠ and cos ⎝ π 6 ⎞ ⎠ are also the same ratio using the same two sides, s and 2s. π 3 ⎛ sin ⎝ π 6 ⎞ ⎛ ⎠ are exactly the same ratio of the same two sides, 3s and 2s. Similarly, cos ⎝ ⎞ ⎠ and π 3 The interrelationship between the sines and cosines of π 6 and π 3 also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, and the right angle is π . That 2 means that a right triangle can be formed with any two angles that add to π 2 the remaining two angles must also add up to π 2 —in other words, any two complementary , angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 7.37. Figure 7.37 Cofunction identity of sine and cosine of complementary angles Using this identity, we can state without calculating, for instance, that the sine of π 12 sine of 5π 12 equals the cosine of π 12 well. . We can also state that if, for a given angle t, cos t = 5 13 equals the cosine of 5π 12 π ⎛ then sin ⎝ 2 , , and that the − t⎞ ⎠ = 5 13 as Cofunction Identities The cofunction identities in radians are listed in Table 7.2. ⎛ cos t = sin ⎝ − t⎞ ⎠ π 2 ⎛ sin t = cos ⎝ − t⎞ ⎠ π 2 tan t = cot ⎛ ⎝ π 2 − t⎞ ⎠ ⎛ cot t = tan ⎝ − t⎞ ⎠ π 2 ⎛ sec t = csc ⎝ − t⎞ ⎠ π 2 ⎛ csc t = sec ⎝ − t⎞ ⎠ π 2 Table 7.2 842 Chapter 7 The Unit Circle: Sine and Cosine Functions Given the sine and cosine of an angle, find the
sine or cosine of its complement. 1. To find the sine of the complementary angle, find the cosine of the original angle. 2. To find the cosine of the complementary angle, find the sine of the original angle. Example 7.15 Using Cofunction Identities If sin t = 5 12 ⎛ , find cos ⎝ − t⎞ ⎠. π 2 Solution According to the cofunction identities for sine and cosine, we have the following. sin t = cos ⎛ ⎝ − t⎞ ⎠ π 2 cos ⎛ ⎝ π 2 − t⎞ ⎠ = 5 12 So 7.15 If csc ⎛ ⎝ π 6 ⎞ ⎠ = 2, find sec ⎛ ⎝ ⎞ ⎠. π 3 Using Trigonometric Functions In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides. Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides. 1. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator. 2. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides. 3. Using the value of the trigonometric function and the known side length, solve for the missing side length. Example 7.16 Finding Missing Side Lengths Using Trigonometric Ratios Find the unknown sides of the triangle in Figure 7.38. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 843 Figure 7.38 Solution We know the angle and the opposite side, so we can use the tangent to find the adjacent side. We rearrange to solve for a. We can use the sine to find the hypotenuse. Again, we rearrange to solve for c. tan(30°) = 7 a a = 7 tan(30°) ≈ 12.1 sin(30°) = 7 c c = 7 sin(30°) ≈ 14 7.16 A right triangle has one angle of π 3 triangle. and a hypotenuse of 20. Find the unknown sides and angle of the Using Right Triangle Trigonometry to Solve Applied Problems Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure 7.39. 844 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.39 Given a tall object, measure its height indirectly. 1. Make a sketch of the problem situation to keep track of known and unknown information. 2. Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible. 3. At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal. 4. Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight. 5. Solve the equation for the unknown height. Example 7.17 Measuring a Distance Indirectly To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° between a line of sight to the top of the tree and the ground, as shown in Figure 7.40. Find the height of the tree. Figure 7.40 Solution We know that the angle of elevation is 57° and the adjacent side is 30 ft long. The opposite side is the unknown height. The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of 57°, letting h be the unknown height. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 845 tan θ = tan(57°) = opposite adjacent h 30 Solve for h. h = 30tan(57°) Multiply. h ≈ 46.2 Use a calculator. The tree is approximately 46 feet tall. How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the 7.17 building making an angle of 5π 12 with the ground? Round to the nearest foot. Access these online resources for additional instruction and practice with right triangle trigonometry. • Finding Trig Functions on Calculator (http://openstaxcollege.org/l/findtrigcal) • Finding Trig Functions Using a Right Triangle (http://openstaxcollege.org/l/trigrttri) • Relate Trig Functions to Sides of a Right Triangle (http://openstaxcollege.org/l/reltrigtri) • Determine Six Trig Functions from a Triangle (http://openstaxcollege.org/l/sixtrigfunc) • Determine Length of Right Triangle Side (http://openstaxcollege.org/l/rttriside) 846 Chapter 7 The Unit Circle: Sine and Cosine Functions 7.2 EXERCISES Verbal For the given right triangle, label the adjacent side, 75. opposite side, and hypotenuse for the indicated angle. sin B = 1 3 , a = 2 89. a = 5, ∡ A = 60° 90. c = 12, ∡ A = 45° Graphical For the following exercises, use Figure 7.41 to evaluate each trigonometric function of angle A. When a right triangle with a hypotenuse of 1 is placed the triangle radius 1, which sides of 76. in a circle of correspond to the x- and y-coordinates? The tangent of an angle compares which sides of the 77. right triangle? Figure 7.41 91. sin A 92. cos A 93. tan A 94. csc A 95. sec A 96. cot A For the following exercises, use Figure 7.42 to evaluate each trigonometric function of angle A. What is the relationship between the two acute angles 78. in a right triangle? 79. Explain the cofunction identity. Algebraic the For complementary angles. following exercises, use cofunctions of 80. 81. 82. 83. cos(34°) = sin(___°) ⎛ cos ⎝ π 3 ⎞ ⎠ = sin(___) csc(21°) = sec(___°) ⎛ tan ⎝ π 4 ⎞ ⎠ = cot(___) For the following exercises, find the lengths of the missing sides if side a is opposite angle A, side b is opposite angle B, and side c is the hypotenuse. 84. 85. 86. cos B = 4 5 , a = 10 sin B = 1 2 , a = 20 tan A = 5 12 , b = 6 87. tan A = 100, b = 100 88. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 847 Figure 7.42 97. sin A 98. cos A 99. tan A 100. csc A 101. sec A 102. cot A For the following exercises, solve for the unknown sides of the given triangle. 103. 104. 105. Technology For the following exercises, use a calculator to find the length of each side to four decimal places. 106. 107. 108. 848 Chapter 7 The Unit Circle: Sine and Cosine Functions 109. 110. 111. b = 15, ∡ B = 15° 112. c = 200, ∡ B = 5° 113. c = 50, ∡ B = 21° 114. a = 30, ∡ A = 27° 115. b = 3.5, ∡ A = 78° Extensions 116. Find x. 117. Find x. This content is available for free at https://cnx.org/content/col11758/1.5 118. Find x. 119. Find x. A radio tower is located 400 feet from a building. 120. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36°, and that the angle of depression to the bottom of the tower is 23°. How tall is the tower? A radio tower is located 325 feet from a building. 121. From a window in the building, a person determines that the angle of elevation to the top of the tower is 43°, and that the angle of depression to the bottom of the tower is 31°. How tall is the tower? 122. Chapter 7 The Unit Circle: Sine and Cosine Functions 849 A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15°, and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument? A 400-foot tall monument is located in the distance. 123. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18°, and that the angle of depression to the bottom of the tower is 3°. How far is the person from the monument? 124. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40°. From the same location, the angle of elevation to the top of the antenna is measured to be 43°. Find the height of the antenna. 125. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be 36°. From the same location, the angle of elevation to the top of the lightning rod is measured to be 38°. Find the height of the lightning rod. Real-World Applications 126. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 127. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 128. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. 129. The angle of elevation to the top of a building in Seat
tle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building. 130. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60°, how far from the base of the tree am I? 850 Chapter 7 The Unit Circle: Sine and Cosine Functions 7.3 \| Unit Circle In this section you will: Learning Objectives ⎛ 7.3.1 Find function values for the sine and cosine of 30° or ⎝ π 6 ⎞ ⎛ ⎠, 45° or ⎝ π 4 ⎛ ⎞ ⎠, and 60∘ or ⎝ ⎞ ⎠. π 3 7.3.2 Identify the domain and range of sine and cosine functions. 7.3.3 Find reference angles. 7.3.4 Use reference angles to evaluate trigonometric functions. Figure 7.43 The Singapore Flyer is the world’s tallest Ferris wheel. (credit: ʺVibin JKʺ/Flickr) Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs. Finding Trigonometric Functions Using the Unit Circle We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. Recall this a unit circle is a circle centered at the origin with radius 1, as shown in Figure 7.44. The angle (in radians) that t intercepts forms an arc of length s. Using the formula s = rt, and knowing that r = 1, we see that for a unit circle, s = t. The x- and y-axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV. For any angle t, we can label the intersection of the terminal side and the unit circle as by its coordinates, (x, y). The coordinates x and y will be the outputs of the trigonometric functions f (t) = cos t and f (t) = sin t, respectively. This means x = cos t and y = sin t. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 851 Figure 7.44 Unit circle where the central angle is t radians Unit Circle A unit circle has a center at (0, 0) and radius 1. In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle t. Let (x, y) be the endpoint on the unit circle of an arc of arc length s. The (x, y) coordinates of this point can be described as functions of the angle. Defining Sine and Cosine Functions from the Unit Circle The sine function relates a real number t to the y-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle t equals the y-value of the endpoint on the unit circle of an arc of length t. In Figure 7.44, the sine is equal to y. Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the y-coordinate of the corresponding point on the unit circle. The cosine function of an angle t equals the x-value of the endpoint on the unit circle of an arc of length t. In Figure 7.45, the cosine is equal to x. Figure 7.45 Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: sin t is the same as sin(t) and cost is the same as cos(t). Likewise, cos2 t is a commonly used shorthand notation for (cos(t))2. Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer. 852 Chapter 7 The Unit Circle: Sine and Cosine Functions Sine and Cosine Functions If t is a real number and a point (x, y) on the unit circle corresponds to a central angle t, then cos t = x sin t = y (7.13) (7.14) Given a point P (x, y) on the unit circle corresponding to an angle of t, find the sine and cosine. 1. The sine of t is equal to the y-coordinate of point P : sin t = y. 2. The cosine of t is equal to the x-coordinate of point P : cos t = x. Example 7.18 Finding Function Values for Sine and Cosine Point P is a point on the unit circle corresponding to an angle of t, as shown in Figure 7.46. Find cos(t) and sin(t). Figure 7.46 Solution We know that cos t is the x-coordinate of the corresponding point on the unit circle and sin t is the y-coordinate of the corresponding point on the unit circle. So: x = cos t = 1 2 y = sin t = 3 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 853 7.18 A certain angle t corresponds to a point on the unit circle at ⎛ ⎝− 2 2 , 2 2 ⎞ as shown in Figure 7.47. Find ⎠ cos t and sin t. Figure 7.47 Finding Sines and Cosines of Angles on an Axis For quadrantral angles, the corresponding point on the unit circle falls on the x- or y-axis. In that case, we can easily calculate cosine and sine from the values of x and y. Example 7.19 Calculating Sines and Cosines along an Axis Find cos(90°) and sin(90°). Solution Moving 90° counterclockwise around the unit circle from the positive x-axis brings us to the top of the circle, where the (x, y) coordinates are (0, 1), as shown in Figure 7.48. 854 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.48 We can then use our definitions of cosine and sine. x = cos t = cos(90°) = 0 y = sin t = sin(90°) = 1 The cosine of 90° is 0; the sine of 90° is 1. 7.19 Find cosine and sine of the angle π. The Pythagorean Identity Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is x2 + y2 = 1. Because x = cos t and y = sin t, we can substitute for x and y to get cos2 t + sin2 t = 1. This equation, cos2 t + sin2 t = 1, is known as the Pythagorean Identity. See Figure 7.49. Figure 7.49 We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 855 Pythagorean Identity The Pythagorean Identity states that, for any real number t, cos2 t + sin2 t = 1 (7.15) Given the sine of some angle t and its quadrant location, find the cosine of t. 1. Substitute the known value of sin t into the Pythagorean Identity. 2. Solve for cos t. 3. Choose the solution with the appropriate sign for the x-values in the quadrant where t is located. Example 7.20 Finding a Cosine from a Sine or a Sine from a Cosine If sin(t) = 3 7 and t is in the second quadrant, find cos(t). Solution If we drop a vertical line from the point on the unit circle corresponding to t, we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See Figure 7.50. Figure 7.50 Substituting the known value for sine into the Pythagorean Identity, cos2(t) + sin2(t) = 1 cos2(t) + 9 49 = 1 cos2(t) = 40 49 cos(t) = ± 40 49 = ± 40 7 = ± 2 10 7 856 Chapter 7 The Unit Circle: Sine and Cosine Functions Because the angle is in the second quadrant, we know the x-value is a negative real number, so the cosine is also negative. cos(t) = − 2 10 7 7.20 If cos(t) = 24 25 and t is in the fourth quadrant, find sin(t). Finding Sines and Cosines of Special Angles We have already learned some properties of the special angles, such as the conversion from radians to degrees, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity. Finding Sines and Cosines of 45° Angles First, we will look at angles of 45° or π 4 so the x- and y-coordinates of the corresponding point on the circle are the same. Because the x- and y-values are the same, the sine and cosine values will also be equal. , as shown in Figure 7.51. A 45° – 45° – 90° triangle is an isosceles triangle, Figure 7.51 At t = π 4 , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line y = x. A unit circle has a radius equal to 1 so the right triangle formed below the line y = x has sides x and y (y = x), and radius = 1. See Figure 7.52. Figure 7.52 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 857 From the Pythagorean Theorem we get We can then substitute y = x. Next we combine like terms. And solving for x, we get In quadrant I, x = 1 2 . At t = π 4 or 45 degrees, x2 + y2 = 1 x2 + x2 = 1 2x2 = 1 x2 = cos t = 1 2 If we then rationalize the denominators, we get (x, y) = (x, x , sin t = 1 2 2 2 2 2 cos t = 1 2 = 2 2 sin t = 1 2 = 2 2 Therefore, the (x, y) coordinates of a point on a circle of radius 1 at an angle of 45° are ⎛ ⎝ 2 2 , 2 2 ⎞ ⎠. Finding Sines and Cosines of 30° and 60° Angles Next, we will find the cosine and sine at an angle of 30°, or π 6 at an angle of 30°, and another at an angle of −30°, as shown in Figure 7.53. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be 60°, as shown in Figure 7.54. . First, we will draw a triangle inside a circle with one side 858 Chapter 7 The Unit Circle: Sine and
Cosine Functions Figure 7.53 Figure 7.54 Because all the angles are equal, the sides are also equal. The vertical line has length 2y, and since the sides are all equal, we can also conclude that r = 2y or y = 1 2 r. Since sin t = y, And since r = 1 in our unit circle, ⎛ sin ⎝ π 6 ⎞ ⎠ = 1 2 r ⎛ sin ⎝ π 6 ⎞ ⎠ = 1 2 (1) = 1 2 Using the Pythagorean Identity, we can find the cosine value. ⎠ + sin2 ⎛ ⎞ ⎝ cos2 ⎛ ⎝ π 6 cos2 ⎛ ⎝ cos2 ⎛ π ⎝ 6 π 6 ⎛ cos ⎝ = Use the square root property. Since y is positive, choose the positive root. The (x, y) coordinates for the point on a circle of radius 1 at an angle of 30° are ⎛ ⎝ 3 2 , 1 2 ⎞ ⎠. At t = π 3 (60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, BAD, as shown in Figure 7.55. Angle A has measure 60°. At point B, we draw an angle ABC with measure of 60°. We know the angles in a triangle sum to 180°, so the measure of angle C is also 60°. Now we have an equilateral triangle. Because each side of the equilateral triangle ABC is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 859 Figure 7.55 The measure of angle ABD is 30°. Angle ABC is double angle ABD, so its measure is 60°. BD is the perpendicular bisector of AC, so it cuts AC in half. This means that AD is 1 2 . Notice that AD is the x-coordinate of the radius, or 1 2 point B, which is at the intersection of the 60° angle and the unit circle. This gives us a triangle BAD with hypotenuse of 1 and side x of length 1 2 . From the Pythagorean Theorem, we get Substituting x = 1 2 , we get Solving for y, we get x2 + y2 = 1 2 ⎞ ⎠ ⎛ ⎝ 1 2 + y2 = 1 + y2 = 1 1 4 y2 = 1 − 1 4 y2 = 3 4 y = ± 3 2 Since t = At t = π 3 has the terminal side in quadrant I where the y-coordinate is positive, we choose y = 3 2 π 3 (60°), the (x, y) coordinates for the point on a circle of radius 1 at an angle of 60° are ⎛ ⎝ the positive value. , 1 2 , 3 2 ⎞ ⎠, so we can find the sine and cosine. ⎛ ⎝ ⎞ ⎠ (x, y , sin t = 3 cos t = 1 2 2 We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. Table 7.3 summarizes these values. 860 Chapter 7 The Unit Circle: Sine and Cosine Functions , or 60° , or 90° π 2 , or 30° π 6 , or 45 Angle Cosine Sine Table 7.3 π 3 1 2 3 2 0 1 Figure 7.56 shows the common angles in the first quadrant of the unit circle. Figure 7.56 Using a Calculator to Find Sine and Cosine To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate cos(30) on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode. Given an angle in radians, use a graphing calculator to find the cosine. 1. If the calculator has degree mode and radian mode, set it to radian mode. 2. Press the COS key. 3. Enter the radian value of the angle and press the close-parentheses key ")". 4. Press ENTER. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 861 Example 7.21 Using a Graphing Calculator to Find Sine and Cosine ⎛ Evaluate cos ⎝ 5π 3 ⎞ ⎠ using a graphing calculator or computer. Solution Enter the following keystrokes: COS( 5 × π ÷ 3 ) ENTER ⎛ cos ⎝ 5π 3 ⎞ ⎠ = 0.5 Analysis We can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sign of 20°, for example, by including the conversion factor to radians as part of the input: SIN( 20 × π ÷ 180 ) ENTER 7.21 ⎛ Evaluate sin ⎝ ⎞ ⎠. π 3 Identifying the Domain and Range of Sine and Cosine Functions Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 and angles larger than 2π can still be graphed on the unit circle and have real values of x, y, and r, there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may be any real number. What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure 7.57. The bounds of the x-coordinate are [−1, 1]. The bounds of the y-coordinate are also [−1, 1]. Therefore, the range of both the sine and cosine functions is [−1, 1]. Figure 7.57 862 Chapter 7 The Unit Circle: Sine and Cosine Functions Finding Reference Angles We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the y-coordinate on the unit circle, the other angle with the same sine will share the same y-value, but have the opposite x-value. Therefore, its cosine value will be the opposite of the first angle’s cosine value. Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same x-value but will have the opposite y-value. Therefore, its sine value will be the opposite of the original angle’s sine value. As shown in Figure 7.58, angle α has the same sine value as angle t; the cosine values are opposites. Angle β has the same cosine value as angle t; the sine values are opposites. sin(t) = sin(α) sin(t) = − sin(β) and and cos(t) = − cos(α) cos(t) = cos(β) Figure 7.58 Recall that an angle’s reference angle is the acute angle, t, formed by the terminal side of the angle t and the horizontal axis. A reference angle is always an angle between 0 and 90°, or 0 and π radians. As we can see from Figure 7.59, for 2 any angle in quadrants II, III, or IV, there is a reference angle in quadrant I. Figure 7.59 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 863 Given an angle between 0 and 2π, find its reference angle. 1. An angle in the first quadrant is its own reference angle. 2. For an angle in the second or third quadrant, the reference angle is \|π − t\| or \|180° − t\|. 3. For an angle in the fourth quadrant, the reference angle is 2π − t or 360° − t. 4. If an angle is less than 0 or greater than 2π, add or subtract 2π as many times as needed to find an equivalent angle between 0 and 2π. Example 7.22 Finding a Reference Angle Find the reference angle of 225° as shown in Figure 7.60. Figure 7.60 Solution Because 225° is in the third quadrant, the reference angle is \|(180° − 225°)\| = \|−45°\| = 45° 7.22 Find the reference angle of 5π 3 . Using Reference Angles Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find (x, y) coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies. 864 Chapter 7 The Unit Circle: Sine and Cosine Functions Using Reference Angles to Evaluate Trigonometric Functions We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the x-values in that quadrant. The sine will be positive or negative depending on the sign of the y-values in that quadrant. Using Reference Angles to Find Cosine and Sine Angles have cosines and sines with the same absolute value as their reference angles. The sign (positive or negative) can be determined from the quadrant of the angle. Given an angle in standard position, find the reference angle, and the cosine and sine of the original angle. 1. Measure the angle between the terminal side of the given angle and the horizontal axis. That is the reference angle. 2. Determine the values of the cosine and sine of the reference angle. 3. Give the cosine the same sign as the x-values in the quadrant of the original angle. 4. Give the sine the same sign as the y-values in the quadrant of the original angle. Example 7.23 Using Reference Angles to Find Sine and Cosine a. Using a reference angle, find the exact value of cos(150°) and sin(150°). b. Using the reference angle, find cos 5π 4 and sin 5π 4 . Solution a. 150° is located in the second quadrant. The angle it makes with the x-axis is 180° − 150° = 30°, so the reference angle is 30°. This tells us that 150° has the same sine and cosine values as 30°, except for the sign. cos(30°) = 3 2 and sin(30°) = 1 2 Since 150° is in the second quadrant, the x-coordinate of the point on the circle is negative, so the cosine value is negative. The y-coordinate is positive, so the sine value is positive. cos(150°) = 3 2 and sin(150°) = 1 2 b. is in th
e third quadrant. Its reference angle is 5π 5π 4 4 In the third quadrant, both x and y are negative, so: − π = π 4 . The cosine and sine of π 4 are both 2 2 . cos5π 4 = − 2 2 and sin5π 4 = − 2 2 7.23 a. Use the reference angle of 315° to find cos(315°) and sin(315°). b. Use the reference angle of − ⎛ to find cos ⎝− π 6 ⎛ ⎞ ⎠ and sin ⎝− π 6 ⎞ ⎠. π 6 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 865 Using Reference Angles to Find Coordinates Now that we have learned how to find the cosine and sine values for special angles in the first quadrant, we can use symmetry and reference angles to fill in cosine and sine values for the rest of the special angles on the unit circle. They are shown in Figure 7.61. Take time to learn the (x, y) coordinates of all of the major angles in the first quadrant. Figure 7.61 Special angles and coordinates of corresponding points on the unit circle In addition to learning the values for special angles, we can use reference angles to find (x, y) coordinates of any point on the unit circle, using what we know of reference angles along with the identities x = cos t y = sin t First we find the reference angle corresponding to the given angle. Then we take the sine and cosine values of the reference angle, and give them the signs corresponding to the y- and x-values of the quadrant. Given the angle of a point on a circle and the radius of the circle, find the (x, y) coordinates of the point. 1. Find the reference angle by measuring the smallest angle to the x-axis. 2. Find the cosine and sine of the reference angle. 3. Determine the appropriate signs for x and y in the given quadrant. Example 7.24 Using the Unit Circle to Find Coordinates Find the coordinates of the point on the unit circle at an angle of 7π 6 . 866 Chapter 7 The Unit Circle: Sine and Cosine Functions Solution We know that the angle 7π 6 is in the third quadrant. First, let’s find the reference angle by measuring the angle to the x-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and π. Next, we will find the cosine and sine of the reference angle. 7π 6 − π = π 6 ⎛ cos ⎝ π 6 ⎞ ⎠ = 3 2 ⎛ sin ⎝ π 6 ⎞ ⎠ = 1 2 We must determine the appropriate signs for x and y in the given quadrant. Because our original angle is in the third quadrant, where both x and y are negative, both cosine and sine are negative. ⎛ cos ⎝ 7π 6 ⎞ ⎠ = − 3 2 sin(7π) = − 1 2 Now we can calculate the (x, y) coordinates using the identities x = cos θ and y = sin θ. The coordinates of the point are ⎛ ⎝− 3 2 , − 1 2 ⎞ on the unit circle. ⎠ 7.24 Find the coordinates of the point on the unit circle at an angle of 5π 3 . Access these online resources for additional instruction and practice with sine and cosine functions. • Trigonometric Functions Using the Unit Circle (http://openstaxcollege.org/l/trigunitcir) • Sine and Cosine from the Unit (http://openstaxcollege.org/l/sincosuc) • Sine and Cosine from the Unit Circle and Multiples of Pi Divided by Six (http://openstaxcollege.org/l/sincosmult) • Sine and Cosine from the Unit Circle and Multiples of Pi Divided by Four (http://openstaxcollege.org/l/sincosmult4) • Trigonometric Functions Using Reference Angles (http://openstaxcollege.org/l/trigrefang) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 867 7.3 EXERCISES Verbal 131. Describe the unit circle. What do the x- and y-coordinates of the points on the 132. unit circle represent? Discuss the difference between a coterminal angle 133. and a reference angle. 134. Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle. 135. Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle. Algebraic For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by t lies. 136. sin(t) < 0 and cos(t) < 0 137. sin(t) > 0 and cos(t) > 0 138. sin(t) > 0 and cos(t) < 0 139. sin(t) > 0 and cos(t) > 0 For the following exercises, find the exact value of each trigonometric function. 140. 141. 142. 143. 144. 145. 146. sin π 2 sin π 3 cos π 2 cos π 3 sin π 4 cos π 4 sin π 6 147. sin π 148. sin 3π 2 149. cos π 150. cos 0 151. cos π 6 152. sin 0 Numeric For the following exercises, state the reference angle for the given angle. 153. 240° 154. −170° 155. 100° 156. −315° 157. 135° 158. 159. 160. 5π 4 2π 3 5π 6 161. −11π 3 162. 163. −7π 4 −π 8 For the following exercises, find the reference angle, the quadrant of the terminal side, and the sine and cosine of each angle. If the angle is not one of the angles on the unit circle, use a calculator and round to three decimal places. 164. 225° 165. 300° 166. 320° 167. 135° 168. 210° 169. 120° 868 170. 250° 171. 150° 172. 173. 174. 175. 176. 177. 178. 179. 5π 4 7π 6 5π 3 3π 4 4π 3 2π 3 5π 6 7π 4 Chapter 7 The Unit Circle: Sine and Cosine Functions 188. State the domain of the sine and cosine functions. 189. State the range of the sine and cosine functions. Graphical For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of t. 190. 191. For the following exercises, find the requested value. 180. If cos(t) = 1 7 sin(t). 181. If cos(t) = 2 9 sin(t). and t is in the fourth quadrant, find and t is in the first quadrant, find 182. If sin(t) = 3 8 cos(t). and t is in the second quadrant, find 183. If sin(t) = − 1 4 cos(t). and t is in the third quadrant, find 192. Find the coordinates of the point on a circle with 184. radius 15 corresponding to an angle of 220°. Find the coordinates of the point on a circle with 185. radius 20 corresponding to an angle of 120°. Find the coordinates of the point on a circle with 186. radius 8 corresponding to an angle of 7π 4 . Find the coordinates of the point on a circle with 187. radius 16 corresponding to an angle of 5π 9 . 193. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 869 194. 197. 195. 198. 196. 199. 870 Chapter 7 The Unit Circle: Sine and Cosine Functions 200. 203. 201. 204. 202. 205. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 871 206. 209. 207. 208. Technology For the following exercises, use a graphing calculator to evaluate. 210. 211. 212. 213. 214. 215. sin 5π 9 cos 5π 9 sin π 10 cos π 10 sin 3π 4 cos 3π 4 216. sin 98° 217. cos 98° 218. Chapter 7 The Unit Circle: Sine and Cosine Functions 872 cos 310° 219. sin 310° Extensions For the following exercises, evaluate. 220. 221. 222. 223. 224. 225. 226. 227. 228. 229. ⎛ sin ⎝ 11π 3 ⎛ ⎞ ⎠ cos ⎝ −5π 6 ⎞ ⎠ ⎛ sin ⎝ 3π 4 ⎛ ⎞ ⎠ cos ⎝ ⎞ ⎠ 5π 3 ⎛ ⎝− 4π sin 3 ⎞ ⎛ ⎠ cos ⎝ ⎞ ⎠ π 2 ⎛ sin ⎝ −9π 4 ⎛ ⎞ ⎠ cos ⎝ −π 6 ⎞ ⎠ ⎛ sin ⎝ π 6 ⎞ ⎛ ⎠ cos ⎝ −π 3 ⎞ ⎠ ⎛ sin ⎝ 7π 4 ⎛ ⎞ ⎠cos ⎝ −2π 3 ⎞ ⎠ ⎛ cos ⎝ 5π 6 ⎞ ⎛ ⎠ cos ⎝ ⎞ ⎠ 2π 3 ⎛ cos ⎝ −π 3 ⎞ ⎛ ⎠cos ⎝ ⎞ ⎠ π 4 ⎛ sin ⎝ −5π 4 ⎛ ⎞ ⎠ sin ⎝ 11π 6 ⎞ ⎠ ⎛ sin(π)sin ⎝ ⎞ ⎠ π 6 Real-World Applications For the following exercises, use this scenario: A child enters a carousel that takes one minute to revolve once around. The child enters at the point (0, 1), that is, on the due north position. Assume the carousel revolves counter clockwise. What are the coordinates of 230. seconds? the child after 45 What are the coordinates of 231. seconds? the child after 90 What are the coordinates of the child after 125 232. seconds? will When coordinates 233. (0.707, –0.707) if the ride lasts 6 minutes? (There are multiple answers.) child have the When have 234. (–0.866, –0.5) if the ride lasts 6 minutes? child will the coordinates This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 873 7.4 \| The Other Trigonometric Functions Learning Objectives In this section you will: 7.4.1 Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent of π 3 , and π 6 π 4 . , 7.4.2 Use reference angles to evaluate the trigonometric functions secant, cotangent. 7.4.3 Use properties of even and odd trigonometric functions. 7.4.4 Recognize and use fundamental identities. 7.4.5 Evaluate trigonometric functions with a calculator. tangent, and A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle with the ground whose tangent is 1 12 or less, regardless of its length. A tangent represents a ratio, so this means that for every 1 inch of rise, the ramp must have 12 inches of run. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions. Finding Exact Values of the Trigonometric Functions Secant, Cosecant, Tangent, and Cotangent We can also define the remaining functions in terms of the unit circle with a point (x, y) corresponding to an angle of t, as shown in Figure 7.62. As with the sine and cosine, we can use the (x, y) coordinates to find the other functions. Figure 7.62 The first function we will define is the tangent. The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. In Figure 7.62, the tangent of angle t is equal to y x, x ≠ 0. Because the ythe tangent of angle t can also be defined value is equal to the sine of t, and the x-value is equal to the cosine of t, as sin t cos t, cos t ≠ 0. The tangent function is abbreviated as tan. The remaining three functions can all be exp
ressed as reciprocals of functions we have already defined. • The secant function is the reciprocal of the cosine function. In Figure 7.62, the secant of angle t is equal to cos t = 1 1 x, x ≠ 0. The secant function is abbreviated as sec. • The cotangent function is the reciprocal of the tangent function. In Figure 7.62, the cotangent of angle t is equal to cos t sin t = x y, y ≠ 0. The cotangent function is abbreviated as cot. 874 Chapter 7 The Unit Circle: Sine and Cosine Functions • The cosecant function is the reciprocal of the sine function. In Figure 7.62, the cosecant of angle t is equal to 1 sin t = 1 y, y ≠ 0. The cosecant function is abbreviated as csc. Tangent, Secant, Cosecant, and Cotangent Functions If t is a real number and (x, y) is a point where the terminal side of an angle of t radians intercepts the unit circle, then y x, x ≠ 0 tan t = sec t = 1 x, x ≠ 0 csc t = 1 y, y ≠ 0 x y, y ≠ 0 cot t = Example 7.25 Finding Trigonometric Functions from a Point on the Unit Circle The point ⎛ ⎝− 3 2 , 1 2 cot t. ⎞ is on the unit circle, as shown in Figure 7.63. Find sin t, cos t, tan t, sec t, csc t, and ⎠ Figure 7.63 Solution Because we know the (x, y) coordinates of the point on the unit circle indicated by angle t, we can use those coordinates to find the six functions: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 875 sin t = y = 1 2 cos t = x = − 3 2 tan t = sec t = 1 csc t = 1 cot ⎛ ⎝− .25 The point ⎛ ⎝ sin t, cos t, tan t, sec t, csc t, and cot t. , − 2 2 ⎞ is ⎠ 2 2 on the unit circle, as shown in Figure 7.64. Find Figure 7.64 Example 7.26 Finding the Trigonometric Functions of an Angle Find sin t, cos t, tan t, sec t, csc t, and cot t. when t = π 6 . Solution We have previously used the properties of equilateral triangles to demonstrate that sin π 6 = 1 2 and cos π 6 = 3 2 . We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values. 876 Chapter 7 The Unit Circle: Sine and Cosine Functions tan π 6 sec π 6 csc π 6 cot π 6 = sin π 6 cos π 6 1 2 3 2 = 1 = cos sin π 6 cos π 6 sin π 6 = = 3 2 1 2 7.26 Find sin t, cos t, tan t, sec t, csc t, and cot t. when t = π 3 . Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting x equal to the cosine and y equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in Table 7.4. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 877 Angle 0 π 6 , or 30° π 4 , or 45° π 3 , or 60° π 2 , or 90° Cosine Sine Tangent Secant 1 0 0 1 Cosecant Undefined Cotangent Undefined Table 7. Undefined Undefined 1 0 Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x- and y-values in the original quadrant. Figure 7.65 shows which functions are positive in which quadrant. To help remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “A,” all of the six trigonometric functions are positive. In quadrant II, “Smart,” only sine and its reciprocal function, cosecant, are positive. In quadrant III, “Trig,” only tangent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Class,” only cosine and its reciprocal function, secant, are positive. 878 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.65 The trigonometric functions are each listed in the quadrants in which they are positive. Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions. 1. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle. 2. Evaluate the function at the reference angle. 3. Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative. Example 7.27 Using Reference Angles to Find Trigonometric Functions Use reference angles to find all six trigonometric functions of − 5π 6 . Solution The angle between this angle’s terminal side and the x-axis is π 6 , so that is the reference angle. Since − 5π 6 is in the third quadrant, where both x and y are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive. ⎛ ⎝− 5π cos 6 ⎛ ⎝− 5π sec ⎛ ⎝− 5π , sin 6 ⎛ ⎝− 5π , csc 2, cot ⎛ 5π , tan ⎝ 6 ⎛ ⎝− 5π .27 Use reference angles to find all six trigonometric functions of − 7π 4 . Using Even and Odd Trigonometric Functions To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 879 Consider the function f (x) = x2, shown in Figure 7.66. The graph of the function is symmetrical about the y-axis. All along the curve, any two points with opposite x-values have the same function value. This matches the result of calculation: (4)2 = (−4)2, (−5)2 = (5)2, and so on. So f (x) = x2 is an even function, a function such that two inputs that are opposites have the same output. That means f (−x) = f (x). Figure 7.66 The function f (x) = x2 is an even function. Now consider the function f (x) = x3, shown in Figure 7.67. The graph is not symmetrical about the y-axis. All along the graph, any two points with opposite x-values also have opposite y-values. So f (x) = x3 is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means f (−x) = − f (x). 880 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.67 The function f (x) = x3 is an odd function. We can test whether a trigonometric function is even or odd by drawing a unit circle with a positive and a negative angle, as in Figure 7.68. The sine of the positive angle is y. The sine of the negative angle is −y. The sine function, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are shown in Table 7.5. Figure 7.68 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 881 sin t = y sin(−t) = −y sin t ≠ sin(−t) cos t = x cos(−t) = x cos t = cos(−t) tan(t) = y x tan(−t) = − y x tan t ≠ tan(−t) sec t = 1 x sec(−t) = 1 x csc t = 1 y csc(−t) = 1 −y cot t = cot(−t) = x y x −y sec t = sec(−t) csc t ≠ csc(−t) cot t ≠ cot(−t) Table 7.5 Even and Odd Trigonometric Functions An even function is one in which f (−x) = f (x). An odd function is one in which f (−x) = − f (x). Cosine and secant are even: Sine, tangent, cosecant, and cotangent are odd: cos(−t) = cos t sec(−t) = sec t sin(−t) = −sin t tan(−t) = −tan t csc(−t) = −csc t cot(−t) = −cot t Example 7.28 Using Even and Odd Properties of Trigonometric Functions If the secant of angle t is 2, what is the secant of −t ? Solution Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle t is 2, the secant of −t is also 2. 7.28 If the cotangent of angle t is 3, what is the cotangent of −t ? Recognizing and Using Fundamental Identities We have explored a number of properties of trigonometric functions. Now, we can take the relationships a step further, and derive some fundamental identities. Identities are statements that are true for all values of the input on which they are defined. Usually, identities can be derived from definitions and relationships we already know. For example, the Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine. 882 Chapter 7 The Unit Circle: Sine and Cosine Functions Fundamental Identities We can derive some useful identities from the six trigonometric functions. The other four trigonometric functions can be related back to the sine and cosine functions using these basic relationships: tan t = sin t cos t sec t = 1 cos t csc t = 1 sin t tan t = cos t sin t cot t = 1 (7.16) (7.17) (7.18) (7.19) Example 7.29 Using Identities to Evaluate Trigonometric Functions a. Given sin(45°) = 2 2 , cos(45°) = 2 2 , evaluate tan(45°). ⎛ b. Given sin ⎝ 5π 6 ⎞ ⎠ = 1 2 ⎛ , cos ⎝ 5π 6 ⎞ ⎠ = − 3 2 ⎛ , evaluate sec ⎝ ⎞ ⎠. 5π 6 Solution Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions. a. b. tan(45°) = sin(45°) cos(45°) 2 2 2 2 = 1 = ⎛ sec ⎝ 5π 6 ⎞ ⎠ = = ⎞ ⎠ 5π 6 1 ⎛ cos ⎝ 1 − 3 2 = −2 3 1 = −2 3 = − 2 3 3 7.29 ⎛ Evaluate csc ⎝ ⎞ ⎠. 7π 6 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 883 Example 7.30 Using Identities to Simplify Trigonometric Expressions Simplify sec t tan t . Solution We can simplify this by rewriting both functions in terms of sine and cosine. sec t tan t = = = 1 cos t sin t cos t cos t cos t 1 sin t 1 sin t = csc t Multiply by the reciprocal. Simplify and use the
identity. By showing that sec t tan t can be simplified to csc t, we have, in fact, established a new identity. sec t tan t = csc t 7.30 Simplify tan t(cos t). Alternate Forms of the Pythagorean Identity We can use these fundamental identities to derive alternate forms of the Pythagorean Identity, cos2 t + sin2 t = 1. One form is obtained by dividing both sides by cos2 t. cos2 t cos2 t = + sin2 t cos2 t 1 cos2 t 1 + tan2 t = sec2 t The other form is obtained by dividing both sides by sin2 t. cos2 t sin2 t = + sin2 t sin2 t 1 sin2 t cot2 t + 1 = csc2 t Alternate Forms of the Pythagorean Identity 1 + tan2 t = sec2 t cot2 t + 1 = csc2 t Example 7.31 Using Identities to Relate Trigonometric Functions 884 Chapter 7 The Unit Circle: Sine and Cosine Functions If cos(t) = 12 13 functions. and t is in quadrant IV, as shown in Figure 7.69, find the values of the other five trigonometric Figure 7.69 Solution We can find the sine using the Pythagorean Identity, cos2 t + sin2 t = 1, and the remaining functions by relating them to sine and cosine. 2 ⎛ ⎝ 12 13 ⎞ ⎠ + sin2 t = 1 2 ⎛ ⎝ sin2 t = 1 − ⎞ 12 ⎠ 13 sin2 t = 1 − 144 169 sin2 t = 25 169 sin t = ± 25 169 sin t = ± 25 169 sin t = ± 5 13 The sign of the sine depends on the y-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y-values are negative, its sine is negative, − 5 13 . The remaining functions can be calculated using identities relating them to sine and cosine. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 885 tan t = sin t cos t = − 5 13 12 13 = − 5 12 sec t = 1 cos t = 1 12 13 csc t = 1 cot t = 1 sin t = 1 − 5 13 tan t = 1 − 5 12 = 13 12 = 13 5 = − 12 5 7.31 If sec(t) = − 17 8 and 0 < t < π, find the values of the other five functions. As we discussed at the beginning of the chapter, a function that repeats its values in regular intervals is known as a periodic function. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or 2π, will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs. Other functions can also be periodic. For example, the lengths of months repeat every four years. If x represents the length time, measured in years, and f (x) represents the number of days in February, then f (x + 4) = f (x). This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that satisfies this condition and is called the period. A period is the shortest interval over which a function completes one full cycle—in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days. Period of a Function The period P of a repeating function f is the number representing the interval such that f (x + P) = f (x) for any value of x. The period of the cosine, sine, secant, and cosecant functions is 2π. The period of the tangent and cotangent functions is π. Example 7.32 Finding the Values of Trigonometric Functions Find the values of the six trigonometric functions of angle t based on Figure 7.70. 886 Chapter 7 The Unit Circle: Sine and Cosine Functions Figure 7.70 Solution sin t = y cos t = x cos cos t = 1 − 1 2 sin t = 1 − 3 2 tan t = 1 3 tan t = sin t sec t = 1 csc t = 1 cot t = 1 = 3 = −.32 Find the values of the six trigonometric functions of angle t based on Figure 7.71. Figure 7.71 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 887 Example 7.33 Finding the Value of Trigonometric Functions If sin(t) = − 3 2 and cos(t) = 1 2 , fin sec(t), csc(t), tan(t), cot(t). Solution sec t = 1 cos t = 1 1 2 = 2 csc t = 1 tan t = sin t sin cos t = − 2 3 3 = − 3 cot t = 1 tan .33 sin(t) = 2 2 and cos(t) = 2 2 , fin sec(t), csc(t), tan(t), and cot(t) Evaluating Trigonometric Functions with a Calculator We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation. Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent. If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor π 180 to convert the degrees to radians. To find the secant of 30°, we could press (for a scientific calcula or): 1 30 × π 180 COS or (for a graphing calculator): 1 ⎛ 30π cos ⎝ 180 ⎞ ⎠ Given an angle measure in radians, use a scientific calculator to find the cosecant. 1. If the calculator has degree mode and radian mode, set it to radian mode. 2. Enter: 1/ 3. Enter the value of the angle inside parentheses. 4. Press the SIN key. 5. Press the = key. 888 Chapter 7 The Unit Circle: Sine and Cosine Functions Given an angle measure in radians, use a graphing utility/calculator to find the cosecant. • If the graphing utility has degree mode and radian mode, set it to radian mode. • Enter: 1/ • Press the SIN key. • Enter the value of the angle inside parentheses. • Press the ENTER key. Example 7.34 Evaluating the Secant Using Technology Evaluate the cosecant of 5π 7 . Solution For a scientific calculator, enter information as follows: 1/(5 × π / 7) SIN = ⎛ csc ⎝ 5π 7 ⎞ ⎠ ≈ 1.279 7.34 Evaluate the cotangent of − π 8 . Access these online resources for additional instruction and practice with other trigonometric functions. • Determing Trig Function Values (http://Openstaxcollege.org/l/trigfuncval) • More Examples of Determining Trig Functions (http://Openstaxcollege.org/l/moretrigfun) • Pythagorean Identities (http://Openstaxcollege.org/l/pythagiden) • Trig Functions on a Calculator (http://Openstaxcollege.org/l/trigcalc) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 889 7.4 EXERCISES Verbal 235. On an interval of [0, 2π), can the sine and cosine values of a radian measure ever be equal? If so, where? What would you estimate the cosine of π degrees to 236. be? Explain your reasoning. For any angle in quadrant II, if you knew the sine of 237. the angle, how could you determine the cosine of the angle? 238. Describe the secant function. Tangent and cotangent have a period of π. What does 239. this tell us about the output of these functions? Algebraic 253. sec 7π 6 254. csc 11π 6 255. cot 13π 6 256. 257. 258. tan 7π 4 sec 3π 4 csc 5π 4 For the following exercises, find the exact value of each expression. 259. cot 11π 4 240. 241. 242. 243. 244. 245. 246. 247. 248. 249. 250. 251. tan π 6 sec π 6 csc π 6 cot π 6 tan π 4 sec π 4 csc π 4 cot π 4 tan π 3 sec π 3 csc π 3 cot π 3 260. 261. 262. 263. tan 8π 3 sec 4π 3 csc 2π 3 cot 5π 3 264. tan 225° 265. sec 300° 266. csc 150° 267. cot 240° 268. tan 330° 269. sec 120° 270. csc 210° 271. cot 315° 272. If sin t = 3 4 cos t, sec t, csc t, tan t, and cot t. , and t is in quadrant II, find the following exercises, use reference angles to For evaluate the expression. 273. If cos t = − 1 3 , and t is in quadrant III, find 252. tan 5π 6 sin t, sec t, csc t, tan t, and cot t. 274. 890 Chapter 7 The Unit Circle: Sine and Cosine Functions If tan t = 12 5 , and 0 ≤ t < , find π 2 sin t, cos t, sec t, csc t, and cot t. 275. If sin t = 3 2 and cos t = 1 2 , find sec t, csc t, tan t, and cot t. If sin 40° ≈ 0.643 and cos 40° ≈ 0.766, find 276. sec 40°, csc 40°, tan 40°, and cot 40°. 277. If sin t = 2 2 , what is the sin(−t) ? 278. If cos t = 1 2 , what is the cos(−t) ? 279. If sec t = 3.1, what is the sec(−t) ? 280. If csc t = 0.34, what is the csc(−t) ? 285. 281. If tan t = −1.4, what is the tan(−t) ? 282. If cot t = 9.23, what is the cot(−t) ? Graphical For the following exercises, use the angle in the unit circle to find the value of the each of the six trigonometric functions. 283. 284. This content is available for free at https://cnx.org/content/col11758/1.5 Technology For the following exercises, use a graphing calculator to evaluate to three decimal places. 286. 287. 288. 289. 290. csc 5π 9 cot 4π 7 sec π 10 tan 5π 8 sec 3π 4 Chapter 7 The Unit Circle: Sine and Cosine Functions 891 291. csc π 4 292. tan 98° 293. cot 33° 294. cot 140° 295. sec 310° Extensions For the following exercises, use identities to evaluate the expression. 296. If tan(t) ≈ 2.7, and sin(t) ≈ 0.94, find cos(t). 297. If tan(t) ≈ 1.3, and cos(t) ≈ 0.61, find sin(t). 298. If csc(t) ≈ 3.2, and cos(t) ≈ 0.95, find tan(t). 299. If cot(t) ≈ 0.58, and cos(t) ≈ 0.5, find csc(t). 300. Determine whether the function f (x) = 2sinx cos x is even, odd, or neither. year. Use the equation to find how many hours of sunlight there are on September 24, the 267th day of the year. State the period of the function. The equation P = 20sin(2πt) + 100 models the 308. blood pressure, P, where t represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures? The height of a piston, h, in inches, can be modeled 309. by the equation y = 2cos x + 6, where x represents the crank angle. Find the height of the piston when the crank angle is 55°. The height of a piston, h, in i
nches, can be modeled 310. by the equation y = 2cos x + 5, where x represents the crank angle. Find the height of the piston when the crank angle is 55°. Determine whether 301. f (x) = 3sin2 x cos x + sec x is even, odd, or neither. the function Determine 302. f (x) = sin x − 2cos2 x is even, odd, or neither. whether the Determine 303. f (x) = csc2 x + sec x is even, odd, or neither. whether the function function For the following exercises, use identities to simplify the expression. 304. csc t tan t 305. sec t csc t Real-World Applications The amount of sunlight 306. ⎛ modeled by the function h = 15cos ⎝ in a certain city can be d⎞ ⎠, where h represents the hours of sunlight, and d is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42nd day of the year. State the period of the function. 1 600 The amount of sunlight 307. ⎛ modeled by the function h = 16cos ⎝ in a certain city can be d⎞ ⎠, where h represents the hours of sunlight, and d is the day of the 1 500 892 Chapter 7 The Unit Circle: Sine and Cosine Functions CHAPTER 7 REVIEW KEY TERMS adjacent side in a right triangle, the side between a given angle and the right angle angle the union of two rays having a common endpoint angle of depression the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned lower than the observer angle of elevation the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned higher than the observer angular speed the angle through which a rotating object travels in a unit of time arc length the length of the curve formed by an arc area of a sector area of a portion of a circle bordered by two radii and the intercepted arc; the fraction θ 2π. multiplied by the area of the entire circle cosecant the reciprocal of the sine function: on the unit circle, csc t = 1 y, y ≠ 0 cosine function the x-value of the point on a unit circle corresponding to a given angle cotangent the reciprocal of the tangent function: on the unit circle, cot t = x y, y ≠ 0 coterminal angles description of positive and negative angles in standard position sharing the same terminal side degree a unit of measure describing the size of an angle as one-360th of a full revolution of a circle hypotenuse the side of a right triangle opposite the right angle identities statements that are true for all values of the input on which they are defined initial side the side of an angle from which rotation begins linear speed the distance along a straight path a rotating object travels in a unit of time; determined by the arc length measure of an angle the amount of rotation from the initial side to the terminal side negative angle description of an angle measured clockwise from the positive x-axis opposite side in a right triangle, the side most distant from a given angle period the smallest interval P of a repeating function f such that f (x + P) = f (x) positive angle description of an angle measured counterclockwise from the positive x-axis Pythagorean Identity a corollary of the Pythagorean Theorem stating that the square of the cosine of a given angle plus the square of the sine of that angle equals 1 quadrantal angle an angle whose terminal side lies on an axis radian the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle radian measure the ratio of the arc length formed by an angle divided by the radius of the circle ray one point on a line and all points extending in one direction from that point; one side of an angle reference angle the measure of the acute angle formed by the terminal side of the angle and the horizontal axis secant the reciprocal of the cosine function: on the unit circle, sec t = 1 x, x ≠ 0 sine function the y-value of the point on a unit circle corresponding to a given angle standard position the position of an angle having the vertex at the origin and the initial side along the positive x-axis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 893 tangent the quotient of the sine and cosine: on the unit circle, tan t = y x, x ≠ 0 terminal side the side of an angle at which rotation ends unit circle a circle with a center at (0, 0) and radius 1 vertex the common endpoint of two rays that form an angle KEY EQUATIONS arc length area of a sector angular speed linear speed s = rθ A = 1 2 θr 2 ω = θ t v = s t linear speed related to angular speed v = rω 894 Chapter 7 The Unit Circle: Sine and Cosine Functions Trigonometric Functions Reciprocal Trigonometric Functions Cofunction Identities Sine sin t = Cosine cos t = Tangent tan t = Secant sec t = Cosecant csc t = Cotangent cot t = opposite hypotenuse adjacent hypotenuse opposite adjacent hypotenuse adjacent hypotenuse opposite adjacent opposite sin t = 1 csc t cos t = 1 sec t tan t = 1 cot t csc t = 1 sin t sec t = 1 cos t cot t = 1 tan t tan t = cot ⎛ cos t = sin ⎝ π 2 π ⎛ sin t = cos ⎝ 2 π 2 π 2 π ⎛ sec t = csc ⎝ 2 ⎛ cot t = tan ⎝ ⎛ ⎝ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ Cosine Sine cos t = x sin t = y Pythagorean Identity cos2 t + sin2 t = 1 Tangent function Secant function Cosecant function tan t = sin t cos t sec t = 1 cos t csc t = 1 sin t Cotangent function cot t = 1 tan t = cos t sin t This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 895 KEY CONCEPTS 7.1 Angles • An angle is formed from the union of two rays, by keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. • An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise. • To draw an angle in standard position, draw the initial side along the positive x-axis and then place the terminal side according to the fraction of a full rotation the angle represents. See Example 7.1. • In addition to degrees, the measure of an angle can be described in radians. See Example 7.2. • To convert between degrees and radians, use the proportion θ 180 θ R π . See Example 7.3 and Example 7.4. = • Two angles that have the same terminal side are called coterminal angles. • We can find coterminal angles by adding or subtracting 360° or 2π. See Example 7.5 and Example 7.6. • Coterminal angles can be found using radians just as they are for degrees. See Example 7.7. • The length of a circular arc is a fraction of the circumference of the entire circle. See Example 7.8. • The area of sector is a fraction of the area of the entire circle. See Example 7.9. • An object moving in a circular path has both linear and angular speed. • The angular speed of an object traveling in a circular path is the measure of the angle through which it turns in a unit of time. See Example 7.10. • The linear speed of an object traveling along a circular path is the distance it travels in a unit of time. See Example 7.11. 7.2 Right Triangle Trigonometry • We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example 7.12. • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example 7.13. • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example 7.14. • Any two complementary angles could be the two acute angles of a right triangle. • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example 7.15. • We can use trigonometric functions of an angle to find unknown side lengths. • Select the trigonometric function representing the ratio of the unknown side to the known side. See Example 7.16. • Right-triangle trigonometry facilitates the measurement of inaccessible heights and distances. • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example 7.17. 7.3 Unit Circle • Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit. • Using the unit circle, the sine of an angle t equals the y-value of the endpoint on the unit circle of an arc of length t whereas the cosine of an angle t equals the x-value of the endpoint. See Example 7.18. 896 Chapter 7 The Unit Circle: Sine and Cosine Functions • The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See Example 7.19. • When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example 7.20. • Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See Example 7.21. • The domain of the sine and cosine functions is all real numbers. • The range of both the sine and cosine functions is [−1, 1]. • The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle. • The signs of the sine and cosine are determined from the x- and y-values in the quadrant of the original angle. • An angle’s reference angle is the size angle, t, formed by the terminal side of the angle t and the horizontal axis. See Example 7.22. • Reference angles can be used to find the sine and cosine of the original angle. See Example 7.23. • Reference angles can also be used to find the coordinates of a point on a circle. See Example 7.24. 7.4 The Other Trigonometric Function
s • The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. • The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function. • The six trigonometric functions can be found from a point on the unit circle. See Example 7.25. • Trigonometric functions can also be found from an angle. See Example 7.26. • Trigonometric functions of angles outside the first quadrant can be determined using reference angles. See Example 7.27. • A function is said to be even if f (−x) = f (x) and odd if f (−x) = − f (x) for all x in the domain of f. • Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd. • Even and odd properties can be used to evaluate trigonometric functions. See Example 7.28. • The Pythagorean Identity makes it possible to find a cosine from a sine or a sine from a cosine. • Identities can be used to evaluate trigonometric functions. See Example 7.29 and Example 7.30. • Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce new identities. See Example 7.31.The trigonometric functions repeat at regular intervals. • The period P of a repeating function f is the smallest interval such that f (x + P) = f (x) for any value of x. • The values of trigonometric functions can be found by mathematical analysis. See Example 7.32 and Example 7.33. • To evaluate trigonometric functions of other angles, we can use a calculator or computer software. See Example 7.34. CHAPTER 7 REVIEW EXERCISES Angles For the following exercises, convert the angle measures to degrees. 311. π 4 312. − 5π 3 For the following exercises, convert the angle measures to radians. 313. −210° This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 897 329. tan π 6 330. ⎛ cos ⎝ π 2 ⎞ ⎠ = sin(___°) 331. csc(18°) = sec(___°) For the following exercises, use the given information to find the lengths of the other two sides of the right triangle. 332. cos B = 3 5 , a = 6 333. tan A = 5 9 , b = 6 For the following exercises, use Figure 7.72 to evaluate each trigonometric function. Figure 7.72 334. sin A 335. tan B For the following exercises, solve for the unknown sides of the given triangle. 336. 337. 314. 180° 315. Find the length of an arc in a circle of radius 7 meters subtended by the central angle of 85°. 316. Find the area of the sector of a circle with diameter 32 feet and an angle of 3π 5 radians. For the following exercises, find the angle between 0° and 360° that is coterminal with the given angle. 317. 420° 318. −80° For the following exercises, find the angle between 0 and 2π in radians that is coterminal with the given angle. 319. − 20π 11 320. 14π 5 For the following exercises, draw the angle provided in standard position on the Cartesian plane. 321. −210° 322. 75° 323. 5π 4 324. − π 3 325. Find the linear speed of a point on the equator of the earth if the earth has a radius of 3,960 miles and the earth rotates on its axis every 24 hours. Express answer in miles per hour. Round to the nearest hundredth. 326. A car wheel with a diameter of 18 inches spins at the rate of 10 revolutions per second. What is the car's speed in miles per hour? Round to the nearest hundredth. Right Triangle Trigonometry For the following exercises, use side lengths to evaluate. 327. 328. cos π 4 cot π 3 898 Chapter 7 The Unit Circle: Sine and Cosine Functions 354. sec 315° 355. If sec(t) = −2.5, what is the sec( − t) ? 356. If tan(t) = −0.6, what is the tan( − t) ? 357. If tan(t) = 1 3 , find tan(t − π). 358. If cos(t) = 2 2 , find sin(t + 2π). 359. Which trigonometric functions are even? 360. Which trigonometric functions are odd? 338. A 15-ft ladder leans against a building so that the angle between the ground and the ladder is 70°. How high does the ladder reach up the side of the building? Find the answer to four decimal places. 339. The angle of elevation to the top of a building in Baltimore is found to be 4 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. Find the answer to four decimal places. Unit Circle 340. Find the exact value of sin π 3 . 341. Find the exact value of cos π 4 . 342. Find the exact value of cos π. 343. State the reference angle for 300°. 344. State the reference angle for 3π 4 . 345. Compute cosine of 330°. 346. Compute sine of 5π 4 . 347. State the domain of the sine and cosine functions. 348. State the range of the sine and cosine functions. The Other Trigonometric Functions For the following exercises, find the exact value of the given expression. cos π 6 349. 350. 351. 352. tan π 4 csc π 3 sec π 4 the following exercises, use reference angles to For evaluate the given expression. 353. sec 11π 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 7 The Unit Circle: Sine and Cosine Functions 899 CHAPTER 7 PRACTICE TEST 361. Convert 5π 6 radians to degrees. distance of 2000 feet from the base of the building. Using this information, find the height of the building. 362. Convert −620° to radians. 373. Find the exact value of sin π 6 . 363. Find the length of a circular arc with a radius 12 centimeters subtended by the central angle of 30°. 374. Compute sine of 240°. 364. Find the area of the sector with radius of 8 feet and an angle of 5π 4 radians. 375. State the domain of the sine and cosine functions. 376. State the range of the sine and cosine functions. Find the angle between 0° and 360° 365. coterminal with 375°. that is 377. Find the exact value of cot π 4 . 366. Find the angle between 0 and 2π in radians that is coterminal with − 4π 7 . 378. Find the exact value of tan π 3 . 379. Use reference angles to evaluate csc 7π 4 . 367. Draw the angle 315° in standard position on the Cartesian plane. 380. Use reference angles to evaluate tan 210°. 368. Draw the angle − π 6 Cartesian plane. in standard position on the 381. If csc t = 0.68, what is the csc( − t) ? 369. A carnival has a Ferris wheel with a diameter of 80 feet. The time for the Ferris wheel to make one revolution is 75 seconds. What is the linear speed in feet per second of a point on the Ferris wheel? What is the angular speed in radians per second? 382. If cos t = 3 2 , find cos(t − 2π). ⎛ 383. Find the missing angle: cos ⎝ ⎞ ⎠ = sin(___) π 6 the missing sides of the triangle 370. Find ABC : sin B = 3 4 , c = 12. 371. Find the missing sides of the triangle. 372. The angle of elevation to the top of a building in Chicago is found to be 9 degrees from the ground at a 900 Chapter 7 The Unit Circle: Sine and Cosine Functions This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 901 8 \| PERIODIC FUNCTIONS Figure 8.1 (credit: "Maxxer_", Flickr) Chapter Outline 8.1 Graphs of the Sine and Cosine Functions 8.2 Graphs of the Other Trigonometric Functions 8.3 Inverse Trigonometric Functions Introduction Each day, the sun rises in an easterly direction, approaches some maximum height relative to the celestial equator, and sets in a westerly direction. The celestial equator is an imaginary line that divides the visible universe into two halves in much the same way Earth’s equator is an imaginary line that divides the planet into two halves. The exact path the sun appears to follow depends on the exact location on Earth, but each location observes a predictable pattern over time. The pattern of the sun’s motion throughout the course of a year is a periodic function. Creating a visual representation of a periodic function in the form of a graph can help us analyze the properties of the function. In this chapter, we will investigate graphs of sine, cosine, and other trigonometric functions. 902 Chapter 8 Periodic Functions 8.1 \| Graphs of the Sine and Cosine Functions Learning Objectives In this section, you will: 8.1.1 Graph variations of y=sin( x ) and y=cos( x ). 8.1.2 Use phase shifts of sine and cosine curves. Figure 8.2 Light can be separated into colors because of its wavelike properties. (credit: "wonderferret"/ Flickr) White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow. Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions (https://cnx.org/content/m49369/latest/) , we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions. Graphing Sine and Cosine Functions Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function. We can create a table of values and use them to sketch a graph. Table 8.1 lists some of the values for the sine function on a unit circle. x sin(x 2π 3 3 2 3π 4 2 2 5π 6 1 2 π 0 Table 8.1 Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure 8.3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 903 Figure 8.3 The sine function Notice how the sine values are positive between 0 and π, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between π and 2π, which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 8.4. Figure 8.4 Plotting values of
the sine function Now let’s take a similar look at the cosine function. Again, we can create a table of values and use them to sketch a graph. Table 8.2 lists some of the values for the cosine function on a unit circle 2π 3 − 1 2 3π 4 5π 6 − 2 2 − 3 2 π −1 x cos(x) 0 1 Table 8.2 As with the sine function, we can plots points to create a graph of the cosine function as in Figure 8.5. Figure 8.5 The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1, 1]. 904 Chapter 8 Periodic Functions In both graphs, the shape of the graph repeats after 2π, which means the functions are periodic with a period of 2π. A periodic function is a function for which a specific horizontal shift, P, results in a function equal to the original function: f (x + P) = f (x) for all values of x in the domain of f . When this occurs, we call the smallest such horizontal shift with P > 0 the period of the function. Figure 8.6 shows several periods of the sine and cosine functions. Figure 8.6 Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. As we can see in Figure 8.7, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because sin(−x) = −sin x. Now we can clearly see this property from the graph. Figure 8.7 Odd symmetry of the sine function Figure 8.8 shows that the cosine function is symmetric about the y-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that cos(−x) = cos x. Figure 8.8 Even symmetry of the cosine function This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 905 Characteristics of Sine and Cosine Functions The sine and cosine functions have several distinct characteristics: • They are periodic functions with a period of 2π. • The domain of each function is (−∞, ∞) and the range is [−1, 1]. • The graph of y = sin x is symmetric about the origin, because it is an odd function. • The graph of y = cos x is symmetric about the y- axis, because it is an even function. Investigating Sinusoidal Functions As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. The general forms of sinusoidal functions are y = Asin(Bx − C) + D and y = Acos(Bx − C) + D (8.1) Determining the Period of Sinusoidal Functions Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period. In the general formula, B is related to the period by P = 2π \|B\| . If \|B\| > 1, then the period is less than 2π and the function undergoes a horizontal compression, whereas if \|B\| < 1, then the period is greater than 2π and the function undergoes a horizontal stretch. For example, f (x) = sin(x), B = 1, so the period is 2π, which we knew. If f (x) = sin(2x), then ⎛ B = 2, so the period is π and the graph is compressed. If f (x) = sin ⎝ x 2 ⎞ ⎠, then B = 1 2 is stretched. Notice in Figure 8.9 how the period is indirectly related to \|B\|. , so the period is 4π and the graph Figure 8.9 Period of Sinusoidal Functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms The period is 2π \|B\| . y = Asin(Bx) y = Acos(Bx) 906 Chapter 8 Periodic Functions Example 8.1 Identifying the Period of a Sine or Cosine Function ⎛ Determine the period of the function f (x) = sin ⎝ x⎞ ⎠. π 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx). In the given equation, B = , so the period will be π 6 P = 2π \|B\| = 2π π 6 = 2π ⋅ 6 π = 12 8.1 ⎛ Determine the period of the function g(x) = cos ⎝ ⎞ ⎠. x 3 Determining Amplitude Returning to the general formula for a sinusoidal function, we have analyzed how the variable B relates to the period. Now let’s turn to the variable A so we can analyze how it is related to the amplitude, or greatest distance from rest. A represents the vertical stretch factor, and its absolute value \|A\| is the amplitude. The local maxima will be a distance \|A\| above the vertical midline of the graph, which is the line x = D; because D = 0 in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If \|A\| > 1, the function is stretched. For example, the amplitude of f (x) = 4 sin x is twice the amplitude of f (x) = 2 sin x. If \|A\| < 1, the function is compressed. Figure 8.10 compares several sine functions with different amplitudes. Figure 8.10 Amplitude of Sinusoidal Functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 907 The amplitude is A, and the vertical height from the midline is \|A\|. In addition, notice in the example that y = Asin(Bx) and y = Acos(Bx) \|A\| = amplitude = 1 2\|maximum − minimum\| Example 8.2 Identifying the Amplitude of a Sine or Cosine Function What is the amplitude of the sinusoidal function f (x) = −4sin(x) ? Is the function stretched or compressed vertically? Solution Let’s begin by comparing the function to the simplified form y = Asin(Bx). In the given function, A = −4, so the amplitude is \|A\| = \|−4\| = 4. The function is stretched. Analysis The negative value of A results in a reflection across the x-axis of the sine function, as shown in Figure 8.11. Figure 8.11 8.2 What is the amplitude of the sinusoidal function f (x) = 1 2 sin(x) ? Is the function stretched or compressed vertically? Analyzing Graphs of Variations of y = sin x and y = cos x Now that we understand how A and B relate to the general form equation for the sine and cosine functions, we will explore the variables C and D. Recall the general form: y = Asin(Bx − C) + D and y = Acos(Bx − C) + D or ⎛ ⎝B⎛ y = Asin ⎝x − C B ⎞ ⎞ ⎛ ⎝B⎛ ⎠ + D and y = Acos ⎠ ⎝x − C B ⎞ ⎞ ⎠ + D ⎠ The value C B for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If C > 0, the graph shifts to the right. If C < 0, the graph shifts to the left. The greater the value of \|C\|, the 908 Chapter 8 Periodic Functions more the graph is shifted. Figure 8.12 shows that the graph of f (x) = sin(x − π) shifts to the right by π units, which is ⎛ more than we see in the graph of f (x) = sin ⎝x − π 4 ⎞ ⎠, which shifts to the right by π 4 units. Figure 8.12 While C relates to the horizontal shift, D indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure 8.13. The function y = cos(x) + D has its midline at y = D. Figure 8.13 Any value of D other than zero shifts the graph up or down. Figure 8.14 compares f (x) = sin x with f (x) = sin x + 2, which is shifted 2 units up on a graph. Figure 8.14 Variations of Sine and Cosine Functions Given an equation in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D, C B is the phase shift and D is the vertical shift. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 909 Example 8.3 Identifying the Phase Shift of a Function ⎛ Determine the direction and magnitude of the phase shift for f (x) = sin ⎝x + ⎞ ⎠ − 2. π 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. In the given equation, notice that B = 1 and C = − . So the phase shift is or π 6 units to the left. Analysis We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation ⎞ ⎞ ⎛ ⎠ − 2. shows a minus sign before C. Therefore f (x) = sin ⎝x + ⎠ ⎞ ⎛ ⎠ − 2 can be rewritten as f (x) = sin ⎝x − ⎛ ⎝− π 6 π 6 If the value of C is negative, the shift is to the left. 8.3 ⎛ Determine the direction and magnitude of the phase shift for f (x) = 3cos ⎝x − ⎞ ⎠. π 2 Example 8.4 Identifying the Vertical Shift of a Function Determine the direction and magnitude of the vertical shift for f (x) = cos(x) − 3. Solution Let’s begin by comparing the equation to the general form y = Acos(Bx − C) + D. In the given equation, D = −3 so the shift is 3 units downward. 8.4 Determine the direction and magnitude of the vertical shift for f (x) = 3sin(x) + 2. 910 Chapter 8 Periodic Functions Given a sinusoidal function in the form f (x) = Asin(Bx − C) + D, identify the midline, amplitude, period, and phase shift. 1. Determine the amplitude as \|A\|. 2. Determine the period as P = 2π \|B\| . 3. Determine the phase shift as C B . 4. Determine the midline as y = D. Example 8.5 Identifying the Variations of a Sinusoidal Function from an Equation Determine the midline, amplitude, period, and phase shift of the function y = 3sin(2x) + 1. Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. A = 3, so the amplitude is \|A\| = 3. Next, B = 2, so the period is P = 2π \|B\| = 2π 2 There is no added constant inside the parentheses, so C = 0 and the phase shift is C = π. B = 0 2 = 0. Finally, D = 1, so the midline is y = 1. Analysis Inspecting the graph, we can determine that the period is π, the midline is y = 1, and the amplitude is 3. See Figure 8.15. Figure 8.15 8.5 Determine the midline, amplitude, period, and phase shift of the function y = 1 2 ⎛ cos ⎝ x 3 − ⎞ ⎠. π 3 This content is available for free at https://cnx.org/content/col117
58/1.5 Chapter 8 Periodic Functions 911 Example 8.6 Identifying the Equation for a Sinusoidal Function from a Graph Determine the formula for the cosine function in Figure 8.16. Figure 8.16 Solution To determine the equation, we need to identify each value in the general form of a sinusoidal function. y = Asin(Bx − C) + D y = Acos(Bx − C) + D The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x = 0, the graph has an extreme point, (0, 0). Since the cosine function has an extreme point for x = 0, let us write our equation in terms of a cosine function. Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y = 0.5. This value, which is the midline, is D in the equation, so D = 0.5. The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So \|A\| = 0.5. Another way we could have determined the amplitude is by recognizing that the difference between the height of local maxima and minima is 1, so \|A\| = 1 2 = 0.5. Also, the graph is reflected about the x-axis so that A = − 0.5. The graph is not horizontally stretched or compressed, so B = 1; and the graph is not shifted horizontally, so C = 0. Putting this all together, g(x) = − 0.5cos(x) + 0.5 8.6 Determine the formula for the sine function in Figure 8.17. Figure 8.17 912 Chapter 8 Periodic Functions Example 8.7 Identifying the Equation for a Sinusoidal Function from a Graph Determine the equation for the sinusoidal function in Figure 8.18. Figure 8.18 Solution With the highest value at 1 and the lowest value at −5, the midline will be halfway between at −2. So D = −2. The distance from the midline to the highest or lowest value gives an amplitude of \|A\| = 3. The period of the graph is 6, which can be measured from the peak at x = 1 to the next peak at x = 7, or from the distance between the lowest points. Therefore, P = 2π \|B\| P = 2π 6 = 6. Using the positive value for B, we find that B = 2π π 3 = ⎛ So far, our equation is either y = 3sin ⎝ π 3 x − C⎞ ⎛ ⎠ − 2 or y = 3cos ⎝ π 3 x − C⎞ ⎠ − 2. For the shape and shift, we have more than one option. We could write this as any one of the following: • a cosine shifted to the right • a negative cosine shifted to the left • a sine shifted to the left • a negative sine shifted to the right While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 913 ⎛ y = 3cos ⎝ or y = − 3cos ⎝ π 3 x + 2π 3 ⎞ ⎠ − 2 Again, these functions are equivalent, so both yield the same graph. 8.7 Write a formula for the function graphed in Figure 8.19. Figure 8.19 Graphing Variations of y = sin x and y = cos x Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations. Instead of focusing on the general form equations y = Asin(Bx − C) + D and y = Acos(Bx − C) + D, we will let C = 0 and D = 0 and work with a simplified form of the equations in the following examples. Given the function y = Asin(Bx), sketch its graph. 1. 2. Identify the amplitude, \|A\|. Identify the period, P = 2π \|B\| . 3. Start at the origin, with the function increasing to the right if A is positive or decreasing if A is negative. 4. At x = π 2\|B\| there is a local maximum for A > 0 or a minimum for A < 0, with y = A. 5. The curve returns to the x-axis at x = π \|B\| . 6. There is a local minimum for A > 0 (maximum for A < 0 ) at x = 3π 2\|B\| with y = – A. 7. The curve returns again to the x-axis at x = π 2\|B\| . 914 Chapter 8 Periodic Functions Example 8.8 Graphing a Function and Identifying the Amplitude and Period ⎛ Sketch a graph of f (x) = − 2sin ⎝ ⎞ ⎠. πx 2 Solution Let’s begin by comparing the equation to the form y = Asin(Bx). Step 1. We can see from the equation that A = − 2, so the amplitude is 2. Step 2. The equation shows that B = \|A\| = 2 , so the period is π 2 P = 2π π 2 = 2π ⋅ 2 π = 4 Step 3. Because A is negative, the graph descends as we move to the right of the origin. Step 4–7. The x-intercepts are at the beginning of one period, x = 0, the horizontal midpoints are at x = 2 and at the end of one period at x = 4. The quarter points include the minimum at x = 1 and the maximum at x = 3. A local minimum will occur 2 units below the midline, at x = 1, and a local maximum will occur at 2 units above the midline, at x = 3. Figure 8.20 shows the graph of the function. Figure 8.20 8.8 Sketch a graph of g(x) = − 0.8cos(2x). Determine the midline, amplitude, period, and phase shift. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 915 Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph. 1. Express the function in the general form y = Asin(Bx − C) + D or y = Acos(Bx − C) + D. 2. 3. 4. Identify the amplitude, \|A\|. Identify the period, P = 2π \|B\| . Identify the phase shift, C B . 5. Draw the graph of f (x) = Asin(Bx) shifted to the right or left by C B and up or down by D. Example 8.9 Graphing a Transformed Sinusoid ⎛ Sketch a graph of f (x) = 3sin ⎝ π 4 x − ⎞ ⎠. π 4 Solution ⎛ Step 1. The function is already written in general form: f (x) = 3sin ⎝ π 4 x − π 4 ⎞ ⎠. This graph will have the shape of a sine function, starting at the midline and increasing to the right. Step 2. \|A\| = \|3\| = 3. The amplitude is 3. Step 3. Since \|B\| = \|π 4\| = , we determine the period as follows. π 4 P = 2π \|B\| = 2π π 4 = 2π ⋅ 4 π = 8 The period is 8. Step 4. Since C = , the phase shift is . The phase shift is 1 unit. Step 5. Figure 8.21 shows the graph of the function. 916 Chapter 8 Periodic Functions Figure 8.21 A horizontally compressed, vertically stretched, and horizontally shifted sinusoid 8.9 ⎛ Draw a graph of g(x) = − 2cos ⎝ π 3 x + π 6 ⎞ ⎠. Determine the midline, amplitude, period, and phase shift. Example 8.10 Identifying the Properties of a Sinusoidal Function x + π⎞ ⎠ + 3, determine the amplitude, period, phase shift, and horizontal shift. Then graph ⎛ Given y = − 2cos ⎝ π 2 the function. Solution Begin by comparing the equation to the general form and use the steps outlined in Example 8.9. y = Acos(Bx − C) + D Step 1. The function is already written in general form. Step 2. Since A = − 2, the amplitude is \|A\| = 2. = 2π π 2 , so the period is P = 2π \|B\| Step 3. \|B\| = π 2 = 2π ⋅ 2 π = 4. The period is 4. Step 4. C = − π, so we calculate the phase shift as C B = −π. The phase shift is − 2. Step 5. D = 3, so the midline is y = 3, and the vertical shift is up 3. Since A is negative, the graph of the cosine function has been reflected about the x-axis. Figure 8.22 shows one cycle of the graph of the function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 917 Figure 8.22 Using Transformations of Sine and Cosine Functions We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter, circular motion can be modeled using either the sine or cosine function. Example 8.11 Finding the Vertical Component of Circular Motion A point rotates around a circle of radius 3 centered at the origin. Sketch a graph of the y-coordinate of the point as a function of the angle of rotation. Solution Recall that, for a point on a circle of radius r, the y-coordinate of the point is y = r sin(x), so in this case, we get the equation y(x) = 3 sin(x). The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in Figure 8.23. Figure 8.23 Analysis 918 Chapter 8 Periodic Functions Notice that the period of the function is still 2π; as we travel around the circle, we return to the point (3, 0) for x = 2π, 4π, 6π, .... Because the outputs of the graph will now oscillate between – 3 and 3, the amplitude of the sine wave is 3. 8.10 What is the amplitude of the function f (x) = 7cos(x) ? Sketch a graph of this function. Example 8.12 Finding the Vertical Component of Circular Motion A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in Figure 8.24. Sketch a graph of the height above the ground of the point P as the circle is rotated; then find a function that gives the height in terms of the angle of rotation. Figure 8.24 Solution Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in Figure 8.25. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 919 Figure 8.25 Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example. Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that y = − 3cos(x) + 4 920 Chapter 8 Periodic Functions 8.11 A weight is attached to a spring that is then hung from a board, as shown in Figure 8.26. As the spring oscillates up an
d down, the position y of the weight relative to the board ranges from –1 in. (at time x = 0) to –7 in. (at time x = π) below the board. Assume the position of y is given as a sinusoidal function of x. Sketch a graph of the function, and then find a cosine function that gives the position y in terms of x. Figure 8.26 Example 8.13 Determining a Rider’s Height on a Ferris Wheel The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center. Passengers board 2 m above ground level, so the center of the wheel must be located 67.5 + 2 = 69.5 m above ground level. The midline of the oscillation will be at 69.5 m. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve. • Amplitude: 67.5, so A = 67.5 • Midline: 69.5, so D = 69.5 • Period: 30, so B = 2π 30 = π 15 • Shape: −cos(t) An equation for the rider’s height would be where t is in minutes and y is measured in meters. ⎛ y = − 67.5cos ⎝ t⎞ ⎠ + 69.5 π 15 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 921 Access these online resources for additional instruction and practice with graphs of sine and cosine functions. • Amplitude and Period of Sine and Cosine (http://openstaxcollege.org/l/ampperiod) • Translations of Sine and Cosine (http://openstaxcollege.org/l/translasincos) • Graphing Sine and Cosine Transformations (http://openstaxcollege.org/l/ transformsincos) • Graphing the Sine Function (http://openstaxcollege.org/l/graphsinefunc) 922 Chapter 8 Periodic Functions 8.1 EXERCISES Verbal Why are the sine and cosine functions called periodic 1. functions? How does the graph of y = sin x compare with the 2. graph of y = cos x ? Explain how you could horizontally translate the graph of y = sin x to obtain y = cos x. 3. For the equation A cos(Bx + C) + D, what constants affect the range of the function and how do they affect the range? How does the range of a translated sine function relate to 4. the equation y = A sin(Bx + C) + D ? How can the unit circle be used to construct the graph of 5. f (t) = sin t ? Graphical For the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum y-values their corresponding x-values on one period for x > 0. Round answers to two decimal places if necessary. and For the following exercises, graph one full period of each function, starting at x = 0. For each function, state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x > 0. State the phase shift and vertical translation, if applicable. Round answers to two decimal places if necessary. 18. 19. 20. 21. 22. ⎛ ⎝t − 5π f (t) = 2sin 6 ⎞ ⎠ ⎛ ⎝t + f (t) = − cos ⎞ ⎠ + 1 π 3 ⎛ f (t) = 4cos ⎝2 ⎛ ⎝t) = − sin ⎝ 1 2 t + 5π 3 ⎞ ⎠ ⎛ f (x) = 4sin ⎝ π 2 ⎞ (x − 3) ⎠ + 7 Determine the amplitude, midline, period, and an 23. equation involving the sine function for the graph shown in Figure 8.27. 6. 7. 8. 9. f (x) = 2sin x f (x) = 2 3 cos x f (x) = − 3sin x f (x) = 4sin x 10. f (x) = 2cos x 11. f (x) = cos(2x) 12. ⎛ f (x) = 2 sin ⎝ x⎞ ⎠ 1 2 13. f (x) = 4 cos(πx) 14. ⎛ f (x) = 3 cos ⎝ x⎞ ⎠ 6 5 15. y = 3 sin(8(x + 4)) + 5 16. y = 2 sin(3x − 21) + 4 17. y = 5 sin(5x + 20) − 2 This content is available for free at https://cnx.org/content/col11758/1.5 Figure 8.27 Determine the amplitude, period, midline, and an 24. equation involving cosine for the graph shown in Figure 8.28. Chapter 8 Periodic Functions 923 Figure 8.28 25. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 8.29. Figure 8.30 27. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 8.31. Figure 8.29 26. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 8.30. Figure 8.31 Determine the amplitude, period, midline, and an 28. equation involving sine for the graph shown in Figure 8.32. Figure 8.32 29. 924 Chapter 8 Periodic Functions Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 8.33. 39. On ⎡ ⎣0, 2π), solve f (x) = 1 2 . 40. On ⎡ ⎣0, 2π), find the x-intercepts of f (x) = cos x. 41. On ⎡ ⎣0, 2π), find the x-values at which the function has a maximum or minimum value. 42. On ⎡ ⎣0, 2π), solve the equation f (x) = 3 2 . Technology 43. Graph h(x) = x + sin x on [0, 2π]. Explain why the graph appears as it does. 44. Graph h(x) = x + sin x on [−100, 100]. Did graph appear as predicted in the previous exercise? the Graph f (x) = x sin x on [0, 2π] and verbalize how 45. the graph varies from the graph of f (x) = sin x. 46. Graph f (x) = x sin x on the window [−10, 10] and explain what the graph shows. 47. Graph f (x) = sin x explain what the graph shows. x on the window ⎡ ⎣−5π, 5π⎤ ⎦ and Real-World Applications A Ferris wheel is 25 meters in diameter and boarded 48. from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t) gives a person’s height in meters above the ground t minutes after the wheel begins to turn. a. Find the amplitude, midline, and period of h(t). b. Find a formula for the height function h(t). c. How high off the ground is a person after 5 minutes? Figure 8.33 Determine the amplitude, period, midline, and an 30. equation involving sine for the graph shown in Figure 8.34. Figure 8.34 Algebraic For the following exercises, let f (x) = sin x. 31. On ⎡ ⎣0, 2π), solve f (x) = 0. 32. 33. 34. On ⎡ ⎣0, 2π), solve f (x) = 1 2 . Evaluate f ⎛ ⎝ ⎞ ⎠. π 2 On [0, 2π), f (x) = 2 2 . Find all values of x. 35. On ⎡ ⎣0, 2π), the maximum value(s) of the function occur(s) at what x-value(s)? 36. On ⎡ ⎣0, 2π), the minimum value(s) of the function occur(s) at what x-value(s)? Show that f (−x) = − f (x). This means that 37. f (x) = sin x is an odd function and possesses symmetry with respect to ________________. For the following exercises, let f (x) = cos x. 38. On ⎡ ⎣0, 2π), solve the equation f (x) = cos x = 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 925 8.2 \| Graphs of the Other Trigonometric Functions Learning Objectives In this section, you will: 8.2.1 Analyze the graph of y=tan x. 8.2.2 Graph variations of y=tan x. 8.2.3 Analyze the graphs of y=sec x and y=csc x. 8.2.4 Graph variations of y=sec x and y=csc x. 8.2.5 Analyze the graph of y=cot x. 8.2.6 Graph variations of y=cot x. We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions. Analyzing the Graph of y = tan x We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that tan x = sin x cos x The period of the tangent function is π because the graph repeats itself on intervals of kπ where k is a constant. If we , we can see the behavior of the graph on one complete cycle. If we look at any graph the tangent function on − π 2 larger interval, we will see that the characteristics of the graph repeat. to π 2 We can determine whether tangent is an odd or even function by using the definition of tangent. sin(−x) tan(−x) = cos(−x) = −sin x cos x = − sin x cos x = − tan x Definition of angent. Sine is an odd function, cosine is even. The quotient of an odd and an even function is odd. Definition of angent. Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table 8.3 tan(x) undefined − 3 – undefined Table 8.3 These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values when π ≈ 1.57, 3 , we can use a table to look for a trend. Because π 3 ≈ 1.05 and π 2 < x < π 2 we will evaluate x at radian measures 1.05 < x < 1.57 as shown in Table 8.4. 926 Chapter 8 Periodic Functions x 1.3 1.5 1.55 1.56 tan x 3.6 14.1 48.1 92.6 Table 8.4 As x approaches π 2 , the outputs of the function get larger and larger. Because y = tan x is an odd function, we see the corresponding table of negative values in Table 8.5. x −1.3 −1.5 −1.55 −1.56 tan x −3.6 −14.1 −48.1 −92.6 Table 8.5 We can see that, as x approaches − π 2 π ⎛ which cos x = 0. For example, cos ⎝ 2 graph of y = tan x has discontinuities at x = , the outputs get smaller and smaller. Remember that there are some values of x for ⎞ ⎠ = 0. At these values, the tangent function is undefin
ed, so the 3π ⎛ ⎞ ⎠ = 0 and cos ⎝ 2 and 3π π . At these values, the graph of the tangent has vertical asymptotes. 2 2 Figure 8.35 represents the graph of y = tan x. The tangent is positive from 0 to π 2 quadrants I and III of the unit circle. and from π to 3π 2 , corresponding to Figure 8.35 Graph of the tangent function Graphing Variations of y = tan x As with the sine and cosine functions, the tangent function can be described by a general equation. y = Atan(Bx) We can identify horizontal and vertical stretches and compressions using values of A and B. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph. Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant A. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 927 Features of the Graph of y = Atan(Bx) • The stretching factor is \|A\|. • The period is P = π \|B\| . • The domain is all real numbers x, where x ≠ π 2\|B\| + π \|B\| k such that k is an integer. • The range is (−∞, ∞). • The asymptotes occur at x = π 2\|B\| + π \|B\| k, where k is an integer. • y = Atan(Bx) is an odd function. Graphing One Period of a Stretched or Compressed Tangent Function We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/ or compressed tangent function of the form f (x) = Atan(Bx). We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our ⎞ limited domain is then the interval ⎛ ⎠, , cross through the origin, and continue to increase as it ⎞ ⎠ and the graph has vertical asymptotes at ± P 2 the graph will come up from the left asymptote at x = − π B. On ⎛ ⎝− where P = ⎝− approaches the right asymptote at x = π 2 . To make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use f ⎛ ⎝ P 4 ⎞ ⎛ ⎝BP ⎠ = Atan 4 ⎞ ⎛ ⎠ = Atan ⎝B π 4B ⎞ ⎠ = A ⎛ because tan ⎝ ⎞ ⎠ = 1. π 4 Given the function f(x) = Atan(Bx), graph one period. 1. 2. Identify the stretching factor, \|A\|. Identify B and determine the period, P = 3. Draw vertical asymptotes at x = − P 2 . π \|B\| and x = P 2 . 4. For A > 0, the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0 ). 5. Plot reference points at ⎛ ⎝ P 4 , A⎞ ⎠, (0, 0), and ⎛ ⎝− P 4 ,−A⎞ ⎠, and draw the graph through these points. Example 8.14 Sketching a Compressed Tangent ⎛ Sketch a graph of one period of the function y = 0.5tan ⎝ x⎞ ⎠. π 2 928 Chapter 8 Periodic Functions Solution First, we identify A and B. Because A = 0.5 and B = π 2 , we can find the stretching/compressing factor and period. The period is π π 2 = 2, so the asymptotes are at x = ± 1. At a quarter period from the origin, we have ⎛ f (0.5) = 0.5tan ⎝ ⎞ ⎠ (8.2) 0.5π 2 ⎞ ⎠ π 4 ⎛ = 0.5tan ⎝ This means the curve must pass through the points (0.5, 0.5), point is at the origin. Figure 8.36 shows the graph of one period of the function. (0, 0), and (−0.5, −0.5). The only inflection = 0.5 Figure 8.36 8.12 ⎛ Sketch a graph of f (x) = 3tan ⎝ x⎞ ⎠. π 6 Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. f (x) = Atan(Bx − C) + D (8.3) The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: Features of the Graph of y = Atan(Bx−C)+D • The stretching factor is \|A\|. • The period is π \|B\| . • The domain is x ≠ C B + π \|B\| k, where k is an integer. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 929 • The range is (−∞, − \|A\|] ∪ [\|A\|, ∞). • The vertical asymptotes occur at x = C B + π 2\|B\| k, where k is an odd integer. • There is no amplitude. • y = A tan(Bx) is and odd function because it is the qoutient of odd and even functions(sin and cosine perspectively). Given the function y = Atan(Bx − C) + D, sketch the graph of one period. 1. Express the function given in the form y = Atan(Bx − C) + D. 2. 3. 4. Identify the stretching/compressing factor, \|A\|. Identify B and determine the period, P = π \|B\| Identify C and determine the phase shift, C B . . 5. Draw the graph of y = Atan(Bx) shifted to the right by C B and up by D. 6. Sketch the vertical asymptotes, which occur at x = C B + π 2\|B\| k, where k is an odd integer. 7. Plot any three reference points and draw the graph through these points. Example 8.15 Graphing One Period of a Shifted Tangent Function Graph one period of the function y = −2tan(πx + π) −1. Solution Step 1. The function is already written in the form y = Atan(Bx − C) + D. Step 2. A = −2, so the stretching factor is \|A\| = 2. Step 3. B = π, so the period is P = π \|B\| Step 4. C = − π, so the phase shift is C = π π = 1. B = −π π = −1. Step 5-7. The asymptotes are at x = − 3 2 and x = − 1 2 and the three recommended reference points are (−1.25, 1), (−1,−1), and (−0.75,−3). The graph is shown in Figure 8.37. 930 Chapter 8 Periodic Functions Figure 8.37 Analysis Note that this is a decreasing function because A < 0. 8.13 How would the graph in Example 8.15 look different if we made A = 2 instead of −2 ? Given the graph of a tangent function, identify horizontal and vertical stretches. 1. Find the period P from the spacing between successive vertical asymptotes or x-intercepts. ⎛ 2. Write f (x) = Atan ⎝ π P x⎞ ⎠. 3. Determine a convenient point (x, f (x)) on the given graph and use it to determine A. Example 8.16 Identifying the Graph of a Stretched Tangent Find a formula for the function graphed in Figure 8.38. Figure 8.38 A stretched tangent function Solution The graph has the shape of a tangent function. Step 1. One cycle extends from –4 to 4, so the period is P = 8. Since P = π \|B\| , we have B = π P = π 8 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 931 ⎛ Step 2. The equation must have the form f (x) = Atan ⎝ x⎞ ⎠. π 8 Step 3. To find the vertical stretch A, we can use the point (2, 2). ⎞ ⎠ ⎛ 2 = Atan ⎝ ⎞ ⎛ ⎠ = Atan ⎝ ⋅ 2 π 8 π 4 ⎛ Because tan ⎝ π 4 ⎞ ⎠ = 1, A = 2. ⎛ This function would have a formula f (x) = 2tan ⎝ x⎞ ⎠. π 8 8.14 Find a formula for the function in Figure 8.39. Figure 8.39 Analyzing the Graphs of y = sec x and y = cscx The secant was defined by the reciprocal identity sec x = 1 leading to vertical asymptotes at π 2 cos x. Notice that the function is undefined when the cosine is 0, , etc. Because the cosine is never more than 1 in absolute value, the secant, being , 3π 2 the reciprocal, will never be less than 1 in absolute value. We can graph y = sec x by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 8.40. The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined. The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant. Note that, because cosine is an even function, secant is also an even function. That is, sec(−x) = sec x. 932 Chapter 8 Periodic Functions Figure 8.40 Graph of the secant function, f (x) = secx = 1 cosx As we did for the tangent function, we will again refer to the constant \|A\| as the stretching factor, not the amplitude. Features of the Graph of y = Asec(Bx) • The stretching factor is \|A\|. • The period is 2π \|B\| . • The domain is x ≠ π 2\|B\| k, where k is an odd integer. • The range is ( − ∞, − \|A\|] ∪ [\|A\|, ∞). π 2\|B\| • The vertical asymptotes occur at x = • There is no amplitude. k, where k is an odd integer. • y = Asec(Bx) is an even function because cosine is an even function. Similar to the secant, the cosecant is defined by the reciprocal identity csc x = 1 sin x. Notice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value. We can graph y = csc x by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure 8.41. The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases. The cosecant graph has vertical asymptotes at each value of x where the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines. Note that, since sine is an odd function, the cosecant function is also an odd function. That is, csc(−x) = −cscx. The graph of cosecant, which is shown in Figure 8.41, is similar to the graph of secant. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 933 Figure 8.41 The graph of the cosecant function, f (x) = cscx = 1 sinx Features of the Graph
of y = Acsc(Bx) • The stretching factor is \|A\|. • The period is 2π \|B\| . • The domain is x ≠ π \|B\| k, where k is an integer. • The range is (−∞, − \|A\|] ∪ [\|A\|, ∞). • The asymptotes occur at x = π \|B\| k, where k is an integer. • y = Acsc(Bx) is an odd function because sine is an odd function. Graphing Variations of y = sec x and y= csc x For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following. y = Asec(Bx − C) + D y = Acsc(Bx − C) + D (8.4) (8.5) Features of the Graph of y = Asec(Bx−C)+D • The stretching factor is \|A\|. • The period is 2π \|B\| . • The domain is x ≠ C B + π 2\|B\| k, where k is an odd integer. • The range is ( − ∞, − \|A\|] ∪ [\|A\|, ∞). 934 Chapter 8 Periodic Functions • The vertical asymptotes occur at x = C B + π 2\|B\| k, where k is an odd integer. • There is no amplitude. • y = Asec(Bx) is an even function because cosine is an even function. Features of the Graph of y = Acsc(Bx−C)+D • The stretching factor is \|A\|. • The period is 2π \|B\| . • The domain is x ≠ C B + π 2\|B\| k, where k is an integer. • The range is ( − ∞, − \|A\|] ∪ [\|A\|, ∞). • The vertical asymptotes occur at x = C B + π \|B\| k, where k is an integer. • There is no amplitude. • y = Acsc(Bx) is an odd function because sine is an odd function. Given a function of the form y = Asec(Bx), graph one period. 1. Express the function given in the form y = Asec(Bx). 2. 3. Identify the stretching/compressing factor, \|A\|. Identify B and determine the period, P = 2π \|B\| . 4. Sketch the graph of y = Acos(Bx). 5. Use the reciprocal relationship between y = cos x and y = sec x to draw the graph of y = Asec(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 8.17 Graphing a Variation of the Secant Function Graph one period of f (x) = 2.5sec(0.4x). Solution Step 1. The given function is already written in the general form, y = Asec(Bx). Step 2. A = 2.5 so the stretching factor is 2.5. Step 3. B = 0.4 so P = 2π 0.4 = 5π. The period is 5π units. Step 4. Sketch the graph of the function g(x) = 2.5cos(0.4x). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 935 Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. Steps 6–7. Sketch two asymptotes at x = 1.25π and x = 3.75π. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5π, −2.5). Figure 8.42 shows the graph. Figure 8.42 8.15 Graph one period of f (x) = − 2.5sec(0.4x). Do the vertical shift and stretch/compression affect the secant’s range? Yes. The range of f (x) = Asec(Bx − C) + D is (−∞, − \|A\| + D] ∪ [\|A\| + D, ∞). Given a function of the form f(x) = Asec(Bx − C) + D, graph one period. 1. Express the function given in the form y = A sec(Bx − C) + D. 2. 3. 4. Identify the stretching/compressing factor, \|A\|. Identify B and determine the period, 2π \|B\| . Identify C and determine the phase shift, C B . 5. Draw the graph of y = A sec(Bx) . but shift it to the right by C B and up by D. 6. Sketch the vertical asymptotes, which occur at x = C B + π 2\|B\| k, where k is an odd integer. 936 Chapter 8 Periodic Functions Example 8.18 Graphing a Variation of the Secant Function ⎛ Graph one period of y = 4sec ⎝ π 3 x − ⎞ ⎠ + 1. π 2 Solution ⎛ Step 1. Express the function given in the form y = 4sec ⎝ π 3 x − ⎞ ⎠ + 1. π 2 Step 2. The stretching/compressing factor is \|A\| = 4. Step 3. The period is Step 4. The phase shift is 2π \|B\| = 2π π 3 = 2π .5 Step 5. Draw the graph of y = Asec(Bx), but shift it to the right by C B = 1.5 and up by D = 6. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 3, and x = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, − 3). Figure 8.43 shows the graph. Figure 8.43 8.16 Graph one period of f (x) = − 6sec(4x + 2) − 8. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 937 The domain of csc x was given to be all x such that x ≠ kπ for any integer k. Would the domain of y = Acsc(Bx − C) + D be x ≠ C + kπ B ? Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input. Given a function of the form y = Acsc(Bx), graph one period. 1. Express the function given in the form y = Acsc(Bx). 2. 3. \|A\|. Identify B and determine the period, P = 2π \|B\| . 4. Draw the graph of y = Asin(Bx). 5. Use the reciprocal relationship between y = sin x and y = csc x to draw the graph of y = Acsc(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 8.19 Graphing a Variation of the Cosecant Function Graph one period of f (x) = −3csc(4x). Solution Step 1. The given function is already written in the general form, y = Acsc(Bx). Step 2. \|A\| = \|−3\| = 3, so the stretching factor is 3. Step 3. B = 4, so P = 2π 4 . The period is π 2 units. π 2 = Step 4. Sketch the graph of the function g(x) = −3sin(4x). Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function. π Steps 6–7. Sketch three asymptotes at x = 0, x = 4 ⎠ and the local minimum at ⎛ ⎞ 3π , −3 ⎝ 8 ⎞ ⎠. Figure 8.44 shows the graph. , 3 maximum at ⎛ ⎝ , and x = π 8 π 2 . We can use two reference points, the local 938 Chapter 8 Periodic Functions Figure 8.44 8.17 Graph one period of f (x) = 0.5csc(2x). Given a function of the form f(x) = Acsc(Bx − C) + D, graph one period. 1. Express the function given in the form y = Acsc(Bx − C) + D. 2. 3. 4. Identify the stretching/compressing factor, \|A\|. Identify B and determine the period, 2π \|B\| . Identify C and determine the phase shift, C B . 5. Draw the graph of y = Acsc(Bx) but shift it to the right by and up by D. 6. Sketch the vertical asymptotes, which occur at x = C B + π \|B\| k, where k is an integer. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 939 Example 8.20 Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant ⎛ Sketch a graph of y = 2csc ⎝ π 2 x⎞ ⎠ + 1. What are the domain and range of this function? Solution ⎛ Step 1. Express the function given in the form y = 2csc ⎝ π 2 Step 2. Identify the stretching/compressing factor, \|A\| = 2. = 2π Step 3. The period is 2π π \|B\| 2 ⋅ 2 π = 4. = 2π 1 x⎞ ⎠ + 1. Step 4. The phase shift is 0 π 2 = 0. Step 5. Draw the graph of y = Acsc(Bx) but shift it up D = 1. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 2, x = 4. The graph for this function is shown in Figure 8.45. Figure 8.45 A transformed cosecant function Analysis The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of ⎛ f (x) = 2sin ⎝ x⎞ ⎠ + 1, shown as the orange dashed wave. π 2 940 Chapter 8 Periodic Functions 8.18 Given the graph ⎛ of f (x) = 2cos ⎝ x⎞ ⎠ + 1 shown π 2 ⎛ g(x) = 2sec ⎝ π 2 x⎞ ⎠ + 1 on the same axes. in Figure 8.46, sketch the graph of Figure 8.46 Analyzing the Graph of y = cot x The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity cot x = 1 tan x. Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers. We can graph y = cot x by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure 8.47. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases. The cotangent graph has vertical asymptotes at each value of x where tan x = 0; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, cot x has vertical asymptotes at all values of x where tan x = 0, and cot x = 0 at all values of x where tan x has its vertical asymptotes. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 941 Figure 8.47 The cotangent function Features of the Graph of y = Acot(Bx) • The stretching factor is \|A\|. • The period is P = • The domain is x ≠ . π \|B\| π \|B\| k, where k is an integer. • The range is (−∞, ∞). • The asymptotes occur at x = π \|B\| k, where k is an integer. • y = Acot(Bx) is an odd function. Graphing Variations of y = cot x We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following. y = Acot(Bx − C) + D (8.6) Properties of the Graph of y = Acot(Bx−C)+D • The stretching factor is \|A\|. • The period is π \|B\| . • The domain is x ≠ C B + π \|B\| k, where k is an integer. • The range is (−∞, − \|A\|] ∪ [\|A\|, ∞). • The vertical asymptotes occur at x = C B + π \|B\| k, where k is an integer. • There is no amplitude. 942 Chapter 8 Periodic Functions • y = Acot(Bx) is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively) Given a
modified cotangent function of the form f(x) = Acot(Bx), graph one period. 1. Express the function in the form f (x) = Acot(Bx). 2. 3. Identify the stretching factor, \|A\|. Identify the period, P = π \|B\| . 4. Draw the graph of y = Atan(Bx). 5. Plot any two reference points. 6. Use the reciprocal relationship between tangent and cotangent to draw the graph of y = Acot(Bx). 7. Sketch the asymptotes. Example 8.21 Graphing Variations of the Cotangent Function Determine the stretching factor, period, and phase shift of y = 3cot(4x), and then sketch a graph. Solution Step 1. Expressing the function in the form f (x) = Acot(Bx) gives f (x) = 3cot(4x). Step 2. The stretching factor is \|A\| = 3. Step 3. The period is P = . π 4 Step 4. Sketch the graph of y = 3tan(4x). Step 5. Plot two reference points. Two such points are ⎛ ⎝ π 16 ⎠ and ⎛ ⎞ , 3 ⎝ ⎞ ⎠. , −3 3π 16 Step 6. Use the reciprocal relationship to draw y = 3cot(4x). Step 7. Sketch the asymptotes, x = 0, x = . π 4 The orange graph in Figure 8.48 shows y = 3tan(4x) and the blue graph shows y = 3cot(4x). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 943 Figure 8.48 Given a modified cotangent function of the form f(x) = Acot(Bx − C) + D, graph one period. 1. Express the function in the form f (x) = Acot(Bx − C) + D. 2. 3. 4. Identify the stretching factor, \|A\|. Identify the period, P = π \|B\| . Identify the phase shift, C B . 5. Draw the graph of y = Atan(Bx) shifted to the right by C B and up by D. 6. Sketch the asymptotes x = C B + π \|B\| k, where k is an integer. 7. Plot any three reference points and draw the graph through these points. Example 8.22 Graphing a Modified Cotangent Sketch a graph of one period of the function f (x) = 4cot ⎛ ⎝ π 8 x − ⎞ ⎠ − 2. π 2 944 Chapter 8 Periodic Functions Solution Step 1. The function is already written in the general form f (x) = Acot(Bx − C) + D. Step 2. A = 4, so the stretching factor is 4. Step 3. B = , so the period is P = = 8. = π 8 π \|B\| π π 8 Step 4. C = π 2 , so the phase shift is C B = π 2 π 8 = 4. ⎛ Step 5. We draw f (x) = 4tan ⎝ π 8 x − ⎞ ⎠ − 2. π 2 Step 6-7. Three points we can use to guide the graph are (6, 2), (8, − 2), and (10, − 6). We use the reciprocal relationship of tangent and cotangent to draw f (x) = 4cot ⎛ ⎝ π 8 x − ⎞ ⎠ − 2. π 2 Step 8. The vertical asymptotes are x = 4 and x = 12. The graph is shown in Figure 8.49. Figure 8.49 One period of a modified cotangent function Using the Graphs of Trigonometric Functions to Solve Real-World Problems Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function. Example 8.23 Using Trigonometric Functions to Solve Real-World Scenarios ⎛ Suppose the function y = 5tan ⎝ t⎞ ⎠ marks the distance in the movement of a light beam from the top of a police car across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car. π 4 a. Find and interpret the stretching factor and period. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 945 b. Graph on the interval ⎡ ⎣0, 5⎤ ⎦. c. Evaluate f (1) and discuss the function’s value at that input. Solution a. We know from the general form of y = Atan(Bt) that \|A\| is the stretching factor and π B is the period. Figure 8.50 We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period. The period is . This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches. b. To graph the function, we draw an asymptote at t = 2 and use the stretching factor and period. See Figure 8.51 Figure 8.51 ⎛ c. period: f (1) = 5tan ⎝ π 4 from the police car. ⎞ ⎠ = 5(1) = 5; after 1 second, the beam of has moved 5 ft from the spot across (1) Access these online resources for additional instruction and practice with graphs of other trigonometric functions. • Graphing the Tangent (http://openstaxcollege.org/l/graphtangent) • Graphing Cosecant and Secant (http://openstaxcollege.org/l/graphcscsec) • Graphing the Cotangent (http://openstaxcollege.org/l/graphcot) 946 Chapter 8 Periodic Functions 8.2 EXERCISES Verbal Explain how the graph of the sine function can be used 49. to graph y = csc x. How can the graph of y = cos x be used to construct 50. the graph of y = sec x ? 51. Explain why the period of tan x is equal to π. Why are there no intercepts on the graph of 52. y = csc x ? How does the period of y = csc x compare with the 53. period of y = sin x ? Algebraic For the following exercises, match each trigonometric function with one of the following graphs. 54. f (x) = tan x 55. f (x) = sec x 56. f (x) = csc x 57. f (x) = cot x For the following exercises, find the period and horizontal shift of each of the functions. 58. f (x) = 2tan(4x − 32) 59. 60. ⎛ h(x) = 2sec ⎝ ⎞ (x + 1) ⎠ π 4 ⎛ m(x) = 6csc ⎝ x + π⎞ ⎠ π 3 61. If tan x = −1.5, find tan(−x). 62. If sec x = 2, find sec(−x). 63. If csc x = −5, find csc(−x). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 947 64. If xsin x = 2, find (−x)sin(−x). 84. f (x) = 5 ⎛ ⎝cot ⎛ ⎝ For the following exercises, find and graph two periods of the periodic function with the given stretching factor, \|A\|, period, and phase shift. 85. A tangent curve, A = 1, period of π 3 ; and phase shift (h, k 86. A tangent curve, A = −2, period of π 4 , and phase shift (h, k) = ⎛ ⎝− π 4 ⎞ , −2 ⎠ For the following exercises, find an equation for the graph of each function. 87. 88. 89. For the following exercises, rewrite each expression such that the argument x is positive. 65. cot(−x)cos(−x) + sin(−x) 66. cos(−x) + tan(−x)sin(−x) Graphical For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes. 67. f (x) = 2tan(4x − 32) 68. 69. 70. 71. ⎛ h(x) = 2sec ⎝ ⎞ ⎠ (x + 1) π 4 ⎛ m(x) = 6csc ⎝ x + π⎞ ⎠ π 3 ⎛ j(x) = tan ⎝ x⎞ ⎠ π 2 ⎛ p(x) = tan ⎝x − ⎞ ⎠ π 2 72. f (x) = 4tan(x) 73. ⎛ f (x) = tan ⎝x + ⎞ ⎠ π 4 74. f (x) = πtan(πx − π) − π 75. f (x) = 2csc(x) 76. f (x) = − 1 4 csc(x) 77. f (x) = 4sec(3x) 78. f (x) = − 3cot(2x) 79. f (x) = 7sec(5x) 80. 81. 82. 83. f (x) = 9 10 csc(πx) ⎛ ⎝x + f (x) = 2csc ⎞ ⎠ − 1 π 4 ⎛ ⎝x − f (x) = − sec ⎞ ⎠ − 2 π 3 f (x) = 7 5 ⎛ ⎝x − csc ⎞ ⎠ π 4 948 Chapter 8 Periodic Functions 93. Technology For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x as 1 sin x. 94. 95. 96. 97. 98. f (x) = \|csc(x)\| f (x) = \|cot(x)\| f (x) = 2 csc(x) f (x) = csc(x) sec(x) Graph f (x) = 1 + sec2 (x) − tan2 (x). What is the function shown in the graph? 99. f (x) = sec(0.001x) 100. f (x) = cot(100πx) 101. f (x) = sin2 x + cos2 x Real-World Applications 102. 90. 91. 92. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 949 x days, for x in the interval 0 to 30 days, is given by ⎛ g(x) = 250,000csc ⎝ x⎞ ⎠. π 30 a. Graph g(x) on the interval ⎡ ⎣0, 35⎤ ⎦. b. Evaluate g(5) and interpret the information. c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? d. Find and discuss the meaning of any vertical asymptotes. A video camera is focused on a rocket on a launching 105. pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x seconds is π 120 x. a. Write a function expressing the altitude h(x), in miles, of the rocket above the ground after x seconds. Ignore the curvature of the Earth. b. Graph h(x) on the interval (0, 60). c. Evaluate and interpret the values h(0) and h(30). d. What happens to the values of h(x) as x approaches 60 seconds? Interpret the meaning of this in terms of the problem. ⎛ The function f (x) = 20tan ⎝ x⎞ π ⎠ marks the distance in the 10 movement of a light beam from a police car across a wall for time x, in seconds, and distance f (x), in feet. a. Graph on the interval ⎡ ⎣0, 5⎤ ⎦. b. Find and interpret the stretching factor, period, and asymptote. c. Evaluate f (1) and f (2.5) and discuss the function’s values at those inputs. 103. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to the right. (See Figure 8.52.) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance d(x), in kilometers, from the fisherman to the boat is given by the function d(x) = 1.5sec(x). a. What is a reasonable domain for d(x) ? b. Graph d(x) on this domain. c. Find and discuss the meaning of any vertical asymptotes on the graph of d(x). d. Calculate and interpret d⎛ ⎝− ⎞ ⎠. Round to the π 3 second decimal place. e. Calculate and interpret d⎛ ⎝ ⎞ ⎠. Round to the second π 6 decimal place. f. What is the minimum distance between the fisherman and the boat? When does this occur? Figure 8.52 A laser rangefinder is locked on a comet approaching 104. Earth. The distance g(x), in kilometers, of the comet after 950 Chapter 8 Periodic Functions 8.3 \| Inverse Trigonometric Functions Learning Objectives In this section, you will: 8.3.1 Understand and use the inverse sine, cosine, a
nd tangent functions. 8.3.2 Find the exact value of expressions involving the inverse sine, cosine, and tangent functions. 8.3.3 Use a calculator to evaluate inverse trigonometric functions. 8.3.4 Find exact values of composite functions with inverse trigonometric functions. For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the inverse trigonometric functions. Understanding and Using the Inverse Sine, Cosine, and Tangent Functions In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 8.53. Figure 8.53 For example, if f (x) = sin x, then we would write f −1(x) = sin−1 x. Be aware that sin−1 x does not mean 1 sinx. The following examples illustrate the inverse trigonometric functions: ⎛ • Since sin ⎝ π 6 ⎞ ⎠ = 1 2 , then π 6 = sin−1 ⎛ ⎝ ⎞ ⎠. 1 2 • Since cos(π) = − 1, then π = cos−1 (−1). ⎛ • Since tan ⎝ π 4 ⎞ ⎠ = 1, then π 4 = tan−1 (1). In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if f (a) = b, then an inverse function would satisfy f −1(b) = a. Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure 8.54 shows the graph of the sine function limited to ⎡ ⎤ ⎦ and the graph ⎣− , π 2 π 2 of the cosine function limited to [0, π]. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 951 Figure 8.54 (a) Sine function on a restricted domain of ⎤ ⎡ π ⎦; (b) Cosine function on a restricted domain of ⎣− 2 [0, π] π 2 , Figure 8.55 shows the graph of the tangent function limited to ⎛ ⎝− π 2 , ⎞ ⎠. π 2 Figure 8.55 Tangent function on a restricted domain of ⎛ ⎝− ⎞ ⎠ , π 2 π 2 These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-toone function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote. On these restricted domains, we can define the inverse trigonometric functions. • The inverse sine function y = sin−1 x means x = sin y. The inverse sine function is sometimes called the arcsine function, and notated arcsinx. y = sin−1 x has domain [−1, 1] and range ⎡ ⎣− π 2 , ⎤ ⎦ π 2 • The inverse cosine function y = cos−1 x means x = cos y. The inverse cosine function is sometimes called the arccosine function, and notated arccos x. y = cos−1 x has domain [−1, 1] and range [0, π] • The inverse tangent function y = tan−1 x means x = tan y. The inverse tangent function is sometimes called the arctangent function, and notated arctan x. 952 Chapter 8 Periodic Functions y = tan−1 x has domain (−∞, ∞) and range ⎛ ⎝− The graphs of the inverse functions are shown in Figure 8.56, Figure 8.57, and Figure 8.58. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that sin−1 x has domain [−1, 1] and range ⎡ ⎤ ⎦, cos−1 x has domain [−1,1] and range [0, π], and tan−1 x has domain of all real numbers and range ⎣− π 2 ⎞ ⎛ ⎠. To find the domain and range of inverse trigonometric functions, switch the domain and range of the original ⎝− functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line y = x. π 2 π 2 , Figure 8.56 The sine function and inverse sine (or arcsine) function Figure 8.57 The cosine function and inverse cosine (or arccosine) function This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 953 Figure 8.58 The tangent function and inverse tangent (or arctangent) function Relations for Inverse Sine, Cosine, and Tangent Functions For angles in the interval ⎡ ⎣− π 2 , π 2 ⎤ ⎦, if sin y = x, then sin−1 x = y. For angles in the interval [0, π], if cos y = x, then cos−1 x = y. For angles in the interval ⎛ ⎝− π 2 , π 2 ⎞ ⎠, if tan y = x, then tan−1 x = y. Example 8.24 Writing a Relation for an Inverse Function ⎛ Given sin ⎝ 5π 12 ⎞ ⎠ ≈ 0.96593, write a relation involving the inverse sine. Solution Use the relation for the inverse sine. If sin y = x, then sin−1 x = y . In this problem, x = 0.96593, and y = 5π 12 . sin−1(0.96593) ≈ 5π 12 8.19 Given cos(0.5) ≈ 0.8776, write a relation involving the inverse cosine. 954 Chapter 8 Periodic Functions Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when (45°), and π we are using the special angles, specifically π 3 6 (60°), and their reflections into other quadrants. (30°), π 4 Given a “special” input value, evaluate an inverse trigonometric function. 1. Find angle x for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function. 2. If x is not in the defined range of the inverse, find another angle y that is in the defined range and has the same sine, cosine, or tangent as x, depending on which corresponds to the given inverse function. Example 8.25 Evaluating Inverse Trigonometric Functions for Special Input Values Evaluate each of the following. a. b. c. d. ⎞ ⎠ sin−1 ⎛ ⎝ 1 2 sin−1 ⎛ ⎝− 2 2 ⎞ ⎠ cos−1 ⎛ ⎝− 3 2 ⎞ ⎠ tan−1 (1) Solution a. Evaluating sin−1 ⎛ ⎝ 1 2 ⎞ ⎠ is the same as determining the angle that would have a sine value of 1 2 . In other words, what angle x would satisfy sin(x) = 1 2 ? There are multiple values that would satisfy this relationship, such as π 6 π 6 will be sin−1 ⎛ ⎝ ⎞ ⎠ = 1 2 one output. and 5π 6 , but we know we need the angle in the interval ⎡ ⎣− π 2 , π 2 ⎤ ⎦, so the answer . Remember that the inverse is a function, so for each input, we will get exactly b. To evaluate sin−1 ⎛ ⎞ ⎠, we know that 5π 4 the interval ⎡ ⎣− ⎝− 2 2 ⎤ ⎦. For that, we need the negative angle coterminal with 7π π 2 4 c. To evaluate cos−1 ⎛ both have a sine value of − 2 2 : sin−1( − 2 2 ⎞ ⎠, we are looking for an angle in the interval [0, π] with a cosine value of , but neither is in ) = − . π 4 and 7π 4 π 2 , ⎝− 3 2 − 3 2 . The angle that satisfies this is cos−1 ⎛ ⎝− 3 2 ⎞ ⎠ = 5π 6 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 955 d. Evaluating tan−1 (1), we are looking for an angle in the interval ⎛ ⎝− π 4 The correct angle is tan−1 (1) = . π 2 , π 2 ⎞ ⎠ with a tangent value of 1. 8.20 Evaluate each of the following. a. b. c. d. sin−1(−1) tan−1 (−1) cos−1 (−1) cos−1 ⎛ ⎝ ⎞ ⎠ 1 2 Using a Calculator to Evaluate Inverse Trigonometric Functions To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN. In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places. In these examples and exercises, the answers will be interpreted as angles and we will use θ as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application. Example 8.26 Evaluating the Inverse Sine on a Calculator Evaluate sin−1(0.97) using a calculator. Solution Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using. In radian mode, sin−1(0.97) ≈ 1.3252. In degree mode, sin−1(0.97) ≈ 75.93°. Note that in calculus and beyond we will use radians in almost all cases. 8.21 Evaluate cos−1 (−0.4) using a calculator. 956 Chapter 8 Periodic Functions Given two sides of a right triangle like the one shown in Figure 8.59, find an angle. Figure 8.59 1. 2. If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use the equation θ = cos−1 ⎛ ⎝ ⎞ ⎠. a h If one give
n side is the hypotenuse of length h and the side of length p opposite to the desired angle is given, use the equation θ = sin−1 ⎛ ⎝ ⎞ ⎠. p h 3. If the two legs (the sides adjacent to the right angle) are given, then use the equation θ = tan−1 ⎛ ⎝ p a ⎞ ⎠. Example 8.27 Applying the Inverse Cosine to a Right Triangle Solve the triangle in Figure 8.60 for the angle θ. Figure 8.60 Solution Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function. cos θ = 9 12 θ = cos−1( 9 12 θ ≈ 0.7227 or about 41.4096° ) Apply definition of he inverse . Evaluate. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 957 8.22 Solve the triangle in Figure 8.61 for the angle θ. Figure 8.61 Finding Exact Values of Composite Functions with Inverse Trigonometric Functions There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f (x) and g(x) be two different trigonometric functions belonging to the set {sin(x), cos(x), tan(x)} and let f −1(y) and g−1(y) be their inverses. Evaluating Compositions of the Form f(f−1(y)) and f−1(f(x)) For any trigonometric function, f ⎛ ⎞ ⎝ f −1 (y) ⎠ = y for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of f −1. However, we have to be a little more careful with expressions of the form f −1 ⎛ ⎝ f (x)⎞ ⎠. Compositions of a trigonometric function and its inverse sin(sin−1 x) = x for − 1 ≤ x ≤ 1 cos(cos−1 x) = x for − 1 ≤ x ≤ 1 tan(tan−1 x) = x for − ∞ < x < ∞ sin−1(sin x) = x only for − π 2 ≤ x ≤ cos−1(cos x) = x only for 0 ≤ x ≤ π tan−1(tan x ) = x only for − < x < π 2 π 2 π 2 Is it correct that sin−1(sin x) = x? No. This equation is correct if x belongs to the restricted domain ⎡ ⎤ ⎦, but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in ⎡ ⎤ ⎦. The situation is similar for cosine and tangent and their inverses. For example, ⎣− π 2 , π 2 ⎣− π 2 , π 2 sin−1 ⎛ ⎝sin⎛ ⎝ 3π 4 ⎞ ⎞ ⎠ ⎠ = π 4 . 958 Chapter 8 Periodic Functions Given an expression of the form f−1(f(θ)) where f(θ) = sin θ, cos θ, or tan θ, evaluate. 1. If θ is in the restricted domain of f , then f −1( f (θ)) = θ. 2. If not, then find an angle ϕ within the restricted domain of f such that f (ϕ) = f (θ). Then f −1 ⎛ ⎝ f (θ)⎞ ⎠ = ϕ. Example 8.28 Using Inverse Trigonometric Functions Evaluate the following: sin−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ 1. π 3 2. 3. 4. sin−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ 2π 3 cos−1 ⎛ ⎛ ⎝cos ⎝ ⎞ ⎞ ⎠ ⎠ 2π 3 cos−1 ⎛ ⎛ ⎝− ⎝cos ⎞ ⎞ ⎠ ⎠ π 3 Solution a. b. c. ⎡ is in ⎣− π 3 π 2 , π 2 ⎤ ⎦, so sin−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ = ⎠ π 3 π 3 . , ⎤ ⎛ ⎦, but sin ⎝ π 2 ⎡ ⎣− is not in π 2 is in [0, π], so cos−1 ⎛ ⎛ ⎝cos ⎝ 2π 3 2π 3 d. − e. π 3 ⎛ is not in [0, π], but cos ⎝− π 3 is in [0, π], so cos−1 ⎛ ⎛ ⎝− ⎝cos ⎞ ⎞ ⎠ = ⎠ π 3 π 3 . 2π 3 ⎛ ⎞ ⎠ = sin ⎝ π 3 ⎠, so sin−1 ⎛ ⎞ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ = ⎠ 2π 3 π 3 . 2π 3 ⎞ ⎞ ⎠ = 2π ⎠ 3 . π 3 ⎛ ⎞ ⎠ = cos ⎝ π 3 ⎞ ⎠ because cosine is an even function. 8.23 Evaluate tan−1 ⎛ ⎛ ⎝tan ⎝ π 8 ⎠ and tan−1 ⎛ ⎞ ⎞ ⎠ ⎛ ⎝tan ⎝ 11π 9 ⎞ ⎞ ⎠. ⎠ Evaluating Compositions of the Form f−1(g(x)) ⎝g(x)⎞ Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form f −1 ⎛ ⎠. For special values of x, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is θ, making the other π − θ. Consider the sine and cosine of each angle of the right triangle in Figure 8.62. 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 959 Figure 8.62 Right triangle illustrating the cofunction relationships π 2 − θ⎞ b ⎛ c = sin ⎝ Because cos θ = π 2 to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract this angle from π 2 − θ if 0 ≤ θ ≤ π. If θ is not in this domain, then we need ⎠, we have sin−1 (cos θ) = ⎠, so cos−1 (sin θ) = . Similarly, sin θ = . These are just a ⎛ c = cos ⎝ the function- − θ if − ≤ θ ≤ − θ⎞ π 2 π 2 π 2 π 2 cofunction relationships presented in another way. Given functions of the form sin−1 (cos x) and cos−1 (sin x), evaluate them. 1. 2. 3. 4. π 2 , ⎡ If x is in ⎣− π 2 If x is not in ⎡ ⎣− If x is in [0, π], then sin−1 (cos x) = π 2 − x. If x is not in [0, π], then find another angle y in [0, π] such that cos y = cos x. sin−1 (cos x) = π 2 − y ⎤ ⎦, then cos−1 (sin x) = π 2 − x. π 2 , π 2 ⎡ ⎤ ⎦, then find another angle y in ⎣− π 2 , π 2 ⎤ ⎦ such that sin y = sin x. cos−1 (sin x) = π 2 − y Example 8.29 Evaluating the Composition of an Inverse Sine with a Cosine Evaluate sin−1 ⎛ ⎛ ⎝cos ⎝ 13π 6 ⎞ ⎞ ⎠ ⎠ a. by direct evaluation. b. by the method described previously. Solution a. Here, we can directly evaluate the inside of the composition. ) = cos( cos(13π 6 π 6 π = cos( 6 + 2π) ) Now, we can evaluate the inverse function as we did earlier. = 3 2 960 Chapter 8 Periodic Functions b. We have x = 13π 6 , y = π 6 , and sin− sin−1 ⎛ ⎛ ⎝cos ⎝ 13π .24 Evaluate cos−1 ⎛ ⎛ ⎝sin ⎝− 11π 4 ⎞ ⎞ ⎠. ⎠ Evaluating Compositions of the Form f(g−1(x)) To evaluate compositions of the form f ⎛ ⎞ ⎝g−1 (x) ⎠, where f and g are any two of the functions sine, cosine, or tangent and ⎛ ⎝cos−1 x⎞ x is any input in the domain of g−1, we have exact formulas, such as sin ⎠ = 1 − x2. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, sin2 x + cos2 x = 1, to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions. Example 8.30 Evaluating the Composition of a Sine with an Inverse Cosine ⎝cos−1 ⎛ ⎛ Find an exact value for sin ⎝ ⎞ ⎞ ⎠. ⎠ 4 5 Solution Beginning with the inside, we can say there is some angle such that θ = cos−1 ⎛ ⎝ and we are looking for sin θ. We can use the Pythagorean identity to do this. 4 5 ⎞ ⎠, which means cos θ = 4 5 , sin2 θ + cos2 θ = 1 sin2 θ + (4 5 = 1 ) 2 Use our known value for cosine. Solve for sine. sin2 θ = 1 − 16 25 sin θ = ± 9 25 = ± 3 5 Since θ = cos−1 ⎛ ⎝ 4 5 ⎞ ⎠ is in quadrant I, sin θ must be positive, so the solution is 3 5 . See Figure 8.63. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 961 Figure 8.63 Right triangle illustrating that if cos θ = 4 5 , then sin θ = 3 5 We know that the inverse cosine always gives an angle on the interval [0, π], so we know that the sine of that ⎞ ⎞ ⎠ = sin θ = 3 ⎠ 5 ⎝cos−1 ⎛ ⎛ angle must be positive; therefore sin ⎝ 4 5 . 8.25 ⎛ ⎝tan−1 ⎛ Evaluate cos ⎝ ⎞ ⎞ ⎠. ⎠ 5 12 Example 8.31 Evaluating the Composition of a Sine with an Inverse Tangent ⎛ ⎝tan−1 ⎛ Find an exact value for sin ⎝ ⎞ ⎞ ⎠. ⎠ 7 4 Solution While we could use a similar technique as in Example 8.29, we will demonstrate a different technique here. From the inside, we know there is an angle such that tan θ = 7 . We can envision this as the opposite and adjacent 4 sides on a right triangle, as shown in Figure 8.64. Figure 8.64 A right triangle with two sides known Using the Pythagorean Theorem, we can find the hypotenuse of this triangle. 962 Chapter 8 Periodic Functions 42 + 72 = hypotenuse 2 hypotenuse = 65 Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse. This gives us our desired composition. sin θ = 7 65 ⎛ ⎝tan−1 ⎛ sin ⎝ 7 4 ⎞ ⎞ ⎠ = sin θ ⎠ = 7 65 = 7 65 65 8.26 ⎝sin−1 ⎛ ⎛ Evaluate cos ⎝ ⎞ ⎞ ⎠. ⎠ 7 9 Example 8.32 Finding the Cosine of the Inverse Sine of an Algebraic Expression ⎝sin−1 ⎛ ⎛ Find a simplified expression for cos ⎝ ⎞ ⎞ ⎠ for − 3 ≤ x ≤ 3. ⎠ x 3 Solution We know there is an angle θ such that sin θ = x 3 . sin2 θ + cos2 + cos2 θ = 1 cos2 θ = 1 − x2 9 cosθ = ± 9 − x2 9 = ± 9 − x2 3 Use the Pythagorean Theorem. Solve for cosine. Because we know that the inverse sine must give an angle on the interval ⎡ ⎣− π 2 , π 2 ⎤ ⎦, we can deduce that the cosine of that angle must be positive. ⎝sin−1 ⎛ ⎛ cos ⎝ x 3 ⎞ ⎞ ⎠ ⎠ = 9 − x2 3 8.27 ⎞ ⎛ ⎠ for − 1 ⎝tan−1 (4x) Find a simplified expression for sin 4 ≤ x ≤ 1 4 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 963 Access this online resource for additional instruction and practice with inverse trigonometric functions. • Evaluate Expressions Involving Inverse Trigonometric Functions (http://openstaxcollege.org/l/evalinverstrig) this website (http://openstaxcollege.org/l/PreCalcLPC06) Visit Learningpod. for additional practice questions from 964 Chapter 8 Periodic Functions 8.3 EXERCISES Verbal 106. Why do the functions f (x) = sin−1 x and g(x) = cos−1 x have different ranges? tan−1 ⎛ ⎝ ⎞ ⎠ −1 3 For the following exercises, use a calculator to evaluate each expression. Express answers to the nearest hundredth. 107. Since the functions y = cos x and y = cos−1 x are 122. cos−1 (−0.4) inverse functions, why is cos−1 ⎛ ⎛ ⎝− ⎝cos − π 6 ? ⎞ ⎞ ⎠ not equal to ⎠ π 6 108. Explain the meaning of π 6 = arcsin(0.5). 109. Most calculators do not have a key to evaluate sec−1 (2). Explain how this can be done using the cosine function or the inverse cosine function. 123. arcsin(0.23) 124. ⎛ arccos ⎝ ⎞ ⎠ 3 5 125. cos−1
(0.8) 126. tan−1 (6) 110. restricted to ⎡ ⎣− Why must the domain of the sine function, sin x, be ⎤ ⎦ for the inverse sine function to exist? , π 2 π 2 111. this Discuss why arccos(cos x) = x for all x. statement is incorrect: Determine whether the following statement is true or answer: explain 112. false and arccos(−x) = π − arccos x. your Algebraic For the following exercises, evaluate the expressions. 113. sin−1 ⎛ ⎝ ⎞ ⎠ 2 2 114. 115. 116. sin−1 ⎛ ⎝− 1 2 ⎞ ⎠ cos−1 ⎛ ⎝ ⎞ ⎠ 1 2 cos−1 ⎛ ⎝− 2 2 ⎞ ⎠ 117. tan−1 (1) 118. tan−1 ⎛ ⎝− 3⎞ ⎠ 119. tan−1 (−1) tan−1 ⎛ ⎝ 3⎞ ⎠ 120. 121. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, find the angle θ in the given right triangle. Round answers to the nearest hundredth. 127. 128. For the following exercises, if possible, without a calculator. If it is not possible, explain why. find the exact value, 129. sin−1 (cos(π)) 130. 131. 132. 133. 134. tan−1 ⎛ ⎝sin(π)⎞ ⎠ cos−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ π 3 tan−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ π 3 sin−1 ⎛ ⎛ ⎝cos ⎝ −π 2 ⎞ ⎞ ⎠ ⎠ tan−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ 4π 3 Chapter 8 Periodic Functions 135. 136. sin−1 ⎛ ⎛ ⎝sin ⎝ ⎞ ⎞ ⎠ ⎠ 5π 6 tan−1 ⎛ ⎛ ⎝sin ⎝ −5π 2 ⎞ ⎞ ⎠ ⎠ 137. ⎝sin−1 ⎛ ⎛ cos ⎝ 138. ⎝cos−1 ⎛ ⎛ sin ⎝ ⎞ ⎞ ⎠ ⎠ 4 5 ⎞ ⎞ ⎠ ⎠ 3 5 139. 140. 141. ⎝tan−1 ⎛ ⎛ sin ⎝ ⎞ ⎞ ⎠ ⎠ 4 3 ⎝tan−1 ⎛ ⎛ cos ⎝ ⎞ ⎞ ⎠ ⎠ 12 5 ⎛ ⎝sin−1 ⎛ cos ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 For the following exercises, find the exact value of the expression in terms of x with the help of a reference triangle. 142. ⎛ ⎞ ⎝sin−1 (x − 1) tan ⎠ 143. ⎛ ⎞ ⎝cos−1 (1 − x) sin ⎠ 144. ⎝sin−1 ⎛ ⎛ cos ⎝ 1 x ⎞ ⎞ ⎠ ⎠ 145. ⎛ ⎞ ⎝tan−1 (3x − 1) cos ⎠ 146. ⎛ ⎝sin−1 ⎛ tan ⎝x + 1 2 ⎞ ⎞ ⎠ ⎠ Extensions the following exercises, evaluate the expression For without using a calculator. Give the exact value. 147. sin−1 ⎛ 1 ⎝ 2 cos−1 ⎛ ⎝ 2 2 ⎞ ⎠ − cos−1 ⎛ ⎝ ⎞ ⎠ − sin−1 ⎛ 3 2 ⎝ ⎞ ⎠ + sin−1 ⎛ 3 ⎝ 2 ⎞ ⎠ + cos−1 ⎛ ⎝ ⎞ ⎠ − cos−1 (1) ⎞ ⎠ − sin−1 (0) 1 2 2 2 For the sin t = following exercises, x x + 1 . find the function if 148. cos t 149. sec t 150. cot t 151. ⎝sin−1 ⎛ ⎛ cos ⎝ x x + 1 ⎞ ⎞ ⎠ ⎠ 965 152. tan−1 ⎛ ⎝ x 2x + 1 ⎞ ⎠ Graphical 153. Graph y = sin−1 x and state the domain and range of the function. 154. Graph y = arccos x and state the domain and range of the function. 155. Graph one cycle of y = tan−1 x and state the domain and range of the function. 156. For what value of x does sin x = sin−1 x ? Use a graphing calculator to approximate the answer. 157. For what value of x does cos x = cos−1 x ? Use a graphing calculator to approximate the answer. Real-World Applications Suppose a 13-foot ladder is leaning against a building, 158. reaching to the bottom of a second-ﬂoor window 12 feet above the ground. What angle, in radians, does the ladder make with the building? Suppose you drive 0.6 miles on a road so that the 159. vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road? An isosceles triangle has two congruent sides of 160. length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side. Without using a calculator, approximate the value of 161. arctan(10,000). Explain why your answer is reasonable. A truss for the roof of a house is constructed from two 162. identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side. 163. The line y = 3 5 x passes through the origin in the x,y- plane. What is the measure of the angle that the line makes with the positive x-axis? 164. The line y = −3 7 x passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the negative x-axis? What percentage grade should a road have if the angle 165. of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade 966 Chapter 8 Periodic Functions means that the road rises 5 feet for every 100 feet of horizontal distance.) A 20-foot 166. the side of a ladder leans up against building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder's angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications? Suppose a 15-foot ladder leans against the side of a 167. house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 967 CHAPTER 8 REVIEW KEY TERMS amplitude the vertical height of a function; the constant A appearing in the definition of a sinusoidal function arccosine another name for the inverse cosine; arccos x = cos−1 x arcsine another name for the inverse sine; arcsin x = sin−1 x arctangent another name for the inverse tangent; arctan x = tan−1 x inverse cosine function the function cos−1 x, which is the inverse of the cosine function and the angle that has a cosine equal to a given number inverse sine function to a given number the function sin−1 x, which is the inverse of the sine function and the angle that has a sine equal inverse tangent function the function tan−1 x, which is the inverse of the tangent function and the angle that has a tangent equal to a given number midline the horizontal line y = D, where D appears in the general form of a sinusoidal function periodic function a function f (x) that satisfies f (x + P) = f (x) for a specific constant P and any value of x phase shift the horizontal displacement of the basic sine or cosine function; the constant C B sinusoidal function any function that can be expressed in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D KEY EQUATIONS Sinusoidal functions f (x) = Asin(Bx − C) + D f (x) = Acos(Bx − C) + D Shifted, compressed, and/or stretched tangent function y = A tan(Bx − C) + D Shifted, compressed, and/or stretched secant function y = A sec(Bx − C) + D Shifted, compressed, and/or stretched cosecant function y = A csc(Bx − C) + D Shifted, compressed, and/or stretched cotangent function y = A cot(Bx − C) + D KEY CONCEPTS 8.1 Graphs of the Sine and Cosine Functions • Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of 2π. 968 Chapter 8 Periodic Functions • The function sin x is odd, so its graph is symmetric about the origin. The function cos x is even, so its graph is symmetric about the y-axis. • The graph of a sinusoidal function has the same general shape as a sine or cosine function. • • In the general formula for a sinusoidal function, the period is P = 2π \|B\| . See Example 8.1. In the general formula for a sinusoidal function, \|A\| represents amplitude. If \|A\| > 1, the function is stretched, whereas if \|A\| < 1, the function is compressed. See Example 8.2. • The value C B in the general formula for a sinusoidal function indicates the phase shift. See Example 8.3. • The value D in the general formula for a sinusoidal function indicates the vertical shift from the midline. See Example 8.4. • Combinations of variations of sinusoidal functions can be detected from an equation. See Example 8.5. • The equation for a sinusoidal function can be determined from a graph. See Example 8.6 and Example 8.7. • A function can be graphed by identifying its amplitude and period. See Example 8.8 and Example 8.9. • A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See Example 8.10. • Sinusoidal functions can be used to solve real-world problems. See Example 8.11, Example 8.12, and Example 8.13. 8.2 Graphs of the Other Trigonometric Functions • The tangent function has period π. • f (x) = Atan(Bx − C) + D is a tangent with vertical and/or horizontal stretch/compression and shift. See Example 8.14, Example 8.15, and Example 8.16. • The secant and cosecant are both periodic functions with a period of 2π. f (x) = Asec(Bx − C) + D gives a shifted, compressed, and/or stretched secant function graph. See Example 8.17 and Example 8.18. • f (x) = Acsc(Bx − C) + D gives a shifted, compressed, and/or stretched cosecant function graph. See Example 8.19 and Example 8.20. • The cotangent function has period π and vertical asymptotes at 0, ± π, ± 2π, .... • The range of cotangent is (−∞, ∞), and the function is decreasing at each point in its range. • The cotangent is zero at ± π 2 , ± 3π 2 , .... • f (x) = Acot(Bx − C) + D is a cotangent with vertical and/or horizontal stretch/compression and shift. See Example 8.21 and Example 8.22. • Real-world scenarios can be solved using graphs of trigonometric functions. See Example 8.23. 8.3 Inverse Trigonometric Functions • An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function. • Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains. • For any trigonometric function f (x), if x = f −1(y), then f (x) = y. However, f (x) = y only implies x = f −1(y) if x is in the restricted domain of f . See Example 8.24. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 969 • Special angles are the outputs of inverse trigonometric functions for special input values; for example, π 4 = tan−1(1) and π 6 = sin−1 ⎛ ⎝ 1 2 ⎞ ⎠. See Example 8.25. • A calculator will return an angle within the restricted domain of the original trigonometric function. See Example 8.26. • • • Inverse functions allow us to find an angle when given two sides of a right triangle. See Example 8.27. In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; ⎞ ⎛ ⎠ = 1 − x2. See Example 8.28.
⎝cos−1 (x) for example, sin If the inside function is a trigonometric function, then the only possible combinations are sin−1 (cos x) = 0 ≤ x ≤ π and cos−1 (sin x) = π 2 − x if − π 2 ≤ x ≤ π 2 . See Example 8.29 and Example 8.30. − x if π 2 • When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See Example 8.31. • When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See Example 8.32. CHAPTER 8 REVIEW EXERCISES Graphs of the Sine and Cosine Functions 176. f (x) = tan x − 4 For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. 168. f (x) = − 3cos x + 3 169. f (x) = 1 4 sin x 170. ⎛ ⎝x + f (x) = 3cos ⎞ ⎠ π 6 171. ⎛ ⎝x − 2π f (x) = − 2sin 3 ⎞ ⎠ 172. ⎛ f (x) = 3sin ⎝x − ⎞ ⎠ − 4 π 4 173. f (x) = 2 ⎛ ⎛ ⎝x − 4π ⎝cos 3 ⎞ ⎞ ⎠ + 1 ⎠ 174. ⎛ ⎝3x − f (x) = 6sin ⎞ ⎠ − 1 π 6 175. f (x) = − 100sin(50x − 20) 177. ⎛ ⎝x − f (x) = 2tan ⎞ ⎠ π 6 178. f (x) = − 3tan(4x) − 2 179. f (x) = 0.2cos(0.1x) + 0.3 For the following exercises, graph two full periods. Identify the period, the phase shift, the amplitude, and asymptotes. 180. f (x) = 1 3 sec x 181. f (x) = 3cot x 182. f (x) = 4csc(5x) 183. ⎛ f (x) = 8sec ⎝ x⎞ ⎠ 1 4 184. f (x) = 2 3 ⎛ csc ⎝ x⎞ ⎠ 1 2 185. f (x) = − csc(2x + π) Graphs of the Other Trigonometric Functions For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. For the following exercises, use this scenario: The population of a city has risen and fallen over a 20-year interval. Its population may be modeled by the following function: y = 12,000 + 8,000sin⎛ the ⎝0.628x), where 970 Chapter 8 Periodic Functions domain is the years since 1980 and the range is the population of the city. 195. sin−1 (1) 186. What is the largest and smallest population the city may have? 187. Graph the function on the domain of [0, 40] . 188. What are the amplitude, period, and phase shift for the function? 189. Over this domain, when does the population reach 18,000? 13,000? 190. What is the predicted population in 2007? 2010? For the following exercises, suppose a weight is attached to a spring and bobs up and down, exhibiting symmetry. 191. Suppose the graph of the displacement function is shown in Figure 8.65, where the values on the x-axis represent the time in seconds and the y-axis represents the displacement in inches. Give the equation that models the vertical displacement of the weight on the spring. 196. cos−1 ⎛ ⎝ ⎞ ⎠ 3 2 197. tan−1 (−1) 198. cos−1 ⎛ ⎝ ⎞ ⎠ 1 2 199. sin−1 ⎛ ⎝ − 3 2 ⎞ ⎠ 200. sin−1 ⎛ ⎛ ⎝cos ⎝ ⎞ ⎞ ⎠ ⎠ π 6 201. cos−1 ⎛ ⎛ ⎝tan ⎝ ⎞ ⎞ ⎠ ⎠ 3π 4 202. ⎝sec−1 ⎛ ⎛ sin ⎝ ⎞ ⎞ ⎠ ⎠ 3 5 203. cot ⎝sin− 204. ⎝cos−1 ⎛ ⎛ tan ⎝ ⎞ ⎞ ⎠ ⎠ 5 13 205. ⎝cos−1 ⎛ ⎛ sin ⎝ x x + 1 ⎞ ⎞ ⎠ ⎠ Graph f (x) = cos x and f (x) = sec x on 206. interval [0, 2π) and explain any observations. the 207. Graph f (x) = sin x and f (x) = csc x and explain any observations. 208. Graph the function f (x) = x 1 − x3 3 ! + x5 5 ! − x7 7 ! on the interval [−1, 1] and compare the graph to the graph of f (x) = sin x on the any same observations. interval. Describe Figure 8.65 192. At time = 0, what is the displacement of the weight? 193. At what equilibrium point equal zero? time does the displacement from the 194. What is the time required for the weight to return to its initial height of 5 inches? In other words, what is the period for the displacement function? Inverse Trigonometric Functions For the following exercises, find the exact value without the aid of a calculator. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 8 Periodic Functions 971 CHAPTER 8 PRACTICE TEST For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. 223. Give in terms of a sine function. 224. Give in terms of a tangent function. 209. f (x) = 0.5sin x 210. f (x) = 5cos x 211. f (x) = 5sin x 212. f (x) = sin(3x) 213. ⎛ ⎝x + f (x) = − cos ⎞ ⎠ + 1 π 3 214. ⎛ f (x) = 5sin ⎝3 ⎛ ⎝ 215. ⎛ f (x) = 3cos ⎝ 1 3 x − 5π 6 ⎞ ⎠ 216. f (x) = tan(4x) 217. ⎛ ⎝x − 7π f (x) = − 2tan 6 ⎞ ⎠ + 2 218. f (x) = πcos(3x + π) 219. f (x) = 5csc(3x) 220. ⎛ f (x) = πsec ⎝ x⎞ ⎠ π 2 221. ⎛ f (x) = 2csc ⎝x + ⎞ ⎠ − 3 π 4 For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function. 222. Give in terms of a sine function. For the following exercises, find the amplitude, period, phase shift, and midline. x + π⎞ 225. ⎛ y = sin ⎝ ⎠ − 3 π 6 226. ⎛ y = 8sin ⎝ 7π 6 x + 7π 2 ⎞ ⎠ + 6 972 Chapter 8 Periodic Functions 227. The outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68°F at midnight and the high and low temperatures during the day are 80°F and 56°F, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t. 228. Water is pumped into a storage bin and empties according to a periodic rate. The depth of the water is 3 feet at its lowest at 2:00 a.m. and 71 feet at its highest, which occurs every 5 hours. Write a cosine function that models the depth of the water as a function of time, and then graph the function for one period. For the following exercises, find the period and horizontal shift of each function. 229. g(x) = 3tan(6x + 42) ⎛ 230. n(x) = 4csc ⎝ 5π 3 x − 20π 3 ⎞ ⎠ 231. Write the equation for the graph in Figure 8.66 in terms of the secant function and give the period and phase shift. Figure 8.66 232. If tan x = 3, find tan(−x). 233. If sec x = 4, find sec(−x). For the following exercises, graph the functions on the specified window and answer the questions. 234. Graph m(x) = sin(2x) + cos(3x) on the viewing window [−10, 10] by [−3, 3]. Approximate the graph’s period. This content is available for free at https://cnx.org/content/col11758/1.5 Graph n(x) = 0.02sin(50πx) on following 235. [0, 1] and [0, 3]. Suppose this function domains in x : models sound waves. Why would these views look so different? the 236. Graph f (x) = sin x observations. x on ⎡ ⎣−0.5, 0.5⎤ ⎦ and explain any For the following exercises, let f (x) = 3 5 cos(6x). 237. What is the largest possible value for f (x) ? 238. What is the smallest possible value for f (x) ? 239. Where is the function increasing on the interval [0, 2π] ? For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift. 240. Sine curve with amplitude 3, period π 3 , and phase shift (h, k 241. Cosine curve with amplitude 2, period π 6 , and phase shift (h, k) = ⎛ ⎝− π 4 ⎞ , 3 ⎠ For the following exercises, graph the function. Describe the graph and, wherever applicable, any periodic behavior, amplitude, asymptotes, or undefined points. 242. f (x) = 5cos(3x) + 4sin(2x) 243. f (x) = esint For the following exercises, find the exact value. 244. sin−1 ⎛ ⎝ ⎞ ⎠ 3 2 245. tan−1 ⎛ ⎝ 3⎞ ⎠ 246. cos−1 ⎛ ⎝− 3 2 ⎞ ⎠ 247. cos−1 ⎛ ⎝sin(π)⎞ ⎠ Chapter 8 Periodic Functions 973 248. cos−1 ⎛ ⎛ ⎝tan ⎝ ⎞ ⎞ ⎠ ⎠ 7π 4 249. ⎞ ⎛ ⎝sin−1 (1 − 2x) cos ⎠ 250. cos−1 (−0.4) 251. ⎝tan−1 ⎛ ⎛ cos ⎞ ⎝x2⎞ ⎠ ⎠ For the following exercises, suppose sin t = x x + 1 . 252. tan t 253. csc t 254. Given Figure 8.67, find the measure of angle θ to three decimal places. Answer in radians. Figure 8.67 the following exercises, determine whether For equation is true or false. the 255. ⎛ ⎛ ⎝sin arcsin ⎝ 5π 6 ⎞ ⎞ ⎠ = 5π ⎠ 6 256. ⎛ ⎛ ⎝cos arccos ⎝ 5π 6 ⎞ ⎞ ⎠ = 5π ⎠ 6 257. The grade of a road is 7%. This means that for every horizontal distance of 100 feet on the road, the vertical rise is 7 feet. Find the angle the road makes with the horizontal in radians. 974 Chapter 8 Periodic Functions This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 975 9 \| TRIGONOMETRIC IDENTITIES AND EQUATIONS Figure 9.1 A sine wave models disturbance. (credit: modification of work by Mikael Altemark, Flickr). Chapter Outline 9.1 Solving Trigonometric Equations with Identities 9.2 Sum and Difference Identities 9.3 Double-Angle, Half-Angle, and Reduction Formulas 9.4 Sum-to-Product and Product-to-Sum Formulas 9.5 Solving Trigonometric Equations Introduction Math is everywhere, even in places we might not immediately recognize. For example, mathematical relationships describe the transmission of images, light, and sound. The sinusoidal graph in Figure 9.1 models music playing on a phone, radio, or computer. Such graphs are described using trigonometric equations and functions. In this chapter, we discuss how to manipulate trigonometric equations algebraically by applying various formulas and trigonometric identities. We will also investigate some of the ways that trigonometric equations are used to model real-life phenomena. 976 Chapter 9 Trigonometric Identities and Equations 9.1 \| Solving Trigonometric Equations with Identities Learning Objectives In this section, you will: 9.1.1 Verify the fundamental trigonometric identities. 9.1.2 Simplify trigonometric expressions using algebra and the identities. Figure 9.2 International passports and travel documents In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when
solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions. Verifying the Fundamental Trigonometric Identities Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways. To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. We will begin with the Pythagorean identities (see Table 9.1), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 977 Pythagorean Identities sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ Table 9.1 The second and third identities can be obtained by manipulating the first. The identity 1 + cot2 θ = csc2 θ is found by rewriting the left side of the equation in terms of sine and cosine. Prove: 1 + cot2 θ = csc2 θ 1 + cot2 θ = = ⎞ ⎛ ⎝1 + cos2 θ ⎠ sin2 θ ⎛ ⎞ ⎛ cos2 θ sin2 θ ⎠ + ⎝ ⎝ sin2 θ sin2 θ = sin2 θ + cos2 θ sin2 θ Rewrite the left side. ⎞ ⎠ Write both terms with the common denominator. = 1 sin2 θ = csc2 θ Similarly, 1 + tan2 θ = sec2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives 1 + tan2 θ = 1 + 2 ⎞ ⎠ ⎛ ⎝ sin θ cos θ 2 sin θ ⎞ = ⎠ cos θ = cos2 θ + sin2 θ cos2 θ cos θ cos θ + ⎛ ⎝ ⎛ ⎝ Rewrite left side. 2 ⎞ ⎠ Write both terms with the common denominator. = 1 cos2 θ = sec2 θ Recall that we determined which trigonometric functions are odd and which are even. The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. (See Table 9.2). Even-Odd Identities tan( − θ) = − tan θ cot( − θ) = − cot θ sin( − θ) = − sin θ csc( − θ) = − csc θ cos( − θ) = cos θ sec( − θ) = sec θ Table 9.2 Recall that an odd function is one in which f (−x)= − f (x) for all x in the domain of f . The sine function is an odd function because sin(−θ) = − sin θ. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of π 2 ⎞ ⎛ ⎠ is opposite the output of sin ⎝− ⎛ . The output of sin ⎝ ⎞ ⎠. Thus, and − π 2 π 2 π 2 978 Chapter 9 Trigonometric Identities and Equations ⎛ sin ⎝ π 2 ⎞ ⎠ = 1 and ⎛ sin ⎝− π 2 ⎞ ⎛ ⎠ = −sin ⎝ = −1 ⎞ ⎠ π 2 This is shown in Figure 9.3. Figure 9.3 Graph of y = sin θ Recall that an even function is one in which f (−x) = f (x) for all x in the domain of f The graph of an even function is symmetric about cos( − θ) = cos θ. For example, consider corresponding inputs π 4 the y-axis. The cosine function is an even function because ⎞ ⎛ . The output of cos ⎠ is the same as the ⎝ and − π 4 π 4 ⎛ output of cos ⎝− ⎞ ⎠. Thus, π 4 See Figure 9.4. ⎛ cos ⎝− π 4 ⎞ ⎛ ⎠ = cos ⎝ π 4 ≈ 0.707 ⎞ ⎠ Figure 9.4 Graph of y = cos θ For all θ in the domain of the sine and cosine functions, respectively, we can state the following: • Since sin(−θ⎞ ⎠ = −sin θ, sine is an odd function. • Since, cos(−θ⎞ ⎠ = cos θ, cosine is an even function. the The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, as can consider identity, tan(−θ⎞ cos θ = − tan θ. Tangent is therefore an odd function, which means that tan(−θ) = − tan(θ) tan(−θ⎞ ⎠ = sin(−θ) cos(−θ⎞ ⎠ ⎠ = −tan θ. We = −sin θ interpret negative tangent tangent angle the of a for all θ in the domain of the tangent function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 979 The cotangent identity, cot(−θ) = − cot θ, also follows from the sine and cosine identities. We can interpret the cotangent −sin θ = − cot θ. Cotangent is therefore an odd function, which means that of a negative angle as cot(−θ) = cos(−θ) sin(−θ) = cos θ cot(−θ) = − cot(θ) for all θ in the domain of the cotangent function. The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as csc(−θ) = −sin θ = − csc θ. The cosecant function is therefore odd. 1 sin(−θ) = 1 Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as sec(−θ) = cos θ = sec θ. The secant function is therefore even. 1 cos(−θ) = 1 To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities. The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 9.3. Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry. Reciprocal Identities sin θ = 1 csc θ csc θ = 1 sin θ cos θ = 1 sec θ sec θ = 1 cos θ tan θ = 1 cot θ cot θ = 1 tan θ Table 9.3 The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 9.4. Quotient Identities tan θ = sin θ cos θ cot θ = cos θ sin θ Table 9.4 The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions. Summarizing Trigonometric Identities The Pythagorean identities are based on the properties of a right triangle. cos2 θ + sin2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ (9.1) (9.2) (9.3) The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan(−θ) = − tan θ (9.4) 980 Chapter 9 Trigonometric Identities and Equations cot(−θ) = − cot θ sin(−θ) = − sin θ csc(−θ) = − csc θ cos(−θ) = cos θ sec(−θ) = sec θ The reciprocal identities define reciprocals of the trigonometric functions. sin θ = 1 csc θ cos θ = 1 sec θ tan θ = 1 cot θ csc θ = 1 sin θ sec θ = 1 cos θ cot θ = 1 tan θ The quotient identities define the relationship among the trigonometric functions. tan θ = sin θ cos θ cot θ = cos θ sin θ Example 9.1 Graphing the Equations of an Identity (9.5) (9.6) (9.7) (9.8) (9.9) (9.10) (9.11) (9.12) (9.13) (9.14) (9.15) (9.16) (9.17) Graph both sides of the identity cot θ = 1 tan θ. In other words, on the graphing calculator, graph y = cot θ and y = 1 tan θ. Solution See Figure 9.5. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 981 Figure 9.5 Analysis We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities. Given a trigonometric identity, verify that it is true. 1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build. 2. Look for opportunities to factor expressions, square a binomial, or add fractions. 3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions. 4. If these steps do not yield the desired result, try converting all terms to sines and cosines. Example 9.2 Verifying a Trigonometric Identity Verify tan θcos θ = sin θ. Solution We will start on the left side, as it is the more complicated side: tan θ cos θ = ⎞ ⎠cos θ ⎞ ⎠cos θ sin θ ⎛ ⎝ cos θ ⎛ sin θ ⎝ cos θ = sin θ = Analysis This identity was fairly simple to verify, as it only required writing tan θ in terms of sin θ and cos θ. 982 Chapter 9 Trigonometric Identities and Equations 9.1 Verify the identity csc θ cos θ tan θ = 1. Example 9.3 Verifying the Equivalency Using the Even-Odd Identities Verify the following equivalency using the even-odd identities: (1 + sin x)⎡ ⎣1 + sin(−x)⎤ ⎦ = cos2 x Solution Working on the left side of the equation, we have (1 + sin x)[1 + sin(−x)] = (1 + sin x)(1 − sin x) Since sin(−x)=−sin x = 1 − sin2 x = cos2 x Diffe ence of squares cos2 x = 1 − sin2 x Example 9.4 Verifying a Trigonometric Identity Involving sec2θ Verify the identity sec2 θ − 1 sec2 θ = sin2 θ Solution As the left side is more complicated, let’s begin there. sec2 θ − 1 sec2 θ = ⎛ ⎞ ⎝tan2 θ + 1 ⎠ − 1 sec2 θ sec2 θ = tan2 θ + 1 = tan2 θ sec2 θ = tan2 θ⎛ 1 ⎝ sec2 θ ⎝cos2 θ⎞ ⎠ = tan2 θ⎛ ⎞ ⎠ ⎞ ⎝cos2 θ⎞ ⎛ ⎠ ⎠ ⎞ ⎝cos2 θ ⎞ ⎟⎛ ⎠ ⎠ = ⎛ sin2 θ ⎝ cos2 θ ⎛ ⎜ sin2 θ cos2 θ ⎝ = sin2 θ = cos2 θ = 1 sec2 θ tan2 θ = sin2 θ cos2 θ There is more than one way to verify an identity. Here is another possibility. Again, we can start
with the left side. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 983 sec2 θ − 1 sec2 θ − 1 sec2 θ = sec2 θ sec2 θ = 1 − cos2 θ = sin2 θ Analysis In the first method, we used the identity sec2 θ = tan2 θ + 1 and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same. 9.2 Show that cot θ csc θ = cos θ. Example 9.5 Creating and Verifying an Identity Create an identity for the expression 2 tan θ sec θ by rewriting strictly in terms of sine. Solution There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression: 2 tan θ sec θ = 2 ⎞ ⎛ ⎝ ⎠ 1 cos θ ⎞ ⎠ sin θ ⎛ ⎝ cos θ = 2 sin θ cos2 θ 2 sin θ 1 − sin2 θ = Thus, Substitute 1 − sin2 θ for cos2 θ. 2 tan θ sec θ = 2 sin θ 1 − sin2 θ Example 9.6 Verifying an Identity Using Algebra and Even/Odd Identities Verify the identity: Solution sin2 (−θ) − cos2 (−θ) sin(−θ) − cos(−θ) = cos θ − sin θ 984 Chapter 9 Trigonometric Identities and Equations Let’s start with the left side and simplify: sin2(−θ) − cos2(−θ) sin(−θ) − cos(−θ) = = = = = [sin(−θ)]2 − [cos(−θ)]2 sin(−θ) − cos(−θ) (−sin θ)2 − (cos θ)2 −sin θ − cos θ (sin θ)2 − (cos θ)2 −sin θ − cos θ (sin θ − cos θ)(sin θ + cos θ) −(sin θ + cos θ) (sin θ − cos θ)(sin θ + cos θ ) −(sin θ + cos θ ) = cos θ − sin θ sin(−x) = −sin x and cos(−x) = cos x Diffe ence of squares 9.3 Verify the identity sin2 θ − 1 tan θ sin θ − tan θ = sin θ + 1 tan θ . Example 9.7 Verifying an Identity Involving Cosines and Cotangents Verify the identity: ⎛ ⎛ ⎝1 − cos2 x⎞ ⎝1 + cot2 x⎞ ⎠ ⎠ = 1. Solution We will work on the left side of the equation. ⎛ ⎛ ⎝1 − cos2 x⎞ ⎝1 + cot2 x1 + cos2 x ⎛ ⎝1 − cos2 x⎞ ⎠ ⎠ sin2 x ⎛ + cos2 x sin2 x ⎛ ⎝1 − cos2 x⎞ ⎠ ⎝ sin2 x sin2 x ⎛ sin2 x + cos2 x ⎛ ⎝1 − cos2 x⎞ ⎠ ⎝ sin2 x ⎞ ⎠ ⎞ ⎠ ⎛ ⎛ ⎝sin2 x⎞ 1 ⎠ ⎝ sin2 x ⎞ ⎠ = 1 Find the common denominator. Using Algebra to Simplify Trigonometric Expressions We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations. For example, the equation (sin x + 1)(sin x − 1) = 0 resembles the equation (x + 1)(x − 1) = 0, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 985 Another example is the difference of squares formula, a2 − b2 = (a − b)(a + b), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve. Example 9.8 Writing the Trigonometric Expression as an Algebraic Expression Write the following trigonometric expression as an algebraic expression: 2cos2 θ + cos θ − 1. Solution Notice that the pattern displayed has the same form as a standard quadratic expression, ax2 + bx + c. Letting cos θ = x, we can rewrite the expression as follows: 2x2 + x − 1 This expression can be factored as (2x + 1)(x − 1). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x. At this point, we would replace x with cos θ and solve for θ. Example 9.9 Rewriting a Trigonometric Expression Using the Difference of Squares Rewrite the trigonometric expression using the difference of squares: 4 cos2 θ − 1. Solution Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. 4 cos2 θ − 1 = (2 cos θ)2 − 1 = (2 cos θ − 1)(2 cos θ + 1) Analysis If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x, rewrite the expression as 4x2 − 1, and factor (2x − 1)(2x + 1). Then replace x with cos θ and solve for the angle. 9.4 Rewrite the trigonometric expression using the difference of squares: 25 − 9 sin2 θ. Example 9.10 986 Chapter 9 Trigonometric Identities and Equations Simplify by Rewriting and Using Substitution Simplify the expression by rewriting and using identities: Solution We can start with the Pythagorean identity. csc2 θ − cot2 θ 1 + cot2 θ = csc2 θ Now we can simplify by substituting 1 + cot2 θ for csc2 θ. We have csc2 θ − cot2 θ = 1 + cot2 θ − cot2 θ = 1 9.5 Use algebraic techniques to verify the identity: cos θ 1 + sin θ = 1 − sin θ cos θ . (Hint: Multiply the numerator and denominator on the left side by 1 − sin θ.) Access these online resources for additional instruction and practice with the fundamental trigonometric identities. • Fundamental Trigonometric Identities (http://openstaxcollege.org/l/funtrigiden) • Verifying Trigonometric Identities (http://openstaxcollege.org/l/verifytrigiden) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 987 9.1 EXERCISES Verbal an even function, We know g(x) = cos x is 1. and f (x) = sin x and h(x) = tan x are odd functions. What about G(x) = cos2 x, F(x) = sin2 x, and H(x) = tan2 x ? Are they even, odd, or neither? Why? Examine the graph of f (x) = sec x on the interval 2. [ − π, π]. How can we tell whether the function is even or odd by only observing the graph of f (x) = sec x ? 3. After examining the reciprocal identity for sec t, explain why the function is undefined at certain points. All of the Pythagorean identities are related. Describe from 4. how to manipulate sin2 t + cos2 t = 1 to the other forms. equations get the to Algebraic For the following exercises, use the fundamental identities to fully simplify the expression. 5. 6. 7. sin x cos x sec x sin(−x) cos(−x) csc(−x) tan x sin x + sec x cos2 x 8. csc x + cos x cot(−x) 9. cot t + tan t sec( − t) 10. 3 sin3 t csc t + cos2 t + 2 cos( − t)cos t 11. −tan(−x)cot(−x) 12. 13. 14. 15. −sin(−x)cos x sec x csc x tan x cot x 1 + tan2 θ csc2 θ + sin2 θ + 1 sec2 θ ⎛ tan x ⎝ csc2 x + tan x sec2 x ⎞ ⎛ ⎝ ⎠ 1 + tan x 1 + cot x ⎞ ⎠ − 1 cos2 x 1 − cos2 x tan2 x + 2 sin2 x For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. 16. 17. 18. 19. 20. tan x + cot x csc x ; cos x sec x + csc x 1 + tan x ; sin x cos x 1 + sin x + tan x; cos x 1 sin xcos x − cot x; cot x 1 − cos x − cos x 1 + cos x; csc x 1 21. (sec x + csc x)(sin x + cos x) − 2 − cot x; tan x 22. 23. 1 csc x − sin x; sec x and tan x 1 + sin x − 1 + sin x 1 − sin x 1 − sin x; sec x and tan x 24. tan x; sec x 25. sec x; cot x 26. sec x; sin x 27. cot x; sin x 28. cot x; csc x For the following exercises, verify the identity. 29. 30. 31. 32. 33. cos x − cos3 x = cos x sin2 x cos x⎛ ⎝tan x − sec(−x)⎞ ⎠ = sin x − 1 1 + sin2 x cos2 x = 1 cos2 x + sin2 x cos2 x = 1 + 2 tan2 x (sin x + cos x)2 = 1 + 2 sin xcos x cos2 x − tan2 x = 2 − sin2 x − sec2 x Extensions For the following exercises, prove or disprove the identity. 34. 1 1 + cos x − 1 1 − cos( − x) = − 2 cot x csc x Chapter 9 Trigonometric Identities and Equations 988 35. csc2 x⎛ ⎝1 + sin2 x⎞ ⎠ = cot2 x 36. ⎛ sec2( − x) − tan2 x tan x ⎝ ⎞ ⎛ ⎝ ⎠ 2 + 2 tan x 2 + 2 cot x ⎞ ⎠ − 2 sin2 x = cos 2x 37. 38. 39. tan x sec xsin(−x) = cos2 x sec(−x) tan x + cot x = − sin(−x) 1 + sin x cos x = cos x 1 + sin(−x) For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. 41. 42. cos2 θ − sin2 θ 1 − tan2 θ = sin2 θ 3 sin2 θ + 4 cos2 θ = 3 + cos2 θ sec θ + tan θ cot θ + cos θ = sec2 θ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 989 9.2 \| Sum and Difference Identities Learning Objectives In this section, you will: 9.2.1 Use sum and difference formulas for cosine. 9.2.2 Use sum and difference formulas for sine. 9.2.3 Use sum and difference formulas for tangent. 9.2.4 Use sum and difference formulas for cofunctions. 9.2.5 Use sum and difference formulas to verify identities. Figure 9.6 Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr) How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances. The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications. In this section, we will learn techniques that will enable us to solve problems s
uch as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity. Using the Sum and Difference Formulas for Cosine Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure 9.7. 990 Chapter 9 Trigonometric Identities and Equations Figure 9.7 The Unit Circle We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles. See Table 9.5. Sum formula for cosine Difference formula for cosine Table 9.5 cos⎛ ⎝α + β⎞ ⎠ = cos α cos β − sin α sin β cos⎛ ⎝α − β⎞ ⎠ = cos α cos β + sin α sin β First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 9.8. Point P is at an angle α from the positive x-axis with coordinates (cos α, sin α) and point Q is at an angle of β from the positive x-axis with coordinates ⎛ ⎠. Note the measure of angle POQ is α − β. ⎝cos β, sin β⎞ ⎠; and Label two more points: A at an angle of ⎛ point B with coordinates (1, 0). Triangle POQ is a rotation of triangle AOB and thus the distance from P to Q is the same as the distance from A to B. ⎠ from the positive x-axis with coordinates ⎛ ⎝α − β⎞ ⎝α − β⎞ ⎝α − β⎞ ⎠, sin⎛ ⎝cos⎛ ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 991 Figure 9.8 We can find the distance from P to Q using the distance formula. dPQ = (cos α − cos β)2 + (sin α − sin β)2 = cos2 α − 2 cos α cos β + cos2 β + sin2 α − 2 sin α sin β + sin2 β Then we apply the Pythagorean identity and simplify. ⎠ + ⎛ ⎝cos2 α + sin2 α⎞ ⎛ ⎝cos2 β + sin2 β⎞ = = 1 + 1 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos α cos β − 2 sin α sin β ⎠ − 2 cos α cos β − 2 sin α sin β Similarly, using the distance formula we can find the distance from A to B. d AB = (cos(α − β) − 1)2 + (sin(α − β) − 0)2 = cos2(α − β) − 2 cos(α − β) + 1 + sin2(α − β) Applying the Pythagorean identity and simplifying we get: ⎛ ⎞ ⎝cos2(α − β) + sin2(α − β) ⎠ − 2 cos(α − β) + 1 = = 1 − 2 cos(α − β) + 1 = 2 − 2 cos(α − β) Because the two distances are the same, we set them equal to each other and simplify. 2 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos(α − β) 2 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos(α − β) Finally we subtract 2 from both sides and divide both sides by −2. cos α cos β + sin α sin β = cos(α − β) Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles. Sum and Difference Formulas for Cosine These formulas can be used to calculate the cosine of sums and differences of angles. cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β (9.18) (9.19) 992 Chapter 9 Trigonometric Identities and Equations Given two angles, find the cosine of the difference between the angles. 1. Write the difference formula for cosine. 2. Substitute the values of the given angles into the formula. 3. Simplify. Example 9.11 Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles ⎛ Using the formula for the cosine of the difference of two angles, find the exact value of cos ⎝ 5π 4 − ⎞ ⎠. π 6 Solution Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values. − π 6 π 6 5π 4 cos(α − β) = cos α cos β + sin α sin β ⎞ ⎛ ⎛ 5π 5π ⎛ ⎞ ⎠ = cos cos ⎠cos ⎝ ⎝ ⎝ 4 4 ⎞ ⎛ ⎞ ⎛ ⎞ 3 ⎝− ⎞ ⎛ ⎠ + sin ⎝ ⎞ ⎛ ⎛ ⎝ ⎠ ⎝ 2 2 1 2 = 4 ⎛ ⎞ ⎠sin ⎝ ⎞ ⎠ π 6 Keep in mind that we can always check the answer using a graphing calculator in radian mode. 9.6 ⎛ Find the exact value of cos ⎝ π 3 − ⎞ ⎠. π 4 Example 9.12 Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine Find the exact value of cos(75°). Solution As 75° = 45° + 30°, we can evaluate cos(75°) as cos(45° + 30°). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 993 cos(α + β) = cos α cos β − sin α sin β cos(45° + 30°) = cos(45°)cos(30°) − sin(45°)sin(30°) ⎞ ⎠ − Keep in mind that we can always check the answer using a graphing calculator in degree mode. Analysis Note that we could have also solved this problem using the fact that 75° = 135° − 60°. cos(α − β) = cos α cos β + sin α sin β cos(135° − 60°) = cos(135°)cos(60°) + sin(135°)sin(60°) ⎞ ⎠ + = ⎛ ⎝ ⎛ ⎞ ⎠ ⎝ 2 2 ⎞ ⎠ 3 2 1 2 ⎞ ⎛ ⎝ ⎠ ⎛ ⎝− 2 2 ⎛ ⎝− .7 Find the exact value of cos(105°). Using the Sum and Difference Formulas for Sine The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas. Sum and Difference Formulas for Sine These formulas can be used to calculate the sines of sums and differences of angles. sin⎛ sin⎛ ⎝α + β⎞ ⎝α − β⎞ ⎠ = sin α cos β + cos α sin β ⎠ = sin α cos β − cos α sin β (9.20) (9.21) Given two angles, find the sine of the difference between the angles. 1. Write the difference formula for sine. 2. Substitute the given angles into the formula. 3. Simplify. Example 9.13 Using Sum and Difference Identities to Evaluate the Difference of Angles 994 Chapter 9 Trigonometric Identities and Equations Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b. a. b. sin(45° − 30°) sin(135° − 120°) Solution a. Let’s begin by writing the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(45° − 30°) = sin(45°)cos(30°) − cos(45°)sin(30°) Next, we need to find the values of the trigonometric expressions. sin(45°) = 2 2 , cos(30°) = 3 2 , cos(45°) = 2 2 , sin(30°) = 1 2 Now we can substitute these values into the equation and simplify. sin(45° − 30°) = . Again, we write the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(135° − 120°) = sin(135°)cos(120°) − cos(135°)sin(120°) Next, we find the values of the trigonometric expressions. sin(135°) = 2 2 , cos(120°) = − 1 2 , cos(135°) = 2 2 , sin(120°) = 3 2 Now we can substitute these values into the equation and simplify. sin(135° − 120°) = 2 2 = − 2 + 6 ⎛ ⎝− sin(135° − 120°) = ⎛ ⎝− 1 2 4 ⎛ ⎝− ⎛ ⎝− Example 9.14 Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function ⎛ ⎝cos−1 1 Find the exact value of sin 2 + sin−1 3 5 ⎞ ⎠. Then check the answer with a graphing calculator. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 995 The pattern displayed in this problem is sin⎛ ⎝α + β⎞ ⎠. Let α = cos−1 1 2 and β = sin−1 3 5 . Then we can write cos α = 1 2 sin We will use the Pythagorean identities to find sin α and cos β. sin α = 1 − cos2 cos β = 1 − sin2 β = 1 − 9 25 = 16 25 = 4 5 Using the sum formula for sine, ⎛ ⎝cos−1 1 sin 2 + sin−1 3 5 ⎞ ⎠ = sin(α + β) = sin α cos β + cos α sin 10 Using the Sum and Difference Formulas for Tangent Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern. Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, tan x = sin x cos x, cos x ≠ 0. Let’s derive the sum formula for tangent. 996 Chapter 9 Trigonometric Identities and Equations Divide the numerator and denominator by cos α cos β. tan(α + β) = = = = = = sin(α + β) cos(α + β) sin α cos β + cos α sin β cos α cos β − sin α sin β sin α cos β + cos α sin β cos α cos β cos α cos β − sin α sin β cos α cos β cos α sin β cos α cos β sin α sin β cos α cos β sin α cos β cos α cos β + cos α cos β cos α cos β − sin β sin α cos α + cos β sin α sin β 1 − cos α cos β tan α + tan β 1 − tan α tan β We can derive the difference formula for tangent in a similar way. Sum and Difference Formulas for Tangent The sum and difference formulas for tangent are: tan⎛ ⎝α + β⎞ ⎠ = tan⎛ ⎝α − β⎞ ⎠ = tan α + tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β (9.22) (9.23) Given two angles, find the tangent of the sum of the angles. 1. Write the sum formula for tangent. 2. Substitute the given angles into the formula. 3. Simplify. Example 9.15 Finding the Exact Value of an Expression Involving Tangent ⎛ Find the exact value of tan ⎝ π 6 + ⎞ ⎠. π 4 Solution Let’s first write the sum formula for tangent and then substitute the given angles into the formula. tan(α + β) = ⎛ tan ⎝ π 6 + ⎞ ⎠ = π 4 tan α + tan β 1 − tan α tan β ⎛ ⎞ ⎛ ⎞ π π ⎠ + tan tan ⎝ ⎠ ⎝ tan ⎝tan ⎝ ⎝ ⎠ ⎠ 4 6 1 − ⎞ ⎞ ⎠ ⎠ Next, we determine the individual function values within the formula: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 997 So we have ⎛ tan ⎝ π 6 ⎞ ⎠ = 1 3 ⎛ , tan ⎝ π 4 ⎞ ⎠ = 1 ⎛ tan ⎝ ⎞ ⎠(1.8 ⎛ Find the exact value of tan ⎝ 2π 3 + ⎞ ⎠. π 4 Example 9.16 Finding Multiple Sums and Differences of Angles Given sin , cos β = − 5 13 , π < β < 3π 2 , find a. b. c. d. sin⎛ ⎝α + β⎞ ⎠ cos⎛ ⎝α + β⎞ ⎠ tan⎛ ⎝α + β⎞ ⎠ tan⎛ ⎝α − β⎞ ⎠ Solution We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas. a. To find sin⎛ ⎝α + β⎞ ⎠, we begin with sin α = 3 5 and 0 < α < π 2 . The side opposite α has length 3, the hypotenuse has length 5, and α is in the first quadrant. See Figure 9.9. Using the Pythagorean Theorem, we can find the length of side a: a2 + 32 = 52 a2 = 16 a = 4 998 Chapter 9 Trigonometric Identities and Equations Figure 9.9 S
ince cos β = − 5 13 and π < β < 3π 2 , the side adjacent to β is −5, the hypotenuse is 13, and β is in the third quadrant. See Figure 9.10. Again, using the Pythagorean Theorem, we have (−5)2 + a2 = 132 25 + a2 = 169 a2 = 144 a = ±12 Since β is in the third quadrant, a = –12. Figure 9.10 The next step is finding the cosine of α and the sine of β. The cosine of α is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 9.10: cos α = 4 5 . We can also find the sine of This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 999 β from the triangle in Figure 9.10, as opposite side over the hypotenuse: sin β = − 12 13 . Now we are ready to evaluate sin⎛ ⎝α + β⎞ ⎠. = sin(α + β) = sin α cos β + cos α sin β ⎛ ⎞ ⎞ ⎛ 3 ⎝− 5 ⎠ ⎝ ⎠ 13 5 − 48 = − 15 65 65 = − 63 65 ⎞ ⎛ ⎝− 12 ⎠ 13 ⎞ ⎠ + 4 5 ⎛ ⎝ b. We can find cos⎛ ⎝α + β⎞ ⎠ in a similar manner. We substitute the values according to the formula. cos(α + β) = cos α cos β − sin α sin β ⎛ ⎞ ⎞ ⎛ ⎝− 5 4 ⎠ ⎝ ⎠ 13 5 + 36 = − 20 65 65 ⎞ ⎛ ⎝− 12 ⎠ 13 ⎞ ⎠ − 3 5 ⎛ ⎝ = c. For tan⎛ ⎝α + β⎞ ⎠, if sin α = 3 5 and cos α = 4 5 , then = 16 65 tan α = 3 5 4 5 = 3 4 If sin β = − 12 13 and cos β = − 5 13 , then Then, tan β = −12 13 −5 13 = 12 5 tan(α + β) = 3 ⎞ ⎠ = 5 ⎛ 12 ⎝ 5 tan α + tan β 1 − tan α tan β 4 + 12 1 − 3 4 63 20 − 16 20 = − 63 16 = d. To find tan⎛ ⎝α − β⎞ ⎠, we have the values we need. We can substitute them in and evaluate. 1000 Chapter 9 Trigonometric Identities and Equations tan(α − β) = 3 ⎞ ⎠ = 5 ⎛ 12 ⎝ 5 tan α − tan β 1 + tan α tan β 4 − 12 1 + 3 4 − 33 20 56 20 = − 33 56 = Analysis A common mistake when addressing problems such as this one is that we may be tempted to think that α and β are angles in the same triangle, which of course, they are not. Also note that tan⎛ ⎝α + β⎞ ⎠ = sin⎛ cos⎛ ⎝α + β⎞ ⎝α + β⎞ ⎠ ⎠ Using Sum and Difference Formulas for Cofunctions Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two positive those two angles are complements, and the sum of the two acute angles in a right triangle is π angles is π , so they are 2 2 , also complements. In Figure 9.11, notice that if one of the acute angles is labeled as θ, labeled ⎛ ⎝ − θ⎞ ⎠. π 2 then the other acute angle must be ⎛ Notice also that sin θ = cos ⎝ ⎠, which is opposite over hypotenuse. Thus, when two angles are complimentary, we can say that the sine of θ equals the cofunction of the complement of θ. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions. − θ⎞ π 2 Figure 9.11 From these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions. Cofunction Identities The cofunction identities are summarized in Table 9.6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1001 ⎛ sin θ = cos ⎝ − θ⎞ ⎠ π 2 ⎛ cos θ = sin ⎝ − θ⎞ ⎠ π 2 tan θ = cot ⎛ ⎝ π 2 − θ⎞ ⎠ ⎛ cot θ = tan ⎝ − θ⎞ ⎠ π 2 ⎛ sec θ = csc ⎝ − θ⎞ ⎠ π 2 ⎛ csc θ = sec ⎝ − θ⎞ ⎠ π 2 Table 9.6 Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using cos⎛ ⎝α − β⎞ ⎠ = cos αcos β + sin αsin β, we can write ⎛ cos ⎝ π 2 Example 9.17 − θ⎞ ⎠ = cos π 2 cos θ + sin π 2 = (0)cos θ + (1)sin θ = sin θ sin θ Finding a Cofunction with the Same Value as the Given Expression in terms of its cofunction. Write tan π 9 Solution The cofunction of tan θ = cot ⎛ ⎝ π 2 − θ⎞ ⎠. Thus, ⎛ tan ⎝ π 9 ⎞ ⎠ = cot ⎛ ⎝ = cot = cot ⎛ ⎝ ⎛ ⎝ ⎞ π − ⎠ 9 − 2π 18 ⎞ ⎠ π 2 9π 18 7π 18 ⎞ ⎠ 9.9 Write sin π 7 in terms of its cofunction. Using the Sum and Difference Formulas to Verify Identities Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity. 1002 Chapter 9 Trigonometric Identities and Equations Given an identity, verify using sum and difference formulas. 1. Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient. 2. Look for opportunities to use the sum and difference formulas. 3. Rewrite sums or differences of quotients as single quotients. 4. If the process becomes cumbersome, rewrite the expression in terms of sines and cosines. Example 9.18 Verifying an Identity Involving Sine Verify the identity sin(α + β) + sin(α − β) = 2 sin α cos β. Solution We see that the left side of the equation includes the sines of the sum and the difference of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β We can rewrite each using the sum and difference formulas. sin(α + β) + sin(α − β) = sin α cos β + cos α sin β + sin α cos β − cos α sin β = 2 sin α cos β We see that the identity is verified. Example 9.19 Verifying an Identity Involving Tangent Verify the following identity. sin(α − β) cos α cos β = tan α − tan β Solution We can begin by rewriting the numerator on the left side of the equation. sin(α − β) cos α cos β = sin α cos β − cos αsin β cos αcos β = sin α cos β cos α cos β − cos α sin β cos α cos β Rewrite using a common denominator. sin β = sin α cos α − cos β = tan α − tan β Cancel. Rewrite in terms of tangent. We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1003 9.10 Verify the identity: tan(π − θ) = − tan θ. Example 9.20 Using Sum and Difference Formulas to Solve an Application Problem Let L1 and L2 denote two non-vertical intersecting lines, and let θ denote the acute angle between L1 and L2. See Figure 9.12. Show that tan θ = m2 − m1 1 + m1 m2 where m1 and m2 are the slopes of L1 and L2 respectively. (Hint: Use the fact tan θ2 = m2. ) that tan θ1 = m1 and Figure 9.12 Solution Using the difference formula for tangent, this problem does not seem as daunting as it might. tan θ = tan⎛ ⎞ ⎠ ⎝θ2 − θ1 tan θ2 − tan θ1 1 + tan θ1 tan θ2 m2 − m1 1 + m1 m2 = = Example 9.21 Investigating a Guy-wire Problem For a climbing wall, a guy-wire R is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire S attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle α between the wires. See Figure 9.13. 1004 Chapter 9 Trigonometric Identities and Equations Figure 9.13 Solution Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that tan β = 47 . We can 50 ⎠ = 40 50 , and tan⎛ ⎝β − α⎞ = 4 5 then use difference formula for tangent. Now, substituting the values we know into the formula, we have tan⎛ ⎝β − α⎞ ⎠ = tan β − tan α 1 + tan βtan α 4 5 = 47 50 − tan α 1 + 47 50tan α − tan α⎞ ⎛ ⎠ ⎝ 47 50 4 ⎛ ⎝1 + 47 50 tan α⎞ ⎠ = 5 Use the distributive property, and then simplify the functions. ⎛ 4(1) + 4 ⎝ 47 50 ⎞ ⎠tan α = 5 4 + 3.76 tan α = 4.7 − 5 tan α ⎞ ⎠ − 5 tan α 47 50 ⎛ ⎝ 5 tan α + 3.76 tan α = 0.7 8.76 tan α = 0.7 tan α ≈ 0.07991 tan−1(0.07991) ≈ .079741 Now we can calculate the angle in degrees. α ≈ 0.079741 ⎛ ⎝ 180 π ⎞ ⎠ ≈ 4.57° Analysis Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given. Access these online resources for additional instruction and practice with sum and difference identities. • Sum and Difference Identities for Cosine (http://openstaxcollege.org/l/sumdifcos) • Sum and Difference Identities for Sine (http://openstaxcollege.org/l/sumdifsin) • Sum and Difference Identities for Tangent (http://openstaxcollege.org/l/sumdiftan) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1005 9.2 EXERCISES Verbal Explain the basis for the cofunction identities and 43. when they apply. 44. ⎛ Is there only one way to evaluate cos ⎝ ⎞ ⎠? Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer. 5π 4 45. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for f (x) = sin(x) and g(x) = cos(x). (Hint: 0 − x = − x ) Algebraic For the following exercises, find the exact value. 46. 47. 48. 49. 50. 51. ⎛ cos ⎝ ⎞ ⎠ 7π 12 ⎛ cos ⎝ ⎞ ⎠ π 12 ⎛ sin ⎝ ⎞ ⎠ 5π 12 ⎛ sin ⎝ 11π 12 ⎞ ⎠ ⎛ tan ⎝− ⎞ ⎠ π 12 ⎛ tan ⎝ 19π 12 ⎞ ⎠ For the following exercises, rewrite in terms of sin x and cos x. 52. 53. 54. 55. ⎛ ⎝x + 11π sin 6 ⎞ ⎠ ⎛ sin ⎝x − 3π 4 ⎞ ⎠ ⎛ ⎝x − 5π cos 6 ⎞ ⎠ ⎛ ⎝x + 2π cos 3 ⎞ ⎠ For the following exercises, simplify the given expression. 56. ⎛ csc ⎝ 57. ⎛ sec ⎝ − t⎞ ⎠ − θ⎞ ⎠ π 2 π 2 58. 59. cot ⎛ ⎝ π 2 − x⎞ ⎠ ⎛ tan ⎝ π 2 − x⎞ ⎠ 60. sin(2x) cos(5x) − sin(5x) cos(2x) 61. 3 2 ⎛ tan ⎝ x⎞ ⎛ x⎞ 7 ⎠ − tan ⎝ ⎠ 5 x⎞ ⎛ x⎞ ⎛ 3 7 ⎠tan 1 + tan ⎝ ⎠ ⎝ 2 5 For the following exercises, find the requested information. 62. Given that sin a = 2 3 interval ⎡ ⎣ the in b both and cos b = − 1 4 , with a and , π⎞ ⎠, find sin(a + b) and π 2 cos(a − b). 63. Given that sin a = 4 5 , and cos b = 1 3 , with a and b both in the cos(a + b). interval ⎡ ⎣0, π 2 ⎞ ⎠,
find sin(a − b) and For the following exercises, find the exact value of each expression. 64. 65. 66. ⎛ ⎝cos−1(0) − cos−1 ⎛ sin ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 ⎛ ⎝cos−1 ⎛ cos ⎝ ⎞ ⎠ + sin−sin−1 ⎛ tan ⎝ 1 2 ⎞ ⎠ − cos−1 ⎛ ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Graphical For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator. 67. ⎛ cos ⎝ π 2 − x⎞ ⎠ 68. sin(π − x) ⎛ tan ⎝ π 3 + x⎞ ⎠ ⎛ sin ⎝ π 3 + x⎞ ⎠ 69. 70. 71. 1006 Chapter 9 Trigonometric Identities and Equations Extensions For the following exercises, prove the identities provided. 89. 90. 91. 92. 93. tan(x + π 4 ) = tan x + 1 1 − tan x tan(a + b) tan(a − b) = sin a cos a + sin b cos b sin a cos a − sin b cos b cos(a + b) cos a cos b = 1 − tan a tan b cos(x + y)cos(x − y) = cos2 x − sin2 y cos(x + h) − cos x h = cos xcos h − 1 h − sin xsin h h the following exercises, prove or disprove the For statements. 94. 95. 96. tan(u + v) = tan u + tan v 1 − tan u tan v tan(u − v) = tan u − tan v 1 + tan u tan v tan(x + y) 1 + tan x tan x = tan x + tan y 1 − tan2 x tan2 y 97. prove or disprove sin⎛ If α, β, and γ are angles in the same triangle, then ⎠ = sin γ. ⎝α + β⎞ If α, β, and y are angles in the same triangle, then 98. prove or disprove tan α + tan β + tan γ = tan α tan β tan γ ⎛ tan ⎝ π 4 − x⎞ ⎠ 72. 73. 74. ⎛ cos ⎝ 7π 6 + x⎞ ⎠ ⎛ sin ⎝ π 4 + x⎞ ⎠ ⎛ cos ⎝ 5π 4 + x⎞ ⎠ For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think 2x = x + x. ) 75. f (x) = sin(4x) − sin(3x)cos x, g(x) = sin x cos(3x) 76. f (x) = cos(4x) + sin x sin(3x), g(x) = − cos x cos(3x) 77. f (x) = sin(3x)cos(6x), g(x) = − sin(3x)cos(6x) 78. f (x) = sin(4x), g(x) = sin(5x)cos x − cos(5x)sin x 79. f (x) = sin(2x), g(x) = 2 sin x cos x 80. 81. f (θ) = cos(2θ), g(θ) = cos2 θ − sin2 θ f (θ) = tan(2θ), g(θ) = tan θ 1 + tan2 θ 82. f (x) = sin(3x)sin x, g(x) = sin2(2x)cos2 x − cos2(2x)sin2 x 83. f (x) = tan( − x), g(x) = tan x − tan(2x) 1 − tan x tan(2x) Technology the following exercises, For find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. 84. sin(75°) 85. sin(195°) 86. 87. 88. cos(165°) cos(345°) tan(−15°) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1007 9.3 \| Double-Angle, Half-Angle, and Reduction Formulas Learning Objectives In this section, you will: 9.3.1 Use double-angle formulas to find exact values. 9.3.2 Use double-angle formulas to verify identities. 9.3.3 Use reduction formulas to simplify an expression. 9.3.4 Use half-angle formulas to find exact values. Figure 9.14 Bicycle ramps for advanced riders have a steeper incline than those designed for novices. Bicycle ramps made for competition (see Figure 9.14) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θ such that tan θ = 5 . The angle is 3 divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one. Using Double-Angle Formulas to Find Exact Values In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α = β. Deriving the double-angle formula for sine begins with the sum formula, sin⎛ ⎝α + β⎞ ⎠ = sin α cos β + cos α sin β If we let α = β = θ, then we have sin(θ + θ) = sin θ cos θ + cos θ sin θ sin(2θ) = 2sin θ cos θ Deriving cos⎛ ⎝α + β⎞ double-angle cosine the ⎠ = cos α cos β − sin α sin β, and letting α = β = θ, we have gives three for us options. First, starting from the sum formula, cos(θ + θ) = cos θ cos θ − sin θ sin θ cos(2θ) = cos2 θ − sin2 θ Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more variations. The first variation is: cos(2θ) = cos2 θ − sin2 θ ⎝1 − sin2 θ⎞ ⎛ = = 1 − 2sin2 θ ⎠ − sin2 θ 1008 Chapter 9 Trigonometric Identities and Equations The second variation is: cos(2θ) = cos2 θ − sin2 θ = cos2 θ − = 2 cos2 θ − 1 ⎝1 − cos2 θ⎞ ⎛ ⎠ Similarly, to derive the double-angle formula for tangent, replacing α = β = θ in the sum formula gives tan α + tan β tan(α + β) = 1 − tan α tan β tan(θ + θ) = tan θ + tan θ 1 − tan θ tan θ 2tan θ 1 − tan2 θ tan(2θ) = Double-Angle Formulas The double-angle formulas are summarized as follows: sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 2 tan θ 1 − tan2 θ tan(2θ) = (9.24) (9.25) (9.26) Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value. 1. Draw a triangle to reflect the given information. 2. Determine the correct double-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Example 9.22 Using a Double-Angle Formula to Find the Exact Value Involving Tangent Given that tan θ = − 3 4 and θ is in quadrant II, find the following: a. b. c. sin(2θ) cos(2θ) tan(2θ) Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given tan θ = − 3 , such that θ is in quadrant II. The tangent of an angle is equal to the opposite 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1009 side over the adjacent side, and because θ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: (−4)2 + (3)2 = c2 16 + 9 = c2 25 = c2 c = 5 Now we can draw a triangle similar to the one shown in Figure 9.15. Figure 9.15 a. Let’s begin by writing the double-angle formula for sine. sin(2θ) = 2 sin θ cos θ We see that we to need to find sin θ and cos θ. Based on Figure 9.15, we see that the hypotenuse equals 5, so sin θ = 3 5 . Substitute these values into the equation, and simplify. , and cos θ = − 4 5 Thus, sin(2θ) = 2 ⎛ ⎞ ⎛ 3 ⎝− 4 ⎝ ⎠ 5 5 = − 24 25 ⎞ ⎠ b. Write the double-angle formula for cosine. Again, substitute the values of the sine and cosine into the equation, and simplify. cos(2θ) = cos2 θ − sin2 θ 2 ⎞ ⎠ ⎛ ⎝ 3 5 cos(2θ) = 2 − ⎛ ⎞ ⎝− 4 ⎠ 5 − 9 = 16 25 25 = 7 25 c. Write the double-angle formula for tangent. tan(2θ) = 2 tan θ 1 − tan2 θ In this formula, we need the tangent, which we were given as tan θ = − 3 4 . Substitute this value into the equation, and simplify. 1010 Chapter 9 Trigonometric Identities and Equations tan(2θ) = 2 ⎞ ⎠ ⎛ ⎝− 3 4 ⎛ ⎝− 16 ⎛ = − 3 16 ⎝ 7 2 = − 24 7 ⎞ ⎠ 9.11 Given sin α = 5 8 , with θ in quadrant I, find cos(2α). Example 9.23 Using the Double-Angle Formula for Cosine without Exact Values Use the double-angle formula for cosine to write cos(6x) in terms of cos(3x). Solution cos(6x) = cos(3x + 3x) = cos 3x cos 3x − sin 3x sin 3x = cos2 3x − sin2 3x Analysis This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function. Using Double-Angle Formulas to Verify Identities Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side. Example 9.24 Using the Double-Angle Formulas to Verify an Identity Verify the following identity using double-angle formulas: 1 + sin(2θ) = (sin θ + cos θ)2 Solution We will work on the right side of the equal sign and rewrite the expression until it matches the left side. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1011 (sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ ⎠ + 2 sin θ cos θ ⎝sin2 θ + cos2 θ⎞ ⎛ = = 1 + 2 sin θ cos θ = 1 + sin(2θ) Analysis This process is not complicated, as long as we recall the perfect square formula from algebra: (a ± b)2 = a2 ± 2ab + b2 where a = sin θ and b = cos θ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent. 9.12 Verify the identity: cos4 θ − sin4 θ = cos(2θ). Example 9.25 Verifying a Double-Angle Identity for Tangent Verify the identity: Solution tan(2θ) = 2 cot θ − tan θ In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation. tan(2θ) = = = = ⎞ ⎠ 1 tan θ 2 tan θ 1 − tan2 θ 2 tan θ⎛ 1 ⎝ tan θ ⎛ (1 − tan2 θ) ⎝ 2 tan θ − tan2 θ 2 cot θ − tan θ tan θ 1 Double-angle formula Multiply by a term that results in desired numerator. ⎞ ⎠ Use reciprocal identity for 1 tan θ. Analysis Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show Let’s work on the right side. 2tan θ 1 − tan2 θ = 2 cot θ − tan θ 1012 Chapter 9 Trigonometric Identities and Equations 2 cot θ − tan θ = 2 1 tan θ − tan θ ⎛ ⎝ tan θ tan θ ⎞ ⎠ = 2 tan θ tan θ (tan θ ) − tan θ(tan θ) 2 tan θ 1 − tan2 θ When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier. = 1 9.13 Verify the identity: cos(2θ)cos
θ = cos3 θ − cos θ sin2 θ. Use Reduction Formulas to Simplify an Expression The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the doubleangle formulas. We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ) = 1 − 2 sin2 θ. Solve for sin2 θ : cos(2θ) = 1 − 2 sin2 θ 2 sin2 θ = 1 − cos(2θ) 1 − cos(2θ) 2 sin2 θ = Next, we use the formula cos(2θ) = 2 cos2 θ − 1. Solve for cos2 θ : cos(2θ) = 2 cos2 θ − 1 1 + cos(2θ) = 2 cos2 θ 1 + cos(2θ) 2 = cos2 θ The last reduction formula is derived by writing tangent in terms of sine and cosine: Substitute the reduction formulas. tan2 θ = sin2 θ cos2 θ 1 − cos(2θ) 2 1 + cos(2θ) 2 = ⎛ ⎞ ⎠ ⎝ 2 1 + cos(2θ) ⎞ ⎠ = = 1 − cos(2θ) ⎛ ⎝ 2 1 − cos(2θ) 1 + cos(2θ) Reduction Formulas The reduction formulas are summarized as follows: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) 1013 (9.27) (9.28) (9.29) Example 9.26 Writing an Equivalent Expression Not Containing Powers Greater Than 1 Write an equivalent expression for cos4 x that does not involve any powers of sine or cosine greater than 1. Solution We will apply the reduction formula for cosine twice. cos4 x = ⎛ ⎝cos2 x + cos(2x) 2 ⎞ ⎠ ⎛ ⎞ ⎝1 + 2cos(2x) + cos2(2x cos(2x) + 1 4 cos(2x) + 1 8 cos(2x) + 1 8 ⎞ ⎠ 1 + cos2(2x) 2 cos(4x) ⎛ ⎝ + 1 8 cos(4x) Substitute reduction formula for cos 2 x. Substitute reduction formula for cos 2 x. Analysis The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra. Example 9.27 Using the Power-Reducing Formulas to Prove an Identity Use the power-reducing formulas to prove sin3 (2x) = ⎤ sin(2x) ⎦ ⎡ ⎣1 − cos(4x)⎤ ⎦ ⎡ ⎣ 1 2 Solution We will work on simplifying the left side of the equation: ⎤ ⎡ sin3(2x) = [sin(2x)] ⎣sin2(2x) ⎦ 1 − cos(4x) ⎡ = sin(2x) ⎣ 2 ⎤ ⎦ ⎛ = sin(2x) ⎝ 1 2 ⎞ ⎠[1 − cos(4x)] = 1 2 [sin(2x)][1 − cos(4x)] Substitute the power-reduction formula. 1014 Chapter 9 Trigonometric Identities and Equations Analysis Note that in this example, we substituted for sin2 (2x). The formula states We let θ = 2x, so 2θ = 4x. 1 − cos(4x) 2 sin2 θ = 1 − cos(2θ) 2 9.14 Use the power-reducing formulas to prove that 10 cos4 x = 15 4 + 5 cos(2x) + 5 4 cos(4x). Using Half-Angle Formulas to Find Exact Values The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θ with α the half-angle formula for sine 2 ⎞ ⎠. Note that the half-angle formulas are preceded by a ± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α 2 ⎛ is found by simplifying the equation and solving for sin ⎝ terminates. α 2 , The half-angle formula for sine is derived as follows: sin2 ⎛ ⎝ α 2 ⎛ sin ⎝ α 2 To derive the half-angle formula for cosine, we have sin2 − cos(2θ) 2 ⎛ ⎝cos2 ⋅ α 2 2 = 1 − cos α 2 ⎠ = ± 1 − cos α ⎞ 2 cos2 ⎛ ⎝ α 2 ⎛ cos ⎝ α 2 For the tangent identity, we have cos2 θ = ⎞ ⎠ ⎞ ⎠ = 1 + cos(2θ) 2 ⎛ ⎝2 ⋅ α 1 + cos 2 2 = 1 + cos α 2 ⎠ = ± 1 + cos α ⎞ 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1015 tan2 θ = ⎞ ⎠ = 1 − cos(2θ) 1 + cos(2θ) ⎛ ⎝2 ⋅ α 1 − cos 2 ⎛ ⎝2 ⋅ α 1 + cos 2 = 1 − cos α 1 + cos α ⎠ = ± 1 − cos α 1 + cos α ⎞ tan2⎛ ⎝ α 2 ⎛ tan ⎝ α 2 Half-Angle Formulas The half-angle formulas are as follows: ⎞ ⎞ 2 2 ⎛ sin ⎝ ⎠ = ± 1 − cos α α 2 ⎠ = ± 1 + cos α α ⎛ cos ⎝ 2 ⎠ = ± 1 − cos α α 1 + cos α 2 ⎛ tan ⎝ = sin α 1 + cos α = 1 − cos α ⎞ sin α ⎞ ⎠ ⎞ ⎠ (9.30) (9.31) (9.32) Example 9.28 Using a Half-Angle Formula to Find the Exact Value of a Sine Function Find sin(15°) using a half-angle formula. Solution Since 15° = 30° 2 , we use the half-angle formula for sine: sin 30° 2 = 1 − cos30 Remember that we can check the answer with a graphing calculator. Analysis Notice that we used only the positive root because sin(15°) is positive. 1016 Chapter 9 Trigonometric Identities and Equations Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle. 1. Draw a triangle to represent the given information. 2. Determine the correct half-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Example 9.29 Finding Exact Values Using Half-Angle Identities Given that tan α = 8 15 and α lies in quadrant III, find the exact value of the following: a. b. c. ⎛ sin ⎝ ⎞ ⎠ α 2 ⎛ cos ⎝ ⎞ ⎠ α 2 ⎛ tan ⎝ ⎞ ⎠ α 2 Solution Using the given information, we can draw the triangle shown in Figure 9.16. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α = − 8 17 and cos α = − 15 17 . Figure 9.16 a. Before we start, we must remember that if α is in quadrant III, α 2 180° 2 < 270° 2 . This means that the terminal side of α 2 , we begin by writing the half-angle formula for sine. Then we substitute the value of the < To find sin α 2 cosine we found from the triangle in Figure 9.16 and simplify. is in quadrant II, since 90° < < 135°. then 180° < α < 270°, so α 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1017 sin α 2 = ± 1 − cos α 1 − = ± 2 ⎛ ⎝− 15 17 2 ⎞ ⎠ ⋅ 1 2 = ± 32 17 2 = ± 32 17 = ± 16 17 = ± 4 17 = 4 17 17 because the angle terminates in quadrant II and sine is positive in , we will write the half-angle formula for cosine, substitute the value of the cosine we found We choose the positive value of sin α 2 quadrant II. b. To find cos α 2 from the triangle in Figure 9.16, and simplify. cos α 2 = ± 1 + cos α 1 + = ± 2 ⎛ ⎝− 15 17 2 ⎞ ⎠ ⋅ 1 2 = ± 2 17 2 = ± 2 17 = ± 1 17 = − 17 17 because the angle is in quadrant II because cosine is negative in , we write the half-angle formula for tangent. Again, we substitute the value of the cosine We choose the negative value of cos α 2 quadrant II. c. To find tan α 2 we found from the triangle in Figure 9.16 and simplify. tan α 2 = ± 1 − cos α 1 + cos α 1 − ( − 15 17) 1 + ( − 15 17) = ± = ± 32 17 2 17 = ± 32 2 = − 16 = −4 1018 Chapter 9 Trigonometric Identities and Equations We choose the negative value of tan α 2 II. because α 2 lies in quadrant II, and tangent is negative in quadrant 9.15 Given that sin α = − 4 5 and α lies in quadrant IV, find the exact value of cos ⎛ ⎝ ⎞ ⎠. α 2 Example 9.30 Finding the Measurement of a Half Angle Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for highlevel competition with an angle of θ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tan θ = 5 3 for higher-level competition, what is the measurement of the angle for novice competition? Solution Since the angle for novice competition measures half the steepness of the angle for the high level competition, and tan θ = 5 for high competition, we can find cos θ from the right triangle and the Pythagorean theorem so 3 that we can use the half-angle identities. See Figure 9.17. 32 + 52 = 34 c = 34 Figure 9.17 We see that cos θ = 3 34 = 3 34 34 . We can use the half-angle formula for tangent: tan θ 2 = 1 − cos θ 1 + cos θ. Since tan θ is in the first quadrant, so is tan θ 2 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1019 tan θ 2 = 1 − 3 34 34 1 + 3 34 34 = 34 − 3 34 34 34 + 3 34 34 = 34 − 3 34 34 + 3 34 ≈ 0.57 We can take the inverse tangent to find the angle: tan−1(0.57) ≈ 29.7°. So the angle of the ramp for novice competition is ≈ 29.7°. Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas. • Double-Angle Identities (http://openstaxcollege.org/l/doubleangiden) • Half-Angle Identities (http://openstaxcollege.org/l/halfangleident) 1020 Chapter 9 Trigonometric Identities and Equations 9.3 EXERCISES Verbal Explain how to determine the reduction identities from 99. the double-angle identity cos(2x) = cos2 x − sin2 x. 100. Explain how to determine the double-angle formula for tan(2x) using the double-angle formulas for cos(2x) and sin(2x). 101. ⎛ tan ⎝ ⎛ by cos ⎝ ⎛ tan ⎝ x 2 ⎛ by dividing the formula for sin ⎝ We can determine the half-angle formula for ⎠ = 1 − cos x x ⎞ ⎞ ⎠ 1 + cos x 2 x 2 ⎞ ⎠ that do not involve any square roots. ⎞ ⎠. Explain how to determine two formulas for x 2 102. ⎛ exercise for tan ⎝ For the half-angle formula given in the previous ⎞ ⎠, explain why dividing by 0 is not a concern. (Hint: examine the values of cos x necessary for the denominator to be 0.) x 2 Algebraic For the following exercises, find the exact values of a) sin(2x), b) cos(2x), and c) tan(2x) without solving for x. 103. If sin x = 1 8 , and x is in quadrant I. 104. If cos x = 2 3 , and x is in quadrant I. 105. If cos x = − 1 2 , and x is in quadrant III. 106. If tan x = −8, and x is in quadrant IV. For the following exercises, find the values of the six trigonometric functions if the conditions provided hold. 107. 108. cos(2θ) = 3 5 and 90° ≤ θ ≤ 180° cos(2θ) = 1 2 and 180° ≤ θ ≤ 270° For the following exercises, simplify to one trigonometric expression. 109. 110. ⎛ 2 sin ⎝ π 4 ⎞ ⎛ ⎠ 2 cos ⎝ ⎞ ⎠ π 4 ⎛ 4 sin ⎝ π 8 ⎛ ⎞ ⎠ cos ⎝ ⎞ ⎠ π 8 This content is av
ailable for free at https://cnx.org/content/col11758/1.5 For the following exercises, find the exact value using halfangle formulas. 111. ⎛ sin ⎝ ⎞ ⎠ π 8 112. ⎛ cos ⎝− 11π 12 ⎞ ⎠ 113. ⎛ sin ⎝ 11π 12 ⎞ ⎠ 114. ⎛ cos ⎝ ⎞ ⎠ 7π 8 115. ⎛ tan ⎝ ⎞ ⎠ 5π 12 116. ⎛ tan ⎝− 3π 12 ⎞ ⎠ 117. ⎛ ⎝− 3π tan 8 ⎞ ⎠ For the following exercises, find the exact values of a) ⎛ without solving for x. sin ⎝ ⎞ ⎛ ⎠, and c) tan ⎝ ⎞ ⎛ ⎠, b) cos ⎝ ⎞ ⎠ x 2 x 2 x 2 118. 119. If tan x = − 4 3 , and x is in quadrant IV. If sin x = − 12 13 , and x is in quadrant III. 120. If csc x = 7, and x is in quadrant II. 121. If sec x = − 4, and x is in quadrant II. For the following exercises, use Figure 9.18 to find the requested half and double angles. Figure 9.18 122. Find sin(2θ), cos(2θ), and tan(2θ). 123. Find sin(2α), cos(2α), and tan(2α). ⎛ Find sin ⎝ θ 2 ⎛ ⎞ ⎠, cos ⎝ θ 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠. θ 2 124. 125. Chapter 9 Trigonometric Identities and Equations 1021 145. sin2 x cos2 x 146. tan2 x sin x 147. tan4 x cos2 x 148. cos2 x sin(2x) 149. cos2 (2x)sin x 150. tan2 ⎛ ⎝ x 2 ⎞ ⎠ sin x the following exercises, For algebraically find an equivalent function, only in terms of sin x and/or cos x, and then check the answer by graphing both functions. 151. sin(4x) 152. cos(4x) Extensions For the following exercises, prove the identities. 153. sin(2x) = 2 tan x 1 + tan2 x 154. cos(2α) = 1 − tan2 α 1 + tan2 α 155. tan(2x) = 2 sin x cos x 2cos2 x − 1 156. ⎛ ⎞ ⎝sin2 x − 1 ⎠ 2 = cos(2x) + sin4 x 157. 158. 159. sin(3x) = 3 sin x cos2 x − sin3 x cos(3x) = cos3 x − 3sin2 x cos x 1 + cos(2t) sin(2t) − cos t = 2 cos t 2 sin t − 1 160. sin(16x) = 16 sin x cos x cos(2x)cos(4x)cos(8x) 161. cos(16x) = ⎛ ⎞ ⎛ ⎞ ⎝cos2 (4x) − sin2 (4x) − sin(8x) ⎝cos2 (4x) − sin2 (4x) + sin(8x) ⎠ ⎠ ⎛ Find sin ⎝ α 2 ⎞ ⎛ ⎠, cos ⎝ α 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠. α 2 For the following exercises, simplify each expression. Do not evaluate. 126. cos2(28°) − sin2(28°) 127. 2cos2(37°) − 1 128. 1 − 2 sin2(17°) 129. cos2(9x) − sin2(9x) 130. 4 sin(8x) cos(8x) 131. 6 sin(5x) cos(5x) For the following exercises, prove the given identity. 132. (sin t − cos t)2 = 1 − sin(2t) 133. sin(2x) = − 2 sin(−x) cos(−x) 134. cot x − tan x = 2 cot(2x) 135. sin(2θ) 1 + cos(2θ) tan2 θ = tan θ For the following exercises, rewrite the expression with an exponent no higher than 1. 136. cos2(5x) 137. cos2(6x) 138. sin4(8x) 139. sin4(3x) 140. cos2 x sin4 x 141. cos4 x sin2 x 142. tan2 x sin2 x Technology For the following exercises, reduce the equations to powers of one, and then check the answer graphically. 143. tan4 x 144. sin2(2x) 1022 Chapter 9 Trigonometric Identities and Equations 9.4 \| Sum-to-Product and Product-to-Sum Formulas Learning Objectives In this section, you will: 9.4.1 Express products as sums. 9.4.2 Express sums as products. Figure 9.19 The UCLA marching band (credit: Eric Chan, Flickr). A band marches down the field creating an amazing sound that bolsters the crowd. That sound travels as a wave that can be interpreted using trigonometric functions. For example, Figure 9.20 represents a sound wave for the musical note A. In this section, we will investigate trigonometric identities that are the foundation of everyday phenomena such as sound waves. Figure 9.20 Expressing Products as Sums We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity. Expressing Products as Sums for Cosine We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1023 cos α cos β + sin α sin β = cos(α − β) + cos α cos β − sin α sin β = cos(α + β) ___________________________________ 2 cos α cos β = cos(α − β) + cos(α + β) Then, we divide by 2 to isolate the product of cosines: cos α cos β = 1 2 [cos(α − β) + cos(α + β)] Given a product of cosines, express as a sum. 1. Write the formula for the product of cosines. 2. Substitute the given angles into the formula. 3. Simplify. Example 9.31 Writing the Product as a Sum Using the Product-to-Sum Formula for Cosine ⎛ Write the following product of cosines as a sum: 2 cos ⎝ 7x 2 ⎠ cos 3x ⎞ 2 . Solution We begin by writing the formula for the product of cosines: cos α cos β = 1 2 ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ + cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ We can then substitute the given angles into the formula and simplify. ⎡ − 3x 7x ⎞ ⎛ 1 ⎣cos ⎠ ⎝ 2 2 2 ⎡ 10x 4x ⎞ ⎛ ⎛ ⎠ + cos = ⎣cos ⎝ ⎝ 2 2 = cos 2x + cos 5x ⎞ ⎛ ⎠ = (2) ⎝ ⎛ 2 cos ⎝ ⎞ ⎛ ⎠cos ⎝ 3x 2 7x 2 ⎞ ⎛ ⎠) + cos ⎝ ⎤ ⎞ ⎦ ⎠ 7x 2 + 3x 2 ⎤ ⎞ ⎦ ⎠ 9.16 Use the product-to-sum formula to write the product as a sum or difference: cos(2θ)cos(4θ). Expressing the Product of Sine and Cosine as a Sum Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine. If we add the sum and difference identities, we get: sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β + _________________________________________ sin(α + β) + sin(α − β) = 2 sin α cos β Then, we divide by 2 to isolate the product of cosine and sine: sin α cos β = 1 2 ⎡ ⎣sin⎛ ⎝α + β⎞ ⎠ + sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ 1024 Chapter 9 Trigonometric Identities and Equations Example 9.32 Writing the Product as a Sum Containing only Sine or Cosine Express the following product as a sum containing only sine or cosine and no products: sin(4θ)cos(2θ). Solution Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify. sin α cos β = 1 2 sin(4θ)cos(2θ) = 1 2 = 1 2 [sin(α + β) + sin(α − β)] [sin(4θ + 2θ) + sin(4θ − 2θ)] [sin(6θ) + sin(2θ)] 9.17 Use the product-to-sum formula to write the product as a sum: sin(x + y)cos(x − y). Expressing Products of Sines in Terms of Cosine Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas: ⎠ = cos α cos β + sin α sin β cos⎛ − cos⎛ ⎝cos α cos β − sin α sin β⎞ ⎠ = − ⎛ ____________________________________________________ ⎠ = 2 sin α sin β ⎠ − cos⎛ cos⎛ ⎝α − β⎞ ⎝α − β⎞ ⎝α + β⎞ ⎝α + β⎞ ⎠ Then, we divide by 2 to isolate the product of sines: sin α sin β = 1 2 ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ − cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas. The Product-to-Sum Formulas The product-to-sum formulas are as follows: [cos(α − β) + cos(α + β)] cos α cos β = 1 2 sin α cos β = 1 [sin(α + β) + sin(α − β)] 2 sin α sin β = 1 2 cos α sin β = 1 2 [sin(α + β) − sin(α − β)] [cos(α − β) − cos(α + β)] (9.33) (9.34) (9.35) (9.36) Example 9.33 Express the Product as a Sum or Difference This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1025 Write cos(3θ) cos(5θ) as a sum or difference. Solution We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify. cos α cos β = 1 2 cos(3θ)cos(5θ) = 1 2 = 1 2 [cos(α − β) + cos(α + β)] [cos(3θ − 5θ) + cos(3θ + 5θ)] [cos(2θ) + cos(8θ)] Use even-odd identity. 9.18 Use the product-to-sum formula to evaluate cos 11π 12 cos π 12 . Expressing Sums as Products Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine. Let u + v = α and u − v = β. 2 2 Then = 2u 2 = u u + v 2 = 2v 2 = v Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have [sin(α + β) + sin(α − β)] sin α cos β = 1 2 ⎞ ⎠ = 1 2 ⎞ ⎠ = sin u + sin v ⎛ ⎞ ⎠cos ⎝ ⎛ ⎞ ⎠cos ⎝ u − v 2 u − v 2 [sin u + sin v] ⎛ sin ⎝ ⎛ 2 sin ⎝ u + v 2 u + v 2 Substitute for(α + β) and (α − β) The other sum-to-product identities are derived similarly. Sum-to-Product Formulas The sum-to-product formulas are as follows: ⎛ sin α − sin β = 2sin ⎝ α + β ⎛ sin α + sin β = 2sin ⎝ 2 α − β 2 α + β 2 ⎛ cos α − cos β = −2sin ⎝ α − β ⎛ ⎞ ⎠cos ⎝ 2 α + β ⎞ ⎞ ⎛ ⎠cos ⎠ ⎝ 2 α − β ⎞ ⎛ ⎠sin ⎝ 2 ⎞ ⎠ ⎞ ⎠ (9.37) (9.38) (9.39) 1026 Chapter 9 Trigonometric Identities and Equations ⎛ cos α + cos β = 2cos ⎝ α + β 2 ⎛ ⎞ ⎠cos ⎝ α − β 2 ⎞ ⎠ (9.40) Example 9.34 Writing the Difference of Sines as a Product Write the following difference of sines expression as a product: sin(4θ) − sin(2θ). Solution We begin by writing the formula for the difference of sines. ⎛ sin α − sin β = 2sin ⎝ α − β 2 ⎛ ⎞ ⎠cos ⎝ α + β 2 ⎞ ⎠ Substitute the values into the formula, and simplify. ⎛ sin(4θ) − sin(2θ) = 2sin ⎝ 4θ − 2θ ⎛ ⎞ ⎠ cos ⎝ 2 6θ 2θ ⎞ ⎞ ⎛ ⎛ ⎠ cos = 2sin ⎠ ⎝ ⎝ 2 2 = 2 sin θ cos(3θ) 4θ + 2θ 2 ⎞ ⎠ 9.19 Use the sum-to-product formula to write the sum as a product: sin(3θ) + sin(θ). Example 9.35 Evaluating Using the Sum-to-Product Formula Evaluate cos(15°) − cos(75°). Check the answer with a graphing calculator. Solution We begin by writing the formula for the difference of cosines. ⎛ cos α − cos β = − 2 sin ⎝ α + β 2 ⎛ ⎞ sin ⎝ ⎠ α − β 2 ⎞ ⎠ Then we substitute the given angles and simplify. ⎛ cos(15°) − cos(75°) = −2sin ⎝ 15° + 75° 2 ⎞ ⎛ ⎠ sin ⎝ = −2sin(45°) sin(−30°) ⎞ ⎛ ⎝− 1 ⎠ 2 ⎛ = −2 ⎝ 2 2 ⎞ ⎠ 15° − 75° 2 ⎞ ⎠ = 2 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1027 Example 9.36 Proving an Identity Prove the identity: Solution cos(4t) − cos(2t) sin(4t) + sin(2t) = − tan t We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side. cos(4t) − cos(2t)
sin(4t) + sin(2t) = = = ⎛ ⎞ 4t − 2t ⎠ sin ⎝ 2 ⎛ ⎞ 4t − 2t ⎠ cos ⎝ 2 ⎞ ⎠ ⎞ ⎠ 4t + 2t ⎛ −2 sin ⎝ 2 4t + 2t ⎛ 2 sin ⎝ 2 −2 sin(3t)sin t 2 sin(3t)cos t − 2 sin(3t)sin t 2 sin(3t)cos t = − sin t cos t = −tan t Analysis Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side. Example 9.37 Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities Verify the identity csc2 θ − 2 = cos(2θ) sin2 θ . Solution For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side. cos(2θ) sin2 θ = 1 − 2 sin2 θ sin2 θ 1 sin2 θ = csc2 θ − 2 = − 2 sin2 θ sin2 θ 9.20 Verify the identity tan θ cot θ − cos2 θ = sin2 θ. 1028 Chapter 9 Trigonometric Identities and Equations Access these online resources for additional instruction and practice with the product-to-sum and sum-to-product identities. • Sum to Product Identities (http://openstaxcollege.org/l/sumtoprod) • Sum to Product and Product to Sum Identities (http://openstaxcollege.org/l/sumtpptsum) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1029 9.4 EXERCISES Verbal Starting with 162. sin α cos β = 1 2 the product [sin(α + β) + sin(α − β)], explain to sum formula how to determine the formula for cos α sin β. Provide 163. calculating two different methods of cos(195°)cos(105°), one of which uses the product to sum. Which method is easier? Describe a situation where we would convert an 164. equation from a sum to a product and give an example. cos(45°)sin(15°) 180. sin(−345°)sin(−15°) 181. sin(195°)cos(15°) 182. sin(−45°)sin(−15°) For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine. 183. cos(23°)sin(17°) Describe a situation where we would convert an 165. equation from a product to a sum, and give an example. 184. 2 sin(100°)sin(20°) Algebraic 185. 2 sin(−100°)sin(−20°) For the following exercises, rewrite the product as a sum or difference. 186. sin(213°)cos(8°) 166. 16 sin(16x)sin(11x) 167. 20 cos(36t)cos(6t) 168. 2 sin(5x)cos(3x) 169. 10 cos(5x)sin(10x) 170. sin(−x)sin(5x) 171. sin(3x)cos(5x) For the following exercises, rewrite the sum or difference as a product. 172. cos(6t) + cos(4t) 173. sin(3x) + sin(7x) 174. cos(7x) + cos(−7x) 175. sin(3x) − sin(−3x) 176. cos(3x) + cos(9x) 177. sin h − sin(3h) For the following exercises, evaluate the product for the following using a sum or difference of two functions. Evaluate exactly. 178. cos(45°)cos(15°) 179. 187. 2 cos(56°)cos(47°) For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine. 188. 189. 190. 191. 192. sin(76°) + sin(14°) cos(58°) − cos(12°) sin(101°) − sin(32°) cos(100°) + cos(200°) sin(−1°) + sin(−2°) For the following exercises, prove the identity. 193. cos(a + b) cos(a − b) = 1 − tan a tan b 1 + tan a tan b 194. 4 sin(3x)cos(4x) = 2 sin(7x) − 2 sinx 195. 6 cos(8x)sin(2x) sin(−6x) = −3 sin(10x)csc(6x) + 3 196. 197. 198. sin x + sin(3x) = 4 sin x cos2 x ⎝cos3 x − cos x sin2 x⎞ ⎛ ⎠ = cos(3x) + cos x 2 2 tan x cos(3x) = sec x⎛ ⎝sin(4x) − sin(2x)⎞ ⎠ 199. cos(a + b) + cos(a − b) = 2 cos a cos b Chapter 9 Trigonometric Identities and Equations Extensions For the following exercises, prove the following sum-toproduct formulas. 215. ⎛ sin x − sin y = 2 sin ⎝ x − y 2 ⎛ ⎞ ⎠cos ⎝ x + y 2 ⎞ ⎠ 216. ⎛ cos x + cos y = 2 cos ⎝ x + y 2 ⎞ ⎛ ⎠cos ⎝ x − y 2 ⎞ ⎠ For the following exercises, prove the identity. 217. sin(6x) + sin(4x) sin(6x) − sin(4x) = tan (5x)cot x 218. cos(3x) + cos x cos(3x) − cos x = − cot (2x)cot x 219. cos(6y) + cos(8y) sin(6y) − sin(4y) = cot y cos (7y)sec (5y) 220. cos⎛ sin⎛ ⎝2y⎞ ⎝2y⎞ ⎠ − cos⎛ ⎠ + sin⎛ ⎝4y⎞ ⎝4y⎞ ⎠ ⎠ = tan y 221. sin(10x) − sin(2x) cos(10x) + cos(2x) = tan(4x) 222. cos x − cos(3x) = 4 sin2 xcos x 223. (cos(2x) − cos(4x))2 + (sin(4x) + sin(2x))2 = 4 sin2(3x) 224. ⎛ tan ⎝ π 4 − t⎞ ⎠ = 1 − tan t 1 + tan t 1030 Numeric For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions. Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 200. cos(58°) + cos(12°) 201. sin(2°) − sin(3°) 202. cos(44°) − cos(22°) 203. cos(176°)sin(9°) 204. sin(−14°)sin(85°) Technology the following exercises, algebraically determine For whether each of the given equation is an identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator. 205. 2 sin(2x)sin(3x) = cos x − cos(5x) 206. cos(10θ) + cos(6θ) cos(6θ) − cos(10θ) = cot(2θ)cot(8θ) 207. sin(3x) − sin(5x) cos(3x) + cos(5x) = tan x 208. 2 cos(2x)cos x + sin(2x)sin x = 2 sin x 209. sin(2x) + sin(4x) sin(2x) − sin(4x) = − tan(3x)cot x For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical. 210. sin(9t) − sin(3t) cos(9t) + cos(3t) 211. 2 sin(8x)cos(6x) − sin(2x) 212. sin(3x) − sin x sin x 213. cos(5x) + cos(3x) sin(5x) + sin(3x) 214. sin x cos(15x) − cos x sin(15x) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1031 9.5 \| Solving Trigonometric Equations Learning Objectives In this section, you will: 9.5.1 Solve linear trigonometric equations in sine and cosine. 9.5.2 Solve equations involving a single trigonometric function. 9.5.3 Solve trigonometric equations using a calculator. 9.5.4 Solve trigonometric equations that are quadratic in form. 9.5.5 Solve trigonometric equations using fundamental identities. 9.5.6 Solve trigonometric equations with multiple angles. 9.5.7 Solve right triangle problems. Figure 9.21 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill) Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. Solving Linear Trigonometric Equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π. In other words, every 2π units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2π: sin θ = sin(θ ± 2kπ) There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with 1032 Chapter 9 Trigonometric Identities and Equations a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. Example 9.38 Solving a Linear Trigonometric Equation Involving the Cosine Function Find all possible exact solutions for the equation cos θ = 1 2 . Solution From the unit circle, we know that cos θ = 1 2 π 3 θ = , 5π 3 These are the solutions in the interval [0, 2π]. All possible solutions are given by θ = π 3 ± 2kπ and θ = 5π 3 ± 2kπ where k is an integer. Example 9.39 Solving a Linear Equation Involving the Sine Function Find all possible exact solutions for the equation sin t = 1 2 . Solution Solving for all possible values of t means that solutions include angles beyond the period of 2π. From Figure . But the problem is asking for all possible values that 9.7, we can see that the solutions are t = π 6 solve the equation. Therefore, the answer is and t = 5π 6 where k is an integer. t = π 6 ± 2πk and t = 5π 6 ± 2πk Given a trigonometric equation, solve using algebra. 1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. 2. Substitute the trigonometric expression with a single variable, such as x or u. 3. Solve the equation the same way an algebraic equation would be solved. 4. Substitute the trigonometric expression back in for
the variable in the resulting expressions. 5. Solve for the angle. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1033 Example 9.40 Solve the Linear Trigonometric Equation Solve the equation exactly: 2 cos θ − 3 = − 5, 0 ≤ θ < 2π. Solution Use algebraic techniques to solve the equation. 2 cos θ − 3 = −5 2 cos θ = −2 cos θ = −1 θ = π 9.21 Solve exactly the following linear equation on the interval [0, 2π) : 2 sin x + 1 = 0. Solving Equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 9.7). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π, not 2π. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π , unless, of course, a problem places its own restrictions on the 2 domain. Example 9.41 Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: 2 sin2 θ − 1 = 0, 0 ≤ θ < 2π. Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ. Then we will find the angles. 2 sin2 θ − 1 = 0 2 sin2 θ = 1 sin2 θ = 1 2 sin2 θ = ± 1 2 sin θ = ± 1 2 , 3π 4 θ = π 4 = ± 2 2 , 7π 4 , 5π 4 1034 Chapter 9 Trigonometric Identities and Equations Example 9.42 Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: csc θ = − 2, 0 ≤ θ < 4π. Solution We want all values of θ for which csc θ = − 2 over the interval 0 ≤ θ < 4π. csc θ = −2 1 sin θ = −2 sin θ = − 1 2 , 11π θ = 7π 6 6 , 19π 6 , 23π 6 Analysis As sin θ = − 1 2 , notice that all four solutions are in the third and fourth quadrants. Example 9.43 Solving an Equation Involving Tangent ⎛ Solve the equation exactly: tan ⎝θ − ⎞ ⎠ = 1, 0 ≤ θ < 2π. π 2 Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of π 4 , the tangent has a value of 1. However, the angle we want is ⎛ ⎝θ − θ − ⎞ ⎛ ⎠. Thus, if tan ⎝ ⎞ ⎠ = 1, π 4 then π 2 π π = 2 4 θ = 3π 4 ± kπ Over the interval [0, 2π), we have two solutions: θ = 3π 4 and θ = 3π 4 + π = 7π 4 9.22 Find all solutions for tan x = 3. Example 9.44 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1035 Identify all Solutions to the Equation Involving Tangent Identify all exact solutions to the equation 2(tan x + 3) = 5 + tan x, 0 ≤ x < 2π. Solution We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign. 2(tan x) + 2(3) = 5 + tan x 2tan x + 6 = 5 + tan x 2tan x − tan x = 5 − 6 tan x = −1 There are two angles on the unit circle that have a tangent value of −1:θ = 3π 4 and θ = 7π 4 . Solve Trigonometric Equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example 9.45 Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation sin θ = 0.8, where θ is in radians. Solution Make sure mode is set to radians. To find θ, use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the sin−1 function. What is shown on the screen is sin−1( . The calculator is ready for the input within the parentheses. For this problem, we enter sin−1 (0.8), and press ENTER. Thus, to four decimals places, The solution is The angle measurement in degrees is sin−1(0.8) ≈ 0.9273 θ ≈ 0.9273 ± 2πk θ ≈ 53.1° θ ≈ 180° − 53.1° ≈ 126.9° Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ. 1036 Chapter 9 Trigonometric Identities and Equations Example 9.46 Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation sec θ = −4, giving your answer in radians. Solution We can begin with some algebra. sec θ = −4 1 cos θ = −4 cos θ = − 1 4 Check that the MODE is in radians. Now use the inverse cosine function. cos−1 ⎛ ⎞ ⎝− 1 ⎠ ≈ 1.8235 4 θ ≈ 1.8235 + 2πk Since π 2 ≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 9.22. Figure 9.22 So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is θ ' ≈ π − 1.8235 ≈ 1.3181. The other solution in quadrant III is θ ' ≈ π + 1.3181 ≈ 4.4597. The solutions are θ ≈ 1.8235 ± 2πk and θ ≈ 4.4597 ± 2πk. 9.23 Solve cos θ = − 0.2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1037 Solving Trigonometric Equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example 9.47 Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: cos2 θ + 3 cos θ − 1 = 0, 0 ≤ θ < 2π. Solution We begin by using substitution and replacing cos θ with x. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x. We have The equation cannot be factored, so we will use the quadratic formula x = −b ± b2 − 4ac 2a . x2 + 3x − 1 = 0 Replace x with cos θ, and solve. x = −3 ± ( − 3)2 − 4(1)( − 1) 2 = −3 ± 13 2 cos θ = −3 ± 13 2 θ = cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ Note that only the + sign is used. This is because we get an error when we solve θ = cos−1 ⎛ ⎝ ⎞ on a ⎠ calculator, since the domain of the inverse cosine function is [−1, 1]. However, there is a second solution: −3 − 13 2 θ = cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ ≈ 1.26 This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is θ = 2π − cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ ≈ 5.02 Example 9.48 1038 Chapter 9 Trigonometric Identities and Equations Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: 2 sin2 θ − 5 sin θ + 3 = 0, 0 ≤ θ ≤ 2π. Solution Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u, or imagine it, as we factor: Now set each factor equal to zero. 2 sin2 θ − 5 sin θ + 3 = 0 (2 sin θ − 3)(sin θ − 1) = 0 2 sin θ − 3 = 0 2 sin θ = 3 sin θ = 3 2 sin θ − 1 = 0 sin θ = 1 Next solve for θ : sin θ ≠ 3 2 π 2 solution θ = . , as the range of the sine function is [−1, 1]. However, sin θ = 1, giving the Analysis Make sure to check all solutions on the given domain as some factors have no solution. 9.24 Solve sin2 θ = 2 cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 9.49 Solving a Trigonometric Equation Using Algebra Solve exactly: 2 sin2 θ + sin θ = 0; 0 ≤ θ < 2π Solution This problem should appear familiar as it is similar to a quadratic. Let sin θ = x. The equation becomes 2x2 + x = 0. We begin by factoring: Set each factor equal to zero. 2x2 + x = 0 x(2x + 1) = 0 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1039 x = 0 (2x + 1) = 0 x = − 1 2 Then, substitute back into the equation the original expression sin θ for x. Thus, sin θ = 0 θ = 0, π sin θ = − 1 2 , 11π θ = 7π 6 6 , 11π 6 The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, 7π 6 . If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. 2 sin2 θ + sin θ = 0 sin θ(2sin θ + 1) = 0 sin θ = 0 θ = 0, π 2 sin θ + 1 = 0 2sin θ = −1 sin θ = − 1 2 , 11π θ = 7π 6 6 Analysis We can see the solutions on the graph in Figure 9.23. On the interval 0 ≤ θ < 2π, four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. the graph crosses the x-axis Figure 9.23 We can verify the solutions on the unit circle in Figure 9.7 as well. Example 9.50 1040 Chapter 9 Trigonometric Identities and Equations Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: 2 sin2 θ − 3 sin θ + 1 = 0, 0 ≤ θ < 2π. Solution We can factor using grouping. Solution values of θ can be found on the unit circle. (2 sin θ − 1)(sin θ − 1) = 0 2 sin θ − 1
= 0 sin θ = 1 2 π 6 sin θ = 1 π θ = 2 θ = , 5π 6 9.25 Solve the quadratic equation 2 cos2 θ + cos θ = 0. Solving Trigonometric Equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation. Example 9.51 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2π. cos x cos(2x) + sin x sin(2x) = 3 2 Solution Notice that the left side of the equation is the difference formula for cosine. cos x cos(2x) + sin x sin(2x) = 3 2 cos(x − 2x) = 3 2 cos( − x) = 3 2 cos x = 3 2 Diffe ence formula for cosine Use the negative angle identity. From the unit circle in Figure 9.7, we see that cos x = 3 2 when x = π 6 , 11π 6 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1041 Example 9.52 Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: cos(2 θ) = cos θ. Solution We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: cos(2θ) = cos θ 2cos2 θ − 1 = cos θ 2 cos2 θ − cos θ − 1 = 0 (2 cos θ + 1)(cos θ − 1) = 0 2 cos θ + 1 = 0 cos θ = − 1 2 cos θ − 1 = 0 cos θ = 1 So, if cos θ = − 1 2 , then θ = 2π 3 ± 2πk and θ = 4π 3 ± 2πk; if cos θ = 1, then θ = 0 ± 2πk. Example 9.53 Solving an Equation Using an Identity Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin2 θ, 0 ≤ θ < 2π. Solution If we rewrite the right side, we can write the equation in terms of cosine: 3 cos θ + 3 = 2sin2 θ ⎝1 − cos2 θ⎞ ⎛ 3 cos θ + 3 = 2 ⎠ 3 cos θ + 3 = 2 − 2cos2 θ 2 cos2 θ + 3 cos θ + 1 = 0 (2 cos θ + 1)(cos θ + 1) = 0 2 cos θ + 1 = 0 cos θ = − 1 2 , 4π θ = 2π 3 3 cos θ + 1 = 0 cos θ = −1 θ = π Our solutions are θ = 2π 3 , 4π 3 , π. 1042 Chapter 9 Trigonometric Identities and Equations Solving Trigonometric Equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x) or cos(3x). When confronted with these equations, recall that y = sin(2x) is a horizontal compression by a factor of 2 of the function y = sin x. On an interval of 2π, we can graph two periods of y = sin(2x), as opposed to one cycle of y = sin x. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x) = 0 compared to sin x = 0. This information will help us solve the equation. Example 9.54 Solving a Multiple Angle Trigonometric Equation Solve exactly: cos(2x) = 1 2 on [0, 2π). Solution We can see that this equation is the standard equation with a multiple of an angle. If cos(α) = 1 2 , we know α is in quadrants I and IV. While θ = cos−1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 will be in quadrants I and IV. Therefore, the possible angles are θ = π 3 x = 5π 6 ⎛ . Does this make sense? Yes, because cos ⎝2 and θ = 5π 3 ⎛ π ⎝ 6 . So, 2x = ⎞ ⎞ ⎛ ⎠ = cos ⎝ ⎠ π 3 or 2x = 5π 3 , which means that x = or π 6 π 3 ⎞ ⎠ = 1 2 . Are there any other possible answers? Let us return to our first step. π 6 In quadrant I, 2x = , so x = π 3 as noted. Let us revolve around the circle again: 2x = + 2π + 6π 3 π 3 π = 3 = 7π 3 so x = 7π 6 . One more rotation yields 2x = + 4π + 12π 3 π 3 π = 3 = 13π 3 x = 13π 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). In quadrant IV, 2x = 5π 3 , so x = 5π 6 as noted. Let us revolve around the circle again: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1043 so x = 11π 6 . One more rotation yields + 2π + 6π 3 2x = 5π 3 = 5π 3 = 11π 3 + 4π + 12π 3 2x = 5π 3 = 5π 3 = 17π 3 x = 17π 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). Our solutions are x = , 5π 6 sin(nx) = c, we must go around the unit circle n times. , and 11π 6 , 7π 6 π 6 . Note that whenever we solve a problem in the form of Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2 + b2 = c2, and model an equation to fit a situation. 1044 Chapter 9 Trigonometric Identities and Equations Example 9.55 Using the Pythagorean Theorem to Model an Equation Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 9.24. Figure 9.24 Solution Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem. a2 + b2 = c2 (23)2 + (69.5)2 ≈ 5359 5359 ≈ 73.2 m The angle of elevation is θ, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places. tan θ = 69.5 23 tan−1 ⎛ ⎝ 69.5 23 ⎞ ⎠ ≈ 1.2522 ≈ 71.69° The angle of elevation is approximately 71.7°, and the length of the cable is 73.2 meters. Example 9.56 Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1045 Solution For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet. See Figure 9.25. Figure 9.25 The side adjacent to θ is a and the hypotenuse is 4a. Thus, cos θ = a 4a = 1 4 ⎞ ⎠ ≈ 75.5° cos−1 ⎛ ⎝ 1 4 The elevation of the ladder forms an angle of 75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: a2 + b2 = (4a)2 b2 = (4a)2 − a2 b2 = 16a2 − a2 b2 = 15a2 b = a 15 Thus, the ladder touches the wall at a 15 feet from the ground. Access these online resources for additional instruction and practice with solving trigonometric equations. • Solving Trigonometric Equations I (http://openstaxcollege.org/l/solvetrigeqI) • Solving Trigonometric Equations II (http://openstaxcollege.org/l/solvetrigeqII) • Solving Trigonometric Equations III (http://openstaxcollege.org/l/solvetrigeqIII) • Solving Trigonometric Equations IV (http://openstaxcollege.org/l/solvetrigeqIV) • Solving Trigonometric Equations V (http://openstaxcollege.org/l/solvetrigeqV) • Solving Trigonometric Equations VI (http://openstaxcollege.org/l/solvetrigeqVI) 1046 Chapter 9 Trigonometric Identities and Equations 9.5 EXERCISES Verbal Will 225. there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not. 226. When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not? 227. When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions? Algebraic For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2π. 228. 2 sin θ = − 2 229. 2 sin θ = 3 230. 2 cos θ = 1 231. 2 cos θ = − 2 232. tan θ = −1 233. tan x = 1 234. cot x + 1 = 0 235. 4 sin2 x − 2 = 0 236. csc2 x − 4 = 0 cos(2θ) = − 3 2 245. 2 sin(πθ) = 1 246. ⎛ 2 cos ⎝ π 5 θ⎞ ⎠ = 3 For the following exercises, find all exact solutions on [0, 2π). 247. sec(x)sin(x) − 2 sin(x) = 0 248. tan(x) − 2 sin(x)tan(x) = 0 249. 2 cos2 t + cos(t) = 1 250. 2 tan2(t) = 3 sec(t) 251. 2 sin(x)cos(x) − sin(x) + 2 cos(x) − 1 = 0 252. cos2 θ = 1 2 253. sec2 x = 1 254. 255. tan2 (x) = −1 + 2 tan(−x) 8 sin2(x) + 6 sin(x) + 1 = 0 256. tan5(x) = tan(x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π). For the following exercises, solve exactly on [0, 2π). 257. sin(3x)cos(6x) − cos(3x)sin(6x) = −0.9 237. 2 cos θ = 2 238. 2 cos θ = −1 239. 2 sin θ = −1 240. 2 sin θ = − 3 241. 2 sin(3θ) = 1 242. 2 sin(2θ) = 3 243. 2 cos(3θ) = − 2 244. This content is available for free at https://cnx.org/content/col11758/1.5 258. sin(6x)cos(11x) − cos(6x)sin(11x) = −0.1 259. cos(2x)cos x + sin(2x)sin x = 1 260. 6 sin(2t) + 9 sin t = 0 261. 9 cos(2θ) = 9 cos2 θ − 4 262. sin(2t) = cos t 263. cos(2t) = sin t 264. cos(6x) − cos(3x) = 0 Chapter 9 Trigonometric Identities and Equations 1047 For the following exercises, solve exactly on the interval [0, 2π). Use the quadratic formula if the equations do not factor. 265. 266. 267. 268. 269. 270. 271. 272. 273. tan2 x − 3 tan x = 0 sin2 x + sin x − 2 = 0 sin2 x − 2 sin x − 4 = 0 5 cos2 x + 3 cos x − 1 = 0 3 cos2 x − 2 cos x − 2 = 0 5 sin2 x + 2 sin x − 1 = 0 tan2 x + 5tan x − 1 = 0 cot2 x = − cot x −tan2 x − tan x − 2 = 0 For
the following exercises, find exact solutions on the interval [0, 2π). Look use trigonometric identities. opportunities for to 274. 275. sin2 x − cos2 x − sin x = 0 sin2 x + cos2 x = 0 276. sin(2x) − sin x = 0 277. cos(2x) − cos x = 0 278. 2 tan x 2 − sec2 x − sin2 x = cos2 x 279. 1 − cos(2x) = 1 + cos(2x) 280. sec2 x = 7 281. 10 sin x cos x = 6 cos x 282. −3 sin t = 15 cos t sin t 4 cos2 x − 4 = 15 cos x 8 sin2 x + 6 sin x + 1 = 0 8 cos2 θ = 3 − 2 cos θ 6 cos2 x + 7 sin x − 8 = 0 283. 284. 285. 286. 287. 12 sin2 t + cos t − 6 = 0 288. tan x = 3 sin x 289. cos3 t = cos t Graphical For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros. 290. 291. 292. 293. 294. 295. 296. 6 sin2 x − 5 sin x + 1 = 0 8 cos2 x − 2 cos x − 1 = 0 100 tan2 x + 20 tan x − 3 = 0 2 cos2 x − cos x + 15 = 0 20 sin2 x − 27 sin x + 7 = 0 2 tan2 x + 7 tan x + 6 = 0 130 tan2 x + 69 tan x − 130 = 0 Technology For the following exercises, use a calculator to find all solutions to four decimal places. 297. sin x = 0.27 298. sin x = −0.55 299. tan x = −0.34 300. cos x = 0.71 the following For equations exercises, algebraically, and then use a calculator to find the values on the interval [0, 2π). Round to four decimal places. solve the 301. 302. 303. 304. 305. 306. tan2 x + 3 tan x − 3 = 0 6 tan2 x + 13 tan x = −6 tan2 x − sec x = 1 sin2 x − 2 cos2 x = 0 2 tan2 x + 9 tan x − 6 = 0 4 sin2 x + sin(2x)sec x − 3 = 0 Chapter 9 Trigonometric Identities and Equations A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him? A 20-foot tall building has a shadow that is 55 feet 324. long. What is the angle of elevation of the sun? A 90-foot tall building has a shadow that is 2 feet 325. long. What is the angle of elevation of the sun? A spotlight on the ground 3 meters from a 2-meter tall 326. man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light? A spotlight on the ground 3 feet from a 5-foot tall 327. woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light? For the following exercises, find a solution to the following word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree. A person does a handstand with his feet touching a 328. wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall? A person does a handstand with her feet touching a 329. wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall? A 23-foot ladder is positioned next to a house. If the 330. ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping? 1048 Extensions For the following exercises, find all solutions exactly to the equations on the interval [0, 2π). 307. 308. 309. 310. 311. 312. 313. 314. 315. 316. csc2 x − 3 csc x − 4 = 0 sin2 x − cos2 x − 1 = 0 sin2 x⎛ ⎝1 − sin2 x⎞ ⎠ + cos2 x⎛ ⎝1 − sin2 x⎞ ⎠ = 0 3 sec2 x + 2 + sin2 x − tan2 x + cos2 x = 0 sin2 x − 1 + 2 cos(2x) − cos2 x = 1 tan2 x − 1 − sec3 x cos x = 0 sin(2x) sec2 x = 0 sin(2x) 2csc2 x = 0 2 cos2 x − sin2 x − cos x − 5 = 0 1 sec2 x + 2 + sin2 x + 4 cos2 x = 4 Real-World Applications An airplane has only enough gas to fly to a city 200 317. miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly? If a loading ramp is placed next to a truck, at a height 318. of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground? If a loading ramp is placed next to a truck, at a height 319. of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground? A woman is watching a launched rocket currently 11 320. miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal? An astronaut is in a launched rocket currently 15 321. miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.) A woman is standing 8 meters away from a 10-meter 322. tall building. At what angle is she looking to the top of the building? 323. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1049 CHAPTER 9 REVIEW KEY TERMS double-angle formulas equal identities derived from the sum formulas for sine, cosine, and tangent in which the angles are even-odd identities set of equations involving trigonometric functions such that if f (−x) = − f (x), the identity is odd, and if f (−x) = f (x), the identity is even half-angle formulas identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions product-to-sum formula a trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions Pythagorean identities set of equations involving trigonometric functions based on the right triangle properties quotient identities pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine reciprocal identities set of equations involving the reciprocals of basic trigonometric definitions reduction formulas function identities derived from the double-angle formulas and used to reduce the power of a trigonometric sum-to-product formula a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions KEY EQUATIONS 1050 Chapter 9 Trigonometric Identities and Equations Pythagorean identities Even-odd identities Reciprocal identities Quotient identities cos2 θ + sin2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ tan( − θ) = −tan θ cot( − θ) = −cot θ sin( − θ) = −sin θ csc( − θ) = −csc θ cos( − θ) = cos θ sec( − θ) = sec θ sin θ = cos θ = tan θ = csc θ = sec θ = cot θ = 1 csc θ 1 sec θ 1 cot θ 1 sin θ 1 cos θ 1 tan θ tan θ = sin θ cos θ cot θ = cos θ sin θ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1051 Sum Formula for Cosine Difference Formula for Cosine Sum Formula for Sine Difference Formula for Sine Sum Formula for Tangent Difference Formula for Tangent Cofunction identities cos⎛ ⎝α + β⎞ ⎠ = cos α cos β − sin αsin β cos⎛ ⎝α − β⎞ ⎠ = cos α cos β + sin α sin β sin⎛ ⎝α + β⎞ ⎠ = sin α cos β + cos α sin β sin⎛ ⎝α − β⎞ ⎠ = sin α cos β − cos α sin β tan⎛ ⎝α + β⎞ ⎠ = tan⎛ ⎝α − β⎞ ⎠ = tan α + tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β tan θ = cot ⎛ cos θ = sin ⎝ π ⎛ sin θ = cos ⎝ ⎛ sec θ = csc ⎝ 2 π 2 ⎛ csc θ = sec ⎝ ⎛ cot θ = tan ⎝ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ 1052 Chapter 9 Trigonometric Identities and Equations Double-angle formulas Reduction formulas sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2sin2 θ = 2cos2 θ − 1 2tan θ 1 − tan2 θ tan(2θ) = sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) sin α 2 cos α 2 tan α 2 Half-angle formulas 2 2 = ± 1 − cos α = ± 1 + cos α = ± 1 − cos α 1 + cos α sin α = 1 + cos α = 1 − cos α sin α Product-to-sum Formulas cos α cos β = 1 2 sin α cos β = 1 2 sin α sin β = 1 2 cos α sin β = 1 2 [cos(α − β) + cos(α + β)] [sin(α + β) + sin(α − β)] [cos(α − β) − cos(α + β)] [sin(α + β) − sin(α − β)] Sum-to-product Formulas ⎛ sin α + sin β = 2 sin ⎝ ⎛ sin α − sin β = 2 sin ⎝ ⎛ cos α − cos β = −2 sin ⎝ 2 α + β 2 ⎛ cos α + cos β = 2 cos ⎝ α − β ⎛ ⎞ ⎞ ⎠cos ⎝ ⎠ 2 α + β ⎞ ⎛ ⎞ ⎠cos ⎠ ⎝ 2 α − β ⎛ ⎞ ⎠sin ⎝ 2 α − β ⎛ ⎞ ⎠cos ⎝ 2 ⎞ ⎠ ⎞ ⎠ KEY CONCEPTS 9.1 Solving Trigonometric Equations with Identities • There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem. • Graphing both sides of an identity will verify it. See Example 9.1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1053 • Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example 9.2 and Example 9.3. • The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example 9.4. • We can create an identity and then verify it. See Example 9.5. • Verifying an identity may involve algebra with the fundamental identities. See Example 9.6 and Example 9.7. • Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example 9.8, Example 9.9, and Example 9.10. 9.2 Sum and Difference Identities • The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles. • The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See Example 9.11 and Example 9.12. • The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the f
irst angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See Example 9.13. • The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See Example 9.14. • The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See Example 9.15. • The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See Example 9.16. • The cofunction identities apply to complementary angles and pairs of reciprocal functions. See Example 9.17. • Sum and difference formulas are useful in verifying identities. See Example 9.18 and Example 9.19. • Application problems are often easier to solve by using sum and difference formulas. See Example 9.20 and Example 9.20. 9.3 Double-Angle, Half-Angle, and Reduction Formulas • Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example 9.22, Example 9.23, Example 9.24, and Example 9.25. • Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example 9.26 and Example 9.27. • Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example 9.28, Example 9.29, and Example 9.30. 9.4 Sum-to-Product and Product-to-Sum Formulas • From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. • We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. See Example 9.31, Example 9.32, and Example 9.33. 1054 Chapter 9 Trigonometric Identities and Equations • We can also derive the sum-to-product identities from the product-to-sum identities using substitution. • We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines. See Example 9.34. • Trigonometric expressions are often simpler to evaluate using the formulas. See Example 9.35. • The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side. See Example 9.36 and Example 9.37. 9.5 Solving Trigonometric Equations • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example 9.38, Example 9.39, and Example 9.40. • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example 9.41, Example 9.42, and Example 9.43, and Example 9.44. • We can also solve trigonometric equations using a graphing calculator. See Example 9.45 and Example 9.46. • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example 9.47, Example 9.48, Example 9.49, and Example 9.50. • We can also use the identities to solve trigonometric equation. See Example 9.51, Example 9.52, and Example 9.53. • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example 9.54. • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example 9.55. CHAPTER 9 REVIEW EXERCISES Solving Trigonometric Equations with Identities For the following exercises, find all solutions exactly that exist on the interval [0, 2π). 331. csc2 t = 3 332. cos2 x = 1 4 333. 2 sin θ = −1 334. tan x sin x + sin(−x) = 0 335. 9 sin ω − 2 = 4 sin2 ω 336. 1 − 2 tan(ω) = tan2(ω) 338. sin3 x + cos2 x sin x the following exercises, determine if For identities are equivalent. the given 339. sin2 x + sec2 x − 1 = ⎛ ⎛ ⎝1 − cos2 x⎞ ⎝1 + cos2 x⎞ ⎠ ⎠ cos2 x 340. tan3 x csc2 x cot2 x cos x sin x = 1 Sum and Difference Identities For the following exercises, find the exact value. 341. ⎛ tan ⎝ ⎞ ⎠ 7π 12 342. ⎛ cos ⎝ 25π 12 ⎞ ⎠ For the following exercises, use basic identities to simplify the expression. 337. sec x cos x + cos x − 1 sec x 343. 344. sin(70°)cos(25°) − cos(70°)sin(25°) cos(83°)cos(23°) + sin(83°)sin(23°) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1055 For the following exercises, prove the identity. 345. cos(4x) − cos(3x)cosx = sin2 x − 4 cos2 x sin2 x For the following exercises, prove the identity. 356. 2cos(2x) sin(2x) = cot x − tan x 346. cos(3x) − cos3 x = − cos x sin2 x − sin x sin(2x) 357. cot x cos(2x) = − sin(2x) + cot x For the following exercise, simplify the expression. 347. 1 2 ⎛ tan ⎝ x⎞ ⎛ x⎞ 1 ⎠ + tan ⎝ ⎠ 8 x⎞ ⎛ x⎞ ⎛ 1 1 ⎠tan 1 − tan ⎠ ⎝ ⎝ 2 8 For the following exercises, rewrite the expression with no powers. 358. cos2 x sin4(2x) For the following exercises, find the exact value. 359. tan2 x sin3 x 348. ⎝sin−1 (0) − cos−1 ⎛ ⎛ cos ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 349. ⎝sin−1 (0) + sin−1 ⎛ ⎛ tan ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Double-Angle, Half-Angle, and Reduction Formulas For the following exercises, find the exact value. 350. Find sin(2θ), cos(2θ), cos θ = − 1 3 and θ is in the interval ⎡ ⎣ and tan(2θ) given , π⎤ π ⎦. 2 351. Find sin(2θ), cos(2θ), and tan(2θ) given sec θ = − 5 3 and θ is in the interval ⎡ ⎣ , π⎤ ⎦. π 2 352. ⎛ sin ⎝ ⎞ ⎠ 7π 8 353. ⎛ sec ⎝ ⎞ ⎠ 3π 8 For the following exercises, use Figure 9.26 to find the desired quantities. Figure 9.26 354. sin(2β), cos(2β), tan(2β), sin(2α), cos(2α), and tan(2α) Sum-to-Product and Product-to-Sum Formulas For the following exercises, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. 360. ⎛ cos ⎝ π 3 ⎞ ⎛ ⎠ sin ⎝ ⎞ ⎠ π 4 ⎛ 361. 2 sin ⎝ 2π 3 ⎛ ⎞ ⎠ sin ⎝ ⎞ ⎠ 5π 6 ⎛ 362. 2 cos ⎝ π 5 ⎞ ⎛ ⎠ cos ⎝ ⎞ ⎠ π 3 For the following exercises, evaluate the sum by using a product formula. Write the exact answer. 363. ⎛ sin ⎝ π 12 ⎞ ⎛ ⎠ − sin ⎝ ⎞ ⎠ 7π 12 364. ⎛ cos ⎝ 5π 12 ⎞ ⎛ ⎠ + cos ⎝ ⎞ ⎠ 7π 12 For the following exercises, change the functions from a product to a sum or a sum to a product. 365. sin(9x)cos(3x) 366. cos(7x)cos(12x) 367. sin(11x) + sin(2x) 368. cos(6x) + cos(5x) β 2 ⎞ ⎛ ⎠, cos ⎝ β 2 ⎞ ⎛ ⎠, tan ⎝ β 2 ⎞ ⎛ ⎠, sin ⎝ α 2 ⎛ ⎞ ⎠, cos ⎝ α 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠ α 2 369. tan x + 1 = 0 Solving Trigonometric Equations For the following exercises, find all exact solutions on the interval [0, 2π). 355. ⎛ sin ⎝ 1056 Chapter 9 Trigonometric Identities and Equations 370. 2 sin(2x) + 2 = 0 For the following exercises, find all exact solutions on the interval [0, 2π). 371. 2 sin2 x − sin x = 0 372. cos2 x − cos x − 1 = 0 373. 2 sin2 x + 5 sin x + 3 = 0 374. cos x − 5 sin(2x) = 0 375. 1 sec2 x + 2 + sin2 x + 4 cos2 x = 0 the following exercises, simplify the equation For algebraically as much as possible. Then use a calculator to find the solutions on the interval [0, 2π). Round to four decimal places. 376. 3 cot2 x + cot x = 1 377. csc2 x − 3 csc x − 4 = 0 For the following exercises, graph each side of the equation to find the approximate solutions on the interval [0, 2π). 378. 20 cos2 x + 21 cos x + 1 = 0 sec2 x − 2 sec x = 15 379. CHAPTER 9 PRACTICE TEST For the following exercises, simplify the given expression. 380. cos(−x)sin x cot x + sin2 x 381. sin(−x)cos(−2x)−sin(−x)cos(−2x) 382. ⎞ ⎛ ⎝sec2 θ − 1 csc(θ)cot(θ) ⎠ 383. ⎞ ⎞ ⎛ ⎛ ⎝1 + tan2 (θ) ⎝1 + cot2 (θ) cos2 (θ)sin2 (θ) ⎠ ⎠ For the following exercises, find the exact value. 384. ⎛ cos ⎝ ⎞ ⎠ 7π 12 385. ⎛ tan ⎝ ⎞ ⎠ 3π 8 This content is available for free at https://cnx.org/content/col11758/1.5 386. ⎝sin−1 ⎛ ⎛ tan ⎝ ⎞ ⎠ + tan−1 3 ⎞ ⎠ 2 2 ⎛ 387. 2sin ⎝ π 4 ⎛ ⎞ ⎠sin ⎝ ⎞ ⎠ π 6 388. ⎛ cos ⎝ 4π 3 + θ⎞ ⎠ 389. ⎛ ⎝− tan π 4 + θ⎞ ⎠ For the following exercises, simplify each expression. Do not evaluate. 390. cos2(32°)tan2(32°) 391. cot ⎞ ⎠ ⎛ ⎝ θ 2 Chapter 9 Trigonometric Identities and Equations 1057 For the following exercises, find all exact solutions to the equation on [0, 2π). 406. tan3 x − tan x sec2 x = tan(−x) 392. cos2 x − sin2 x − 1 = 0 407. sin(3x) − cos x sin(2x) = cos2 x sin x − sin3 x 393. cos2 x = cos x 4 sin2 x + 2 sin x − 3 = 0 408. sin(2x) sin x − cos(2x) cos x = sec x 394. cos(2x) + sin2 x = 0 395. 2 sin2 x − sin x = 0 396. Rewrite the expression as a product instead of a sum: cos(2x) + cos(−8x). For the following exercise, rewrite the product as a sum or difference. 397. 8cos(15x)sin(3x) For the following exercise, rewrite the sum or difference as a product. 398. 2⎛ ⎝sin(8θ) − sin(4θ)⎞ ⎠ 399. Find all solutions of tan(x) − 3 = 0. 400. Find the solutions of sec2 x − 2 sec x = 15 on the interval [0, 2π) algebraically; then graph both sides of the equation to determine the answer. For the following exercises, find all solutions exactly on the interval 0 ≤ θ ≤ π ⎛ 401. 2cos ⎝ ⎞ ⎠ = 1 θ 2 402. 3cot(y) = 1 403. Find sin(2θ), cos(2θ), and tan(2θ) given cot θ = − 3 4 and θ is on the interval ⎡ ⎣ , π⎤ ⎦. π 2 404. cos θ = 7 25 ⎛ Find sin ⎝ θ 2 and θ is in quadrant IV. ⎛ ⎞ ⎠, cos ⎝ θ 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠ given θ 2 405. Rewrite the expression sin4 x with no powers greater than 1. For the fol
lowing exercises, prove the identity. 409. Plot the points and find a function of the form y = Acos(Bx + C) + D that fits the given data. x y 0 −2 1 2 2 −2 3 2 4 −2 5 2 410. The displacement h(t) in centimeters of a mass suspended by a spring is modeled by the function h(t) = 1 sin(120πt), where t is measured in seconds. 4 Find the amplitude, period, and frequency of displacement. this 411. A woman is standing 300 feet away from a 2000-foot building. If she looks to the top of the building, at what angle above horizontal is she looking? A bored worker looks down at her from the 15th floor (1500 feet above her). At what angle is he looking down at her? Round to the nearest tenth of a degree. governed 412. Two frequencies of sound are played on an equation instrument by n(t) = 8 cos(20πt)cos(1000πt). What are the period and frequency of the “fast” and “slow” oscillations? What is the amplitude? the 413. The average monthly snowfall in a small village in the Himalayas is 6 inches, with the low of 1 inch occurring in July. Construct a function that models this behavior. During what period is there more than 10 inches of snowfall? 414. A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring t seconds after being released. At what time will the spring come to rest? In this case, use 1 cm amplitude as rest. 415. Water levels near a glacier currently average 9 feet, varying seasonally by 2 inches above and below the average and reaching their highest point in January. Due 1058 Chapter 9 Trigonometric Identities and Equations to global warming, the glacier has begun melting faster than normal. Every year, the water levels rise by a steady 3 inches. Find a function modeling the depth of the water t months from now. If the docks are 2 feet above current water levels, at what point will the water first rise above the docks? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1059 10 \| FURTHER APPLICATIONS OF TRIGONOMETRY Figure 10.1 General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr) Chapter Outline 10.1 Non-right Triangles: Law of Sines 10.2 Non-right Triangles: Law of Cosines 10.3 Polar Coordinates 10.4 Polar Coordinates: Graphs 10.5 Polar Form of Complex Numbers 10.6 Parametric Equations 10.7 Parametric Equations: Graphs 10.8 Vectors 1060 Chapter 10 Further Applications of Trigonometry Introduction The world’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California.[1] Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, in Trigonometric Functions including finding (https://cnx.org/content/m49369/latest/) and investigate applications more deeply and meaningfully. the height of a tree. We extend topics we introduced 10.1 \| Non-right Triangles: Law of Sines Learning Objectives In this section, you will: 10.1.1 Use the Law of Sines to solve oblique triangles. 10.1.2 Find the area of an oblique triangle using the sine function. 10.1.3 Solve applied problems using the Law of Sines. Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 10.2 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. Figure 10.2 Using the Law of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 10.3. Figure 10.3 2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 10.4. 1. Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1061 Figure 10.4 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 10.5. Figure 10.5 Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 10.6. Figure 10.6 Using the right triangle relationships, we know that sin α = and sin β = h a. Solving both equations for h gives two h b different expressions for h. We then set the expressions equal to each other. h = bsin α and h = asin β bsin α = asin β ⎛ ⎛ ⎞ 1 ⎠(bsin α) = (asin β) ⎝ ⎝ ab sin α a = sin β b ⎞ ⎠ 1 ab Multiply both sides by 1 ab. Similarly, we can compare the other ratios. sin α a = sin γ c and sin β b = sin γ c Collectively, these relationships are called the Law of Sines. sin α a = sin β b = sin λ c Note the standard way of labeling triangles: angle α (alpha) is opposite side a; angle β (beta) is opposite side b; and angle γ (gamma) is opposite side c. See Figure 10.7. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. 1062 Chapter 10 Further Applications of Trigonometry Figure 10.7 Law of Sines Given a triangle with angles and opposite sides labeled as in Figure 10.7, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ (10.1) (10.2) To solve an oblique triangle, use any pair of applicable ratios. Example 10.1 Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in Figure 10.8 to the nearest tenth. Figure 10.8 Solution The three angles must add up to 180 degrees. From this, we can determine that β = 180° − 50° − 30° = 100° To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1063 sin(50°) 10 csin(50°) 10 = sin(30°) c = sin(30°) c = sin(30 ° ) c ≈ 6.5 Multiply both sides by c. 10 sin(50°) Multiply by the reciprocal to isolate c. Similarly, to solve for b, we set up another proportion. sin(50°) 10 = sin(100°) b bsin(50°) = 10sin(100°) 10sin(100°) sin(50°) b = b ≈ 12.9 Therefore, the complete set of angles and sides is Multiply both sides by b. Multiply by the reciprocal to isolate b. α = 50° a = 10 β = 100° b ≈ 12.9 γ = 30° c ≈ 6.5 10.1 Solve the triangle shown in Figure 10.9 to the nearest tenth. Figure 10.9 Using The Law of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. Possible Outcomes for SSA Triangles Oblique triangles in the category SSA may have four different outcomes. Figure 10.10 illustrates the solutions with the known sides a and b and known angle α. 1064 Chapter 10 Further Applications of Trigonometry Figure 10.10 Example 10.2 Solving an Oblique SSA Triangle Solve the triangle in Figure 10.11 for the missing side and find the missing angle measures to the nearest tenth. Figure 10.11 Solution Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we have the proportion = sin α a = sin β b sin β sin(35°) 8 6 8sin(35°) = sin β 6 0.7648 ≈ sin β sin−1(0.7648) ≈ 49.9° β ≈ 49.9° This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1065 However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β ? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right
angle perspective, we have Figure 10.12. It appears that there may be a second triangle that will fit the given criteria. Figure 10.12 The angle supplementary to β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1°. (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ, we have We can then use these measurements to solve the other triangle. Since γ′ is supplementary to γ, we have γ = 180° − 35° − 130.1° ≈ 14.9° Now we need to find c and c′. We have Finally, γ′ = 180° − 35° − 49.9° ≈ 95.1° c sin(14.9°) = c = 6 sin(35°) 6sin(14.9°) sin(35°) ≈ 2.7 c′ sin(95.1°) = c′ = 6 sin(35°) 6sin(95.1°) sin(35°) ≈ 10.4 To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 10.13. 1066 Chapter 10 Further Applications of Trigonometry Figure 10.13 However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first triangle (a) in Figure 10.13. 10.2 Given α = 80°, a = 120, and b = 121, find the missing side and angles. If there is more than one possible solution, show both. Example 10.3 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure 10.14, solve for the unknown side and angles. Round your answers to the nearest tenth. Figure 10.14 Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85°, and its corresponding side c = 12, and we know side b = 9. We will use this proportion to solve for β. sin(85°) 12 9sin(85°) 12 = sin β 9 = sin β Isolate the unknown. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1067 To find β, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. ⎞ ⎠ β = sin−1 ⎛ ⎝ 9sin(85°) 12 β ≈ sin−1(0.7471) β ≈ 48.3° In this case, there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7°. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives if we subtract β from 180°, we find that which is impossible, and so β ≈ 48.3°. α = 180° − 85° − 131.7° ≈ − 36.7°, To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7°. Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions. = sin(46.7 ° ) a = sin(46.7 ° ) sin(85 ° ) 12 asin(85 ° ) 12 a = 12sin(46.7 ° ) sin(85 ° ) ≈ 8.8 The complete set of solutions for the given triangle is α ≈ 46.7° a ≈ 8.8 β ≈ 48.3° b = 9 γ = 85° c = 12 10.3 Given α = 80°, a = 100, b = 10, find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. Example 10.4 Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Solution Using the given information, we can solve for the angle opposite the side of length 10. See Figure 10.15. 1068 Chapter 10 Further Applications of Trigonometry = sin(50°) 4 sin α 10 10sin(50°) sin α = 4 sin α ≈ 1.915 Figure 10.15 We can stop here without finding the value of α. Because the range of the sine function is [−1, 1], it is impossible for the sine value to be 1.915. In fact, inputting sin−1 (1.915) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. 10.4 Determine the number of triangles possible given a = 31, b = 26, β = 48°. Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as Area = 1 bh, where b is base and h is 2 height. For oblique triangles, we must find h before we can use the area formula. Observing the two triangles in Figure 10.16, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = h c or csin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α′, or 180 − α. Figure 10.16 Thus, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1069 Similarly, Area = 1 2 (base)⎛ ⎝height⎞ ⎠ = 1 2 b(csin α) Area = 1 2 ⎝bsin γ⎞ a⎛ ⎠ = 1 2 ⎝csin β⎞ a⎛ ⎠ Area of an Oblique Triangle The formula for the area of an oblique triangle is given by Area = 1 2 = 1 2 = 1 2 bcsin α acsin β absin γ (10.3) This is equivalent to one-half of the product of two sides and the sine of their included angle. Example 10.5 Finding the Area of an Oblique Triangle Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102°. Round the area to the nearest integer. Solution Using the formula, we have absin γ Area = 1 2 Area = 1 (90)(52)sin(102°) 2 Area ≈ 2289 square units 10.5 Find the area of the triangle given β = 42°, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth. Solving Applied Problems Using the Law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. Example 10.6 Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 10.17. Round the altitude to the nearest tenth of a mile. 1070 Chapter 10 Further Applications of Trigonometry Figure 10.17 Solution To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a, and then use right triangle relationships to find the height of the aircraft, h. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. sin(130°) 20 = sin(35°) a asin(130°) = 20sin(35°) 20sin(35°) sin(130°) a = a ≈ 14.98 The distance from one station to the aircraft is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for h. sin(15°) = sin(15°) = opposite hypotenuse h a sin(15°) = h 14.98 h = 14.98sin(15°) h ≈ 3.88 The aircraft is at an altitude of approximately 3.9 miles. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1071 10.6 The diagram shown in Figure 10.18 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B, is 62°, and the distance between the viewing points of the two end zones is 145 yards. Figure 10.18 Access these online resources for additional instruction and practice with trigonometric applications. • Law of Sines: The Basics (http://openstaxcollege.org/l/sinesbasic) • Law of Sines: The Ambiguous Case (http://openstaxcollege.org/l/sinesambiguous) 1072 Chapter 10 Further Applications of Trigonometry 10.1 EXERCISES Verbal 1. Describe the altitude of a triangle. 2. Compare right triangles and oblique triangles. When can you use the Law of Sines to find a missing 3. angle? 17. a = 12, c = 17, α = 35° 18. a = 20.5, b = 35.0, β = 25° 19. a = 7, c = 9, α = 43° 20. a = 7, b = 3, β = 24° In the Law of Sines, what is the relationship between the 4. angle in the numerator and the side in the denominator? 21. b = 13, c = 5, γ = 10° 5. What type of triangle results in an ambiguous case? 22. a = 2.3, c = 1.8, γ = 28° Algebraic 23. β = 119°, b = 8.2, a = 11.3 For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 6. 7. α = 43°, γ = 69°, a = 20 α = 35°, γ = 73°, c = 20 8. α = 60°, β = 60°, γ = 60° 9. a = 4, α = 60°, β = 100° 10. b = 10, β = 95°, γ = 30° For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. 11. Find side b when A = 37°, B = 49°, c = 5. 12. Find side a when A = 132°, C = 23°, b = 10. 13. Find side c when B = 37°, C = 21, b = 23. For the following exercises, assume α is opposite side a, β is opposite side c. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. side b, and γ is opposite 14. α = 119°, a = 14, b = 26 15. γ = 113°, b = 10, c = 32 16. b = 3.5, c = 5.3, γ = 80° This content is available for free at https://cnx.org/content/col11758/1.5 24. Find angle A when a = 24, b = 5, B = 22°. 25. Find angle A when a = 13, b = 6, B = 20°. 26. Find angle B when A = 12°, a = 2, b = 9. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a = 5, c = 6, β = 35° 28. b = 11, c = 8, α = 28° 29. a = 32, b = 24, γ = 75° 30. a = 7.2, b = 4.5, γ = 43° Graphical For the following exercises, find the length of side x. Round to the nearest tenth. 31. 32
. Chapter 10 Further Applications of Trigonometry 1073 33. 34. 35. 36. 39. 40. 41. Notice that x is an obtuse angle. 42. For the following exercises, find the measure of angle x, if possible. Round to the nearest tenth. 37. 38. For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. 43. 1074 Chapter 10 Further Applications of Trigonometry 44. 45. 46. 47. 48. This content is available for free at https://cnx.org/content/col11758/1.5 49. Extensions Find the radius of the circle in Figure 10.19. Round to 50. the nearest tenth. Figure 10.19 Find the diameter of the circle in Figure 10.20. 51. Round to the nearest tenth. Figure 10.20 Find m ∠ ADC in Figure 10.21. Round to the 52. nearest tenth. Chapter 10 Further Applications of Trigonometry 1075 Figure 10.21 Figure 10.24 53. Find AD in Figure 10.22. Round to the nearest tenth. 56. Solve the triangle in Figure 10.25. (Hint: Draw a perpendicular from H to JK). Round each answer to the nearest tenth. Figure 10.25 Figure 10.22 Solve both triangles in Figure 10.23. Round each 54. answer to the nearest tenth. 57. Solve the triangle in Figure 10.26. (Hint: Draw a perpendicular from N to LM). Round each answer to the nearest tenth. Figure 10.26 Figure 10.23 Find AB in the parallelogram shown in Figure 55. 10.24. In Figure 10.27, ABCD is not a parallelogram. 58. ∠ m is obtuse. Solve both triangles. Round each answer to the nearest tenth. 1076 Chapter 10 Further Applications of Trigonometry Figure 10.29 61. Figure 10.30 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A and B, which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 86.2° and 83.9°, respectively. How far is the satellite from station A and how high is the satellite above the ground? Round answers to the nearest whole mile. Figure 10.27 Real-World Applications A pole leans away from the sun at an angle of 7° to the 59. vertical, as shown in Figure 10.28. When the elevation of the sun is 55°, the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth. Figure 10.28 Figure 10.30 60. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 10.29. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. 62. A communications tower is located at the top of a steep hill, as shown in Figure 10.31. The angle of inclination of the hill is 67°. A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. Find the length of the cable required for the guy wire to the nearest whole meter. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1077 be 32° and 56°, as shown in Figure 10.34. Find the distance of the plane from point A to the nearest tenth of a kilometer. Figure 10.31 The roof of a house is at a 20° angle. An 8-foot solar 63. panel is to be mounted on the roof and should be angled 38° relative to the horizontal for optimal results. (See Figure 10.32). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. Figure 10.32 64. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° and 44°, as shown in Figure 10.33. Find the distance of the plane from point A to the nearest tenth of a kilometer. Figure 10.33 A pilot is flying over a straight highway. He determines 65. the angles of depression to two mileposts, 4.3 km apart, to Figure 10.34 In order to estimate the height of a building, two 66. students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. In order to estimate the height of a building, two 67. students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. Points A and B are on opposite sides of a lake. Point 68. C is 97 meters from A. The measure of angle BAC is determined to be 101°, and the measure of angle ACB is determined to be 53°. What is the distance from A to B, rounded to the nearest whole meter? 69. A man and a woman standing 31 2 miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. Two search teams spot a stranded climber on a 70. mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. A street light is mounted on a pole. A 6-foot-tall man is 71. standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is 1078 Chapter 10 Further Applications of Trigonometry standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. Three cities, A, B, and C, are located so that city A 72. is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 miles from city B, how far is city A from city B ? Round the distance to the nearest tenth of a mile. 73. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. Brian’s house is on a corner lot. Find the area of the 74. front yard if the edges measure 40 and 56 feet, as shown in Figure 10.35. Figure 10.35 75. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. A yield sign measures 30 inches on all three sides. 76. What is the area of the sign? Naomi bought a modern dining table whose top is in 77. the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 10.36. Figure 10.36 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1079 10.2 \| Non-right Triangles: Law of Cosines Learning Objectives In this section, you will: 10.2.1 Use the Law of Cosines to solve oblique triangles. 10.2.2 Solve applied problems using the Law of Cosines. 10.2.3 Use Heron’s formula to ﬁnd the area of a triangle. Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 10.37. How far from port is the boat? Figure 10.37 Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases. Using the Law of Cosines to Solve Oblique Triangles The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level. Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure 10.38. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted. 1080 Chapter 10 Further Applications of Trigonometry Figure 10.38 We can drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that cos θ = x(adjacent) b(hypotenuse) and sin θ =
y(opposite) b(hypotenuse) In terms of θ, x = bcos θ and y = bsin θ. The (x, y) point located at C has coordinates (bcos θ, bsin θ). Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagorean Theorem. Thus, a2 = (x − c)2 + y2 = (bcos θ − c)2 + (bsin θ)2 ⎝b2 cos2 θ − 2bccos θ + c2⎞ ⎛ ⎠ + b2 sin2 θ = = b2 cos2 θ + b2 sin2 θ + c2 − 2bccos θ = b2 ⎛ ⎠ + c2 − 2bccos θ ⎝cos2 θ + sin2 θ⎞ a2 = b2 + c2 − 2bccos θ Substitute (bcos θ) for x and (bsin θ) for y. Expand the perfect square. Group terms noting that cos2 θ + sin2 θ = 1. Factor out b2. The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve. Law of Cosines The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 10.39, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations. a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ (10.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1081 Figure 10.39 To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle. cos α = cos β = cos γ = b2 + c2 − a2 2bc a2 + c2 − b2 2ac a2 + b2 − c2 2ab Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle. 1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles. 2. Apply the Law of Cosines to find the length of the unknown side or angle. 3. Apply the Law of Sines or Cosines to find the measure of a second angle. 4. Compute the measure of the remaining angle. Example 10.7 Finding the Unknown Side and Angles of a SAS Triangle Find the unknown side and angles of the triangle in Figure 10.40. Figure 10.40 Solution 1082 Chapter 10 Further Applications of Trigonometry First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines. Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b, as we know the measurement of the opposite angle β. b2 = a2 + c2 − 2accos β b2 = 102 + 122 − 2(10)(12)cos(30∘) b2 = 100 + 144 − 240 ⎛ ⎝ ⎞ ⎠ 3 2 b2 = 244 − 120 3 b = 244 − 120 3 b ≈ 6.013 Substitute the measurements for the known quantities. Evaluate the cosine and begin to simplify. Use the square root property. Because we are solving for a length, we use only the positive square root. Now that we know the length b, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α, we have sin α a = sin β b = sin α 10 sin α = sin(30°) 6.013 10sin(30°) 6.013 α = sin−1 ⎛ ⎝ α ≈ 56.3° 10sin(30°) 6.013 Multiply both sides of the equation by 10. ⎞ ⎠ Find the inverse sine of 10sin(30°) 6.013 . The other possibility for α would be α = 180° – 56.3° ≈ 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0° and 180°. Proceeding with α ≈ 56.3°, we can then find the third angle of the triangle. The complete set of angles and sides is γ = 180° − 30° − 56.3° ≈ 93.7° α ≈ 56.3° β = 30° γ ≈ 93.7° a = 10 b ≈ 6.013 c = 12 10.7 Find the missing side and angles of the given triangle: α = 30°, b = 12, c = 24. Example 10.8 Solving for an Angle of a SSS Triangle Find the angle α for the given triangle if side a = 20, side b = 25, and side c = 18. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1083 Solution For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α, we have a2 = b2 + c2 −2bccos α 202 = 252 + 182 −2(25)(18)cos α 400 = 625 + 324 − 900cos α 400 = 949 − 900cos α −549 = −900cos α −549 = cos α −900 0.61 ≈ cos α cos−1(0.61) ≈ α α ≈ 52.4° Substitute the appropriate measurements. Simplify in each step. Isolate cos α. Find the inverse cosine. See Figure 10.41. Figure 10.41 Analysis Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method. 10.8 Given a = 5, b = 7, and c = 10, find the missing angles. Solving Applied Problems Using the Law of Cosines Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few. Example 10.9 Using the Law of Cosines to Solve a Communication Problem On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart 1084 Chapter 10 Further Applications of Trigonometry along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway. Solution For simplicity, we start by drawing a diagram similar to Figure 10.42 and labeling our given information. Figure 10.42 Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let a = 2420, b = 5050, and c = 6000. Thus, θ corresponds to the opposite side a = 2420. a2 = b2 + c2 − 2bccos θ (2420)2 = (5050)2 + (6000)2 − 2(5050)(6000)cos θ (2420)2 − (5050)2 − (6000)2 = − 2(5050)(6000)cos θ (2420)2 − (5050)2 − (6000)2 −2(5050)(6000) = cos θ cos θ ≈ 0.9183 θ ≈ cos−1(0.9183) θ ≈ 23.3° To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 10.43. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem. Figure 10.43 Using the angle θ = 23.3° and the basic trigonometric identities, we can find the solutions. Thus This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1085 cos(23.3°) = x 5050 x = 5050cos(23.3°) x ≈ 4638.15 feet sin(23.3°) = y 5050 y = 5050sin(23.3°) y ≈ 1997.5 feet The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway. Example 10.10 Calculating Distance Traveled Using a SAS Triangle Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 10.44. Figure 10.44 Solution turned 20 degrees, so the obtuse angle of the non-right The boat triangle is the supplemental angle, 180° − 20° = 160°. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port. 1086 Chapter 10 Further Applications of Trigonometry x2 = 82 + 102 − 2(8)(10)cos(160°) x2 = 314.35 x = 314.35 x ≈ 17.7 miles The boat is about 17.7 miles from port. Using Heron’s Formula to Find the Area of a Triangle We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known. Heron’s Formula Heron’s formula finds the area of oblique triangles in which sides a, b, and c are known. Area = s(s − a)(s − b)(s − c) (10.5) is one half of the perimeter of the triangle, sometimes called the semi-perimeter. where s = (a + b + c) 2 Example 10.11 Using Heron’s Formula to Find the Area of a Given Triangle Find the area of the triangle in Figure 10.45 using Heron’s formula. Figure 10.45 Solution First, we calculate s. Then we apply the formula. s = (a + b + c) 2 s = (10 + 15 + 7) 2 = 16 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1087 Area = s(s − a)(s − b)(s − c) Area = 16(16 − 10)(16 − 15)(16 − 7) Area ≈ 29.4 The area is approximately 29.4 square units. Use Heron’s formula to find the area of a triangle with sides of lengths a = 29.7 ft, b = 42.3 ft, and 10.9 c = 38.4 ft. Example 10.12 Applying Heron
’s Formula to a Real-World Problem A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 10.46 for a view of the city property. Figure 10.46 Solution Find the measurement for s, which is one-half of the perimeter. (62.4 + 43.5 + 34.1) 2 s = s = 70 m 1088 Chapter 10 Further Applications of Trigonometry Apply Heron’s formula. Area = 70(70 − 62.4)(70 − 43.5)(70 − 34.1) Area = 506,118.2 Area ≈ 711.4 The developer has about 711.4 square meters. 10.10 Find the area of a triangle given a = 4.38 ft , b = 3.79 ft, and c = 5.22 ft. Access these online resources for additional instruction and practice with the Law of Cosines. • Law of Cosines (http://openstaxcollege.org/l/lawcosines) • Law of Cosines: Applications (http://openstaxcollege.org/l/cosineapp) • Law of Cosines: Applications 2 (http://openstaxcollege.org/l/cosineapp2) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1089 10.2 EXERCISES Verbal 97. a = 108, b = 132, c = 160; find angle C. If you are looking for a missing side of a triangle, what 78. do you need to know when using the Law of Cosines? For the following exercises, solve the triangle. Round to the nearest tenth. If you are looking for a missing angle of a triangle, 79. what do you need to know when using the Law of Cosines? 98. A = 35°, b = 8, c = 11 80. Explain what s represents in Heron’s formula. Explain the relationship between the Pythagorean 81. Theorem and the Law of Cosines. When must you use the Law of Cosines instead of the 82. Pythagorean Theorem? Algebraic is opposite side b, and γ For the following exercises, assume α is opposite side a, β is opposite side c. If possible, solve each triangle for the unknown side. Round to the nearest tenth. 83. γ = 41.2°, a = 2.49, b = 3.13 84. α = 120°, b = 6, c = 7 85. β = 58.7°, a = 10.6, c = 15.7 86. γ = 115°, a = 18, b = 23 87. α = 119°, a = 26, b = 14 88. γ = 113°, b = 10, c = 32 89. β = 67°, a = 49, b = 38 90. α = 43.1°, a = 184.2, b = 242.8 91. α = 36.6°, a = 186.2, b = 242.2 92. β = 50°, a = 105, b = 45 For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth. 93. a = 42, b = 19, c = 30; find angle A. 94. a = 14, b = 13, c = 20; find angle C. 95. a = 16, b = 31, c = 20; find angle B. 96. a = 13, b = 22, c = 28; find angle A. 99. B = 88°, a = 4.4, c = 5.2 100. C = 121°, a = 21, b = 37 101. a = 13, b = 11, c = 15 102. a = 3.1, b = 3.5, c = 5 103. a = 51, b = 25, c = 29 For the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth. Find the area of a triangle with sides of length 18 in, 104. 21 in, and 32 in. Round to the nearest tenth. Find the area of a triangle with sides of length 20 cm, 105. 26 cm, and 37 cm. Round to the nearest tenth. 106. a = 1 2 m, b = 1 3 m, c = 1 4 m 107. a = 12.4 ft, b = 13.7 ft, c = 20.2 ft 108. a = 1.6 yd, b = 2.6 yd, c = 4.1 yd Graphical For the following exercises, find the length of side x. Round to the nearest tenth. 109. 110. 111. 1090 Chapter 10 Further Applications of Trigonometry 112. 113. 114. 116. 117. For the following exercises, find the measurement of angle A. 118. 115. This content is available for free at https://cnx.org/content/col11758/1.5 119. Chapter 10 Further Applications of Trigonometry 1091 Find the measure of each angle in the triangle shown in Figure 10.47. Round to the nearest tenth. 124. Figure 10.47 For the following exercises, solve for the unknown side. Round to the nearest tenth. 120. 121. 122. 123. 125. 126. 127. 128. For the following exercises, find the area of the triangle. Round to the nearest hundredth. Chapter 10 Further Applications of Trigonometry 1092 Extensions 129. A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal. 130. The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal. 131. The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal. A regular octagon is inscribed in a circle with a radius 132. of 8 inches. (See Figure 10.48.) Find the perimeter of the octagon. 137. 138. Figure 10.48 Real-World Applications A regular pentagon is inscribed in a circle of radius 12 133. cm. (See Figure 10.49.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter. A surveyor has taken the measurements shown in 139. Figure 10.50. Find the distance across the lake. Round answers to the nearest tenth. Figure 10.49 Figure 10.50 A satellite calculates the distances and angle shown in 140. Figure 10.51 (not to scale). Find the distance between the two cities. Round answers to the nearest tenth. the following that For x2 = 25 + 36 − 60cos(52) represents the relationship of three sides of a triangle and the cosine of an angle. exercises, suppose 134. Draw the triangle. 135. Find the length of the third side. For the following exercises, find the area of the triangle. 136. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1093 Figure 10.51 141. An airplane flies 220 miles with a heading of 40°, and then flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth. A 113-foot tower is located on a hill that is inclined 142. 34° to the horizontal, as shown in Figure 10.52. A guywire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed. Figure 10.52 Two ships left a port at the same time. One ship 143. traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Find the distance between the two ships after 10 hours of travel. 144. The graph in Figure 10.53 represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours. Figure 10.53 A triangular swimming pool measures 40 feet on one 145. side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)? A pilot flies in a straight path for 1 hour 30 min. She 146. then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position? Los Angeles is 1,744 miles from Chicago, Chicago is 147. 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle. Philadelphia is 140 miles from Washington, D.C., 148. Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle. Two planes leave the same airport at the same time. 149. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours? Two airplanes take off in different directions. One 150. travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude? A parallelogram has sides of length 15.4 units and 9.8 151. units. Its area is 72.9 square units. Find the measure of the longer diagonal. The four sequential sides of a quadrilateral have 152. lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral? 1094 Chapter 10 Further Applications of Trigonometry 153. The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral? Find the area of a triangular piece of land that 154. measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot. Find the area of a triangular piece of land that 155. measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1095 10.3 \| Polar Coordinates Learning Objectives In this section, you will: 10.3.1 Plot points using polar coordinates. 10.3.2 Convert from polar coordinates to rectangular coordinates. 10.3.3 Convert from rectangular coordinates to polar coordinates. 10.3.4 Transform equations between polar and rectangular forms. 10.3.5 Identify and graph polar equations by converting to rectangular equations. Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind (see Figure 10.54). How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid. Figure 10.54 Plotting Points Using Polar Coordinates When we think about plotting points in the plane, we usually think of rectangular coordinates (x, y) in the Cartesian coordinate plane. However, there
are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled (r, θ) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane. The polar grid is scaled as the unit circle with the positive x-axis now viewed as the polar axis and the origin as the pole. The first coordinate r is the radius or length of the directed line segment from the pole. The angle θ, measured in radians, indicates the direction of r. We move counterclockwise from the polar axis by an angle of θ, and measure a directed line segment the length of r in the direction of θ. Even though we measure θ first and then r, the polar point is written with the r-coordinate first. For example, to plot the point ⎛ units in the counterclockwise direction and ⎝2, π 4 ⎞ ⎠, we would move π 4 then a length of 2 from the pole. This point is plotted on the grid in Figure 10.55. 1096 Chapter 10 Further Applications of Trigonometry Figure 10.55 Example 10.13 Plotting a Point on the Polar Grid Plot the point ⎛ ⎝3, ⎞ ⎠ on the polar grid. π 2 Solution The angle π 2 is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the π 2 direction, as shown in Figure 10.56. Figure 10.56 10.11 Plot the point ⎛ ⎝2, π 3 ⎞ ⎠ in the polar grid. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1097 Example 10.14 Plotting a Point in the Polar Coordinate System with a Negative Component Plot the point ⎛ ⎝−2, π 6 ⎞ ⎠ on the polar grid. Solution We know that π 6 is located in the first quadrant. However, r = −2. We can approach plotting a point with a negative r in two ways: 1. Plot the point ⎛ ⎝2, π 6 ⎞ ⎠ by moving π 6 in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant; 2. Move π 6 in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant. See Figure 10.57(a). Compare this to the graph of the polar coordinate ⎛ ⎝2, ⎞ ⎠ shown in Figure 10.57(b). π 6 Figure 10.57 10.12 Plot the points ⎛ ⎝3, − ⎞ ⎠ π 6 and ⎛ ⎝2, 9π 4 ⎞ ⎠ on the same polar grid. Converting from Polar Coordinates to Rectangular Coordinates When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables x, y, r, and θ. cos θ = sin θ = x r → x = rcos θ y r → y = rsin θ 1098 Chapter 10 Further Applications of Trigonometry Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 10.58. An easy way to remember the equations above is to think of cos θ as the adjacent side over the hypotenuse and sin θ as the opposite side over the hypotenuse. Figure 10.58 Converting from Polar Coordinates to Rectangular Coordinates To convert polar coordinates (r, θ) to rectangular coordinates (x, y), let cos θ = sin θ = x r → x = rcos θ y r → y = rsin θ Given polar coordinates, convert to rectangular coordinates. 1. Given the polar coordinate (r, θ), write x = rcos θ and y = rsin θ. 2. Evaluate cos θ and sin θ. 3. Multiply cos θ by r to find the x-coordinate of the rectangular form. 4. Multiply sin θ by r to find the y-coordinate of the rectangular form. Example 10.15 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates ⎛ ⎝3, ⎞ ⎠ as rectangular coordinates. π 2 Solution Use the equivalent relationships. x = rcos θ x = 3cos π 2 y = rsin θ y = 3sin π 2 = 0 = 3 The rectangular coordinates are (0, 3). See Figure 10.59. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1099 Figure 10.59 Example 10.16 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates (−2, 0) as rectangular coordinates. Solution See Figure 10.60. Writing the polar coordinates as rectangular, we have x = rcos θ x = −2cos(0) = −2 y = rsin θ y = −2sin(0) = 0 The rectangular coordinates are also (−2, 0). 1100 Chapter 10 Further Applications of Trigonometry Figure 10.60 10.13 Write the polar coordinates ⎛ ⎝−1, 2π 3 ⎞ ⎠ as rectangular coordinates. Converting from Rectangular Coordinates to Polar Coordinates To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point. Converting from Rectangular Coordinates to Polar Coordinates Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 10.61. cos θ = x r or x = rcos θ y r or y = rsin θ sin θ = r 2 = x2 + y2 y x tan θ = (10.6) Figure 10.61 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1101 Example 10.17 Writing Rectangular Coordinates as Polar Coordinates Convert the rectangular coordinates (3, 3) to polar coordinates. Solution We see that the original point (3, 3) is in the first quadrant. To find θ, use the formula tan θ = y x. This gives tan θ = 3 3 tan θ = 1 π tan−1(1) = 4 To find r, we substitute the values for x and y into the formula r = x2 + y2. We know that r must be positive, as π 4 is in the first quadrant. Thus r = 32 + 32 r = 9 + 9 r = 18 = 3 2 So, r = 3 2 and θ= , giving us the polar point ⎛ ⎝3 2, π 4 ⎞ ⎠. See Figure 10.62. π 4 Figure 10.62 Analysis There are other sets of polar coordinates that will be the same as our first solution. For example, the points ⎝−3 2, 5π ⎝−3 2, 5π ⎝3 2, π ⎞ ⎠ will coincide with the original solution of ⎛ ⎛ ⎞ ⎠ 4 4 4 indicates a move further counterclockwise by π, which is directly opposite π . The radius is expressed as 4 ⎝3 2, − 7π 4 ⎠. The point ⎛ ⎞ ⎠ and ⎛ ⎞ − 3 2. However, the angle 5π 4 is located in the third quadrant and, as r is negative, we extend the directed 1102 Chapter 10 Further Applications of Trigonometry line segment in the opposite direction, into the first quadrant. This is the same point as ⎛ ⎝3 2, π 4 ⎞ ⎠. The point ⎝3 2, − 7π ⎛ 4 ⎞ ⎠ is a move further clockwise by − 7π 4 , from π 4 . The radius, 3 2, is the same. Transforming Equations between Polar and Rectangular Forms We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation. Given an equation in polar form, graph it using a graphing calculator. 1. Change the MODE to POL, representing polar form. 2. Press the Y= button to bring up a screen allowing the input of six equations: r1, r2, . . . , r6. 3. Enter the polar equation, set equal to r. 4. Press GRAPH. Example 10.18 Writing a Cartesian Equation in Polar Form Write the Cartesian equation x2 + y2 = 9 in polar form. Solution The goal is to eliminate x and y from the equation and introduce r and θ. Ideally, we would write the equation r as a function of θ. To obtain the polar form, we will use the relationships between (x, y) and (r, θ). Since x = rcos θ and y = rsin θ, we can substitute and solve for r. (rcos θ)2 + (rsin θ)2 = 9 r 2 cos2 θ + r 2 sin2 θ = 9 r 2(cos2 θ + sin2 θ) = 9 r 2(1) = 9 r = ± 3 Substitute cos2 θ + sin2 θ = 1. Use the square root property. Thus, x2 + y2 = 9, r = 3, and r = − 3 should generate the same graph. See Figure 10.63. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1103 Figure 10.63 (a) Cartesian form x2 + y2 = 9 (b) Polar form r = 3 To graph a circle in rectangular form, we must first solve for y. x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x2 Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form Y1 = 9 − x2 and Y2 = − 9 − x2. Press GRAPH. Example 10.19 Rewriting a Cartesian Equation as a Polar Equation Rewrite the Cartesian equation x2 + y2 = 6y as a polar equation. Solution This equation appears similar to the previous example, but it requires different steps to convert the equation. We can still follow the same procedures we have already learned and make the following substitutions: r 2 = 6y r 2 = 6rsin θ r 2 − 6rsin θ = 0 r(r − 6sin θ) = 0 r = 0 or r = 6sin θ Use x2 + y2 = r 2. Substitute y = rsin θ. Set equal to 0. Factor and solve. We reject r = 0, as it only represents one point, (0, 0). 1104 Chapter 10 Further Applications of Trigonometry Therefore, the equations x2 + y2 = 6y and r = 6sin θ should give us the same graph. See Figure 10.64. Figure 10.64 (a) Cartesian form x2 + y2 = 6y (b) polar form r = 6sin θ The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical. Example 10.20 Rewriting a Cartesian Equation in Polar Form Rewrite the Cartesian equation y = 3x + 2 as a polar equation. Solution We will use the relationships x = rcos θ and y = rsin θ. y = 3x + 2 rsin θ = 3rcos θ + 2 rsin θ − 3rcos θ = 2 r(sin θ − 3cos θ) = 2 r = 2 sin θ − 3cos θ Isolate r. Solve for r. 10.14 Rewrite the Cartesian equation y2 = 3 − x2 in polar form. Identify and Graph Polar Equations by Converting to Rectangular Equ
ations We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1105 Example 10.21 Graphing a Polar Equation by Converting to a Rectangular Equation Covert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding graph. Solution The conversion is r = 2sec θ r = 2 rcos θ = 2 x = 2 cos θ Notice that the equation r = 2sec θ drawn on the polar grid is clearly the same as the vertical line x = 2 drawn on the rectangular grid (see Figure 10.65). Just as x = c is the standard form for a vertical line in rectangular form, r = csec θ is the standard form for a vertical line in polar form. Figure 10.65 (a) Polar grid (b) Rectangular coordinate system A similar discussion would demonstrate that the graph of the function r = 2csc θ will be the horizontal line y = 2. In fact, r = ccsc θ is the standard form for a horizontal line in polar form, corresponding to the rectangular form y = c. Example 10.22 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = 3 1 − 2cos θ as a Cartesian equation. Solution 1106 Chapter 10 Further Applications of Trigonometry The goal is to eliminate θ and r, and introduce x and y. We clear the fraction, and then use substitution. In order to replace r with x and y, we must use the expression x2 + y2 = r 2. 3 1 − 2cos θ r = r(1 − 2cos θ − 2x = 3 r⎛ ⎝1 − 2⎛ ⎝ Use cos θ = x r to eliminate θ. r = 3 + 2x r 2 = (3 + 2x)2 x2 + y2 = (3 + 2x)2 Isolate r. Square both sides. Use x2 + y2 = r 2. The Cartesian equation is x2 + y2 = (3 + 2x)2. However, to graph it, especially using a graphing calculator or computer program, we want to isolate y. x2 + y2 = (3 + 2x)2 y2 = (3 + 2x)2 − x2 y = ± (3 + 2x)2 − x2 When our entire equation has been changed from r and θ to x and y, we can stop, unless asked to solve for y or simplify. See Figure 10.66. Figure 10.66 The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1107 In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation. x2 + y2 = (3 + 2x)2 x2 + y2 − (3 + 2x)2 = 0 x2 + y2 − (9 + 12x + 4x2) = 0 x2 + y2 − 9 − 12x − 4x2 = 0 − 3x2 − 12x + y2 = 9 3x2 + 12x − y2 = − 9 3(x2 + 4x + ) − y2 = − 9 3(x2 + 4x + 4) − y2 = − 9 + 12 3(x + 2)2 − y2 = 3 (x + 2)2 − y2 3 = 1 Multiply through by −1. Organize terms to complete the square for x. 10.15 Rewrite the polar equation r = 2sin θ in Cartesian form. Example 10.23 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = sin(2θ) in Cartesian form. Solution r = sin(2θ) r = 2sin θcos θ r = 2 2xy r 2 r 3 = 2xy ⎝ x2 + y2⎞ ⎛ ⎠ = 2xy 3 Use the double angle identity for sine. x Use cos θ = r and sin θ = y r . Simplify. Multiply both sides by r 2. As x2 + y2 = r 2, r = x2 + y2. This equation can also be written as 3 2 ⎝x2 + y2⎞ ⎛ ⎠ = 2xy or x2 + y2 = ⎛ ⎝2xy⎞ ⎠ 2 3 Access these online resources for additional instruction and practice with polar coordinates. • Introduction to Polar Coordinates (http://openstaxcollege.org/l/intropolar) • Comparing Polar and Rectangular Coordinates (http://openstaxcollege.org/l/polarrect) 1108 Chapter 10 Further Applications of Trigonometry 10.3 EXERCISES Verbal How are polar coordinates different from rectangular 156. coordinates? How are the polar axes different from the x- and y- 157. axes of the Cartesian plane? 158. Explain how polar coordinates are graphed. 159. How are the points ⎛ ⎝3, ⎠ and ⎛ ⎞ ⎝−3, π 2 ⎞ ⎠ related? π 2 160. Explain why the points ⎛ ⎝−3, ⎞ ⎠ and ⎛ ⎝3, − π 2 ⎞ ⎠ are π 2 the same. Algebraic the following exercises, convert the given polar For to Cartesian coordinates with r > 0 and coordinates 0 ≤ θ ≤ 2π. Remember to consider the quadrant in which is located when determining θ for the the given point point. 161. ⎛ ⎝7, 7π 6 ⎞ ⎠ 162. (5, π) 163. ⎛ ⎝6, − ⎞ ⎠ π 4 164. ⎛ ⎝−3, ⎞ ⎠ π 6 165. ⎛ ⎝4, 7π 4 ⎞ ⎠ For the following exercises, convert the given Cartesian coordinates to polar coordinates with r > 0, 0 ≤ θ < 2π. Remember to consider the quadrant in which the given point is located. 166. (4, 2) 167. (−4, 6) 168. (3, −5) 169. (−10, −13) 170. (8, 8) For the following exercises, convert the given Cartesian equation to a polar equation. This content is available for free at https://cnx.org/content/col11758/1.5 171. x = 3 172. y = 4 173. y = 4x2 174. y = 2x4 175. x2 + y2 = 4y 176. x2 + y2 = 3x 177. x2 − y2 = x 178. x2 − y2 = 3y 179. x2 + y2 = 9 180. x2 = 9y 181. y2 = 9x 182. 9xy = 1 the following exercises, convert For the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented. 183. r = 3sin θ 184. r = 4cos θ 185. 186. r = 4 sin θ + 7cos θ r = 6 cos θ + 3sin θ 187. r = 2sec θ 188. r = 3csc θ 189. r = rcos θ + 2 190. r 2 = 4sec θ csc θ 191. r = 4 192. r 2 = 4 193. Chapter 10 Further Applications of Trigonometry 1109 r = 1 4cos θ − 3sin θ 194. r = 3 cos θ − 5sin θ Graphical For the following exercises, find the polar coordinates of the point. 195. 196. 197. 198. 199. For the following exercises, plot the points. 200. ⎛ ⎝−2, ⎞ ⎠ π 3 201. ⎛ ⎝−1, − ⎞ ⎠ π 2 202. ⎛ ⎝3.5, 7π 4 ⎞ ⎠ 203. ⎛ ⎝−4, ⎞ ⎠ π 3 204. ⎛ ⎝5, ⎞ ⎠ π 2 205. ⎛ ⎝4, −5π 4 ⎞ ⎠ 206. ⎛ ⎝3, 5π 6 ⎞ ⎠ 207. ⎛ ⎝−1.5, 7π 6 ⎞ ⎠ 208. ⎛ ⎝−2, ⎞ ⎠ π 4 1110 209. ⎛ ⎝1, 3π 2 ⎞ ⎠ For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis. 210. 5x − y = 6 211. 2x + 7y = − 3 Chapter 10 Further Applications of Trigonometry Use a graphing calculator to find the polar 228. coordinates of (−2, 0) in radians. Round to the nearest hundredth. Extensions 229. Describe the graph of r = asec θ; a > 0. 230. Describe the graph of r = asec θ; a < 0. x2 + ⎛ ⎝y − 1⎞ ⎠ 2 = 1 231. Describe the graph of r = acsc θ; a > 0. 2 = 13 232. Describe the graph of r = acsc θ; a < 0. 233. What polar equations will give an oblique line? For the following exercise, graph the polar inequality. 212. 213. (x + 2)2 + ⎛ ⎝y + 3⎞ ⎠ 214. x = 2 215. x2 + y2 = 5y 216. x2 + y2 = 3x 234. r < 4 235. 0 ≤ θ ≤ π 4 236. θ = 2373 238 239. − For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane. 217. r = 6 218. r = − 4 219. θ = − 2π 3 220. θ = π 4 221. r = sec θ 222. r = −10sin θ 223. r = 3cos θ Technology 224. coordinates of ⎛ Use a graphing calculator to find the rectangular ⎞ ⎠. Round to the nearest thousandth. ⎝2, − π 5 225. coordinates of ⎛ Use a graphing calculator to find the rectangular ⎞ ⎠. Round to the nearest thousandth. ⎝−3, 3π 7 Use a graphing calculator to find the polar 226. coordinates of (−7, 8) in degrees. Round to the nearest thousandth. Use a graphing calculator to find the polar 227. coordinates of (3, − 4) in degrees. Round to the nearest hundredth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1111 10.4 \| Polar Coordinates: Graphs Learning Objectives In this section you will: 10.4.1 Test polar equations for symmetry. 10.4.2 Graph polar equations by plotting points. The planets move through space in elliptical, periodic orbits about the sun, as shown in Figure 10.67. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as (r, θ). We interpret r as the distance from the sun and θ as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. Figure 10.67 Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA/JPL-Caltech) Testing Polar Equations for Symmetry Just as a rectangular equation such as y = x2 describes the relationship between x and y on a Cartesian grid, a polar equation describes a relationship between r and θ on a polar grid. Recall that the coordinate pair (r, θ) indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of θ, and extend a ray from the pole (origin) r units in the direction of θ. All points that satisfy the polar equation are on the graph. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of r) to determine the graph of a polar equation. In the first test, we consider symmetry with respect to the line θ = (y-axis). We replace (r, θ) with (−r, − θ) to π 2 determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation r = 2sin θ; 1112 Chapter 10 Further Applications of Trigonometry r = 2sin θ −r = 2sin(−θ) −r = −2sin θ r = 2sin θ Replace (r, θ) with (−r, −θ). Identity: sin(−θ
) = −sin θ. Multiply both sides by−1. This equation exhibits symmetry with respect to the line θ = π 2 . In the second test, we consider symmetry with respect to the polar axis ( x -axis). We replace (r, θ) with (r, − θ) or (−r, π − θ) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation r = 1 − 2cos θ. r = 1 − 2cos θ r = 1 − 2cos(−θ) r = 1 − 2cos θ Replace (r, θ) with (r, −θ). Even/Odd identity The graph of this equation exhibits symmetry with respect to the polar axis. In the third test, we consider symmetry with respect to the pole (origin). We replace (r, θ) with (−r, θ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation r = 2sin(3θ). r = 2sin(3θ) −r = 2sin(3θ) The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests , the polar axis, or the pole. In these does not necessarily indicate that a graph will not be symmetric about the line θ = π 2 instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. Symmetry Tests A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 10.68. Figure 10.68 (a) A graph is symmetric with respect to the line θ = (y-axis) if replacing (r, θ) with ( − r, − θ) π 2 yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing (r, θ) with (r, − θ) or (−r, π−θ) yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing (r, θ) with (−r, θ) yields an equivalent equation. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1113 Given a polar equation, test for symmetry. 1. Substitute the appropriate combination of components for (r, θ) : (−r, − θ) for θ = (r, − θ) for polar axis symmetry; and (−r, θ) for symmetry with respect to the pole. symmetry; π 2 2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry. Example 10.24 Testing a Polar Equation for Symmetry Test the equation r = 2sin θ for symmetry. Solution Test for each of the three types of symmetry. 1) Replacing (r, θ) with ( − r, − θ) yields the same result. Thus, the graph is symmetric with respect to the line θ = . π 2 2) Replacing θ with − θ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. −r = 2sin(−θ) −r = −2sin θ Even-odd identity r = 2sin θ Multiply by −1 Passed r = 2sin(−θ) r = −2sin θ r = −2sin θ ≠ 2sin θ Failed Even-odd identity 3) Replacing r with – r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. −r = 2sin θ r = −2sin θ ≠ 2sin θ Failed Table 10.1 Analysis Using a graphing calculator, we can see that the equation r = 2sin θ is a circle centered at (0, 1) with radius . We can also see that the graph is not symmetric with the polar r = 1 and is indeed symmetric to the line θ = π 2 axis or the pole. See Figure 10.69. 1114 Chapter 10 Further Applications of Trigonometry Figure 10.69 10.16 Test the equation for symmetry: r = − 2cos θ. Graphing Polar Equations by Plotting Points To graph in the rectangular coordinate system we construct a table of x and y values. To graph in the polar coordinate system we construct a table of θ and r values. We enter values of θ into a polar equation and calculate r. However, using the properties of symmetry and finding key values of θ and r means fewer calculations will be needed. Finding Zeros and Maxima To find the zeros of a polar equation, we solve for the values of θ that result in r = 0. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for x. We use the same process for polar equations. Set r = 0, and solve for θ. For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of θ into the equation that result in the maximum value of the trigonometric functions. Consider r = 5cos θ; the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when θ = 0, so our polar equation is 5cos θ, and the value θ = 0 will yield the maximum \|r\|. Similarly, the maximum value of the sine function is 1 when θ = , and if our polar equation is r = 5sin θ, the value π 2 will yield the maximum \|r\|. We may find additional information by calculating values of r when θ = 0. These θ = π 2 points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation. Example 10.25 Finding Zeros and Maximum Values for a Polar Equation Using the equation in Example 10.24, find the zeros and maximum \|r\| and, if necessary, the polar axis intercepts of r = 2sin θ. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1115 Solution To find the zeros, set r equal to zero and solve for θ. 2sin θ = 0 sin θ = 0 θ = sin−1 0 θ = nπ where n is an integer Substitute any one of the θ values into the equation. We will use 0. r = 2sin(0) r = 0 The points (0, 0) and (0, ± nπ) are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept. To find the maximum value of the equation, look at the maximum value of the trigonometric function sin θ, which occurs when θ = π 2 ⎛ ± 2kπ resulting in sin ⎝ π 2 ⎞ ⎠ = 1. Substitute π 2 for θ. ⎞ ⎠ π 2 ⎛ r = 2sin ⎝ r = 2(1) r = 2 Analysis The point ⎛ ⎝2, π 2 ⎞ ⎠ will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See Table 10.1 and Figure 10.70. 1116 Chapter 10 Further Applications of Trigonometry θ 0 π 6 π 3 π 2 2π 3 5π 6 π r = 2sin θ r = 2sin(0) = 0 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 π 6 ⎛ r = 2sin ⎝ π 3 ⎞ ⎠ ≈ 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 2 π 2 r 0 1 1.73 2 ⎛ r = 2sin ⎝ 2π 3 ⎞ ⎠ ≈ 1.73 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 5π 6 r = 2sin(π) = 0 1 0 Table 10.1 Figure 10.70 Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros 10.17 and maximum values of \|r\| : r = 3cos θ. Investigating Circles Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1117 There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations. Formulas for the Equation of a Circle Some of the formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ, where a is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is \|a\| ⎠. Figure 2 , or one-half the diameter. For r = acos θ, the center is ⎛ ⎝ ⎞ ⎠. For r = asin θ, , 0 the center is ⎛ ⎝ , π⎞ a 2 a 2 10.71 shows the graphs of these four circles. Figure 10.71 Example 10.26 Sketching the Graph of a Polar Equation for a Circle Sketch the graph of r = 4cos θ. Solution First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the ± kπ. zeros and maximum \|r\| for r = 4cos θ. First, set r = 0, and solve for θ . Thus, a zero occurs at θ = π 2 A key point to plot is ⎛ ⎝0, ⎞ ⎠ . π 2 To find the maximum value of r, note that the maximum value of the cosine function is 1 when θ = 0 ± 2kπ. Substitute θ = 0 into the equation: r = 4cos θ r = 4cos(0) r = 4(1) = 4 The maximum value of the equation is 4. A key point to plot is (4, 0). As r = 4cos θ is symmetric with respect to the polar axis, we only need to calculate r-values for θ over the interval [0, π]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to Table 10.2. The graph is shown in Figure 10.72. 1118 Chapter 10 Further Applications of Trigonometry π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π 3.46 2.83 2 0 −2 −2.83 −3.46 4 θ r 0 4 Table 10.2 Figure 10.72 Investigating Cardioids While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own. Formulas for a Cardioid The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a b = 1. The cardioid graph passes through the pole, as we can see in Figure 10.73. Figure 10.73 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1119 Given the polar equation of a cardioid, sketch its graph. 1. Check equation for the three types of symmetry. 2. Find the zeros. Set r = 0. 3. Find the maximum value of the equation according to the maximum value of the trigonometric expression. 4. Make a table of values for r and θ. 5. Plot the points
and sketch the graph. Example 10.27 Sketching the Graph of a Cardioid Sketch the graph of r = 2 + 2cos θ. Solution First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting r = 0, we have θ = π + 2kπ. The zero of the equation is located at (0, π). The graph passes through this point. The maximum value of r = 2 + 2cos θ occurs when cos θ is a maximum, which is when cos θ = 1 or when θ = 0. Substitute θ = 0 into the equation, and solve for r. r = 2 + 2cos(0) r = 2 + 2(1) = 4 The point (4, 0) is the maximum value on the graph. We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [0, π]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in Table 10.3, and then we plot the points and draw the graph. See Figure 10.74. θ r 0 π 4 π 2 2π 3 4 3.41 2 1 π 0 Table 10.3 1120 Chapter 10 Further Applications of Trigonometry Figure 10.74 Investigating Limaçons The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled b ≥ 2. limaçons when 1 < b < 2 and convex limaçons when a a Formulas for One-Loop Limaçons The formulas that produce the graph of a dimpled one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ a b < 2. All four graphs are shown in Figure 10.75. where a > 0, b > 0, and 1< Figure 10.75 Dimpled limaçons This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1121 Given a polar equation for a one-loop limaçon, sketch the graph. 1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted. 2. Find the zeros. 3. Find the maximum values according to the trigonometric expression. 4. Make a table. 5. Plot the points and sketch the graph. Example 10.28 Sketching the Graph of a One-Loop Limaçon Graph the equation r = 4 − 3sin θ. Solution First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a , yet it fails all the three symmetry tests. A graph that clearly displays symmetry with respect to the line θ = π 2 graphing calculator will immediately illustrate the graph’s reflective quality. Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting r = 0 results in θ being undefined. What does this mean? How could θ be undefined? The angle θ is undefined for any value of sin θ > 1. Therefore, θ is undefined because there is no value of θ for which sin θ > 1. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating r when θ = 0. r(0) = 4 − 3sin(0 So, there is at least one polar axis intercept at (4, 0). Next, as the maximum value of the sine function is 1 when θ = , we will substitute θ = π 2 into the equation π 2 and solve for r. Thus, r = 1. Make a table of the coordinates similar to Table 10.4 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4 Table 10.4 The graph is shown in Figure 10.76. 1122 Chapter 10 Further Applications of Trigonometry Figure 10.76 One-loop limaçon Analysis This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving sin θ is likely symmetric with respect to the line θ = , evaluating more points helps to verify that the graph is correct. π 2 10.18 Sketch the graph of r = 3 − 2cos θ. Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it. Formulas for Inner-Loop Limaçons The formulas that generate the inner-loop limaçons are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a < b. The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 10.77 for the graphs. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1123 Figure 10.77 Example 10.29 Sketching the Graph of an Inner-Loop Limaçon Sketch the graph of r = 2 + 5cos θ. Solution Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when r = 0, θ = 1.98. The maximum \|r\| is found when cos θ = 1 or when θ = 0. Thus, the maximum is found at the point (7, 0). Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge. See Table 10.5 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7 Table 10.5 As expected, the values begin to repeat after θ = π. The graph is shown in Figure 10.78. 1124 Chapter 10 Further Applications of Trigonometry Figure 10.78 Inner-loop limaçon Investigating Lemniscates The lemniscate is a polar curve resembling the infinity symbol ∞ or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition. Formulas for Lemniscates The formulas that generate the graph of a lemniscate are given by r 2 = a2 cos 2θ and r 2 = a2 sin 2θ where a ≠ 0. The formula r 2 = a2 sin 2θ is symmetric with respect to the pole. The formula r 2 = a2 cos 2θ is symmetric with respect to the pole, the line θ = , and the polar axis. See Figure 10.79 for the graphs. π 2 Figure 10.79 Example 10.30 Sketching the Graph of a Lemniscate This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1125 Sketch the graph of r 2 = 4cos 2θ. Solution The equation exhibits symmetry with respect to the line θ = , the polar axis, and the pole. π 2 Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution u = 2θ. 0 = 4cos 2θ 0 = 4cos u 0 = cos u cos− = 2θ = θ = Substitute 2θ back in for u. So, the point ⎛ ⎝0, ⎞ ⎠ π 4 is a zero of the equation. Now let’s find the maximum value. Since the maximum of cos u = 1 when u = 0, the maximum cos 2θ = 1 when 2θ = 0. Thus, r 2 = 4cos(0) r 2 = 4(1) = 4 r = ± 4 = 2 We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line θ = , and the π 2 polar axis, we only need to plot points in the first quadrant. Make a table similar to Table 10.6 Table 10.6 Plot the points on the graph, such as the one shown in Figure 10.80. Figure 10.80 Lemniscate 1126 Chapter 10 Further Applications of Trigonometry Analysis Making a substitution such as u = 2θ is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown. Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of r. This is because there are no real square roots for these values of θ. In other words, the corresponding r-values of 4cos(2θ) are complex numbers because there is a negative number under the radical. Investigating Rose Curves The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern. Rose Curves The formulas that generate the graph of a rose curve are given by r = acos nθ and r = asin nθ where a ≠ 0. If n is even, the curve has 2n petals. If n is odd, the curve has n petals. See Figure 10.81. Figure 10.81 Example 10.31 Sketching the Graph of a Rose Curve (n Even) Sketch the graph of r = 2cos 4θ. Solution Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line θ = and the pole. π 2 Now we will find the zeros. First make the substitution u = 4θ. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1127 0 = 2cos 4θ 0 = cos 4θ 0 = cos u cos− = 4θ = The zero is θ = . The point ⎛ ⎝0, π 8 π 8 ⎞ ⎠ is on the curve. Next, we find the maximum \|r\|. We know that the maximum value of cos u = 1 when θ = 0. Thus, r = 2cos(4 ⋅ 0) r = 2cos(0) r = 2(1) = 2 The point (2, 0) is on the curve. The graph of the rose curve has unique properties, which are revealed in Table 10.7. π 8 π 4 3π 8 π 2 5π 8 3π 4 0 −2 0 2 0 −2 θ r 0 2 Table 10.7 As r = 0 when θ = π 8 , it makes sense to divide values in the table by π 8 units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at r = 0, each petal extends out a distance of r = 2, and then turns back to zero 2n times for a total of eight petals. See the graph in Figure 10.82. Figure 10.82 Rose curve, n even Analysis 1128 Chapter 10 Further Applications of Trigonometry When thes
e curves are drawn, it is best to plot the points in order, as in the Table 10.7. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn. 10.19 Sketch the graph of r = 4sin(2θ). Example 10.32 Sketching the Graph of a Rose Curve (n Odd) Sketch the graph of r = 2sin(5θ). Solution The graph of the equation shows symmetry with respect to the line θ = . Next, find the zeros and maximum. π 2 We will want to make the substitution u = 5θ. 0 = 2sin(5θ) 0 = sin u sin−1 0 = 0 u = 0 5θ = 0 θ = 0 The maximum value is calculated at the angle where sin θ is a maximum. Therefore, ⎛ r = 2sin ⎝5 ⋅ ⎞ ⎠ π 2 r = 2(1) = 2 Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for n odd yields the same number of petals as n, there will be five petals on the graph. See Figure 10.83. Create a table of values similar to Table 10.8. π 6 π 3 π 2 2π 3 5π 6 1 −1.73 2 −1.73 1 π 0 θ r 0 0 Table 10.8 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1129 Figure 10.83 Rose curve, n odd 10.20 Sketch the graph of r = 3cos(3θ). Investigating the Archimedes’ Spiral The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics. Archimedes’ Spiral The formula that generates the graph of the Archimedes’ spiral is given by r = θ for θ ≥ 0. As θ increases, r increases at a constant rate in an ever-widening, never-ending, spiraling path. See Figure 10.84. Figure 10.84 1130 Chapter 10 Further Applications of Trigonometry Given an Archimedes’ spiral over [0, 2π], sketch the graph. 1. Make a table of values for r and θ over the given domain. 2. Plot the points and sketch the graph. Example 10.33 Sketching the Graph of an Archimedes’ Spiral Sketch the graph of r = θ over [0, 2π]. Solution As r is equal to θ, the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted. Create a table such as Table 10.9. θ r π 4 π 2 π 3π 2 7π 4 2π 0.785 1.57 3.14 4.71 5.50 6.28 Table 10.9 Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 10.85. Figure 10.85 Archimedes’ spiral Analysis The domain of this polar curve is [0, 2π]. In general, however, the domain of this function is (−∞, ∞). Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1131 10.21 Sketch the graph of r = − θ over the interval [0, 4π]. Summary of Curves We have explored a number of seemingly complex polar curves in this section. Figure 10.86 and Figure 10.87 summarize the graphs and equations for each of these curves. Figure 10.86 Figure 10.87 Access these online resources for additional instruction and practice with graphs of polar coordinates. • Graphing Polar Equations Part 1 (http://openstaxcollege.org/l/polargraph1) • Graphing Polar Equations Part 2 (http://openstaxcollege.org/l/polargraph2) • Animation: The Graphs of Polar Equations (http://openstaxcollege.org/l/polaranim) • Graphing Polar Equations on the TI-84 (http://openstaxcollege.org/l/polarTI84) 1132 Chapter 10 Further Applications of Trigonometry 10.4 EXERCISES Verbal 240. Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane. 241. Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the xaxis, y-axis, and origin? What are the steps to follow when graphing polar 242. equations? Describe the shapes of the graphs of cardioids, 243. limaçons, and lemniscates. What part of the equation determines the shape of the 244. graph of a polar equation? Graphical 261. r = 3 + 2sin θ 262. r = 7 + 4sin θ 263. r = 4 + 3cos θ 264. r = 5 + 4cos θ 265. r = 10 + 9cos θ 266. r = 1 + 3sin θ 267. r = 2 + 5sin θ 268. r = 5 + 7sin θ 269. r = 2 + 4cos θ For the following exercises, test the equation for symmetry. 270. r = 5 + 6cos θ 245. r = 5cos 3θ 246. r = 3 − 3cos θ 247. r = 3 + 2sin θ 248. r = 3sin 2θ 249. r = 4 250. r = 2θ 251. r = 4cos θ 2 252. r = 2 θ 253. r = 3 1 − cos2 θ 254. r = 5sin 2θ For the following exercises, graph the polar equation. Identify the name of the shape. 255. r = 3cos θ 256. r = 4sin θ 257. r = 2 + 2cos θ 258. r = 2 − 2cos θ 259. r = 5 − 5sin θ 260. r = 3 + 3sin θ This content is available for free at https://cnx.org/content/col11758/1.5 271. r 2 = 36cos(2θ) 272. r 2 = 10cos(2θ) 273. r 2 = 4sin(2θ) 274. r 2 = 10sin(2θ) 275. r = 3sin(2θ) 276. r = 3cos(2θ) 277. r = 5sin(3θ) 278. r = 4sin(4θ) 279. r = 4sin(5θ) 280. r = −θ 281. r = 2θ 282. r = − 3θ Technology For the following exercises, use a graphing calculator to sketch the graph of the polar equation. 283. r = 1 θ 284. r = 1 θ Chapter 10 Further Applications of Trigonometry 1133 285. r = 2sin θtan θ, a cissoid 286. r = 2 1 − sin2 θ , a hippopede On a graphing utility, graph each polar equation. 302. Explain the similarities and differences you observe in the graphs. r1 = 3θ r2 = 2θ r3 = θ Extensions For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. 303. r1 = 3 + 2sin θ, r2 = 2 304. r1 = 6 − 4cos θ, r2 = 4 305. r1 = 1 + sin θ, r2 = 3sin θ 306. r1 = 1 + cos θ, r2 = 3cos θ 307. r1 = cos(2θ), r2 = sin(2θ) 308. 309. 310. r1 = sin2 (2θ), r2 = 1 − cos(4θ) r1 = 3, r2 = 2sin(θ) r1 2 = sin θ, r2 2 = cos θ 311. r1 = 1 + cos θ, r2 = 1 − sin θ 287. r = 5 + cos(4θ) 288. r = 2 − sin(2θ) 289. r = θ 2 290. r = θ + 1 291. r = θsin θ 292. r = θcos θ For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of [0, 4π] and then explain the differences shown in the graphs. 293. r = θ, r = − θ 294. r = θ, r = θ + sin θ 295. r = sin θ + θ, r = sin θ − θ 296. 297. 298. ⎛ r = 2sin ⎝ θ 2 ⎞ ⎛ ⎠, r = θsin ⎝ ⎞ ⎠ θ 2 r = sin⎛ ⎝cos(3θ)⎞ ⎠ r = sin(3θ) On a graphing utility, [0, 4π], [0, 8π], [0, 12π], and ⎡ effect of increasing the width of the domain. ⎣0, 16π⎤ ⎛ graph r = sin ⎝ 16 5 ⎦. Describe θ⎞ ⎠ on the 299. On a r = sin θ + graphing 3 ⎛ ⎛ ⎝sin ⎝ ⎞ θ⎞ ⎠ ⎠ 5 2 on [0, 4π]. utility, graph and sketch 300. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. r1 = 3sin(3θ) r2 = 2sin(3θ) r3 = sin(3θ) 301. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. r1 = 3 + 3cos θ r2 = 2 + 2cos θ r3 = 1 + cos θ 1134 Chapter 10 Further Applications of Trigonometry 10.5 \| Polar Form of Complex Numbers Learning Objectives In this section, you will: 10.5.1 Plot complex numbers in the complex plane. 10.5.2 Find the absolute value of a complex number. 10.5.3 Write complex numbers in polar form. 10.5.4 Convert a complex number from polar to rectangular form. 10.5.5 Find products of complex numbers in polar form. 10.5.6 Find quotients of complex numbers in polar form. 10.5.7 Find powers of complex numbers in polar form. 10.5.8 Find roots of complex numbers in polar form. “God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science. We first encountered complex numbers in Complex Numbers. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem. Plotting Complex Numbers in the Complex Plane Plotting a complex number a + bi is similar to plotting a real number, except that the horizontal axis represents the real part of the number, a, and the vertical axis represents the imaginary part of the number, bi. Given a complex number a + bi, plot it in the complex plane. 1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis. 2. Plot the point in the complex plane by moving a units in the horizontal direction and b units in the vertical direction. Example 10.34 Plotting a Complex Number in the Complex Plane Plot the complex number 2 − 3i in the complex plane. Solution From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See Figure 10.88. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1135 Figure 10.88 10.22 Plot the point 1 + 5i in the complex plane. Finding the Absolute Value of a Complex Number The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or \|z\|. It measures the distance from the origin to a point in the plane. For example, the graph of z = 2 + 4i, in Figure 10.89, shows \|z\|. Figure 10.89 Absolute Value of a Complex Number Given z = x + yi, a complex number, the absolute value of z is defined as It is the distance from the origin to the point (x, y). \|z\| = x2 + y2 1136 Chapt
er 10 Further Applications of Trigonometry Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, (0, 0). Example 10.35 Finding the Absolute Value of a Complex Number with a Radical Find the absolute value of z = 5 − i. Solution Using the formula, we have See Figure 10.90. 2 + (−1)2 \|z\| = x2 + y2 \|z\| = ⎛ ⎝ 5⎞ \|z\| = 5 + 1 \|z\| = 6 ⎠ Figure 10.90 10.23 Find the absolute value of the complex number z = 12 − 5i. Example 10.36 Finding the Absolute Value of a Complex Number Given z = 3 − 4i, find \|z\|. Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1137 \|z\| = x2 + y2 \|z\| = (3)2 + (−4)2 \|z\| = 9 + 16 \|z\| = 25 \|z\| = 5 The absolute value z is 5. See Figure 10.91. Figure 10.91 10.24 Given z = 1 − 7i, find \|z\|. Writing Complex Numbers in Polar Form The polar form of a complex number expresses a number in terms of an angle θ and its distance from the origin r. Given a complex number in rectangular form expressed as z = x + yi, we use the same conversion formulas as we do to write the number in trigonometric form: We review these relationships in Figure 10.92. x = rcos θ y = rsin θ r = x2 + y2 1138 Chapter 10 Further Applications of Trigonometry Figure 10.92 We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point (x, y). The modulus, then, is the same as r, the radius in polar form. We use θ to indicate the angle of direction (just as with polar coordinates). Substituting, we have z = x + yi z = rcos θ + (rsin θ)i z = r(cos θ + isin θ) Polar Form of a Complex Number Writing a complex number in polar form involves the following conversion formulas: Making a direct substitution, we have x = rcos θ y = rsin θ r = x2 + y2 z = x + yi z = (rcos θ) + i(rsin θ) z = r(cos θ + isin θ) where r is the modulus and θ is the argument. We often use the abbreviation rcis θ to represent r(cos θ + isin θ). Example 10.37 Expressing a Complex Number Using Polar Coordinates Express the complex number 4i using polar coordinates. Solution On the complex plane, the number z = 4i is the same as z = 0 + 4i. Writing it in polar form, we have to calculate r first. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1139 r = x2 + y2 r = 02 + 42 r = 16 r = 4 Next, we look at x. If x = rcos θ, and x = 0, then θ = the complex number z = 0 + 4i can be written as z = 4 ⎛ ⎛ ⎝cos ⎝ π 2 ⎞ ⎛ ⎠ + isin ⎝ π 2 π . In polar coordinates, 2 ⎝ π ⎞ ⎞ ⎛ ⎠ or 4cis ⎠ 2 ⎞ ⎠. See Figure 10.93. Figure 10.93 10.25 Express z = 3i as r cis θ in polar form. Example 10.38 Finding the Polar Form of a Complex Number Find the polar form of − 4 + 4i. Solution First, find the value of r. r = x2 + y2 r = (−4)2 + r = 32 r = 4 2 ⎛ ⎝42⎞ ⎠ Find the angle θ using the formula: 1140 Chapter 10 Further Applications of Trigonometry x r cos θ = cos θ = −4 4 2 cos θ = − 1 2 θ = cos−1 ⎛ ⎝− 1 2 ⎞ ⎠ = 3π 4 ⎛ Thus, the solution is 4 2cis ⎝ ⎞ ⎠. 3π 4 10.26 Write z = 3 + i in polar form. Converting a Complex Number from Polar to Rectangular Form Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given z = r(cos θ + isin θ), first evaluate the trigonometric functions cos θ and sin θ. Then, multiply through by r. Example 10.39 Converting from Polar to Rectangular Form Convert the polar form of the given complex number to rectangular form: z = 12 ⎛ ⎛ ⎝cos ⎝ π 6 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠ ⎠ π 6 Solution We begin by evaluating the trigonometric expressions. ⎛ cos ⎝ π 6 ⎞ ⎠ = 3 2 ⎛ and sin ⎝ π 6 ⎞ ⎠ = 1 2 After substitution, the complex number is z = 12 ⎛ ⎝ 3 2 + 1 2 i⎞ ⎠ We apply the distributive property: z = 12 ⎛ ⎝ 3 2 = (12) 3 2 i⎞ ⎠ + 1 2 + (12)1 2 i The rectangular form of the given point in complex form is 6 3 + 6i. = 6 3 + 6i This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1141 Example 10.40 Finding the Rectangular Form of a Complex Number Find the rectangular form of the complex number given r = 13 and tan θ = 5 12 . Solution If tan θ = 5 , and tan θ = 12 y r . and sin θ = y x, we first determine r = x2 + y2 = 122 + 52 = 13. We then find cos θ = x r z = 13(cos θ + isin θ) 12 = 13 13 = 12 + 5i + 5 13 i⎞ ⎠ ⎛ ⎝ The rectangular form of the given number in complex form is 12 + 5i. 10.27 Convert the complex number to rectangular form: ⎛ ⎝cos11π 6 z = 4 + isin11π 6 ⎞ ⎠ Finding Products of Complex Numbers in Polar Form Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments. Products of Complex Numbers in Polar Form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the product of these numbers is given as: ⎡ ⎣cos⎛ z1 z2 = r1 r2 z1 z2 = r1 r2 cis⎛ ⎝θ1 + θ2 ⎝θ1 + θ2 ⎞ ⎠ ⎞ ⎠ + isin⎛ ⎝θ1 + θ2 ⎤ ⎞ ⎦ ⎠ Notice that the product calls for multiplying the moduli and adding the angles. Example 10.41 Finding the Product of Two Complex Numbers in Polar Form Find the product of z1 z2, given z1 = 4(cos(80°) + isin(80°)) and z2 = 2(cos(145°) + isin(145°)). Solution Follow the formula 1142 Chapter 10 Further Applications of Trigonometry z1 z2 = 4 ⋅ 2[cos(80° + 145°) + isin(80° + 145°)] z1 z2 = 8[cos(225°) + isin(225°)] ⎡ 5π ⎞ ⎛ ⎛ z1 z2 = 8 ⎠ + isin ⎣cos ⎝ ⎝ 4 ⎡ ⎤ ⎞ + i⎛ ⎣− 2 ⎝− 2 ⎦ ⎠ 2 2 z1 z2 = − 4 2 − 4i 2 z1 z2 = 8 5π 4 ⎤ ⎞ ⎦ ⎠ Finding Quotients of Complex Numbers in Polar Form The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments. Quotients of Complex Numbers in Polar Form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the quotient of these numbers is z1 z2 z1 z2 = = r1 r2 r1 r2 ⎡ ⎣cos⎛ ⎝θ1 − θ2 ⎞ ⎠ + isin⎛ ⎝θ1 − θ2 ⎤ ⎞ ⎠ ⎦, z2 ≠ 0 cis⎛ ⎝θ1 − θ2 ⎞ ⎠, z2 ≠ 0 Notice that the moduli are divided, and the angles are subtracted. Given two complex numbers in polar form, find the quotient. 1. Divide r1 r2 . 2. Find θ1 − θ2. 3. Substitute the results into the formula: z = r(cos θ + isin θ). Replace r with r1 r2 , and replace θ with θ1 − θ2. 4. Calculate the new trigonometric expressions and multiply through by r. Example 10.42 Finding the Quotient of Two Complex Numbers Find the quotient of z1 = 2(cos(213°) + isin(213°)) and z2 = 4(cos(33°) + isin(33°)). Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1143 [cos(213° − 33°) + isin(213° − 33°)] [cos(180°) + isin(180°)] z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 = + 0i] + 0i 10.28 Find the product and the quotient of z1 = 2 3(cos(150°) + isin(150°)) and z2 = 2(cos(30°) + isin(30°)). Finding Powers of Complex Numbers in Polar Form Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the argument by n. It is the standard method used in modern mathematics. De Moivre’s Theorem If z = r(cos θ + isin θ) is a complex number, then zn zn where n is a positive integer. Example 10.43 = r n ⎡ = r n ⎣cos(nθ) + isin(nθ)⎤ cis(nθ) ⎦ Evaluating an Expression Using De Moivre’s Theorem Evaluate the expression (1 + i)5 using De Moivre’s Theorem. Solution Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write (1 + i) in polar form. Let us find r. r = x2 + y2 r = (1)2 + (1)2 r = 2 Then we find θ. Using the formula tan θ = y x gives 1144 Chapter 10 Further Applications of Trigonometry tan θ = 1 1 tan θ = 1 π θ = 4 Use De Moivre’s Theorem to evaluate the expression. n = r n (a + bi) (1 + i)5 = ( 2)5 ⎡ [cos(nθ) + isin(nθ)] π ⎞ ⎛ ⎛ ⎠ + isin ⎣cos ⎝5 ⋅ ⎝5 ⋅ 4 ⎤ ⎡ 5π 5π ⎞ ⎞ ⎛ ⎛ (1 + i)5 = 4 2 ⎠ + isin ⎣cos ⎦ ⎠ ⎝ ⎝ 4 4 ⎡ ⎤ ⎞ + i⎛ ⎣− 2 ⎝− 2 (1 + i)5 = 4 2 ⎦ ⎠ 2 2 (1 + i)5 = − 4 − 4i ⎤ ⎞ ⎦ ⎠ π 4 Finding Roots of Complex Numbers in Polar Form To find the nth root of a complex number in polar form, we use the nth Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding nth roots of complex numbers in polar form. The nth Root Theorem To find the nth root of a complex number in polar form, use the formula given as 1 n z 1 n ⎡ ⎛ ⎣cos ⎝ = r n + 2kπ θ n ⎞ ⎛ ⎠ + isin ⎝ n + 2kπ θ n ⎤ ⎞ ⎦ ⎠ where k = 0, 1, 2, 3, . . . , n − 1. We add 2kπ n to θ n in order to obtain the periodic roots. Example 10.44 Finding the nth Root of a Complex Number Evaluate the cube roots of z = 8 ⎛ ⎛ ⎝cos ⎝ 2π 3 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠. ⎠ 2π 3 Solution We have cos ⎜ ⎣ ⎝ 2π 3 3 + 2kπ 3 ⎛ ⎞ ⎜ ⎟ + isin ⎝ ⎠ 2π 3 3 + 2kπ cos ⎝ 2π 9 + 2kπ 3 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 2kπ 3 ⎤ ⎞ ⎦ ⎠ There will be three roots: k = 0, 1, 2. When k = 0, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1145 1 3 = 2 z ⎛ ⎛ ⎝cos ⎝ 2π 9 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠ ⎠ 2π 9 When k = 1, we have cos ⎝ 2π 9 + 6π 9 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 6π 9 ⎤ ⎞ ⎦ ⎠ Add 2(1)π 3 to each angle. ⎛ ⎛ ⎝cos ⎝ 8π 9 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠ ⎠ 8π 9 When k = 2, we have cos ⎝ 2π 9 + 12π 9 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 12π 9 ⎤ ⎞ ⎦ ⎠ Add 2(2)π 3 to each angle. ⎛ ⎛ ⎝cos ⎝ 14π 9 ⎛ ⎞ ⎠ + isin ⎝ 14π 9 ⎞ ⎞ ⎠ ⎠ Remember to find the common denominator to simplify fractions in situations like this one. For k = 1, the angle simplification is 2(1)π 3 ⎞ ⎠ ⎛ ⎝ 3 3 + 2π 2(1)π
= 2π 3 3 3 3 = 2π 9 = 8π 9 ⎞ ⎛ 1 ⎠ + ⎝ 3 + 6π 9 10.29 Find the four fourth roots of 16(cos(120°) + isin(120°)). Access these online resources for additional instruction and practice with polar forms of complex numbers. • The Product and Quotient of Complex Numbers in Trigonometric Form (http://openstaxcollege.org/l/prodquocomplex) • De Moivre’s Theorem (http://openstaxcollege.org/l/demoivre) 1146 Chapter 10 Further Applications of Trigonometry 10.5 EXERCISES Verbal 312. A complex number is a + bi. Explain each part. 332. z = 3cis(240°) 333. z = 2cis(100°) What does the absolute value of a complex number 313. represent? For the following exercises, find z1 z2 in polar form. 314. How is a complex number converted to polar form? How do we find the product of 315. numbers? two complex 316. What is De Moivre’s Theorem and what is it used for? Algebraic For the following exercises, find the absolute value of the given complex number. 317. 5 + 3i 318. −7 + i 319. −3 − 3i 320. 2 − 6i 321. 2i 322. 2.2 − 3.1i 334. 335. 336. 337. 338. 339. z1 = 2 3cis(116°); z2 = 2cis(82°) z1 = 2cis(205°); z2 = 2 2cis(118°) z1 = 3cis(120°); z2 = 1 4 cis(60°) ⎛ z1 = 3cis ⎝ π 4 ⎛ ⎞ ⎠; z2 = 5cis ⎝ ⎞ ⎠ π 6 ⎛ z1 = 5cis ⎝ 5π 8 ⎞ ⎛ ⎠; z2 = 15cis ⎝ ⎞ ⎠ π 12 ⎛ z1 = 4cis ⎝ π 2 ⎞ ⎛ ⎠; z2 = 2cis ⎝ ⎞ ⎠ π 4 For the following exercises, find z1 z2 in polar form. 340. z1 = 21cis(135°); z2 = 3cis(65°) 341. z1 = 2cis(90°); z2 = 2cis(60°) For the following exercises, write the complex number in polar form. 342. z1 = 15cis(120°); z2 = 3cis(40°) 323. 2 + 2i 324. 8 − 4i 325. − 1 2 − 1 2 i 326. 3 + i 327. 3i For the following exercises, convert the complex number from polar to rectangular form. 328. ⎛ z = 7cis ⎝ 329. ⎛ z = 2cis ⎝ ⎞ ⎠ π 6 ⎞ ⎠ π 3 330. ⎛ z = 4cis ⎝ ⎞ ⎠ 7π 6 331. z = 7cis(25°) This content is available for free at https://cnx.org/content/col11758/1.5 343. ⎛ z1 = 6cis ⎝ π 3 ⎞ ⎛ ⎠; z2 = 2cis ⎝ ⎞ ⎠ π 4 344. ⎛ z1 = 5 2cis(π); z2 = 2cis ⎝ ⎞ ⎠ 2π 3 345. ⎛ z1 = 2cis ⎝ 3π 5 ⎞ ⎛ ⎠; z2 = 3cis ⎝ ⎞ ⎠ π 4 For the following exercises, find the powers of each complex number in polar form. 346. 347. 348. 349. Find z3 when z = 5cis(45°). Find z4 when z = 2cis(70°). Find z2 when z = 3cis(120°). ⎛ Find z2 when z = 4cis ⎝ ⎞ ⎠. π 4 350. ⎛ Find z4 when z = cis ⎝ ⎞ ⎠. 3π 16 Chapter 10 Further Applications of Trigonometry 1147 351. ⎛ Find z3 when z = 3cis ⎝ ⎞ ⎠. 5π 3 Use the polar to rectangular feature on the graphing 371. calculator to change 2cis(45°) to rectangular form. For the following exercises, evaluate each root. 352. Evaluate the cube root of z when z = 27cis(240°). Use the polar to rectangular feature on the graphing 372. calculator to change 5cis(210°) to rectangular form. 353. Evaluate z = 16cis(100°). the square root of z when 354. ⎛ Evaluate the cube root of z when z = 32cis ⎝ ⎞ ⎠. 2π 3 355. Evaluate the square root of z when z = 32cis(π). 356. ⎛ Evaluate the cube root of z when z = 8cis ⎝ ⎞ ⎠. 7π 4 Graphical For the following exercises, plot the complex number in the complex plane. 357. 2 + 4i 358. −3 − 3i 359. 5 − 4i 360. −1 − 5i 361. 3 + 2i 362. 2i 363. −4 364. 6 − 2i 365. −2 + i 366. 1 − 4i Technology For the following exercises, find all answers rounded to the nearest hundredth. Use the rectangular to polar feature on the graphing 367. calculator to change 5 + 5i to polar form. Use the rectangular to polar feature on the graphing 368. calculator to change 3 − 2i to polar form. Use the rectangular to polar feature on the graphing 369. calculator to change −3 − 8i to polar form. Use the polar to rectangular feature on the graphing 370. calculator to change 4cis(120°) to rectangular form. 1148 Chapter 10 Further Applications of Trigonometry 10.6 \| Parametric Equations Learning Objectives In this section, you will: 10.6.1 Parameterize a curve. 10.6.2 Eliminate the parameter. 10.6.3 Find a rectangular equation for a curve defined parametrically. 10.6.4 Find parametric equations for curves defined by rectangular equations. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 10.94. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time. Figure 10.94 In this section, we will consider sets of equations given by x(t) and y(t) where t is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of t, the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations. Parameterizing a Curve When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both x and y vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both x and y are dependent functions. In the example in the section opener, the parameter is time, t. The x position of the moon at time, t, is represented as the function x(t), and the y position of the moon at time, t, is represented as the function y(t). Together, x(t) and y(t) are called parametric equations, and generate an ordered pair ⎛ ⎠. Parametric equations primarily describe motion and direction. ⎝x(t), y(t)⎞ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1149 When we parameterize a curve, we are translating a single equation in two variables, such as x and y , into an equivalent pair of equations in three variables, x, y, and t. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time. When we graph parametric equations, we can observe the individual behaviors of x and of y. There are a number of shapes that cannot be represented in the form y = f (x), meaning that they are not functions. For example, consider the graph of a circle, given as r 2 = x2 + y2. Solving for y gives y = ± r 2 − x2, or two equations: y1 = r 2 − x2 and y2 = − r 2 − x2. If we graph y1 and y2 together, the graph will not pass the vertical line test, as shown in Figure 10.95. Thus, the equation for the graph of a circle is not a function. Figure 10.95 However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward. Parametric Equations Suppose t is a number on an interval, I. The set of ordered pairs, ⎛ a plane curve based on the parameter t. The equations x = f (t) and y = g(t) are the parametric equations. ⎠, where x = f (t) and y = g(t), forms ⎝x(t), y(t)⎞ Example 10.45 Parameterizing a Curve Parameterize the curve y = x2 − 1 letting x(t) = t. Graph both equations. Solution If x(t) = t, then to find y(t) we replace the variable x with the expression given in x(t). In other words, y(t) = t 2 − 1. Make a table of values similar to Table 10.10, and sketch the graph. 1150 Chapter 10 Further Applications of Trigonometry t x(t) y(t) −4 −4 y(−4) = (−4)2 − 1 = 15 −3 −3 y(−3) = (−3)2 − 1 = 8 −2 −2 y(−2) = (−2)2 − 1 = 3 −1 −1 y(−1) = (−1)(0) = (0)2 − 1 = − 1 y(1) = (1)2 − 1 = 0 y(2) = (2)2 − 1 = 3 y(3) = (3)2 − 1 = 8 y(4) = (4)2 − 1 = 15 Table 10.10 See the graphs in Figure 10.96. It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as t increases. Figure 10.96 (a) Parametric y(t) = t 2 − 1 (b) Rectangular y = x2 − 1 Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1151 The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of y = x2 − 1. Construct 10.30 x(t) = t − 3, y(t) = 2t + 4; − 1 ≤ t ≤ 2. table of a values and plot the parametric equations: Example 10.46 Finding a Pair of Parametric Equations Find a pair of parametric equations that models the graph of y = 1 − x2, using the parameter x(t) = t. Plot some points and sketch the graph. Solution If x(t) = t and we substitute t for x into the y equation, then y(t) = 1 − t 2. Our pair of parametric equations is x(t) = t y(t) = 1 − t 2 To graph the equations, first we construct a table of values like that in Table 10.11. We can choose values around t = 0, from t = − 3 to t = 3. The values in the x(t) column will be the same as those in the t column because x(t) = t. Calculate values for the column y(t). 1152 Chapter 10 Further Applications of Trigonometry t x(t) = t y(t) = 1 − t 2 −3 −3 y(−3) = 1 − (−3)2 = − 8 −2 −2 y(−2) = 1 − (−2)2 = − 3 −1 −1 y(−1) = 1 − (−1)(0) = 1 − 0 = 1 y(1) = 1 − (1)2 = 0 y(2) = 1 − (2)2 = − 3 y(3) = 1 − (3)2 = − 8 Table 10.11 The graph of y = 1 − t 2 is a parabola facing downward, as shown in Figure 10.97. We have mapped the curve over the interval [−3, 3], shown as a solid line with arrows indicating the orientation of the curve according to t. O
rientation refers to the path traced along the curve in terms of increasing values of t. As this parabola is symmetric with respect to the line x = 0, the values of x are reflected across the y-axis. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1153 Figure 10.97 10.31 Parameterize the curve given by x = y3 − 2y. Example 10.47 Finding Parametric Equations That Model Given Criteria An object travels at a steady rate along a straight path (−5, 3) to (3, −1) in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object. Solution The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x-value of the object starts at −5 meters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of 8 m , or 2 m / s. We can write the x-coordinate as a linear function 4 s with respect to time as x(t) = 2t − 5. In the linear function template y = mx + b, 2t = mx and − 5 = b. Similarly, the y-value of the object starts at 3 and goes to −1, which is a change in the distance y of −4 meters in 4 seconds, which is a rate of −4 m , or − 1m / s. We can also write the y-coordinate as the linear function 4 s y(t) = − t + 3. Together, these are the parametric equations for the position of the object, where x and y are expressed in meters and t represents time: 1154 Chapter 10 Further Applications of Trigonometry x(t) = 2t − 5 y(t) = − t + 3 Using these equations, we can build a table of values for t, x, and y (see Table 10.12). In this example, we limited values of t to non-negative numbers. In general, any value of t can be used. t 0 1 2 3 4 x(t) = 2t − 5 y(t) = − t + 3 x = 2(00) + 3 = 3 x = 2(11) + 3 = 2 x = 2(22) + 3 = 1 x = 2(3) − 5 = 1 y = − (3) + 3 = 0 x = 2(4) − 5 = 3 y = − (4) + 3 = − 1 Table 10.12 From this table, we can create three graphs, as shown in Figure 10.98. Figure 10.98 (a) A graph of x vs. t, representing the horizontal position over time. (b) A graph of y vs. t, representing the vertical position over time. (c) A graph of y vs. x, representing the position of the object in the plane at time t. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1155 Again, we see that, in Figure 10.98(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows. Eliminating the Parameter In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x and y. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter t from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations. Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for t. We substitute the resulting expression for t into the second equation. This gives one equation in x and y. Example 10.48 Eliminating the Parameter in Polynomials Given x(t) = t 2 + 1 and y(t) = 2 + t, eliminate the parameter, and write the parametric equations as a Cartesian equation. Solution We will begin with the equation for y because the linear equation is easier to solve for t. Next, substitute y − 2 for t in x(t). y − 2)2 + 1 x = y2 − 4y + 4 + 1 x = y2 − 4y + 5 x = y2 − 4y + 5 The Cartesian form is x = y2 − 4y + 5. Substitute the expression for t into x. Analysis This is an equation for a parabola in which, in rectangular terms, x is dependent on y. From the curve’s vertex at (1, 2), the graph sweeps out to the right. See Figure 10.99. In this section, we consider sets of equations given by the functions x(t) and y(t), where t is the independent variable of time. Notice, both x and y are functions of time; so in general y is not a function of x. 1156 Chapter 10 Further Applications of Trigonometry Figure 10.99 10.32 Given the equations below, eliminate the parameter and write as a rectangular equation for y as a function of x. Example 10.49 x(t) = 2t 2 + 6 y(t) = 5 − t Eliminating the Parameter in Exponential Equations Eliminate the parameter and write as a Cartesian equation: x(t) = e−t and y(t) = 3et , t > 0. Solution Isolate et . Substitute the expression into y(t). The Cartesian form is y = 3 x. Analysis x = e−t et = 1 x y = 3et ⎛ ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1157 The graph of the parametric equation is shown in Figure 10.100(a). The domain is restricted to t > 0. The Cartesian equation, y = 3 x is shown in Figure 10.100(b) and has only one restriction on the domain, x ≠ 0. Figure 10.100 Example 10.50 Eliminating the Parameter in Logarithmic Equations Eliminate the parameter and write as a Cartesian equation: x(t) = t + 2 and y(t) = log(t). Solution Solve the first equation for tx − 2)2 = t Square both sides. Then, substitute the expression for t into the y equation. y = log(t) y = log(x − 2)2 The Cartesian form is y = log(x − 2)2. Analysis 1158 Chapter 10 Further Applications of Trigonometry To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on x = t + 2 to t > 0; we restrict the domain on x to x > 2. The domain for the parametric equation y = log(t) is restricted to t > 0; we limit the domain on y = log(x − 2)2 to x > 2. 10.33 Eliminate the parameter and write as a rectangular equation. x(t) = t 2 y(t) = ln t t > 0 Eliminating the Parameter from Trigonometric Equations Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. First, we use the identities: Solving for cos t and sin t, we have Then, use the Pythagorean Theorem: Substituting gives Example 10.51 x(t) = acos t y(t) = bsin t x a = cos t y b = sin t cos2 t + sin2 t = 1 cos2 t + sin2 Eliminating the Parameter from a Pair of Trigonometric Parametric Equations Eliminate the parameter from the given pair of trigonometric equations where 0 ≤ t ≤ 2π and sketch the graph. x(t) = 4cos t y(t) = 3sin t Solution Solving for cos t and sin t, we have x = 4cos t x = cos t 4 y = 3sin t y = sin t 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1159 Next, use the Pythagorean identity and make the substitutions. cos2 t + sin2 y2 9 + = 1 The graph for the equation is shown in Figure 10.101. x2 16 Figure 10.101 Analysis Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify x2 16 t = y2 9 the coordinates are (0, 3). This shows the orientation of the curve with increasing values of t. = 1 as an ellipse centered at (0, 0). Notice that when t = 0 the coordinates are (4, 0), and when + π 2 Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: 10.34 x(t) = 2cos t and y(t) = 3sin t. Finding Cartesian Equations from Curves Defined Parametrically When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as x(t) = t. In this case, y(t) can be any expression. For example, consider the following pair of equations. x(t) = t y(t) = t 2 − 3 Rewriting this set of parametric equations is a matter of substituting x for t. Thus, the Cartesian equation is y = x2 − 3. Example 10.52 1160 Chapter 10 Further Applications of Trigonometry Finding a Cartesian Equation Using Alternate Methods Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. x(t) = 3t − 2 y(t) = t + 1 Solution Method 1. First, let’s solve the x equation for t. Then we can substitute the result into the y equation. x = 3t − 2 x + 2 = 3t x + 2 3 = t Now substitute the expression for t into the y equation Method 2. Solve the y equation for t and substitute this expression in the x equation. Make the substitution and then solve for y(y − 1) − 2 x = 3y − 3 − 2 x = 3y − 5 x + 5 = 3y 10.35 Write the given parametric equations as a Cartesian equation: x(t) = t 3 and y(t) = t 6. Finding Parametric Equations for Curves Defined by Rectangular Equations Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent x, and then substitute it into the y equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for x as the domain of the rectangular equation, then the graphs will be different. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1161 Example 10.53 Finding a Set of Par
ametric Equations for Curves Defined by Rectangular Equations Find a set of equivalent parametric equations for y = (x + 3)2 + 1. Solution An obvious choice would be to let x(t) = t. Then y(t) = (t + 3)2 + 1. But let’s try something more interesting. What if we let x = t + 3 ? Then we have y = (x + 3)2 + 1 y = ((t + 3) + 3)2 + 1 y = (t + 6)2 + 1 x(t) = t + 3 y(t) = (t + 6)2 + 1 The set of parametric equations is See Figure 10.102. Figure 10.102 Access these online resources for additional instruction and practice with parametric equations. • Introduction to Parametric Equations (http://openstaxcollege.org/l/introparametric) • Converting Parametric Equations to Rectangular Form (http://openstaxcollege.org/l/ convertpara) 1162 Chapter 10 Further Applications of Trigonometry 10.6 EXERCISES Verbal 373. What is a system of parametric equations? 389. x(t) = e2t ⎧ ⎨ y(t) = e6t ⎩ 374. Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter. 390. ⎧ x(t) = t 5 ⎨ y(t) = t 10 ⎩ Explain how to eliminate a parameter given a set of 375. parametric equations. What is a benefit of writing a system of parametric 376. equations as a Cartesian equation? 377. What is a benefit of using parametric equations? Why are there many sets of parametric equations to 378. represent on Cartesian function? Algebraic For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 379. x(t) = 5 − t ⎧ ⎨ y(t) = 8 − 2t ⎩ 380. x(t) = 6 − 3t ⎧ ⎨ y(t) = 10 − t ⎩ 381. x(t) = 2t + 1 ⎧ ⎨ y(t) = 3 t ⎩ 382. x(t) = 3t − 1 ⎧ ⎨ y(t) = 2t 2 ⎩ 383. x(t) = 2et ⎧ ⎨ y(t) = 1 − 5t ⎩ 384. x(t) = e−2t ⎧ ⎨ y(t) = 2e−t ⎩ 385. x(t) = 4log(t) ⎧ ⎨ y(t) = 3 + 2t ⎩ 386. x(t) = log(2t) ⎧ ⎨ y(t) = t − 1 ⎩ 387. 388. ⎧ x(t) = t 3 − t ⎨ y(t) = 2t ⎩ ⎧ x(t) = t − t 4 ⎨ y(t) = t + 2 ⎩ This content is available for free at https://cnx.org/content/col11758/1.5 391. x(t) = 4cos t ⎧ ⎨ y(t) = 5sin t ⎩ 392. x(t) = 3sin t ⎧ ⎨ y(t) = 6cos t ⎩ 393. ⎧ x(t) = 2cos2 t ⎨ y(t) = − sin t ⎩ 394. x(t) = cos t + 4 ⎧ ⎨ y(t) = 2sin2 t ⎩ 395. x(t) = t − 1 ⎧ ⎨ y(t) = t 2 ⎩ 396. x(t) = − t ⎧ ⎨ y(t) = t 3 + 1 ⎩ 397. x(t) = 2t − 1 ⎧ ⎨ y(t) = t 3 − 2 ⎩ For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x-y table. 398. x(t) = 2t − 1 ⎧ ⎨ y(t) = t + 4 ⎩ 399. x(t) = 4 − t ⎧ ⎨ y(t) = 3t + 2 ⎩ 400. x(t) = 2t − 1 ⎧ ⎨ y(t) = 5t ⎩ 401. x(t) = 4t − 1 ⎧ ⎨ y(t) = 4t + 2 ⎩ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting x(t) = t or by setting y(t) = t. 402. y(x) = 3x2 + 3 Chapter 10 Further Applications of Trigonometry 1163 403. y(x) = 2sin x + 1 404. x(y) = 3log(y) + y 405. x(y) = y + 2y For the following exercises, parameterize (write parametric equations using by each Cartesian x(t) = acos t and y(t) = bsin t. Identify the curve. equation for) 406. 407. x2 4 + y2 9 = 1 x2 16 + y2 36 = 1 408. x2 + y2 = 16 409. x2 + y2 = 10 Parameterize the line from (3, 0) to (−2, −5) so the line is at (3, 0) at t = 0, and at (−2, −5) at 410. that t = 1. Parameterize the line from (−1, 0) to (3, −2) so the line is at (−1, 0) at t = 0, and at (3, −2) at 411. that t = 1. Parameterize the line from (−1, 5) to (2, 3) so that 412. the line is at (−1, 5) at t = 0, and at (2, 3) at t = 1. Parameterize the line from (4, 1) to (6, −2) so that 413. the line is at (4, 1) at t = 0, and at (6, −2) at t = 1. Technology For the following exercises, use the table feature in the the graphs graphing calculator intersect. to determine whether 414. 415. x1(t) = 3t ⎧ ⎨ y1(t) = 2t − 1 ⎩ and x2(t) = t + 3 ⎧ ⎨ y2(t) = 4t − 4 ⎩ ⎧ x1(t) = t 2 ⎨ y1(t) = 2t − 1 ⎩ and x2(t) = − t + 6 ⎧ ⎨ y2(t) = t + 1 ⎩ For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 416. ⎧ x1 (t) = 3t 2 − 3t + 7 ⎨ y1 (t) = 2t + 3 ⎩ x y t –1 0 1 417. ⎧ x1 (t) = t 2 − 4 ⎨ y1 (t) = 2t 418. ⎧ x1 (t) = t 4 ⎨ y1 (t1 0 1 2 Extensions Find two different sets of parametric equations for 419. y = (x + 1)2. Find two different sets of parametric equations for 420. y = 3x − 2. Find two different sets of parametric equations for 421. y = x2 − 4x + 4. 1164 Chapter 10 Further Applications of Trigonometry 10.7 \| Parametric Equations: Graphs Learning Objectives In this section you will: 10.7.1 Graph plane curves described by parametric equations by plotting points. 10.7.2 Graph parametric equations. It is the bottom of the ninth inning, with two outs and two men on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and at an angle of approximately 45° to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations. In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems. Figure 10.103 Parametric equations can model the path of a projectile. (credit: Paul Kreher, Flickr) Graphing Parametric Equations by Plotting Points In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable. Given a pair of parametric equations, sketch a graph by plotting points. 1. Construct a table with three columns: t, x(t), and y(t). 2. Evaluate x and y for values of t over the interval for which the functions are defined. 3. Plot the resulting pairs (x, y). Example 10.54 Sketching the Graph of a Pair of Parametric Equations by Plotting Points Sketch the graph of the parametric equations x(t) = t 2 + 1, y(t) = 2 + t. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1165 Construct a table of values for t, x(t), and y(t), as in Table 10.13, and plot the points in a plane. t −5 −4 −3 −2 −1 0 1 2 3 4 5 x(t) = t 2 + 1 y(t) = 2 + t 26 17 10 5 2 1 2 5 10 17 26 −3 −2 −1 0 1 2 3 4 5 6 7 Table 10.13 The graph is a parabola with vertex at the point (1, 2), opening to the right. See Figure 10.104. 1166 Chapter 10 Further Applications of Trigonometry Figure 10.104 Analysis As values for t progress in a positive direction from 0 to 5, the plotted points trace out the top half of the parabola. As values of t become negative, they trace out the lower half of the parabola. There are no restrictions on the domain. The arrows indicate direction according to increasing values of t. The graph does not represent a function, as it will fail the vertical line test. The graph is drawn in two parts: the positive values for t, and the negative values for t. 10.36 Sketch the graph of the parametric equations x = t, y = 2t + 3, 0 ≤ t ≤ 3. Example 10.55 Sketching the Graph of Trigonometric Parametric Equations Construct a table of values for the given parametric equations and sketch the graph: x = 2cos t y = 4sin t Solution Construct a table like that in Table 10.14 using angle measure in radians as inputs for t, and evaluating x and y. Using angles with known sine and cosine values for t makes calculations easier. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1167 t x = 2cos t y = 4sin t 0 π 6 π 3 π 2 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π x = 2cos(0) = 2 y = 4sin(0) = 0 ⎛ x = 2cos ⎝ ⎞ ⎠ = 3 π 6 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 π 6 ⎛ x = 2cos ⎝ ⎞ ⎠ = 1 π 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 3 π 3 ⎛ x = 2cos ⎝ ⎞ ⎠ = 0 π 2 ⎛ y = 4sin ⎝ ⎞ ⎠ = 4 π 2 ⎛ x = 2cos ⎝ ⎞ ⎠ = − 1 2π 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 3 2π 3 ⎛ x = 2cos ⎝ 5π 6 ⎞ ⎠ = − 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 5π 6 x = 2cos(π) = − 2 y = 4sin(π) = 0 ⎛ x = 2cos ⎝ 7π 6 ⎞ ⎠ = − 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = − 2 7π 6 ⎛ x = 2cos ⎝ ⎞ ⎠ = − 1 4π 3 ⎛ y = 4sin ⎝ 4π 3 ⎞ ⎠ = − 2 3 ⎛ x = 2cos ⎝ ⎞ ⎠ = 0 3π 2 ⎛ y = 4sin ⎝ ⎞ ⎠ = − 4 3π 2 ⎛ x = 2cos ⎝ ⎞ ⎠ = 1 5π 3 ⎛ y = 4sin ⎝ 5π 3 ⎞ ⎠ = − 2 3 ⎛ x = 2cos ⎝ 11π 6 ⎞ ⎠ = 3 ⎛ y = 4sin ⎝ 11π 6 ⎞ ⎠ = − 2 x = 2cos(2π) = 2 y = 4sin(2π) = 0 Table 10.14 Figure 10.105 shows the graph. 1168 Chapter 10 Further Applications of Trigonometry Figure 10.105 By the symmetry shown in the values of x and y, we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing t values. Analysis We have seen that parametric equations can be graphed by plotting points. However, a graphing calculator will save some time and reveal nuances in a graph that may be too tedious to discover using only hand calculations. Make sure to change the mode on the calculator to parametric (PAR). To confirm, the Y = window should show X1T = Y1T = instead of Y1 = . 10.37 Graph the parametric equations: x = 5cos t, y = 3sin t. Example 10.56 Graphing Parametric Equations and Rectangular Form Together Graph the parametric equations x = 5cos t and y = 2sin t. First, construct the graph using data points generated from the parametric form. Then graph the rectangular form of the equation. Compare the two graphs. Solution Construct a table of values like that in Table 10.15. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1169 t x = 5cos t y = 2sin t 0 1 2 3 4 5 −1 −2 −3 −4 −5 x = 5cos(0) = 5 y = 2sin(0) = 0 x = 5cos(1) ≈ 2.7 y = 2sin(1) ≈ 1.7 x = 5cos(2) ≈ −2.1 y = 2sin(2) ≈ 1.8 x = 5cos(3) ≈ −4.95 y = 2sin(3) ≈ 0.28 x = 5cos(4) ≈ −3.3 y = 2sin(4) ≈ −1.5 x = 5cos(5) ≈ 1.4 y = 2sin(5) ≈ −1.9 x = 5cos(−1) ≈ 2.7 y = 2sin(−1) ≈ −1.7 x = 5cos(−2) ≈ −2.1 y = 2sin(−2) ≈ −1.8 x = 5cos(−3) ≈ −4.95 y = 2sin(−3) ≈ −0.28 x = 5cos(−4) ≈ −3.3 y = 2sin(−4) ≈ 1.5 x = 5cos(−5) ≈ 1.4 y = 2sin(−5) ≈ 1.9 Table 10.15 Plot the (x, y) values from the table. See Figure 10
.106. 1170 Chapter 10 Further Applications of Trigonometry Figure 10.106 Next, translate the parametric equations to rectangular form. To do this, we solve for t in either x(t) or y(t), and then substitute the expression for t in the other equation. The result will be a function y(x) if solving for t as a function of x, or x(y) if solving for t as a function of y. x = 5cos t x = cos t 5 y = 2sin t y = sin t 2 Solve for cos t. Solve for sin t. Then, use the Pythagorean Theorem. cos2 t + sin2 y2 4 x2 25 + = 1 Analysis In Figure 10.107, the data from the parametric equations and the rectangular equation are plotted together. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Clearly, both forms produce the same graph. Figure 10.107 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1171 Example 10.57 Graphing Parametric Equations and Rectangular Equations on the Coordinate System Graph the parametric equations x = t + 1 and y = t, t ≥ 0, and the rectangular equivalent y = x − 1 on the same coordinate system. Solution Construct a table of values for the parametric equations, as we did in the previous example, and graph y = t, t ≥ 0 on the same grid, as in Figure 10.108. Figure 10.108 Analysis With the domain on t restricted, we only plot positive values of t. The parametric data is graphed in blue and the graph of the rectangular equation is dashed in red. Once again, we see that the two forms overlap. 10.38 Sketch the graph of the parametric equations x = 2cos θ and y = 4sin θ, along with the rectangular equation on the same grid. Applications of Parametric Equations Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in x and y give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of x and y change depending on t, as the location of a moving object at a particular time. A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of θ to the horizontal, with an initial speed of v0, and at a height h above the horizontal. The path of an object propelled at an inclination of θ to the horizontal, with initial speed v0, and at a height h above the horizontal, is given by x = (v0 cosθ)t y = − 1 2 gt 2 + (v0 sinθ)t + h 1172 Chapter 10 Further Applications of Trigonometry where g accounts for the effects of gravity and h is the initial height of the object. Depending on the units involved in the problem, use g = 32 ft / s2 or g = 9.8 m / s2. The equation for x gives horizontal distance, and the equation for y gives the vertical distance. Given a projectile motion problem, use parametric equations to solve. 1. The horizontal distance is given by x = ⎛ ⎝v0 cos θ⎞ ⎠t. Substitute the initial speed of the object for v0. 2. The expression cos θ indicates the angle at which the object is propelled. Substitute that angle in degrees for cos θ. 3. The vertical distance is given by the formula y = − 1 2 the effect of gravity. Depending on units involved, use g = 32 ft/s2 or g = 9.8 m/s2. Again, substitute the initial speed for v0, and the height at which the object was propelled for h. ⎠t + h. The term − 1 2 gt 2 represents ⎝v0 sin θ⎞ gt 2 + ⎛ 4. Proceed by calculating each term to solve for t. Example 10.58 Finding the Parametric Equations to Describe the Motion of a Baseball Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of 45° to the horizontal, making contact 3 feet above the ground. a. Find the parametric equations to model the path of the baseball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? d. Is it a home run? Solution a. Use the formulas to set up the equations. The horizontal position is found using the parametric equation for x. Thus, The vertical position is found using the parametric equation for y. Thus, x = (v0 cos θ)t x = (140cos(45°))t b. Substitute 2 into the equations to find the horizontal and vertical positions of the ball. y = − 16t 2 + (v0 sin θ)t + h y = − 16t 2 + (140sin(45°))t + 3 x = (140cos(45°))(2) x = 198 feet y = − 16(2)2 + (140sin(45°))(2) + 3 y = 137 feet After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1173 c. To calculate how long the ball is in the air, we have to find out when it will hit ground, or when y = 0. Thus, ⎝140sin(45∘)⎞ ⎠t + 3 y = − 16t 2 + ⎛ y = 0 t = 6.2173 Set y(t) = 0 and solve the quadratic. When t = 6.2173 seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.) d. We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for t, and input t into y. ⎠t ⎝140cos(45°)⎞ ⎠t ⎝140cos(45°)⎞ x = ⎛ 400 = ⎛ t = 4.04 y = − 16(4.04)2 + ⎛ y = 141.8 ⎝140sin(45°)⎞ ⎠(4.04) + 3 The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See Figure 10.109. Figure 10.109 Access the following online resource for additional instruction and practice with graphs of parametric equations. • Graphing Parametric Equations on the TI-84 (http://openstaxcollege.org/l/graphpara84) 1174 Chapter 10 Further Applications of Trigonometry 10.7 EXERCISES Verbal What are two methods used to graph parametric 422. equations? What is one difference in point-plotting parametric 423. equations compared to Cartesian equations? 424. Why are some graphs drawn with arrows? Name a few common types of graphs of parametric 425. equations. t x y −3 −2 −1 0 1 2 Why are 426. understanding projectile motion? parametric graphs important in Graphical For the following exercises, graph each set of parametric equations by making a table of values. Include the orientation on the graph. 427. x(t) = t ⎧ ⎨ y(t 429. x(t) = 2 + t ⎧ ⎨ y(t) = 3 − 2t ⎩ −2 −1 0 1 2 3 t x y −3 −2 −1 0 1 2 3 428. x(t) = t − 1 ⎧ ⎨ y(t) = t 2 ⎩ 430. x(t) = − 2 − 2t ⎧ ⎨ y(t) = 3 + t ⎩ −3 −2 −1 0 1 t x y 431. ⎧ x(t) = t 3 ⎨ y(t) = t + 2 ⎩ −2 −1 0 1 2 t x y This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1175 432. ⎧ x(t) = t 2 ⎨ y(t) = t + 3 ⎩ −2 −1 0 1 2 t x y For the following exercises, sketch the curve and include the orientation. 444. x(t) = t − 1 ⎧ ⎨ y(t) = − t 2 ⎩ 445. ⎧ x(t) = t 3 ⎨ y(t) = t + 3 ⎩ 446. x(t) = 2cos t ⎧ ⎨ y(t) = − sin t ⎩ 447. x(t) = 7cos t ⎧ ⎨ y(t) = 7sin t ⎩ 448. x(t) = e2t ⎧ ⎨ y(t) = − e t ⎩ 433. x(t) = t ⎧ ⎨ y(t) = t ⎩ 434. x(t) = − t ⎧ ⎨ y(t) = t ⎩ 435. x(t) = 5 − \|t\| ⎧ ⎨ y(t) = t + 2 ⎩ 436. x(t) = − t + 2 ⎧ ⎨ y(t) = 5 − \|t\| ⎩ 437. x(t) = 4sin t ⎧ ⎨ y(t) = 2cos t ⎩ 438. x(t) = 2sin t ⎧ ⎨ y(t) = 4cos t ⎩ 439. 440. ⎧ x(t) = 3cos2 t ⎨ y(t) = −3sin t ⎩ ⎧ x(t) = 3cos2 t ⎨ y(t) = −3sin2 t ⎩ 441. x(t) = sec t ⎧ ⎨ y(t) = tan t ⎩ 442. x(t) = sec t ⎧ ⎨ y(t) = tan2 t ⎩ 443. ⎧ x(t) = 1 e2t ⎨ y(t) = e− t ⎩ For the following exercises, graph the equation and include the orientation. Then, write the Cartesian equation. For the following exercises, graph the equation and include the orientation. 449. 450. 451. x = t 2, y = 3t, 0 ≤ t ≤ 5 x = 2t, y = t 2, y = 25 − t 2, 0 < t ≤ 5 452. x(t) = − t, y(t) = t, t ≥ 0 453. x = − 2cos t, y = 6 sin t, 0 ≤ t ≤ π 454. x = − sec t, y = tan t, − π 2 < t < π 2 For the following exercises, use the parametric equations for integers a and b: x(t) = acos((a + b)t) y(t) = acos((a − b)t) Graph on the domain [−π, 0], where a = 2 and 455. b = 1, and include the orientation. Graph on the domain [−π, 0], where a = 3 and 456. b = 2 , and include the orientation. Graph on the domain [−π, 0], where a = 4 and 457. b = 3 , and include the orientation. Graph on the domain [−π, 0], where a = 5 and 458. b = 4 , and include the orientation. If a is 1 more than b, describe the effect the values 459. of a and b have on the graph of the parametric equations. 460. Describe the graph if a = 100 and b = 99. 1176 Chapter 10 Further Applications of Trigonometry What happens if b is 1 more than a ? Describe the 461. graph. 462. If the parametric equations x(t) = t 2 and y(t) = 6 − 3t have the graph of a horizontal parabola opening to the right, what would change the direction of the curve? For the following exercises, describe the graph of the set of parametric equations. 463. 464. 465. x(t) = − t 2 and y(t) is linear y(t) = t 2 and x(t) is linear y(t) = − t 2 and x(t) is linear Write the parametric equations of a circle with center 466. (0, 0), radius 5, and a counterclockwise orientation. 475. Write the parametric equations of an ellipse with 467. center (0, 0), major axis of length 10, minor axis of length 6, and a counterclockwise orientation. the window [−3, 3] by [−3, 3] on For the following exercises, use a graphing utility to graph on domain [0, 2π) for the following values of a and b , and include the orientation. the x(t) = sin(at) ⎧ ⎨ y(t) = sin(bt) ⎩ 468. a = 1, b = 2 469. a = 2, b = 1 470. a = 3, b = 3 471. a = 5, b = 5 472. a = 2, b = 5 473. a = 5, b = 2 Technology 476. by equations parametric For the following exerci
ses, look at the graphs that were form created x(t) = acos(bt) ⎧ ⎨ y(t) = csin(dt) ⎩ graphing calculator to find the values of a, b, c, to achieve each graph. . Use the parametric mode on the and d the of 474. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1177 477. and then eliminate time to write height as a function of horizontal position. be 485. A skateboarder riding on a level surface at a constant speed of 9 ft/s throws a ball in the air, the height of which can equation y(t) = − 16t 2 + 10t + 5. Write parametric equations for the ball’s position, and then eliminate time to write height as a function of horizontal position. described the by For the following exercises, use this scenario: A dart is thrown upward with an initial velocity of 65 ft/s at an angle of elevation of 52°. Consider the position of the dart at any time t. Neglect air resistance. Find parametric equations that model the problem 486. situation. Find all possible values of x that 487. situation. represent the 488. When will the dart hit the ground? 489. Find the maximum height of the dart. 490. At what time will the dart reach maximum height? For the following exercises, look at the graphs of each of the four parametric equations. Although they look unusual and beautiful, they are so common that they have names, as indicated in each exercise. Use a graphing utility to graph each on the indicated domain. 491. x(t) = 14cos t − cos(14t) ⎧ An epicycloid: ⎨ y(t) = 14sin t + sin(14t) ⎩ on the domain [0, 2π] . 492. A hypocycloid: x(t) = 6sin t + 2sin(6t) ⎧ ⎨ y(t) = 6cos t − 2cos(6t) ⎩ on the domain [0, 2π] . For the following exercises, use a graphing utility to graph the given parametric equations. a. b. c. x(t) = cost − 1 ⎧ ⎨ y(t) = sint + t ⎩ x(t) = cost + t ⎧ ⎨ y(t) = sint − 1 ⎩ x(t) = t − sint ⎧ ⎨ y(t) = cost − 1 ⎩ Graph all three sets of parametric equations on the 478. domain [0, 2π]. Graph all three sets of parametric equations on the 479. domain [0, 4π]. Graph all three sets of parametric equations on the 480. domain ⎡ ⎣−4π, 6π⎤ ⎦. The graph of each set of parametric equations appears 481. to “creep” along one of the axes. What controls which axis the graph creeps along? 493. A hypotrochoid: x(t) = 2sin t + 5cos(6t) ⎧ ⎨ y(t) = 5cos t − 2sin(6t) ⎩ on the domain [0, 2π] . 494. x(t) = 5sin(2t)sint ⎧ A rose: ⎨ y(t) = 5sin(2t)cost ⎩ on the domain [0, 2π] . Explain the effect on the graph of the parametric 482. equation when we switched sin t and cos t . Explain the effect on the graph of the parametric 483. equation when we changed the domain. Extensions An object is thrown in the air with vertical velocity of 484. 20 ft/s and horizontal velocity of 15 ft/s. The object’s height can be described by the equation y(t) = − 16t 2 + 20t , while the object moves horizontally with constant velocity 15 ft/s. Write parametric equations for the object’s position, 1178 Chapter 10 Further Applications of Trigonometry 10.8 \| Vectors In this section you will: Learning Objectives 10.8.1 View vectors geometrically. 10.8.2 Find magnitude and direction. 10.8.3 Perform vector addition and scalar multiplication. 10.8.4 Find the component form of a vector. 10.8.5 Find the unit vector in the direction of v . 10.8.6 Perform operations with vectors in terms of i and j . 10.8.7 Find the dot product of two vectors. An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 10.110. What are the ground speed and actual bearing of the plane? Figure 10.110 Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors. A Geometric View of Vectors A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities: • Lower case, boldfaced type, with or without an arrow on top such as v, u, w, v→ , u→ , w→ . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1179 • Given initial point P and terminal point Q, a vector can be represented as PQ → . The arrowhead on top is what indicates that it is not just a line, but a directed line segment. • Given an initial point of (0, 0) and terminal point (a, b), a vector may be represented as 〈 a, b 〉 . This last symbol 〈 a, b 〉 has special significance. It is called the standard position. The position vector has an initial point (0, 0) and a terminal point 〈 a, b 〉 . To change any vector into the position vector, we think about the change in the → is C(x1, y1) and the terminal x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector CD point is D(x2, y2), then the position vector is found by calculating → = 〈 x2 − x1, y2 − y1 〉 AB = 〈 a, b 〉 In Figure 10.111, we see the original vector CD → → and the position vector AB . Figure 10.111 Properties of Vectors A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at (0, 0) and is identified by its terminal point 〈 a, b 〉 . Example 10.59 Find the Position Vector Consider the vector whose initial point is P(2, 3) and terminal point is Q(6, 4). Find the position vector. Solution The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus v = 〈 6 − 2, 4 − 3 〉 = 〈 4, 1 〉 The position vector begins at (0, 0) and terminates at (4, 1). The graphs of both vectors are shown in Figure 10.112. 1180 Chapter 10 Further Applications of Trigonometry Figure 10.112 We see that the position vector is 〈 4, 1 〉 . Example 10.60 Drawing a Vector with the Given Criteria and Its Equivalent Position Vector Find the position vector given that vector v has an initial point at (−3, 2) and a terminal point at (4, 5), then graph both vectors in the same plane. Solution The position vector is found using the following calculation: v = 〈 4 − ( − 3), 5 − 2 〉 = 〈 7, 3 〉 Thus, the position vector begins at (0, 0) and terminates at (7, 3). See Figure 10.113. Figure 10.113 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1181 10.39 Draw a vector v that connects from the origin to the point (3, 5). Finding Magnitude and Direction To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function. Magnitude and Direction of a Vector the magnitude is found by \|v\| = a2 + b2. The direction is equal to the angle Given a position vector v = 〈 a, b 〉 , formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tan θ = ⎞ ⎠, as illustrated in Figure 10.114. ⎠ ⇒ θ = tan− Figure 10.114 Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal. Example 10.61 Finding the Magnitude and Direction of a Vector Find the magnitude and direction of the vector with initial point P(−8, 1) and terminal point Q(−2, − 5). Draw the vector. Solution First, find the position vector. We use the Pythagorean Theorem to find the magnitude. u = 〈 −2, − (−8), −5−1 〉 = 〈 6, − 6 〉 \|u\| = (6)2 + ( − 6)2 = 72 = 6 2 The direction is given as 1182 Chapter 10 Further Applications of Trigonometry = −1 ⇒ θ = tan−1(−1) tan θ = −6 6 = − 45° However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, − 45° + 360° = 315°. See Figure 10.115. Figure 10.115 Example 10.62 Showing That Two Vectors Are Equal Show that vector v with initial point at (5, −3) and terminal point at (−1, 2) is equal to vector u with initial point at (−1, −3) and terminal point at (−7, 2). Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector. Solution As shown in Figure 10.116, draw the vector v starting at initial (5, −3) and terminal point (−1, 2). Draw the vector u with initial point (−1, −3) and terminal point (−7, 2). Find the standard position for each. Next, find and sketch the position vector for v and u. We have v = 〈 −1 − 5, 2 − ( − 3) 〉 = 〈 −6, 5 〉 u = 〈 −7 − (−1), 2 − (−3) 〉 = 〈 −6, 5 〉 Since the position vectors are the same, v and u are the same. An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1183 \|v\| = (−1 − 5)2 + (2 − (−3))2 = (−6)2 + (5)2 = 36 + 25 = 61 \|u\| = (−7 − (−1))2 + (2 − (−3))2 = (−6)2 + (5)2 = 36 + 25 = 61 As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives tan θ = − 5 6 = −
39.8° ⇒ θ = tan−1 ⎛ ⎝− 5 6 ⎞ ⎠ However, we can see that the position vector terminates in the second quadrant, so we add 180°. Thus, the direction is − 39.8° + 180° = 140.2°. Figure 10.116 Performing Vector Addition and Scalar Multiplication Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u = 〈 x, y 〉 as an arrow or directed line segment from the origin to the point (x, y), vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector. To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 10.117. 1184 Chapter 10 Further Applications of Trigonometry Figure 10.117 Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 10.118 for a visual that compares vector addition and vector subtraction using parallelograms. Figure 10.118 Example 10.63 Adding and Subtracting Vectors Given u = 〈 3, − 2 〉 and v = 〈 −1, 4 〉 , find two new vectors u + v, and u − v. Solution To find the sum of two vectors, we add the components. Thus, u + v = 〈 3, − 2 〉 + 〈 −1), − 2 + 4 〉 = 〈 2, 2 〉 See Figure 10.119(a). To find the difference of two vectors, add the negative components of v to u. Thus, u + ( − v) = 〈 3, − 2 〉 + 〈 1, − 4 〉 = 〈 3 + 1, − 2 + ( − 4) 〉 = 〈 4, − 6 〉 See Figure 10.119(b). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1185 Figure 10.119 (a) Sum of two vectors (b) Difference of two vectors Multiplying By a Scalar While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector. Scalar Multiplication Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply v = 〈 a, b 〉 by k , we have Only the magnitude changes, unless k is negative, and then the vector reverses direction. kv = 〈 ka, kb 〉 Example 10.64 Performing Scalar Multiplication Given vector v = 〈 3, 1 〉 , find 3v, 1 2 v, and −v. Solution See Figure 10.120 for a geometric interpretation. If v = 〈 3, 1 〉 , then 3v = 〈 3 ⋅ 3, 3 ⋅ 1 〉 = 〈 9v = 〈 −3, −1 〉 ⋅ 3, 1 2 , 1 2 〉 ⋅ 1 〉 1186 Chapter 10 Further Applications of Trigonometry Figure 10.120 Analysis Notice that the vector 3v is three times the length of v, 1 2 but in the opposite direction. v is half the length of v, and –v is the same length of v, 10.40 Find the scalar multiple 3 u given u = 〈 5, 4 〉 . Example 10.65 Using Vector Addition and Scalar Multiplication to Find a New Vector Given u = 〈 3, − 2 〉 and v = 〈 −1, 4 〉 , find a new vector w = 3u + 2v. Solution First, we must multiply each vector by the scalar. 3u = 3 〈 3, − 2 〉 = 〈 9, − 6 〉 2v = 2 〈 −1, 4 〉 = 〈 −2, 8 〉 w = 3u + 2v = 〈 9, − 6 〉 + 〈 −2, 8 〉 = 〈 9 − 2, − 6 + 8 〉 = 〈 7, 2 〉 Then, add the two together. So, w = 〈 7, 2 〉 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1187 Finding Component Form In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the x direction, and the vertical component is the y direction. For example, we can see in the graph in Figure 10.121 that the position vector 〈 2, 3 〉 comes from adding the vectors v1 and v2. We have v1 with initial point (0, 0) and terminal point (2, 0). We also have v2 with initial point (0, 0) and terminal point (0, 3). v1 = 〈 2 − 0, 0 − 0 〉 = 〈 2, 0 〉 Therefore, the position vector is v2 = 〈 0 − 0, 3 − 0 〉 = 〈 0, 3 〉 v = 〈 2 + 0, 3 + 0 〉 = 〈 2, 3 〉 Using the Pythagorean Theorem, the magnitude of v1 is 2, and the magnitude of v2 is 3. To find the magnitude of v, use the formula with the position vector. \|v\| = \|v1\|2 + \|v2\|2 = 22 + 32 = 13 The magnitude of v is 13. To find the direction, we use the tangent function tan θ = y x. tan θ = v2 v1 tan θ = 3 2 θ = tan−1 ⎛ ⎝ ⎞ ⎠ = 56.3° 3 2 Figure 10.121 Thus, the magnitude of v is 13 and the direction is 56.3∘ off the horizontal. Example 10.66 Finding the Components of the Vector 1188 Chapter 10 Further Applications of Trigonometry Find the components of the vector v with initial point (3, 2) and terminal point (7, 4). Solution First find the standard position. See the illustration in Figure 10.122. v = 〈 7 − 3, 4 − 2 〉 = 〈 4, 2 〉 Figure 10.122 The horizontal component is v1 = 〈 4, 0 〉 and the vertical component is v2 = 〈 0, 2〉. Finding the Unit Vector in the Direction of v In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations. Unit vectors are defined in terms of components. The horizontal unit vector is written as i = 〈 1, 0 〉 and is directed along the positive horizontal axis. The vertical unit vector is written as j = 〈 0, 1 〉 and is directed along the positive vertical axis. See Figure 10.123. Figure 10.123 The Unit Vectors If v is a nonzero vector, then v \|v\| is a unit vector in the direction of v. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1189 Example 10.67 Finding the Unit Vector in the Direction of v Find a unit vector in the same direction as v = 〈 −5, 12〉. Solution First, we will find the magnitude. \|v\| = ( − 5)2 + (12)2 = 25 + 144 = 169 = 13 Then we divide each component by \|v\|, which gives a unit vector in the same direction as v: v \|v\| = − 5 13 i + 12 13 j or, in component form See Figure 10.124. v \|v\| = 〈 − 5 13 , 12 13 〉 Figure 10.124 1190 Chapter 10 Further Applications of Trigonometry Verify that the magnitude of the unit vector equals 1. The magnitude of − 5 13 i + 12 13 j is given as 2 ⎛ ⎝− 5 13 ⎞ ⎠ + ⎛ ⎝ 12 13 ⎞ ⎠ 2 = 25 169 = 169 169 + 144 169 = 1 The vector u = 5 13 i + 12 13 j is the unit vector in the same direction as v = 〈 −5, 12 〉 . Performing Operations with Vectors in Terms of i and j So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j. Vectors in the Rectangular Plane Given a vector v with initial point P = (x1, y1) and terminal point Q = (x2, y2 ), v is written as v = (x2 − x1)i + (y1 − y2) j The position vector from (0, 0) to (a, b), where (x2 − x1) = a and (y2 − y1) = b, is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j. The magnitude of v = ai + bj is given as \|v\| = a2 + b2. See Figure 10.125. Figure 10.125 Example 10.68 Writing a Vector in Terms of i and j Given a vector v with initial point P = (2, −6) and terminal point Q = (−6, 6), write the vector in terms of i and j. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1191 Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j = ( − 6 − 2)i + (6 − ( − 6)) j = − 8i + 12 j Example 10.69 Writing a Vector in Terms of i and j Using Initial and Terminal Points Given initial point P1 = (−1, 3) and terminal point P2 = (2, 7), write the vector v in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j v = (2 − ( − 1))i + (7 − 3) j = 3i + 4 j Write the vector u with initial point P = (−1, 6) and terminal point Q = (7, − 5) in terms of i and 10.41 j. Performing Operations on Vectors in Terms of i and j When vectors are written in terms of i and j, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components. Adding and Subtracting Vectors in Rectangular Coordinates Given v = ai + bj and u = ci + dj, then v + u = (a + c)i + (b + d) j v − u = (a − c)i + (b − d) j Example 10.70 Finding the Sum of the Vectors Find the sum of v1 = 2i − 3 j and v2 = 4i + 5 j. Solution According to the formula, we have 1192 Chapter 10 Further Applications of Trigonometry v1 + v2 = (2 + 4)i + ( − 3 + 5) j = 6i + 2 j Calculating the Component Form of a Vector: Direction We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using i and j. For any of t
hese vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction. Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with \|v\| replacing r. Vector Components in Terms of Magnitude and Direction Given a position vector v = 〈 x, y 〉 and a direction angle θ, cos θ = x \|v\| x = \|v\|cos θ and sin θ = y \|v\| y = \|v\|sin θ Thus, v = xi + y j = \|v\|cos θi + \|v\|sin θ j, and magnitude is expressed as \|v\| = x2 + y2. Example 10.71 Writing a Vector in Terms of Magnitude and Direction Write a vector with length 7 at an angle of 135° to the positive x-axis in terms of magnitude and direction. Solution Using the conversion formulas x = \|v\|cos θi and y = \|v\|sin θ j, we find that x = 7cos(135°)i = − 7 2 2 y = 7sin(135°) j = 7 2 2 This vector can be written as v = 7cos(135°)i + 7sin(135°) j or simplified as 10.42 A vector travels from the origin to the point (3, 5). Write the vector in terms of magnitude and direction. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1193 Finding the Dot Product of Two Vectors As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses. The dot product of two vectors involves multiplying two vectors together, and the result is a scalar. Dot Product The dot product of two vectors v = 〈 a, b 〉 and u = 〈 c, d 〉 is the sum of the product of the horizontal components and the product of the vertical components. v ⋅ u = ac + bd To find the angle between the two vectors, use the formula below. u \|u\| cos θ = v \|v\| ⋅ Example 10.72 Finding the Dot Product of Two Vectors Find the dot product of v = 〈 5, 12 〉 and u = 〈 −3, 4 〉 . Solution Using the formula, we have v ⋅ u = 〈 5, 12 〉 ⋅ 〈 −3, 4 〉 = 5 ⋅ ( − 3) + 12 ⋅ 4 = − 15 + 48 = 33 Example 10.73 Finding the Dot Product of Two Vectors and the Angle between Them Find the dot product of v1 = 5i + 2j and v2 = 3i + 7j. Then, find the angle between the two vectors. Solution Finding the dot product, we multiply corresponding components. v1 ⋅ v2 = 〈 5, 2 〉 ⋅ 〈 3 = 15 + 14 = 29 To find the angle between them, we use the formula cos θ = v \|v\| ⋅ u \|u\| . 1194 Chapter 10 Further Applications of Trigonometry v \|v\| ⋅ u \|u\| 〉 ⋅ 〈 3 58 + 7 58 〉 + 2 29 + 2 29 = 〈 5 29 ⋅ 3 58 + 14 = 5 29 = 15 1682 1682 ⋅ 7 58 = 29 1682 = 0.707107 cos−1(0.707107) = 45° See Figure 10.126. Figure 10.126 Example 10.74 Finding the Angle between Two Vectors Find the angle between u = 〈 −3, 4 〉 and v = 〈 5, 12 〉 . Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1195 ⋅ ⋅ ⎛ ⎝ ⎞ ⎠ = v u \|v\| \|u\| = θ = cos−1 ⎛ u ⎝ \|u\| −3i + 4 j 5 ⎛ ⎝− 3 5 = − 15 65 ⎞ ⋅ 5 ⎠ + 13 + 48 65 ⋅ ⎞ v ⎠ \|v\| 5i + 12 j 13 4 5 ⋅ 12 13 ⎛ ⎝ ⎞ ⎠ = 33 65 θ = cos−1 ⎛ ⎝ = 59.5∘ ⎞ ⎠ 33 65 See Figure 10.127. Figure 10.127 Example 10.75 Finding Ground Speed and Bearing Using Vectors We now have the tools to solve the problem we introduced in the opening of the section. 1196 Chapter 10 Further Applications of Trigonometry An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 10.128. Figure 10.128 Solution The ground speed is represented by x in the diagram, and we need to find the angle α in order to calculate the adjusted bearing, which will be 140° + α . Notice in Figure 10.128, that angle BCO must be equal to angle AOC by the rule of alternating interior angles, so angle BCO is 140°. We can find x by the Law of Cosines: x2 = (16.2)2 + (200)2 − 2(16.2)(200)cos(140°) x2 = 45, 226.41 x = 45, 226.41 x = 212.7 The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines. = sin α 16.2 sin α = sin(140°) 212.7 16.2sin(140°) 212.7 = 0.04896 sin−1(0.04896) = 2.8° Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour. Access these online resources for additional instruction and practice with vectors. • Introduction to Vectors (http://openstaxcollege.org/l/introvectors) • Vector Operations (http://openstaxcollege.org/l/vectoroperation) • The Unit Vector (http://openstaxcollege.org/l/unitvector) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1197 10.8 EXERCISES Verbal What are the characteristics of the letters that are 495. commonly used to represent vectors? 496. How is a vector more specific than a line segment? 497. What are i and j, and what do they represent? 498. What is component form? When a unit vector is expressed as 〈 a, b 〉 , which 499. letter is the coefficient of the i and which the j ? Algebraic Given a vector with initial point (5, 2) and terminal 500. point (−1, − 3), find an equivalent vector whose initial form point is (0, 0). Write the vector in component 〈 a, b 〉 . Given a vector with initial point (−4, 2) and 501. terminal point (3, − 3), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈 a, b 〉 . Given a vector with initial point (7, − 1) and 502. terminal point (−1, − 7), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈 a, b 〉 . P1 = (8, 3), P2 = (6, 5), P3 = (11, 8), and P4 = (9, 10) Given initial point P1 = (−3, 1) and terminal point 508. P2 = (5, 2), write the vector v in terms of i and j. Given initial point P1 = (6, 0) and terminal point 509. P2 = (−1, − 3), write the vector v in terms of i and j. For the following exercises, use the vectors u = i + 5j, v = −2i− 3j, and w = 4i − j. 510. Find u + (v − w) 511. Find 4v + 2u For the following exercises, use the given vectors to compute u + v, u − v, and 2u − 3v. 512. u = 〈 2, − 3 〉 , v = 〈 1, 5 〉 513. u = 〈 −3, 4 〉 , v = 〈 −2, 1 〉 Let v = −4i + 3j. Find a vector that is half the length 514. and points in the same direction as v. Let v = 5i + 2j. Find a vector that is twice the length 515. and points in the opposite direction as v. For the following exercises, find a unit vector in the same direction as the given vector. For the following exercises, determine whether the two vectors u and v are equal, where u has an initial point P1 and a terminal point P2 and v has an initial point P3 and a terminal point P4 . 516. a = 3i + 4j 517. b = −2i + 5j 518. c = 10i – j P1 = (5, 1), P2 = (3, − 2), P3 = (−1, 3), and 503. P4 = (9, − 4) P1 = (2, − 3), P2 = (5, 1), P3 = (6, − 1), and 504. P4 = (9, 3) P1 = (−1, − 1), P2 = (−4, 5), P3 = (−10, 6), 505. and P4 = (−13, 12) P1 = (3, 7), P2 = (2, 1), P3 = (1, 2), and 506. P4 = (−1, − 4) 507. 519 520. u = 100i + 200j 521. u = −14i + 2j the following exercises, For direction of the vector, 0 ≤ θ < 2π. find the magnitude and 522. 〈 0, 4 〉 523. 〈 6, 5 〉 524. 〈 2, −5 〉 1198 Chapter 10 Further Applications of Trigonometry 525. 〈 −4, −6 〉 For the following exercises, use the vectors shown to sketch 2u + v. 526. Given u = 3i − 4j and v = −2i + 3j, calculate u ⋅ v. 536. 527. Given u = −i − j and v = i + 5j, calculate u ⋅ v. Given u = 〈 −2, 4 〉 and v = 〈 −3, 1 〉 , 528. calculate u ⋅ v. u = 〈 −1, 6 〉 and v = 〈 6, − 1 〉 , 529. Given calculate u ⋅ v. Graphical For the following exercises, given v, draw v, 3v and 1 2 v. 530. 〈 2, −1 〉 531. 〈 −1, 4 〉 532. 〈 −3, −2 〉 537. For the following exercises, use the vectors shown to sketch u + v, u − v, and 2u. 533. 534. 535. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, use the vectors shown to sketch u − 3v. 538. 539. Chapter 10 Further Applications of Trigonometry 1199 A 60-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. A 25-pound box is resting on a ramp that is inclined 550. 8°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. Find the magnitude of the horizontal and vertical 551. components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth. Find the magnitude of the horizontal and vertical 552. components of the vector with magnitude 4 pounds pointed in a direction of 127° above the horizontal. Round to the nearest hundredth. Find the magnitude of the horizontal and vertical 553. components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth. For the following exercises, write the vector shown in component form. 540. 541. Given initial point P1 = (2, 1) and terminal point 542. P2 = (−1, 2), write the vector v in terms of i and j, then draw the vector on the graph. Find the magnitude of the horizontal and vertical 554. components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth. Given initial point P1 = (4, − 1) and terminal point 543. P2 = (−3, 2), write the vector v in terms of i and j. Draw the points and the vector on the graph. Given initial point P1 = (3, 3) and terminal point 544. P2 = (−3, 3), wr
ite the vector v in terms of i and j. Draw the points and the vector on the graph. Extensions For the following exercises, use the given magnitude and direction in standard position, write the vector in component form. 545. \|v\| = 6, θ = 45 ° 546. \|v\| = 8, θ = 220° 547. \|v\| = 2, θ = 300° 548. \|v\| = 5, θ = 135° 549. Real-World Applications A woman leaves home and walks 3 miles west, then 2 555. miles southwest. How far from home is she, and in what direction must she walk to head directly home? A boat leaves the marina and sails 6 miles north, then 556. 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina? A man starts walking from home and walks 4 miles 557. east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk? A woman starts walking from home and walks 4 558. miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? A man starts walking from home and walks 3 miles at 559. 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? 1200 Chapter 10 Further Applications of Trigonometry 560. A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? Suppose a body has a force of 10 pounds acting on it 571. to the right, 25 pounds acting on it ─135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body? The condition of equilibrium is when the sum of the 572. forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Suppose a body has a force of 3 pounds acting on it to 573. the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector. An airplane is heading north at an airspeed of 600 km/ 561. hr, but there is a wind blowing from the southwest at 80 km/ hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? An airplane is heading north at an airspeed of 500 km/ 562. hr, but there is a wind blowing from the northwest at 50 km/ hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? An airplane needs to head due north, but there is a 563. wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? An airplane needs to head due north, but there is a 564. wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? 565. As part of a video game, the point (5, 7) is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point. 566. As part of a video game, the point (7, 3) is rotated counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point. 567. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? Two children are throwing a ball back and forth 568. straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? A 50-pound object rests on a ramp that is inclined 569. 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound. Suppose a body has a force of 10 pounds acting on it 570. to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1201 CHAPTER 10 REVIEW KEY TERMS altitude a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles ambiguous case a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle Archimedes’ spiral a polar curve given by r = θ. When multiplied by a constant, the equation appears as r = aθ. As r = θ, the curve continues to widen in a spiral path over the domain. argument the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis cardioid a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as r = a ± bcos θ and r = a ± bsin θ, where a b = 1 convex limaҫon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that a b ≥ 2 De Moivre’s Theorem formula used to find the nth power or nth roots of a complex number; states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the angles by n dimpled limaҫon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that 1 < a b < 2 dot product given two vectors, the sum of the product of the horizontal components and the product of the vertical components Generalized Pythagorean Theorem used for SAS and SSS triangles initial point the origin of a vector an extension of the Law of Cosines; relates the sides of an oblique triangle and is inner-loop limaçon a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by r = a ± bcos θ and r = a ± bsin θ where a < b Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle Law of Sines states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side lemniscate a polar curve resembling a figure 8 and given by the equation r 2 = a2 cos 2θ and r 2 = a2 sin 2θ, a ≠ 0 magnitude the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem modulus the absolute value of a complex number, or the distance from the origin to the point (x, y); also called the amplitude oblique triangle any triangle that is not a right triangle one-loop limaҫon a polar curve represented by r = a ± bcos θ and r = a ± bsin θ such that a > 0, b > 0, and a b > 1; may be dimpled or convex; does not pass through the pole parameter a variable, often representing time, upon which x and y are both dependent polar axis on the polar grid, the equivalent of the positive x-axis on the rectangular grid 1202 Chapter 10 Further Applications of Trigonometry polar coordinates on the polar grid, the coordinates of a point labeled (r, θ), where θ indicates the angle of rotation from the polar axis and r represents the radius, or the distance of the point from the pole in the direction of θ polar equation an equation describing a curve on the polar grid. polar form of a complex number a complex number expressed in terms of an angle θ and its distance from the origin r; can be found by using conversion formulas x = rcos θ, y = rsin θ, and r = x2 + y2 pole the origin of the polar grid resultant a vector that results from addition or subtraction of two vectors, or from scalar multiplication rose curve a polar equation resembling a flower, given by the equations r = acos nθ and r = asin nθ; when n is even there are 2n petals, and the curve is highly symmetrical; when n is odd there are n petals. scalar a quantity associated with magnitude but not direction; a constant scalar multiplication the product of a constant and each component of a vector standard position the placement of a vector with the initial point at (0, 0) and the terminal point (a, b), represented by the change in the x-coordinates and the change in the y-coordinates of the original vector terminal point the end point of a vector, usually represented by an arrow indicating its direction unit vector a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the x-axis and is defined as v1 = 〈 1, 0 〉 the vertical unit vector runs along the y-axis and is defined as v2 = 〈 0, 1 〉 . vector a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point) vector addition the sum of two vectors, found by adding corresponding components KEY EQUATIONS Law of Sines Area for oblique triangles sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ Area = 1 2 = 1 2 = 1 2 bcsin α acsin β absin γ Law of Cosines a2 = b2 + c2 − 2bccos α b2 = a2 + c2 − 2accos β c2 = a2 + b2 − 2abcos γ Heron’s formula Area
= s(s − a)(s − b)(s − c) (a + b + c) 2 where s = This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1203 Conversion formulas cos θ = x r → x = rcos θ y r → y = rsin θ sin θ = r 2 = x2 + y2 y tan θ = x KEY CONCEPTS 10.1 Non-right Triangles: Law of Sines • The Law of Sines can be used to solve oblique triangles, which are non-right triangles. • According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. • There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See Example 10.1. • The ambiguous case arises when an oblique triangle can have different outcomes. • There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See Example 10.2 and Example 10.3. • The Law of Sines can be used to solve triangles with given criteria. See Example 10.4. • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See Example 10.5. • There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See Example 10.6. 10.2 Non-right Triangles: Law of Cosines • The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles. • The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example 10.7 and Example 10.8. • The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example 10.9 and Example 10.10. • Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 10.11 and See Example 10.12. 10.3 Polar Coordinates • The polar grid is represented as a series of concentric circles radiating out from the pole, or origin. • To plot a point in the form (r, θ), θ > 0, move in a counterclockwise direction from the polar axis by an angle of θ, and then extend a directed line segment from the pole the length of r in the direction of θ. If θ is negative, move in a clockwise direction, and extend a directed line segment the length of r in the direction of θ. See Example 10.13. • If r is negative, extend the directed line segment in the opposite direction of θ. See Example 10.14. • To convert from polar coordinates to rectangular coordinates, use the formulas x = rcos θ and y = rsin θ. See Example 10.15 and Example 10.16. 1204 Chapter 10 Further Applications of Trigonometry • To convert x r , sin θ = cos θ = y r , tan θ = y x, and r = x2 + y2. See Example 10.17. from rectangular coordinates to polar coordinates, use one or more of the formulas: • Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See Example 10.18, Example 10.19, and Example 10.20. • Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. See Example 10.21, Example 10.22, and Example 10.23. 10.4 Polar Coordinates: Graphs • It is easier to graph polar equations if we can test the equations for symmetry with respect to the line θ = , the π 2 polar axis, or the pole. • There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See Example 10.24. • Polar equations may be graphed by making a table of values for θ and r. • The maximum value of a polar equation is found by substituting the value θ that leads to the maximum value of the trigonometric expression. • The zeros of a polar equation are found by setting r = 0 and solving for θ. See Example 10.25. • Some formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ. See Example 10.26. • The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ, for a > 0, b > 0, and a b = 1. See Example 10.27. • The formulas that produce the graphs of a one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for 1 < a b < 2. See Example 10.28. • The formulas that produce the graphs of an inner-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for a > 0, b > 0, and a < b. See Example 10.29. • The formulas that produce the graphs of a lemniscates are given by r 2 = a2 cos 2θ and r 2 = a2 sin 2θ, where a ≠ 0. See Example 10.30. • The formulas that produce the graphs of rose curves are given by r = acos nθ and r = asin nθ, where a ≠ 0; if n is even, there are 2n petals, and if n is odd, there are n petals. See Example 10.31 and Example 10.32. • The formula that produces the graph of an Archimedes’ spiral is given by r = θ, θ ≥ 0. See Example 10.33. 10.5 Polar Form of Complex Numbers • Complex numbers in the form a + bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the x-axis as the real axis and the y-axis as the imaginary axis. See Example 10.34. • The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: \|z\| = a2 + b2. See Example 10.35 and Example 10.36. • To write complex numbers in polar form, we use the formulas x = rcos θ, y = rsin θ, and r = x2 + y2. Then, z = r(cos θ + isin θ). See Example 10.37 and Example 10.38. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1205 • To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by r. See Example 10.39 and Example 10.40. • To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property. See Example 10.41. • To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles. See Example 10.42. • To find the power of a complex number zn 10.43. , raise r to the power n, and multiply θ by n. See Example • Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See Example 10.44. 10.6 Parametric Equations • Parameterizing a curve involves translating a rectangular equation in two variables, x and y, into two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See Example 10.45, Example 10.46, and Example 10.47. • Sometimes equations are simpler to graph when written in rectangular form. By eliminating t, an equation in x and y is the result. • To eliminate t, solve one of the equations for t, and substitute the expression into the second equation. See Example 10.48, Example 10.49, Example 10.50, and Example 10.51. • Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for t in one of the equations, and substitute the expression into the second equation. See Example 10.52. • There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation. • Find an expression for x such that the domain of the set of parametric equations remains the same as the original rectangular equation. See Example 10.53. 10.7 Parametric Equations: Graphs • When there is a third variable, a third parameter on which x and y depend, parametric equations can be used. • To graph parametric equations by plotting points, make a table with three columns labeled t, x(t), and y(t). Choose values for t in increasing order. Plot the last two columns for x and y. See Example 10.54 and Example 10.55. • When graphing a parametric curve by plotting points, note the associated t-values and show arrows on the graph indicating the orientation of the curve. See Example 10.56 and Example 10.57. • Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion. See Example 10.58. • Projectile motion depends on two parametric equations: x = (v0 cos θ)t and y = − 16t 2 + (v0 sin θ)t + h. Initial velocity is symbolized as v0. θ represents the initial angle of the object when thrown, and h represents the height at which the object is propelled. 10.8 Vectors • The position vector has its initial point at the origin. See Example 10.59. • If the position vector is the same for two vectors, they are equal. See Example 10.60. 1206 Chapter 10 Further Applications of Trigonometry • Vectors are defined by their magnitude and direction. See Example 10.61. • If two vectors have the same magnitude and direction, they are equal. See Example 10.62. • Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example 10.63. • Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example 10.64 and Example 10.65. • Vectors are comprised of two components: the horizontal component along the positive x-
axis, and the vertical component along the positive y-axis. See Example 10.66. • The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude. • The magnitude of a vector in the rectangular coordinate system is \|v\| = a2 + b2. See Example 10.67. • In the rectangular coordinate system, unit vectors may be represented in terms of i and j where i represents the horizontal component and j represents the vertical component. Then, v = ai + bj is a scalar multiple of v by real numbers a and b. See Example 10.68 and Example 10.69. • Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See Example 10.70. • A vector v = ai + bj is written in terms of magnitude and direction as v = \|v\|cos θi + \|v\|sin θ j. See Example 10.71. • The dot product of two vectors is the product of the i terms plus the product of the j terms. See Example 10.72. • We can use the dot product to find the angle between two vectors. Example 10.73 and Example 10.74. • Dot products are useful for many types of physics applications. See Example 10.75. CHAPTER 10 REVIEW EXERCISES Non-right Triangles: Law of Sines For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 574. β = 50°, a = 105, b = 45 575. α = 43.1°, a = 184.2, b = 242.8 576. Solve the triangle. 577. Find the area of the triangle. 578. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25° and 49°, as shown in Figure 10.129. Find the distance of the plane from point A and the elevation of the plane. Figure 10.129 Non-right Triangles: Law of Cosines 579. Solve the triangle, rounding to the nearest tenth, assuming α is opposite side a, β is opposite side b, and γ s opposite side c : a = 4, b = 6, c = 8. 580. Solve the triangle in Figure 10.130, rounding to the nearest tenth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1207 591. x2 + y2 = − 2y the following exercises, convert For equation to a Cartesian equation. the given polar 592. r = 7cos θ 593. r = −2 4cos θ + sin θ For the following exercises, convert to rectangular form and graph. 594. θ = 3π 4 595. r = 5sec θ Polar Coordinates: Graphs the following exercises, For symmetry. 596. r = 4 + 4sin θ 597. r = 7 test each equation for 598. Sketch a graph of the polar equation r = 1 − 5sin θ. Label the axis intercepts. 599. Sketch a graph of the polar equation r = 5sin(7θ). 600. Sketch a graph of the polar equation r = 3 − 3cos θ Polar Form of Complex Numbers For the following exercises, find the absolute value of each complex number. 601. −2 + 6i 602. 4 − 3i Figure 10.130 581. Find the area of a triangle with sides of length 8.3, 6.6, and 9.1. 582. To find the distance between two cities, a satellite calculates the distances and angle shown in Figure 10.131 (not to scale). Find the distance between the cities. Round answers to the nearest tenth. Figure 10.131 Polar Coordinates 583. Plot the point with polar coordinates ⎛ ⎝3, ⎞ ⎠. π 6 584. Plot the point with polar coordinates ⎛ ⎝5, − 2π 3 ⎞ ⎠ 585. Convert ⎛ ⎝6, − 3π 4 ⎞ ⎠ to rectangular coordinates. 586. Convert ⎛ ⎝−2, 3π 2 ⎞ ⎠ to rectangular coordinates. 587. Convert (7, − 2) to polar coordinates. Write the complex number in polar form. 588. Convert (−9, − 4) to polar coordinates. For the following exercises, convert the given Cartesian equation to a polar equation. 589. x = − 2 590. x2 + y2 = 64 603. 5 + 9i 604. 1 2 − 3 2 i For the following exercises, convert the complex number from polar to rectangular form. 605. ⎛ z = 5cis ⎝ ⎞ ⎠ 5π 6 1208 Chapter 10 Further Applications of Trigonometry 606. z = 3cis(40°) For the following exercises, find the product z1 z2 in polar form. 607. z1 = 2cis(89°) z2 = 5cis(23°) 608. ⎛ z1 = 10cis ⎝ ⎞ ⎠ π 6 ⎛ z2 = 6cis ⎝ ⎞ ⎠ π 3 For the following exercises, find the quotient form. 609. z1 = 12cis(55°) z2 = 3cis(18°) 610. ⎛ z1 = 27cis ⎝ ⎞ ⎠ 5π 3 ⎛ z2 = 9cis ⎝ ⎞ ⎠ π 3 618. x(t) = − cos t ⎧ ⎨ y(t) = 2sin2 t ⎩ 619. Parameterize (write a parametric equation for) each using x(t) = acos t and by Cartesian equation y(t) = bsin t for x2 25 + y2 16 = 1. 620. Parameterize the line from ( − 2, 3) to (4, 7) so that the line is at ( − 2, 3) at t = 0 and (4, 7) at t = 1. Parametric Equations: Graphs z1 z2 in polar For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation. 621. ⎧ x(t) = 3t 2 ⎨ y(t) = 2t − 1 ⎩ 622. x(t) = et ⎧ ⎨ y(t) = − 2e5 t ⎩ 623. x(t) = 3cos t ⎧ ⎨ y(t) = 2sin t ⎩ For the following exercises, find the powers of each complex number in polar form. 611. Find z4 when z = 2cis(70°) ⎛ 612. Find z2 when z = 5cis ⎝ ⎞ ⎠ 3π 4 624. A ball is launched with an initial velocity of 80 feet per second at an angle of 40° to the horizontal. The ball is released at a height of 4 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 3 seconds? c. How long is the ball in the air? For the following exercises, evaluate each root. 613. Evaluate the cube root of z when z = 64cis(210°). Vectors ⎛ 614. Evaluate the square root of z when z = 25cis ⎝ ⎞ ⎠. 3π 2 For the following exercises, plot the complex number in the complex plane. 615. 6 − 2i 616. −1 + 3i Parametric Equations For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 617. x(t) = 3t − 1 ⎧ ⎨ y(t) = t ⎩ This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, determine whether the two vectors, u and v, are equal, where u has an initial point P1 and a terminal point P2, and v has an initial point P3 and a terminal point P4. 625. P4 = (9, 2) P1 = (−1, 4), P2 = (3, 1), P3 = (5, 5) and P1 = (6, 11), P2 = (−2, 8), P3 = (0, − 1) and 626. P4 = (−8, 2) the following vectors exercises, For u = 2i − j,v = 4i − 3 j, and w = − 2i + 5 j to evaluate the expression. use the 627. u − v Chapter 10 Further Applications of Trigonometry 1209 628. 2v − u + w For the following exercises, find a unit vector in the same direction as the given vector. 629. a = 8i − 6j 630. b = −3i − j the following exercises, For direction of the vector. find the magnitude and 631. 〈 6, −2 〉 632. 〈 −3, −3 〉 For the following exercises, calculate u ⋅ v. 633. u = −2i + j and v = 3i + 7j 634. u = i + 4j and v = 4i + 3j 635. Given v = 〈−3, 4 〉 draw v, 2v, and 1 2 v. 636. Given the vectors shown in Figure 10.132, sketch u + v, u − v and 3v. Figure 10.132 637. Given initial point P1 = (3, 2) and terminal point P2 = (−5, − 1), write the vector v in terms of i and j. Draw the points and the vector on the graph. CHAPTER 10 PRACTICE TEST 638. Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve the triangle, if possible, and round each answer tenth, given β = 68°, b = 21, c = 16. to the nearest 639. Find the area of the triangle in Figure 10.133. Round each answer to the nearest tenth. Figure 10.133 1210 Chapter 10 Further Applications of Trigonometry 640. A pilot flies in a straight path for 2 hours. He then makes a course correction, heading 15° to the right of his original course, and flies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position? 656. Eliminate the parameter t to rewrite the following equation: parametric x(t) = t + 1 ⎧ ⎨ y(t) = 2t 2 . ⎩ equations Cartesian as a 641. Convert (2, 2) the point. to polar coordinates, and then plot 657. Parameterize (write a parametric equation for) the following Cartesian equation by using x(t) = acos t and 642. Convert ⎛ ⎝2, π 3 ⎞ ⎠ to rectangular coordinates. y(t) = bsin t : x2 36 + y2 100 = 1. 643. Convert the polar equation to a Cartesian equation: x2 + y2 = 5y. 658. Graph the set of parametric equations and find the x(t) = − 2sin t ⎧ Cartesian equation: ⎨ y(t) = 5cos t ⎩ . 659. A ball is launched with an initial velocity of 95 feet per second at an angle of 52° to the horizontal. The ball is released at a height of 3.5 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? For the following exercises, use the vectors u = i − 3j and v = 2i + 3j. 660. Find 2u − 3v. 661. Calculate u ⋅ v. 662. Find a unit vector in the same direction as v. 663. Given vector v has an initial point P1 = (2, 2) and terminal point P2 = (−1, 0), write the vector v in terms of i and j. On the graph, draw v, and − v. Convert 644. r = − 3csc θ. to rectangular form and graph: 645. Test the equation for symmetry: r = − 4sin⎛ ⎝2θ). 646. Graph r = 3 + 3cos θ. 647. Graph r = 3 − 5sin θ. 648. Find the absolute value of the complex number 5 − 9i. 649. Write the complex number in polar form: 4 + i. 650. Convert ⎛ rectangular form: z = 5cis ⎝ the complex number 2π 3 ⎞ ⎠. from polar to Given z1 = 8cis(36°) and z2 = 2cis(15°), evaluate each expression. 651. z1 z2 652. z1 z2 653. (z2)3 654. z1 655. Plot the complex number −5 − i in the complex plane. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1211 11 \| SYSTEMS OF EQUATIONS AND INEQUALITIES Figure 11.1 Enigma machines like this one, once owned by Italian dictator Benito Mussolini, were used by government and military officials for enciphering and deciphering top-secret communications during World War II. (credit: Dave Addey, Flickr) Chapter Outline 11.1 Systems of Linear Equations: Two Variables 11.2 Systems of Linear Equations: Three Variables 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables 11.4 Partial Fractions 11.5 Matric
es and Matrix Operations 11.6 Solving Systems with Gaussian Elimination 11.7 Solving Systems with Inverses 11.8 Solving Systems with Cramer's Rule 1212 Chapter 11 Systems of Equations and Inequalities Introduction By 1943, it was obvious to the Nazi regime that defeat was imminent unless it could build a weapon with unlimited destructive power, one that had never been seen before in the history of the world. In September, Adolf Hitler ordered German scientists to begin building an atomic bomb. Rumors and whispers began to spread from across the ocean. Refugees and diplomats told of the experiments happening in Norway. However, Franklin D. Roosevelt wasn’t sold, and even doubted British Prime Minister Winston Churchill’s warning. Roosevelt wanted undeniable proof. Fortunately, he soon received the proof he wanted when a group of mathematicians cracked the “Enigma” code, proving beyond a doubt that Hitler was building an atomic bomb. The next day, Roosevelt gave the order that the United States begin work on the same. The Enigma is perhaps the most famous cryptographic device ever known. It stands as an example of the pivotal role cryptography has played in society. Now, technology has moved cryptanalysis to the digital world. Many ciphers are designed using invertible matrices as the method of message transference, as finding the inverse of a matrix is generally part of the process of decoding. In addition to knowing the matrix and its inverse, the receiver must also know the key that, when used with the matrix inverse, will allow the message to be read. In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear, and then partial fractions. We will not be breaking any secret codes here, but we will lay the foundation for future courses. 11.1 \| Systems of Linear Equations: Two Variables Learning Objectives In this section, you will: 11.1.1 Solve systems of equations by graphing. 11.1.2 Solve systems of equations by substitution. 11.1.3 Solve systems of equations by addition. 11.1.4 Identify inconsistent systems of equations containing two variables. 11.1.5 Express the solution of a system of dependent equations containing two variables. Figure 11.2 (credit: Thomas Sørenes) A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions. Introduction to Systems of Equations In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1213 infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables. 2x + y = 15 3x – y = 5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists. 2(4) + (7) = 15 True 3(4) − (7) = 5 True In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system. Types of Linear Systems There are three types of systems of linear equations in two variables, and three types of solutions. • An independent system has exactly one solution pair (x, y). The point where the two lines intersect is the only solution. • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations. Figure 11.3 compares graphical representations of each type of system. Figure 11.3 1214 Chapter 11 Systems of Equations and Inequalities Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution. 1. Substitute the ordered pair into each equation in the system. 2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. Example 11.1 Determining Whether an Ordered Pair Is a Solution to a System of Equations Determine whether the ordered pair (5, 1) is a solution to the given system of equations. x + 3y = 8 2x − 9 = y Solution Substitute the ordered pair (5, 1) into both equations. (5) + 3(1) = 8 8 = 8 True 2(5) − 9 = (1) 1=1 True The ordered pair (5, 1) satisfies both equations, so it is the solution to the system. Analysis We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 11.4. Figure 11.4 11.1 Determine whether the ordered pair (8, 5) is a solution to the following system. 5x−4y = 20 2x + 1 = 3y This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1215 Solving Systems of Equations by Graphing There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes. Example 11.2 Solving a System of Equations in Two Variables by Graphing Solve the following system of equations by graphing. Identify the type of system. Solution Solve the first equation for y. Solve the second equation for y. 2x + y = −8 x − y = −1 2x + y = −8 y = −2x−8 x − y = −1 y = x + 1 Graph both equations on the same set of axes as in Figure 11.5. Figure 11.5 The lines appear to intersect at the point (−3,−2). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations. 2(−3) + (−2) = −8 −8 = −8 True (−3) − (−2) = −1 −1 = −1 True The solution to the system is the ordered pair (−3,−2), so the system is independent. 1216 Chapter 11 Systems of Equations and Inequalities 11.2 Solve the following system of equations by graphing. 2x − 5y = −25 −4x + 5y = 35 Can graphing be used if the system is inconsistent or dependent? Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system. Solving Systems of Equations by Substitution Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical. Given a system of two equations in two variables, solve using the substitution method. 1. Solve one of the two equations for one of the variables in terms of the other. 2. Substitute the expression for this variable into the second equation, then solve for the remaining variable. 3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an or
dered pair. 4. Check the solution in both equations. Example 11.3 Solving a System of Equations in Two Variables by Substitution Solve the following system of equations by substitution. Solution First, we will solve the first equation for y. − x + y = −5 2x − 5y = 1 −x + y = −5 y = x−5 Now we can substitute the expression x−5 for y in the second equation. 2x − 5y = 1 2x − 5(x − 5) = 1 2x − 5x + 25 = 1 − 3x = −24 x = 8 Now, we substitute x = 8 into the first equation and solve for y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1217 −(8) + y = −5 y = 3 Our solution is (8, 3). Check the solution by substituting (8, 3) into both equations8) + (3) = − 5 2x − 5y = 1 2(8) − 5(3) = 1 True True 11.3 Solve the following system of equations by substitution. x = y + 3 4 = 3x−2y Can the substitution method be used to solve any linear system in two variables? Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions. Solving Systems of Equations in Two Variables by the Addition Method A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition. Given a system of equations, solve using the addition method. 1. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right. 2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable. 3. Solve the resulting equation for the remaining variable. 4. Substitute that value into one of the original equations and solve for the second variable. 5. Check the solution by substituting the values into the other equation. Example 11.4 Solving a System by the Addition Method Solve the given system of equations by addition. 1218 Chapter 11 Systems of Equations and Inequalities x + 2y = −1 −x + y = 3 Solution Both equations are already set equal to a constant. Notice that the coefficient of x in the second equation, –1, is the opposite of the coefficient of x in the first equation, 1. We can add the two equations to eliminate x without needing to multiply by a constant. x + 2y = − 1 −x + y = 3 3y = 2 Now that we have eliminated x, we can solve the resulting equation for y. 3y = 2 y = 2 3 Then, we substitute this value for y into one of the original equations and solve for x The solution to this system is ⎛ ⎝− 7 3 ⎞ ⎠. , 2 3 Check the solution in the first equation. x + 2y = −1 ⎛ ⎝− 1 = −1 True Analysis We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 11.6 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1219 Figure 11.6 Example 11.5 Using the Addition Method When Multiplication of One Equation Is Required Solve the given system of equations by the addition method. 3x + 5y = −11 x − 2y = 11 Solution Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3x in it and the second equation has x. So if we multiply the second equation by −3, the x-terms will add to zero. x−2y = 11 −3(x−2y) = −3(11) Multiply both sides by −3. −3x + 6y = −33 Use the distributive property. Now, let’s add them. 3x + 5y = −11 −3x + 6y = −33 _______________ 11y = −44 y = −4 For the last step, we substitute y = −4 into one of the original equations and solve for x. 3x + 5y = − 11 3x + 5( − 4) = − 11 3x − 20 = − 11 3x = 9 x = 3 Our solution is the ordered pair (3, −4). See Figure 11.7. Check the solution in the original second equation. x − 2y = 11 (3) − 2( − 4) = 3 + 8 = 11 True 1220 Chapter 11 Systems of Equations and Inequalities Figure 11.7 11.4 Solve the system of equations by addition. 2x−7y = 2 3x + y = −20 Example 11.6 Using the Addition Method When Multiplication of Both Equations Is Required Solve the given system of equations in two variables by addition. 2x + 3y = −16 5x−10y = 30 Solution One equation has 2x and the other has 5x. The least common multiple is 10x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x by multiplying the first equation by −5 and the second equation by 2. Then, we add the two equations together. − 5(2x + 3y) = − 5(−16) − 10x − 15y = 80 2(5x − 10y) = 2(30) 10x − 20y = 60 −10x−15y = 80 10x−20y = 60 ________________ −35y = 140 y = −4 Substitute y = −4 into the original first equation. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1221 2x + 3(−4) = −16 2x − 12 = −16 2x = −4 x = −2 The solution is (−2, −4). Check it in the other equation. 5x−10y = 30 5(−2)−10(−4) = 30 −10 + 40 = 30 30 = 30 See Figure 11.8. Figure 11.8 Example 11.7 Using the Addition Method in Systems of Equations Containing Fractions Solve the given system of equations in two variables by addition Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. 6 + x 3 ⎞ ⎠ = 6(3) y ⎛ ⎝ 6 2x + y = 18 y ⎞ ⎛ ⎠ = 4(1) ⎝ 4 2x − y = 4 x 2 − 4 1222 Chapter 11 Systems of Equations and Inequalities Now multiply the second equation by −1 so that we can eliminate the x-variable. −1(2x − y) = −1(4) −2x + y = −4 Add the two equations to eliminate the x-variable and solve the resulting equation. Substitute y = 7 into the first equation. 2x + y = 18 −2x + y = −4 _____________ 2y = 14 y = 7 2x + (7) = 18 2x = 11 x = 11 2 = 7.5 The solution is ⎛ ⎝ 11 2 ⎞ ⎠. Check it in the other equation. , 7 x 2 11 2 2 11 11.5 Solve the system of equations by addition. 2x + 3y = 8 3x + 5y = 10 Identifying Inconsistent Systems of Equations Containing Two Variables Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12 = 0. Example 11.8 Solving an Inconsistent System of Equations Solve the following system of equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1223 x = 9−2y x + 2y = 13 Solution We can approach this problem in two ways. Because one equation is already solved for x, is to use substitution. the most obvious step x + 2y = 13 (9 − 2y) + 2y = 13 9 + 0y = 13 9 = 13 Clearly, this statement is a contradiction because 9 ≠ 13. Therefore, the system has no solution. The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows. x = 9−2y 2y = − We then convert the second equation expressed to slope-intercept form. x + 2y = 13 2y = − x + 13 y = − 1 2 x + 13 2 Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect + 13 2 Analysis Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 11.9. Figure 11.9 1224 Chapter 11 Systems of Equations and Inequalities 11.6 Solve the following system of equations in two variables. 2y−2x = 2 2y−2x = 6 Expressing the Solution of a System of Dependent Equations Containing Two Variables Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0 = 0. Example 11.9 Finding a Solution to a Dependent System of Linear Equations Find a solution to the system of equations using the addition method. x + 3y = 2 3x + 9y = 6 Solution With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x. If we multiply both sides of the first equation by −3, then we will be able to eliminate the x -variable. Now add the equations. x + 3y = 2 (−3)(x + 3y) = (−3)(2) −3x − 9y = − 6 − 3x − 9y = −6 + 3x + 9y = 6 ______________ 0 = 0 We can see that there will be an infinite number of solutions that satisfy both equations. Analysis If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form. x + 2 3 x + 3y = 2 3y = − x + 2 y = − 1 3 3x + 9y = 6 9y = −3x + See Figure 11.10. Notice the results are the same. The general solution to the system is ⎛ ⎝x, −1 3 x + 2 3 ⎞ ⎠. This content is available for free at https://cnx.org/content/col11758/1.5 C
hapter 11 Systems of Equations and Inequalities 1225 Figure 11.10 11.7 Solve the following system of equations in two variables. y−2x = 5 −3y + 6x = −15 Using Systems of Equations to Investigate Profits Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R = xp, where x = quantity and p = price. The revenue function is shown in orange in Figure 11.11. The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 11.11. The x -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars. Figure 11.11 The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money. 1226 Chapter 11 Systems of Equations and Inequalities The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P(x) = R(x) − C(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses. Example 11.10 Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C(x) = 0.85x + 35,000 and the revenue function R(x) = 1.55x, find the break-even point and the profit function. Solution Write the system of equations using y to replace function notation. y = 0.85x + 35,000 y = 1.55x Substitute the expression 0.85x + 35,000 from the first equation into the second equation and solve for x. 0.85x + 35,000 = 1.55x 35,000 = 0.7x 50,000 = x Then, we substitute x = 50,000 into either the cost function or the revenue function. 1.55(50,000) = 77,500 The break-even point is (50,000, 77,500). The profit function is found using the formula P(x) = R(x) − C(x). P(x) = 1.55x − (0.85x + 35, 000) = 0.7x − 35, 000 The profit function is P(x) = 0.7x−35,000. Analysis The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11.12. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1227 Figure 11.12 We see from the graph in Figure 11.13 that the profit function has a negative value until x = 50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss. Figure 11.13 Example 11.11 Writing and Solving a System of Equations in Two Variables 1228 Chapter 11 Systems of Equations and Inequalities The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? Solution Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. c + a = 2,000 The revenue from all children can be found by multiplying $25.00 by the number of children, 25c. The revenue from all adults can be found by multiplying $50.00 by the number of adults, 50a. The total revenue is $70,000. We can use this to write an equation for the revenue. We now have a system of linear equations in two variables. 25c + 50a = 70,000 c + a = 2,000 25c + 50a = 70,000 In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either c or a. We will solve for a. Substitute the expression 2,000 − c in the second equation for a and solve for c. c + a = 2,000 a = 2,000 − c 25c + 50(2,000 − c) = 70,000 25c + 100,000 − 50c = 70,000 − 25c = −30,000 c = 1,200 Substitute c = 1,200 into the first equation to solve for a. 1,200 + a = 2,000 a = 800 We find that 1,200 children and 800 adults bought tickets to the circus that day. Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were 11.8 bought for a total of $14,200, how many children and how many adults bought meal tickets? Access these online resources for additional instruction and practice with systems of linear equations. • Solving Systems of Equations Using Substitution (http://openstaxcollege.org/l/syssubst) • Solving Systems of Equations Using Elimination (http://openstaxcollege.org/l/syselim) • Applications of Systems of Equations (http://openstaxcollege.org/l/sysapp) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1229 11.1 EXERCISES Verbal Can a system of linear equations have exactly two 1. solutions? Explain why or why not. 2. If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins. 3. If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company? 4. If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point? Given a system of equations, explain at 5. different methods of solving that system. least two Algebraic For the following exercises, determine whether the given ordered pair is a solution to the system of equations. 15. 16. 17. 18. 19. 20. −2x + 3y = 1.2 −3x − 6y = 1.8 x−0.2y = 1 −10x + 2y = 5 3x + 5y = 9 30x + 50y = −90 −3x + y = 2 12x−4y = − = 16 = 11 y = 3 6. 7. 8. 9. 5x − y = 4 x + 6y = 2 and (4, 0) −3x − 5y = 13 − x + 4y = 10 and (−6, 1) 3x + 7y = 1 2x + 4y = 0 and (2, 3) −2x + 5y = 7 2x + 9y = 7 and (−1, 1) 10. x + 8y = 43 3x−2y = −1 and (3, 5) the following exercises, For substitution. solve each system by 11. 12. 13. 14. x + 3y = 5 2x + 3y = 4 3x−2y = 18 5x + 10y = −10 4x + 2y = −10 3x + 9y = 0 2x + 4y = −3.8 9x−5y = 1.3 For the following exercises, solve each system by addition. 21. 22. 23. 24. 25. 26. 27. 28. −2x + 5y = −42 7x + 2y = 30 6x−5y = −34 2x + 6y = 4 5x − y = −2.6 −4x−6y = 1.4 7x−2y = 3 4x + 5y = 3.25 −x + 2y = −1 5x−10y = 6 7x + 6y = 2 −28x−24y = − = − 43 120 1230 29. 30. −0.2x + 0.4y = 0.6 x−2y = −3 −0.1x + 0.2y = 0.6 5x−10y = 1 For the following exercises, solve each system by any method. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 5x + 9y = 16 x + 2y = 4 6x−8y = −0.6 3x + 2y = 0.9 5x−2y = 2.25 7x−4y = 3 x − 5 12 −6x + 5 2 y = − 55 12 y = 55 2 7x−4y = 7 6 2x + 4y = 1 3 3x + 6y = 11 2x + 4y = − 21 12 6 y = 2 y = −.2x + 1.3y = −0.1 4.2x + 4.2y = 2.1 0.1x + 0.2y = 2 0.35x−0.3y = 0 Graphical For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions. 41. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 3x − y = 0.6 x−2y = 1.3 42. −x + 2y = 4 2x−4y = 1 43. 44. 45. x + 2y = 7 2x + 6y = 12 3x−5y = 7 x−2y = 3 3x−2y = 5 −9x + 6y = −15 Technology For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth. 46. 47. 48. 49. 50. 0.1x + 0.2y = 0.3 −0.3x + 0.5y = 1 −0.01x + 0.12y = 0.62 0.15x + 0.20y = 0.52 0.5x + 0.3y = 4 0.25x−0.9y = 0.46 0.15x + 0.27y = 0.39 −0.34x + 0.56y = 1.8 −0.71x + 0.92y = 0.13 0.83x + 0.05y = 2.1 Extensions For the following exercises, solve each system in terms of A, B, C, D, E, and F where A – F are nonzero numbers. Note that A ≠ B and AE ≠ BD. 51. 52. 53 + Ay = 1 x + By = 1 Ax + y = 0 Bx + y = 1 54. Ax + By = C x + y = 1 Chapter 11 Systems of Equations and Inequalities 1231 55. Ax + By = C Dx + Ey = F Real-World Applications For the following exercises, solve for the desired quantity. 56. A stuffed animal business has a total cost of production C = 12x + 30 and a revenue function R = 20x. Find the break-even point. A fast-food restaurant has a cost of production 57. C(x) = 11x + 120 and a revenue function R(x) = 5x. When does the company start to turn a profit? A cell phone factory has a cost of production function 58. C(x) = 150x + 10, 000 and revenue R(x) = 200x. What is the break-even point? a 59. A musician charges C(x) = 64x + 20,000, where x is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? a cost factory A guitar production has 60. C(x) = 75x + 50,000. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function. of For the following exercises, use a system of equations with two variables and two equations to solve. linear Find two numbers whose sum is 28 and difference is 61. 13. A number is 9 more than another number. Twice the 62. sum of the two numbers is 10. Find the two numbers. The startup cost for a restaurant is $120,000, and each 63. meal costs $10 for the restaurant to
make. If each meal is then sold for $15, after how many meals does the restaurant break even? A moving company charges a flat rate of $150, and an 64. additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost? A total of 1,595 first- and second-year college students 65. gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15. How many freshmen and sophomores were in attendance? 276 students enrolled in a freshman-level chemistry 66. class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed. There were 130 faculty at a conference. If there were 67. 18 more women than men attending, how many of each gender attended the conference? A jeep and BMW enter a highway running east-west at 68. the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control. If a scientist mixed 10% saline solution with 60% 69. saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed? An investor earned triple the profits of what she earned 70. last year. If she made $500,000.48 total for both years, how much did she earn in profits each year? An investor who dabbles in real estate invested 1.1 71. million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals? If an investor invests a total of $25,000 into two bonds, 72. one that pays 3% simple interest, and the other that pays 27 % interest, and the investor earns $737.50 annual 8 interest, how much was invested in each account? If an investor invests $23,000 into two bonds, one that 73. pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account? CDs cost $5.96 more than DVDs at All Bets Are Off 74. Electronics. How much would 6 CDs and 2 DVDs cost if 5 CDs and 2 DVDs cost $127.73? A store clerk sold 60 pairs of sneakers. The high-tops 75. sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold? A concert manager counted 350 ticket receipts the day 76. after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold? Admission into an amusement park for 4 children and 2 77. adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and 1232 Chapter 11 Systems of Equations and Inequalities adults, what is the price of the child’s ticket and the price of the adult ticket? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1233 11.2 \| Systems of Linear Equations: Three Variables Learning Objectives In this section, you will: 11.2.1 Solve systems of three equations in three variables. 11.2.2 Identify inconsistent systems of equations containing three variables. 11.2.3 Express the solution of a system of dependent equations containing three variables. Figure 11.14 (credit: “Elembis,” Wikimedia Commons) John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund? Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. Solving Systems of Three Equations in Three Variables In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution (x, y, z), which we call an ordered triple. A system in upper triangular form looks like the following: Ax + By + Cz = D Ey + Fz = G Hz = K The third equation can be solved for z, and then we back-substitute to find y and x. To write the system in upper triangular form, we can perform the following operations: 1. Interchange the order of any two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a nonzero multiple of one equation to another equation. 1234 Chapter 11 Systems of Equations and Inequalities The solution set to a three-by-three system is an ordered triple ⎧ ⎬. Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes. ⎨(x, y, z)⎫ ⎩ ⎭ Number of Possible Solutions Figure 11.15 and Figure 11.16 illustrate possible solution scenarios for three-by-three systems. • Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ⎬. Graphically, the ordered triple defines a point that is the intersection of three planes ordered triple ⎧ ⎨(x, y, z)⎫ ⎭ ⎩ in space. • Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as 0 = 0. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space. • Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as 3 = 0. Graphically, a system with no solution is represented by three planes with no point in common. Figure 11.15 (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions. Figure 11.16 All three figures represent three-by-three systems with no solution. (a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. (c) All three planes are parallel, so there is no point of intersection. Example 11.12 Determining Whether an Ordered Triple Is a Solution to a System Determine whether the ordered triple (3, −2, 1) is a solution to the system. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1235 x + y + z = 2 6x − 4y + 5z = 31 5x + 2y + 2z = 13 Solution We will check each equation by substituting in the values of the ordered triple for x, y, and z. x + y + z = 2 (3) + (−2) + (1) = 2 True 6x−4y + 5z = 31 6(3)−4(−2) + 5(1) = 31 18 + 8 + 5 = 31 True 5x + 2y + 2z = 13 5(3) + 2(−2) + 2(1) = 13 15−4 + 2 = 13 True The ordered triple (3, −2, 1) is indeed a solution to the system. Given a linear system of three equations, solve for three unknowns. 1. Pick any pair of equations and solve for one variable. 2. Pick another pair of equations and solve for the same variable. 3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system. 4. Back-substitute known variables into any one of the original equations and solve for the missing variable. Example 11.13 Solving a System of Three Equations in Three Variables by Elimination Find a solution to the following system: x−2y + 3z = 9 (1) − x + 3y − z = −6 (2) 2x−5y + 5z = 17 (3) Solution There will always be several choices as to where to begin, but the most obvious first step here is to eliminate x by adding equations (1) and (2). x − 2y + 3z = 9 (1) − x + 3y − z = −6 (2) y + 2z = 3 (3) The second step is multiplying equation (1) by −2 and adding the result to equation (3). These two steps will eliminate the variable x. −2x + 4y − 6z = −18 (1) multiplied by − 2 2x − 5y + 5z = 17 (3) ____________________________________ − y − z = −1 (5) In equations (4) and (5), we have created a new two-by-two system. We can solve for z by adding the two equations. 1236 Chapter 11 Systems of Equations and Inequalities y + 2z = 3 (4) −y − z = − 1 (5) z = 2 (6) Choosing one equation from each new system, we obtain the upper triangular form: x−2y + 3z = 9 (1) y + 2z = 3 (4) z = 2 (6) Next, we back-substitute z = 2 into equation (4) and solve for y. y + 2(21 Finally, we can back-substitute z = 2 and y = −
1 into equation (1). This will yield the solution for x. x−2(−1) + 3(2 The solution is the ordered triple (1, −1, 2). See Figure 11.17. Figure 11.17 Example 11.14 Solving a Real-World Problem Using a System of Three Equations in Three Variables In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund? Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1237 To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts: x = amount invested in money-market fund y = amount invested in municipal bonds z = amount invested in mutual funds The first equation indicates that the sum of the three principal amounts is $12,000. x + y + z = 12,000 We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds. z = y + 4,000 The third equation shows that the total amount of interest earned from each fund equals $670. Then, we write the three equations as a system. 0.03x + 0.04y + 0.07z = 670 x + y + z = 12,000 − y + z = 4,000 0.03x + 0.04y + 0.07z = 670 To make the calculations simpler, we can multiply the third equation by 100. Thus, x + y + z = 12,000 (1) − y + z = 4,000 (2) 3x + 4y + 7z = 67,000 (3) Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up. x + y + z = 12,000 3x + 4y + 7z = 67,000 − y + z = 4,000 Step 2. Multiply equation (1) by −3 and add to equation (2). Write the result as row 2. x + y + z = 12,000 y + 4z = 31,000 − y + z = 4,000 Step 3. Add equation (2) to equation (3) and write the result as equation (3). x + y + z = 12,000 y + 4z = 31,000 5z = 35,000 Step 4. Solve for z in equation (3). Back-substitute that value in equation (2) and solve for y. Then, backsubstitute the values for z and y into equation (1) and solve for x. 1238 Chapter 11 Systems of Equations and Inequalities 5z = 35,000 z = 7,000 y + 4(7,000) = 31,000 y = 3,000 x + 3,000 + 7,000 = 12,000 x = 2,000 John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds. 11.9 Solve the system of equations in three variables. 2x + y−2z = −1 3x−3y − z = 5 x−2y + 3z = 6 Identifying Inconsistent Systems of Equations Containing Three Variables Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as 3 = 7 or some other contradiction. Example 11.15 Solving an Inconsistent System of Three Equations in Three Variables Solve the following system. x−3y + z = 4 (1) − x + 2y−5z = 3 (2) 5x−13y + 13z = 8 (3) Solution Looking at the coefficients of x, we can see that we can eliminate x by adding equation (1) to equation (2). x−3y + z = 4 (1) −x + 2y−5z = 3 (2) − y−4z = 7 (4) Next, we multiply equation (1) by −5 and add it to equation (3). −5x + 15y − 5z = −20 5x − 13y + 13z = 8 ______________________________________ 2y + 8z = −12 (1) multiplied by −5 (3) (5) Then, we multiply equation (4) by 2 and add it to equation (5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1239 −2y − 8z = 14 (4) multiplied by 2 2y + 8z = − 12 (5) _______________________________________ 0 = 2 The final equation 0 = 2 is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution. Analysis In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent. 11.10 Solve the system of three equations in three variables. x + y + z = 2 y−3z = 1 2x + y + 5z = 0 Expressing the Solution of a System of Dependent Equations Containing Three Variables We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line. Example 11.16 Finding the Solution to a Dependent System of Equations Find the solution to the given system of three equations in three variables. 2x + y−3z = 0 (1) 4x + 2y−6z = 0 (2) x − y + z = 0 (3) Solution First, we can multiply equation (1) by −2 and add it to equation (2). −4x−2y + 6z = 0 equation (1) multiplied by −2 4x + 2y−6z = 0 (2) ____________________________________________ 0 = 0 We do not need to proceed any further. The result we get is an identity, 0 = 0, which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by −2, and adding it to equation (1). We then perform the same steps as above and find the same result, 0 = 0. When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have 1240 Chapter 11 Systems of Equations and Inequalities We then solve the resulting equation for z. 2x + y−3z = 0 x − y + z = 0 _____________ 3x−2z = 0 3x−2z = 0 z = 3 2 x We back-substitute the expression for z into one of the equations and solve for y. x⎞ ⎠ = 0 ⎛ 2x + y − 3 ⎝ 3 2 2x + − 2x x So the general solution is ⎛ x, 3 2 dependent on the value selected for x. ⎝x, 5 2 x⎞ ⎠. In this solution, x can be any real number. The values of y and z are Analysis As shown in Figure 11.18, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations. Figure 11.18 Does the generic solution to a dependent system always have to be written in terms of x? No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of x and if needed x and y. 11.11 Solve the following system. x + y + z = 7 3x − 2y − z = 4 x + 6y + 5z = 24 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1241 Access these online resources for additional instruction and practice with systems of equations in three variables. • Ex 1: System of Three Equations with Three Unknowns Using Elimination (http://openstaxcollege.org/l/systhree) • Ex. 2: System of Three Equations with Three Unknowns Using Elimination (http://openstaxcollege.org/l/systhelim) 1242 Chapter 11 Systems of Equations and Inequalities 11.2 EXERCISES Verbal Can a linear system of three equations have exactly two 78. solutions? Explain why or why not 79. If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique. 80. If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example. Using the method of addition, is there only one way to 81. solve the system? 82. Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not. Algebraic For the following exercises, determine whether the ordered triple given is the solution to the system of equations. 83. 84. 85. 86. 2x−6y + 6z = −12 x + 4y + 5z = −1 −x + 2y + 3z = −1 and (0, 1, −1) 6x − y + 3z = 6 3x + 5y + 2z = 0 x + y = 0 and (3, −3, −5) 6x−7y + z = 2 −x − y + 3z = 4 2x + y − z = 1 and (4, 2, −61 and (4, 4, −1) 87. −x − y + 2z = 3 5x + 8y−3z = 4 −x + 3y−5z = −5 and (4, 1, −7) the following exercises, For substitution. solve each system by 3x−4y + 2z = −15 2x + 4y + z = 16 2x + 3y + 5z = 20 88. 89. This content is available for free at https://cnx.org/content/col11758/1.5 5x−2y + 3z = 20 2x−4y−3z = −9 x + 6y−8z = 21 90. 91. 92. 5x + 2y + 4z = 9 −3x + 2y + z = 10 4x−3y + 5z = −3 4x−3y + 5z = 31 −x + 2y + 4z = 20 x + 5y−2z = −29 5x−2y + 3z = 4 −4x + 6y−7z = −1 3x + 2y − z = 4 93. 4x + 6y + 9z = 0 −5x + 2y−6z = 3 7x−4y + 3z = −3 For the following exercises, solve each system by Gaussian elimination. 2x − y + 3z = 17 −5x + 4y−2z = −46 2y + 5z = −7 5x−6y + 3z = 50 − x + 4y = 10 2x − z = 10 2x + 3y−6z = 1 −4x−6y + 12z = −2 x + 2y + 5z = 10 4x + 6y−2z = 8 6x + 9y−3z = 12 −2x−3y + z = −4 2x + 3y−4z = 5 −3x + 2y + z = 11 −x + 5y + 3z = 4 10x + 2y−14z = 8 −x−2y−4z = −1 −12x−6y + 6z = −12 94. 95. 96. 97. 98. 99. 100. Chapter 11 Systems of Equations and Inequalities 1243 x + y + z = 14 2y + 3z = −14 −16y−24z = −112 101. 102. 103. 104. 5x−3y + 4z = −1 −4x + 2y−3z = 0 −x + 5y + 7z = −11 x + y + z = 0 2x − y + 3z = 0 x − z = 0 3x + 2y−5z = 6 5x−4y + 3z = −12 4x + 5y−2z = 15 x + y + z = 0 2x − y + 3z = 0 x − z = 1 105. 3x − 1 2 y − z = − 1 2 4x + 106. 107. 6x−5y + 6z = 38 4x − 74 = − 13 10 z = − 7 20 z = − 5 4 108 10 2 1 2 1 4 1 8 109. 1108 z = −5 111 = − 23 3 z = 0 112 = 55 12 z = 5 3 113. 114. 115. 116. 117. 118. 119. 1 40 − 1 2 x + 3 12 x + 1 y + 1 80 60 16 3 8 z = 1 100 z = − 1 5 z = 3 20 0.1x−0.2y + 0.3z = 2 0.5
x−0.1y + 0.4z = 8 0.7x−0.2y + 0.3z = 8 0.2x + 0.1y−0.3z = 0.2 0.8x + 0.4y−1.2z = 0.1 1.6x + 0.8y−2.4z = 0.2 1.1x + 0.7y−3.1z = −1.79 2.1x + 0.5y−1.6z = −0.13 0.5x + 0.4y−0.5z = −0.07 0.5x−0.5y + 0.5z = 10 0.2x−0.2y + 0.2z = 4 0.1x−0.1y + 0.1z = 2 0.1x + 0.2y + 0.3z = 0.37 0.1x−0.2y−0.3z = −0.27 0.5x−0.1y−0.3z = −0.03 0.5x−0.5y−0.3z = 0.13 0.4x−0.1y−0.3z = 0.11 0.2x−0.8y−0.9z = −0.32 120. 0.5x + 0.2y−0.3z = 1 0.4x−0.6y + 0.7z = 0.8 0.3x−0.1y−0.9z = 0.6 1244 121. 122. 0.3x + 0.3y + 0.5z = 0.6 0.4x + 0.4y + 0.4z = 1.8 0.4x + 0.2y + 0.1z = 1.6 0.8x + 0.8y + 0.8z = 2.4 0.3x−0.5y + 0.2z = 0 0.1x + 0.2y + 0.3z = 0.6 Extensions For the following exercises, solve the system for x, y, and z. 123−1 2 x−2 3 y−−3 3 = 2 3 = 0 + + + 124. 5x−3y − z + 1 2 = 1 2 + 2z = −3 y−9 2 −4y + z = 4 6x + x + 8 2 125. 126. 127. x + 4 7 x−− 12 −−5 2 y−3 2 + − − + z−− 4x + 3y−2z = 11 0.02x + 0.015y−0.01z = 0.065 = 1 Real-World Applications Three even numbers sum up to 108. The smaller is 128. half the larger and the middle number is 3 4 the larger. What are the three numbers? Three numbers sum up to 147. The smallest number is 129. half the middle number, which is half the largest number. What are the three numbers? Chapter 11 Systems of Equations and Inequalities attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance? An animal shelter has a total of 350 animals 131. comprised of cats, dogs, and rabbits. If the number of rabbits is 5 less than one-half the number of cats, and there are 20 more cats than dogs, how many of each animal are at the shelter? Your roommate, Sarah, offered to buy groceries for 132. you and your other roommate. The total bill was $82. She forgot to save the individual receipts but remembered that your groceries were $0.05 cheaper than half of her groceries, and that your other roommate’s groceries were $2.10 more than your groceries. How much was each of your share of the groceries? Your roommate, John, offered to buy household 133. supplies for you and your other roommate. You live near the border of three states, each of which has a different sales tax. The total amount of money spent was $100.75. Your supplies were bought with 5% tax, John’s with 8% tax, and your third roommate’s with 9% sales tax. The total amount of money spent without taxes is $93.50. If your supplies before tax were $1 more than half of what your third roommate’s supplies were before tax, how much did each of you spend? Give your answer both with and without taxes. Three coworkers work for the same employer. Their 134. jobs are warehouse manager, office manager, and truck driver. The sum of the annual salaries of the warehouse manager and office manager is $82,000. The office manager makes $4,000 more than the truck driver annually. The annual salaries of the warehouse manager and the truck driver total $78,000. What is the annual salary of each of the co-workers? At a carnival, $2,914.25 in receipts were taken at the 135. end of the day. The cost of a child’s ticket was $20.50, an adult ticket was $29.75, and a senior citizen ticket was $15.25. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold? A local band sells out for their concert. They sell all 136. 1,175 tickets for a total purse of $28,112.50. The tickets were priced at $20 for student tickets, $22.50 for children, and $29 for adult tickets. If the band sold twice as many adult as children tickets, how many of each type was sold? In a bag, a child has 325 coins worth $19.50. There 137. were three types of coins: pennies, nickels, and dimes. If the bag contained the same number of nickels as dimes, how many of each type of coin was in the bag? 130. consisting of children, parents, and grandparents, At a family reunion, there were only blood relatives, in 138. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1245 production, and Saudi Arabia produced 2% more oil than Russia. What percent of the world oil production did Saudi Arabia, the United States, and Russia produce?[2] The top three sources of oil imports for the United 144. States in the same year were Saudi Arabia, Mexico, and Canada. The three top countries accounted for 47% of oil imports. The United States imported 1.8% more from Saudi Arabia than they did from Mexico, and 1.7% more from Saudi Arabia than they did from Canada. What percent of imports were from these three the United States oil countries?[3] The top three oil producers in the United States in a 145. certain year are the Gulf of Mexico, Texas, and Alaska. The three regions were responsible for 64% of the United States oil production. The Gulf of Mexico and Texas combined for 47% of oil production. Texas produced 3% more than Alaska. What percent of United States oil production came from these regions?[4] At one time, in the United States, 398 species of 146. animals were on the endangered species list. The top groups were mammals, birds, and fish, which comprised 55% of the endangered species. Birds accounted for 0.7% more than fish, and fish accounted for 1.5% more than mammals. What percent of the endangered species came from mammals, birds, and fish? Meat consumption in the United States can be broken 147. into three categories: red meat, poultry, and fish. If fish makes up 4% less than one-quarter of poultry consumption, and red meat consumption is 18.2% higher than poultry of meat consumption, what consumption?[5] percentages the are Last year, at Haven’s Pond Car Dealership, for a particular model of BMW, Jeep, and Toyota, one could purchase all three cars for a total of $140,000. This year, due to inflation, the same cars would cost $151,830. The cost of the BMW increased by 8%, the Jeep by 5%, and the Toyota by 12%. If the price of last year’s Jeep was $7,000 less than the price of last year’s BMW, what was the price of each of the three cars last year? A recent college graduate took advantage of his 139. business education and invested in three investments immediately after graduating. He invested $80,500 into three accounts, one that paid 4% simple interest, one that % simple interest, and one that paid 21 paid 31 % simple 2 8 interest. He earned $2,670 interest at the end of one year. If the amount of the money invested in the second account was four times the amount invested in the third account, how much was invested in each account? You inherit one million dollars. You invest it all in 140. three accounts for one year. The first account pays 3% compounded annually, the second account pays 4% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $34,000 in interest. If you invest four times the money into the account that pays 3% compared to 2%, how much did you invest in each account? You inherit one hundred thousand dollars. You invest 141. it all in three accounts for one year. The first account pays 4% compounded annually, the second account pays 3% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $3,650 in interest. If you invest five times the money in the account that pays 4% compared to 3%, how much did you invest in each account? The top three countries in oil consumption in a certain 142. year are as follows: the United States, Japan, and China. In millions of barrels per day, the three top countries consumed 39.8% of the world’s consumed oil. The United States consumed 0.7% more than four times China’s consumption. The United States consumed 5% more than triple Japan’s consumption. What percent of the world oil consumption did the United States, Japan, and China consume?[1] The top three countries in oil production in the same 143. year are Saudi Arabia, the United States, and Russia. In millions of barrels per day, the top three countries produced 31.4% of the world’s produced oil. Saudi Arabia and the the world’s United States combined for 22.1% of 1. “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 2. “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 3. “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 4. “USA: The coming global oil crisis,” accessed April 6, 2014, http://www.oilcrisis.com/us/. 5. “The United States Meat Industry at a Glance,” accessed April 6, 2014, http://www.meatami.com/ht/d/sp/i/47465/pid/ 47465. 1246 Chapter 11 Systems of Equations and Inequalities 11.3 \| Systems of Nonlinear Equations and Inequalities: Two Variables Learning Objectives In this section, you will: 11.3.1 Solve a system of nonlinear equations using substitution. 11.3.2 Solve a system of nonlinear equations using elimination. 11.3.3 Graph a nonlinear inequality. 11.3.4 Graph a system of nonlinear inequalities. Halley’s Comet (Figure 11.19) orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear. Figure 11.19 Halley’s Comet (credit: "NASA Blueshift"/Flickr) In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations. Solving a System of Nonlinear Equations Using Substitution A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is
not linear. Recall that a linear equation can take the form Ax + By + C = 0. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes. Intersection of a Parabola and a Line There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line. Possible Types of Solutions for Points of Intersection of a Parabola and a Line Figure 11.20 illustrates possible solution sets for a system of equations involving a parabola and a line. • No solution. The line will never intersect the parabola. • One solution. The line is tangent to the parabola and intersects the parabola at exactly one point. • Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1247 Figure 11.20 Given a system of equations containing a line and a parabola, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the parabola equation. 3. Solve for the remaining variable. 4. Check your solutions in both equations. Example 11.17 Solving a System of Nonlinear Equations Representing a Parabola and a Line Solve the system of equations. x − y = −1 y = x2 + 1 Solution Solve the first equation for x and then substitute the resulting expression into the second equation. x − y = −1 x = y−1 y = x2 + 1 y = (y−1)2 + 1 Solve for x. Substitute expression for x. Expand the equation and set it equal to zero. y = (y−1)2 = (y2 −2y + 1) + 1 = y2 −2y + 2 0 = y2 −3y + 2 = (y−2)(y−1) 1248 Chapter 11 Systems of Equations and Inequalities Solving for y gives y = 2 and y = 1. Next, substitute each value for y into the first equation to solve for x. Always substitute the value into the linear equation to check for extraneous solutions. x − y = −1 x − (2) = −1 x = 1 x − (1) = −1 x = 0 The solutions are (1, 2) and (0, 1), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 11.21. Figure 11.21 Could we have substituted values for y into the second equation to solve for x in Example 11.17? Yes, but because x is squared in the second equation this could give us extraneous solutions for x. For y = 1 This gives us the same value as in the solution. For y = 2 y = x2 + 1 y = x2 + 1 x2 = 0 x = ± 0 = 0 y = x2 + 1 2 = x2 + 1 x2 = 1 x = ± 1 = ± 1 Notice that −1 is an extraneous solution. 11.12 Solve the given system of equations by substitution. 3x − y = −2 2x2 − y = 0 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1249 Intersection of a Circle and a Line Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line. Possible Types of Solutions for the Points of Intersection of a Circle and a Line Figure 11.22 illustrates possible solution sets for a system of equations involving a circle and a line. • No solution. The line does not intersect the circle. • One solution. The line is tangent to the circle and intersects the circle at exactly one point. • Two solutions. The line crosses the circle and intersects it at two points. Figure 11.22 Given a system of equations containing a line and a circle, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the equation for the circle. 3. Solve for the remaining variable. 4. Check your solutions in both equations. Example 11.18 Finding the Intersection of a Circle and a Line by Substitution Find the intersection of the given circle and the given line by substitution. x2 + y2 = 5 y = 3x−5 Solution One of the equations has already been solved for y. We will substitute y = 3x−5 into the equation for the circle. x2 + (3x−5)2 = 5 x2 + 9x2 −30x + 25 = 5 10x2 −30x + 20 = 0 1250 Chapter 11 Systems of Equations and Inequalities Now, we factor and solve for x. 10(x2 − 3x + 2) = 0 10(x − 2)(x − 1) = 0 x = 2 x = 1 Substitute the two x-values into the original linear equation to solve for y. y = 3(2)−5 = 1 y = 3(1)−5 = −2 The line intersects the circle at (2, 1) and (1, −2), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 11.23. Figure 11.23 11.13 Solve the system of nonlinear equations. x2 + y2 = 10 x−3y = −10 Solving a System of Nonlinear Equations Using Elimination We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse. Possible Types of Solutions for the Points of Intersection of a Circle and an Ellipse Figure 11.24 illustrates possible solution sets for a system of equations involving a circle and an ellipse. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1251 • No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other. • One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point. • Two solutions. The circle and the ellipse intersect at two points. • Three solutions. The circle and the ellipse intersect at three points. • Four solutions. The circle and the ellipse intersect at four points. Figure 11.24 Example 11.19 Solving a System of Nonlinear Equations Representing a Circle and an Ellipse Solve the system of nonlinear equations. x2 + y2 = 26 (1) 3x2 + 25y2 = 100 (2) Solution Let’s begin by multiplying equation (1) by −3, and adding it to equation (2). ( − 3)(x2 + y2) = ( − 3)(26) − 3x2 − 3y2 = − 78 3x2 + 25y2 = 100 22y2 = 22 After we add the two equations together, we solve for y. Substitute y = ± 1 into one of the equations and solve for x. y2 = 1 y = ± 1 = ± 1 1252 Chapter 11 Systems of Equations and Inequalities x2 + (1)2 = 26 x2 + 1 = 26 x2 = 25 x = ± 25 = ± 5 x2 + (−1)2 = 26 x2 + 1 = 26 x2 = 25 = ± 5 There are four solutions: (5, 1), (−5, 1), (5, −1), and (−5, −1). See Figure 11.25. Figure 11.25 11.14 Find the solution set for the given system of nonlinear equations. 4x2 + y2 = 13 x2 + y2 = 10 Graphing a Nonlinear Inequality All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality. Recall that when the inequality is greater than, y > a, or less than, y < a, the inequality is greater than or equal to, y ≥ a, or less than or equal to, y ≤ a, graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See Figure 11.26. the graph is drawn with a dashed line. When the graph is drawn with a solid line. The This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1253 Figure 11.26 (a) an example of y > a; (b) an example of y ≥ a; (c) an example of y < a; (d) an example of y ≤ a Given an inequality bounded by a parabola, sketch a graph. 1. Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set. 2. 3. If the boundary is included in the region (the operator is ≤ or ≥ ), the parabola is graphed as a solid line. If the boundary is not included in the region (the operator is < or >), the parabola is graphed as a dashed line. 4. Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line. 5. Shade the region representing the solution set. Example 11.20 Graphing an Inequality for a Parabola Graph the inequality y > x2 + 1. Solution First, graph the corresponding equation y = x2 + 1. Since y > x2 + 1 has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points (0, 2) and (2, 0). One point is clearly inside the parabola and the other point is clearly outside. 1254 Chapter 11 Systems of Equations and Inequalities y > x2 + 1 2 > (0)2 + 1 2 > 1 True 0 > (2)2 + 1 0 > 5 False The graph is shown in Figure 11.27. We can see that the solution set consists of all points inside the parabola, but not on the graph itself. Figure 11.27 Graphing a System of Nonlinear Inequalities Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities
is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region. Given a system of nonlinear inequalities, sketch a graph. 1. Find the intersection points by solving the corresponding system of nonlinear equations. 2. Graph the nonlinear equations. 3. Find the shaded regions of each inequality. 4. Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality. Example 11.21 Graphing a System of Inequalities Graph the given system of inequalities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1255 x2 − y ≤ 0 2x2 + y ≤ 12 Solution These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable y will be eliminated. Then we solve for x. x2 − y = 0 2x2 + y = 12 ____________ 3x2 = 12 x2 = 4 x = ± 2 Substitute the x-values into one of the equations and solve for y. x2 − y = 0 (2) (−2) The two points of intersection are (2, 4) and (−2, 4). Notice that the equations can be rewritten as follows. x2 − y ≤ 0 x2 ≤ y y ≥ x2 2x2 + y ≤ 12 y ≤ −2x2 + 12 Graph each inequality. See Figure 11.28. The feasible region is the region between the two equations bounded by 2x2 + y ≤ 12 on the top and x2 − y ≤ 0 on the bottom. 1256 Chapter 11 Systems of Equations and Inequalities Figure 11.28 11.15 Graph the given system of inequalities. y ≥ x2 − 1 x − y ≥ − 1 Access these online resources for additional instruction and practice with nonlinear equations. • Solve a System of Nonlinear Equations Using Substitution (http://openstaxcollege.org/l/ nonlinsub) • Solve a System of Nonlinear Equations Using Elimination (http://openstaxcollege.org/l/ nonlinelim) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1257 11.3 EXERCISES Verbal 148. Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers. 149. When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution? 150. When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region? If you graph a revenue and cost function, explain how 151. to determine in what regions there is profit. 152. If you perform your break-even analysis and there is more than one solution, explain how you would determine which x-values are profit and which are not. Algebraic For the following exercises, solve the system of nonlinear equations using substitution. 153. x + y = 4 x2 + y2 = 9 154. y = x−3 x2 + y2 = 9 155. 156. y = x x2 + y2 = 9 y = − x x2 + y2 = 9 157. x = 2 x2 − y2 = 9 For the following exercises, solve the system of nonlinear equations using elimination. 4x2 −9y2 = 36 4x2 + 9y2 = 36 x2 + y2 = 25 x2 − y2 = 1 158. 159. 160. 2x2 + 4y2 = 4 2x2 −4y2 = 25x−10 161. 162. y2 − x2 = 9 3x2 + 2y2 = 8 x2 + y2 + 1 16 y = 2x2 = 2500 For the following exercises, use any method to solve the system of nonlinear equations. 163. −2x2 + y = −5 6x − y = 9 164. −x2 + y = 2 − x + y = 2 165. 166. 167. 168. 169. 170. x2 + y2 = 1 y = 20x2 −1 x2 + y2 = 1 y = − x2 2x3 − x2 = y y = 1 2 − x 9x2 + 25y2 = 225 (x−6)2 + y2 = 1 x4 − x2 = y x2 + y = 0 2x3 − x2 = y x2 + y = 0 For the following exercises, use any method to solve the nonlinear system. 171. x2 + y2 = 9 y = 3 − x2 172. x2 − y2 = 9 x = 3 1258 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. x2 − y2 = 9 y = 3 x2 − y2 = 9 x − y = 0 −x2 + y = 2 −4x + y = −1 −x2 + y = 2 2y = − x x2 + y2 = 25 x2 − y2 = 36 x2 + y2 = 1 y2 = x2 16x2 −9y2 + 144 = 0 y2 + x2 = 16 3x2 − y2 = 12 (x−1)2 + y2 = 1 3x2 − y2 = 12 (x−1)2 + y2 = 4 3x2 − y2 = 12 x2 + y2 = 16 x2 − y2 − 6x − 4y − 11 = 0 − x2 + y2 = 5 x2 + y2 −6y = 7 x2 + y = 1 185. x2 + y2 = 6 xy = 1 Graphical For the following exercises, graph the inequality. 186. x2 + y < 9 187. x2 + y2 < 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities the following exercises, graph the system of For inequalities. Label all points of intersection. 188. x2 + y < 1 y > 2x 189. x2 + y < −5 y > 5x + 10 190. 191. 192. x2 + y2 < 25 3x2 − y2 > 12 x2 − y2 > −4 x2 + y2 < 12 x2 + 3y2 > 16 3x2 − y2 < 1 Extensions For the following exercises, graph the inequality. 193. y ≥ e x y ≤ ln(x) + 5 194. y ≤ − log(x) y ≤ e x For the following exercises, find the solutions to the nonlinear equations with two variables. 195. 196. 197. 198. 4 x2 + 1 x2 − 2 5 y2 = 24 y2 + 4 = 0 6 x2 − 1 x2 − 6 y2 = 8 y2 = 1 8 1 x2 − xy + y2 −2 = 0 x + 3y = 4 x2 − xy−2y2 −6 = 0 x2 + y2 = 1 199. x2 + 4xy−2y2 −6 = 0 x = y + 2 Chapter 11 Systems of Equations and Inequalities 1259 Technology the following exercises, system of For inequalities. Use a calculator to graph the system to confirm the answer. solve the 200. xy < 1 y > x 201. x2 + y < 3 y > 2x Real-World Applications For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. Two numbers add up to 300. One number is twice the 202. square of the other number. What are the numbers? The squares of two numbers add to 360. The second 203. number is half the value of the first number squared. What are the numbers? A laptop company has discovered their cost and 204. revenue functions for each day: C(x) = 3x2 −10x + 200 and R(x) = −2x2 + 100x + 50. If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit. 205. revenue A cell phone company has the following cost and functions: C(x) = 8x2 −600x + 21,500 and R(x) = −3x2 + 480x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. 1260 Chapter 11 Systems of Equations and Inequalities 11.4 \| Partial Fractions Learning Objectives In this section, you will: 11.4.1 Decompose P( x ) Q( x ) , where Q( x ) has only nonrepeated linear factors. 11.4.2 Decompose P( x ) Q( x ) , where Q( x ) has repeated linear factors. 11.4.3 Decompose P( x ) Q( x ) , where Q( x ) has a nonrepeated irreducible quadratic factor. 11.4.4 Decompose P( x ) Q( x ) , where Q( x ) has a repeated irreducible quadratic factor. Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions. Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression. Decomposing P(x) Q(x) Where Q(x) Has Only Nonrepeated Linear Factors Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions. For example, suppose we add the following fractions: We would first need to find a common denominator, (x + 2)(x−3). Next, we would write each expression with this common denominator and find the sum of the terms. 2 x−3 + −1 ⎝ x + 2 x − 3 2x + 4 − x + 3 (x + 2)(x − 3) = x + 7 x2 − x − 6 Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions. x + 7 x2 − x−6 Simplifie sum = 2 x−3 + −1 x + 2 Partial fraction decomposition We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator. When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of x2 − x−6 are (x−3)(x + 2), the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1261 Partial Fraction Decomposition of P(x) Q(x) : Q(x) Has Nonrepeated Linear Factors The partial fraction decomposition of P(x) Q(x) than the degree of Q(x) is when Q(x) has nonrepeated linear factors and the degree of P(x) is less P(x) Q(x) = A1 ⎝a1 x + b1 ⎛ ⎞ ⎠ + A2 ⎝a2 x
+ b2 ⎛ ⎞ ⎠ + A3 ⎝a3 x + b3 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An ⎝an x + bn ⎛ . ⎞ ⎠ Given a rational expression with distinct linear factors in the denominator, decompose it. 1. Use a variable for the original numerators, usually A, B, or C, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use An for each numerator A1 ⎝a1 x + b1 2. Multiply both sides of the equation by the common denominator to eliminate fractions. A2 ⎝a2 x + b2 An ⎝an x + bn P(x) Q(x. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.22 Decomposing a Rational Function with Distinct Linear Factors Decompose the given rational expression with distinct linear factors. 3x (x + 2)(x−1) Solution We will separate the denominator factors and give each numerator a symbolic label, like A, B , or C. 3x (x + 2)(x−1) = A (x + 2) + B (x−1) Multiply both sides of the equation by the common denominator to eliminate the fractions: ⎡ (x + 2)(x−1) ⎣ 3x (x + 2)(x−1) ⎡ ⎤ ⎦ = (x + 2)(x−1) ⎣ A (x + 2) ⎤ ⎦ + (x + 2)(x−1) ⎡ ⎣ B (x−1) ⎤ ⎦ The resulting equation is Expand the right side of the equation and collect like terms. 3x = A(x−1) + B(x + 2) 3x = Ax − A + Bx + 2B 3x = (A + B)x − A + 2B Set up a system of equations associating corresponding coefficients + 2B 1262 Chapter 11 Systems of Equations and Inequalities Add the two equations and solve for B + 2B 3 = 0 + 3B 1 = B Substitute B = 1 into one of the original equations in the system. 3 = A + 1 2 = A Thus, the partial fraction decomposition is 3x (x + 2)(x−1) = 2 (x + 2) + 1 (x−1) Another method to use to solve for A or B is by considering the equation that resulted from eliminating the fractions and substituting a value for x that will make either the A- or B-term equal 0. If we let x = 1, the A- term becomes 0 and we can simply solve for B. 3x = A(x − 1) + B(x + 2) 3(1) = A[(1) − 1] + B[(1) + 2] 3 = 0 + 3B 1 = B Next, either substitute B = 1 into the equation and solve for A, or make the B-term 0 by substituting x = −2 into the equation. 3x = A(x − 1) + B(x + 2) 3( − 2) = A[( − 2) − 1] + B[( − 2) + 2] − 6 = − 3A + 0 = A −6 −3 2 = A We obtain the same values for A and B using either method, so the decompositions are the same using either method. 3x (x + 2)(x−1) = 2 (x + 2) + 1 (x−1) Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics. 11.16 Find the partial fraction decomposition of the following expression. x (x−3)(x−2) Decomposing P(x) Q(x) Where Q(x) Has Repeated Linear Factors Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1263 Partial Fraction Decomposition of P(x) Q(x) : Q(x) Has Repeated Linear Factors The partial fraction decomposition of P(x) Q(x) of P(x) is less than the degree of Q(x), is , when Q(x) has a repeated linear factor occurring n times and the degree P(x) Q(x) = A1 (ax + b) + A2 (ax + b)2 + A3 (ax + b)3 + ⋅ ⋅ ⋅ + An (ax + b) n Write the denominator powers in increasing order. Given a rational expression with repeated linear factors, decompose it. 1. Use a variable like A, B, or C for the numerators and account for increasing powers of the denominators. P(x) Q(x) = A1 (ax + b) + A2 (ax + b)2 + . . . + An (ax + b) n 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.23 Decomposing with Repeated Linear Factors Decompose the given rational expression with repeated linear factors. −x2 + 2x + 4 x3 −4x2 + 4x Solution The denominator factors are x(x−2)2. To allow for the repeated factor of (x−2), the decomposition will include three denominators: x, (x−2), and (x−2)2. Thus, −x2 + 2x + 4 x3 −4x2 + 4x = A x + B (x−2) + C (x−2)2 Next, we multiply both sides by the common denominator. x(x−2)2 ⎡ ⎢−x2 + 2x + 4 x(x−2)2 ⎣ ⎡ ⎢A x + B ⎣ − x2 + 2x + 4 = A(x−2)2 + Bx(x−2) + Cx C (x−2)2 ⎤ ⎥ = ⎦ (x−2) + ⎤ ⎥x(x−2)2 ⎦ On the right side of the equation, we expand and collect like terms. −x2 + 2x + 4 = A(x2 − 4x + 4) + B(x2 − 2x) + Cx = Ax2 − 4Ax + 4A + Bx2 − 2Bx + Cx = (A + B)x2 + ( − 4A − 2B + C)x + 4A 1264 Chapter 11 Systems of Equations and Inequalities Next, we compare the coefficients of both sides. This will give the system of equations in three variables: −x2 + 2x + 4 = (A + B)x2 + (−4A−2B + C)x + 4A A + B = −1 (1) −4A−2B + C = 2 (2) 4A = 4 (3) Solving for A , we have Substitute A = 1 into equation (1). 4A = 4 A = 1 A + B = −1 (1) + B = −1 B = −2 Then, to solve for C, substitute the values for A and B into equation (2). −4A−2B + C = 2 −4(1)−2(−2) + C = 2 − Thus, −x2 + 2x + 4 x3 −4x2 + 4x = 1 x − 2 (x−2) + 2 (x−2)2 11.17 Find the partial fraction decomposition of the expression with repeated linear factors. 6x−11 (x−1)2 , Where Q(x) Has a Nonrepeated Irreducible Decomposing P(x) Q(x) Quadratic Factor So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators A, B, or C representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as Ax + B, Bx + C, etc. Decomposition of P(x) Q(x) : Q(x) Has a Nonrepeated Irreducible Quadratic Factor The partial fraction decomposition of P(x) Q(x) such that Q(x) has a nonrepeated irreducible quadratic factor and the degree of P(x) is less than the degree of Q(x) is written as P(x) Q(x) = A1 x + B1 ⎝a1 x2 + b1 x + c1 ⎛ ⎞ ⎠ + A2 x + B2 ⎝a2 x2 + b2 x + c2 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An x + Bn ⎛ ⎝an x2 + bn x + cn ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1265 The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: A, B, C, and so on. Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it. 1. Use variables such as A, B, or C for the constant numerators over linear factors, and linear expressions such as A1 x + B1, A2 x + B2, etc., for the numerators of each quadratic factor in the denominator. P(x) Q(x) = A ax + b + A1 x + B1 ⎝a1 x2 + b1 x + c1 ⎛ ⎞ ⎠ + A2 x + B2 ⎝a2 x2 + b2 x + c2 ⎛ ⎞ ⎠ + ⋅ ⋅ ⋅ + An x + Bn ⎛ ⎝an x2 + bn x + cn ⎞ ⎠ 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.24 Decomposing P(x) Q(x) Factor When Q(x) Contains a Nonrepeated Irreducible Quadratic Find a partial fraction decomposition of the given expression. 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ Solution We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus, 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ = A (x + 3) + Bx + C ⎛ ⎞ ⎝x2 + x + 2 ⎠ We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator. ⎡ ⎢ 8x2 + 12x − 20 (x + 3)(x2 + x + 2) (x + 3)(x2 + x + 2x + 3) ⎣ + Bx + C (x2 + x + 2) ⎤ ⎥(x + 3)(x2 + x + 2) ⎦ 8x2 + 12x − 20 = A(x2 + x + 2) + (Bx + C)(x + 3) Notice we could easily solve for A by choosing a value for x that will make the Bx + C term equal 0. Let x = −3 and substitute it into the equation. 8x2 + 12x − 20 = A(x2 + x + 2) + (Bx + C)(x + 3) 8( − 3)2 + 12( − 3) − 20 = A(( − 3)2 + ( − 3) + 2) + (B( − 3) + C)(( − 3) + 3) 16 = 8A A = 2 1266 Chapter 11 Systems of Equations and Inequalities Now that we know the value of A, substitute it back into the equation. Then expand the right side and collect like terms. 8x2 + 12x−20 = 2(x2 + x + 2) + (Bx + C)(x + 3) 8x2 + 12x−20 = 2x2 + 2x + 4 + Bx2 + 3B + Cx + 3C 8x2 + 12x−20 = (2 + B)x2 + (2 + 3B + C)x + (4 + 3C) Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations. 2 + B = 8 (1) 2 + 3B + C = 12 (2) 4 + 3C = −20 (3) Solve for B using equation (1) and solve for C using equation (3). (1 + 3C = −20 (3) 3C = −24 C = −8 Thus, the partial fraction decomposition of the expression is 8x2 + 12x−20 ⎛ ⎞ ⎝x2 + x + 2 (x + 3) ⎠ = 2 (x + 3) + 6x−8 ⎛ ⎞ ⎝x2 + x + 2 ⎠ Could we have just set up a system of equations to solve Example 11.24? Yes, we could have solved it by setting up a system of equations without solving for A first. The expansion on the right would be: 8x2 + 12x−20 = Ax2 + Ax + 2A + Bx2 + 3B + Cx + 3C 8x2 + 12x−20 = (A + B)x2 + (A + 3B + C)x + (2A + 3C) So the system of equations would be: A + B = 8 A + 3B + C = 12 2A + 3C = −20 Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic 11.18 factor. 5x2 −6x + 7 ⎞ ⎛ ⎝x2 + 1 (x−1) ⎠ Decomposing P(x) Q(x) When Q(x) Has a Repeated Irreducible Quadratic Factor Now that we can decompose a simplified rational expressio
n with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1267 decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers. Decomposition of P(x) Q(x) When Q(x) Has a Repeated Irreducible Quadratic Factor The partial fraction decomposition of P(x) Q(x) P(x) is less than the degree of Q(x), is , when Q(x) has a repeated irreducible quadratic factor and the degree of P(x) ⎝ax2 + bx + c⎞ ⎛ ⎠ n = A1 x + B1 ⎝ax2 + bx + c⎞ ⎛ ⎠ + A2 x + B2 ⎝ax2 + bx + c⎞ ⎛ ⎠ 2 + A3 x + B3 ⎝ax2 + bx + c⎞ ⎛ ⎠ 3 + ⋅ ⋅ ⋅ + An x + Bn ⎝ax2 + bx + c⎞ ⎛ ⎠ n Write the denominators in increasing powers. Given a rational expression that has a repeated irreducible factor, decompose it. 1. Use variables like A, B, or C for the constant numerators over linear factors, and linear expressions such as A1 x + B1, A2 x + B2, etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as P(x) Q(x) = A ax + b + A1 x + B1 (ax2 + bx + c) + A2 x + B2 (ax2 + bx + c)2 + ⋯ + An + Bn (ax2 + bx + c) n 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 11.25 Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator Decompose the given expression that has a repeated irreducible factor in the denominator. x4 + x3 + x2 − x + 1 ⎞ x⎛ ⎝x2 + 1 ⎠ 2 Solution The factors of the denominator are x, (x2 + 1), and (x2 + 1)2. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form Ax + B. So, let’s begin the decomposition. x4 + x3 + x2 − x + 1 ⎞ x⎛ ⎝x2 + 1 ⎠ 2 = A x + Bx + C ⎞ ⎛ ⎝x2 + 1 ⎠ + Dx + E 2 ⎞ ⎛ ⎝x2 + 1 ⎠ We eliminate the denominators by multiplying each term by x⎛ ⎞ ⎝x2 + 1 ⎠ 2 . Thus, 1268 Chapter 11 Systems of Equations and Inequalities x4 + x3 + x2 − x + 1 = A⎛ ⎞ ⎝x2 + 1 ⎠ 2 ⎞ ⎛ ⎝x2 + 1 ⎠ + (Dx + E)(x) + (Bx + C)(x) Expand the right side. x4 + x3 + x2 − x + 1 = A(x4 + 2x2 + 1) + Bx4 + Bx2 + Cx3 + Cx + Dx2 + Ex = Ax4 + 2Ax2 + A + Bx4 + Bx2 + Cx3 + Cx + Dx2 + Ex Now we will collect like terms. x4 + x3 + x2 − x + 1 = (A + B)x4 + (C)x3 + (2A + B + D)x2 + (C + E)x + A Set up the system of equations matching corresponding coefficients on each side of the equal sign. A + B = 1 C = 1 2A + 1 A = 1 We can use substitution from this point. Substitute A = 1 into the first equation. Substitute A = 1 and B = 0 into the third equation. 1 + B = 1 B = 0 Substitute C = 1 into the fourth equation. 2(11 1 + E = −1 E = −2 Now we have solved for all of the unknowns on the right side of the equal sign. We have A = 1, B = 0, C = 1, D = −1, and E = −2. We can write the decomposition as follows: x4 + x3 + x2 − x + 1 x⎛ ⎞ ⎝x2 + x2 + 1 ⎠ − x + 2 ⎛ ⎞ ⎝x2 + 1 ⎠ 2 11.19 Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor. x3 −4x2 + 9x−5 2 ⎞ ⎛ ⎝x2 −2x + 3 ⎠ Access these online resources for additional instruction and practice with partial fractions. • Partial Fraction Decomposition (http://openstaxcollege.org/l/partdecomp) • Partial Fraction Decomposition With Repeated Linear Factors (http://openstaxcollege.org/ l/partdecomprlf) • Partial Fraction Decomposition With Linear and Quadratic Factors (http://openstaxcollege.org/l/partdecomlqu) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1269 11.4 EXERCISES Verbal 206. Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction Can you explain why a partial fraction decomposition 207. is unique? (Hint: Think about it as a system of equations.) Can you explain how to verify a partial fraction 208. decomposition graphically? 209. You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could doublecheck your answer. 220. 221. 222. 223. 224. 6x x2 −4 2x−3 x2 −6x + 5 4x−1 x2 − x−6 4x + 3 x2 + 8x + 15 3x−1 x2 −5x + 6 For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. eventually 225. −5x−19 (x + 4)2 210. Once you have a system of equations generated by the partial fraction decomposition, can you explain another had method solve you if to 7x + 13 3x2 + 8x + 15 simplify it? For + B example , we 3x + 5 = A x + 1 to 7x + 13 = A(3x + 5) + B(x + 1). Explain how you could intelligently choose an x -value that will eliminate either A or B and solve for A and B. Algebraic For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors. 211. 212. 213. 214. 215. 216. 217. 218. 219. 5x + 16 x2 + 10x + 24 3x−79 x2 −5x−24 −x−24 x2 −2x−24 10x + 47 x2 + 7x + 10 x 6x2 + 25x + 25 32x−11 20x2 −13x + 2 x + 1 x2 + 7x + 10 5x x2 −9 10x x2 −25 226. x (x−2)2 227. 7x + 14 (x + 3)2 228. −24x−27 (4x + 5)2 229. −24x−27 (6x−7)2 230. 5 − x (x−7)2 231. 232. 233. 234. 235. 5x + 14 2x2 + 12x + 18 5x2 + 20x + 8 2x(x + 1)2 4x2 + 55x + 25 5x(3x + 5)2 54x3 + 127x2 + 80x + 16 2x2 (3x + 2)2 x3 −5x2 + 12x + 144 ⎞ x2 ⎛ ⎝x2 + 12x + 36 ⎠ 1270 Chapter 11 Systems of Equations and Inequalities For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor. x3 + 6x2 + 5x + 9 ⎞ ⎛ ⎝x2 + 1 ⎠ 2 236. 237. 238. 239. 240. 241. 242. 243. 244. 245. 246. 247. 248. 4x2 + 6x + 11 ⎞ ⎛ ⎝x2 + x + 3 (x + 2) ⎠ 4x2 + 9x + 23 ⎞ ⎛ ⎝x2 + 6x + 11 (x−1) ⎠ −2x2 + 10x + 4 ⎛ ⎞ ⎝x2 + 3x + 8 (x−1) ⎠ x2 + 3x + 1 ⎞ ⎛ ⎝x2 + 5x−2 (x + 1) ⎠ 4x2 + 17x−1 ⎛ ⎞ ⎝x2 + 6x + 1 (x + 3) ⎠ 4x2 ⎛ ⎞ ⎝x2 + 7x−5 (x + 5) ⎠ 4x2 + 5x + 3 x3 −1 −5x2 + 18x−4 x3 + 8 3x2 −7x + 33 x3 + 27 x2 + 2x + 40 x3 −125 4x2 + 4x + 12 8x3 −27 −50x2 + 5x−3 125x3 −1 −2x3 −30x2 + 36x + 216 x4 + 216x For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. 3x3 + 2x2 + 14x + 15 ⎞ ⎛ ⎝x2 + 4 ⎠ 2 249. 250. This content is available for free at https://cnx.org/content/col11758/1.5 251. 252. 253. 254. 255. x3 − x2 + x−1 ⎞ ⎛ ⎝x2 −3 ⎠ 2 x2 + 5x + 5 (x + 2)2 x3 + 2x2 + 4x 2 ⎞ ⎛ ⎝x2 + 2x + 9 ⎠ x2 + 25 ⎞ ⎛ ⎝x2 + 3x + 25 ⎠ 2 2x3 + 11x + 7x + 70 ⎛ ⎞ ⎝2x2 + x + 14 ⎠ 2 256. 5x + 2 x⎛ ⎞ ⎝x2 + 4 ⎠ 2 257. x4 + x3 + 8x2 + 6x + 36 x⎛ ⎞ ⎝x2 + 6 ⎠ 2 258. 2x−9 ⎛ ⎝x2 − x⎞ ⎠ 2 259. 5x3 −2x + 1 2 ⎛ ⎝x2 + 2x⎞ ⎠ Extensions the following exercises, For expansion. find the partial fraction 260. 261. x2 + 4 (x + 1)3 x3 −4x2 + 5x + 4 (x−2)3 For the following exercises, perform the operation and then find the partial fraction decomposition. 7 x + 8 + 5 x−2 − x−1 x2 −6x−16 262. 263. Chapter 11 Systems of Equations and Inequalities 1271 1 x−4 − 3 x + 6 − 2x + 7 x2 + 2x−24 264. 2x x2 −16 − 1−2x x2 + 6x + 8 − x−5 x2 −4x 1272 Chapter 11 Systems of Equations and Inequalities 11.5 \| Matrices and Matrix Operations Learning Objectives In this section, you will: 11.5.1 Find the sum and difference of two matrices. 11.5.2 Find scalar multiples of a matrix. 11.5.3 Find the product of two matrices. Figure 11.29 (credit: “SD Dirk,” Flickr) Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. Table 11.1 shows the needs of both teams. Wildcats Mud Cats Goals Balls Jerseys Table 11.1 6 30 14 10 24 20 A goal costs $300; a ball costs $10; and a jersey costs $30. How can we find the total cost for the equipment needed for each team? In this section, we discover a method in which the data in the soccer equipment table can be displayed and used for calculating other information. Then, we will be able to calculate the cost of the equipment. Finding the Sum and Difference of Two Matrices To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1273 vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named A, B, and C are shown below. A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, B = ⎡ 1 2 7 ⎢ 0 −5 6 ⎣ 7 8 2 ⎤ ⎥, C = ⎦ ⎡ − Describing Matrices A matrix is often referred to by its size or dimensions: m × n indicating m rows and n columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix A identified as ai j, we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a23. A = a11 a12 a13 ⎡ a21 a22 a23 ⎢ ⎣ a31 a32 a33 ⎤ ⎥ ⎦ A square matrix is a matrix with dimensions n × n, meaning that it has the same number of rows as columns. The 3×3 matrix above is an example of a square matrix. A row matrix is a matrix consisting of one row with dimensions 1 × n. [a11 a12 a13] A column matrix is a matrix consisting of one column with dimensions m × 1. a11 ⎡ a21 ⎢ ⎣ a31 ⎤ ⎥ ⎦ A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations. Ma
trices A matrix is a rectangular array of numbers that is usually named by a capital letter: A, B, C, and so on. Each entry in a matrix is referred to as ai j, such that i represents the row and j represents the column. Matrices are often referred to by their dimensions: m × n indicating m rows and n columns. Example 11.26 Finding the Dimensions of the Given Matrix and Locating Entries Given matrix A : a. What are the dimensions of matrix A ? b. What are the entries at a31 and a22 ? 2 ⎤ ⎥ ⎦ Solution a. The dimensions are 3 × 3 because there are three rows and three columns. 1274 Chapter 11 Systems of Equations and Inequalities b. Entry a31 is the number at row 3, column 1, which is 3. The entry a22 is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column. Adding and Subtracting Matrices We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries. In order to do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions. We can add or subtract a 3 × 3 matrix and another 3 × 3 matrix, but we cannot add or subtract a 2 × 3 matrix and a 3 × 3 matrix because some entries in one matrix will not have a corresponding entry in the other matrix. Adding and Subtracting Matrices Given matrices A and B of like dimensions, addition and subtraction of A and B will produce matrix C or matrix D of the same dimension. Matrix addition is commutative. It is also associative. A + B = C such that ai j + bi j = ci j A − B = D such that ai j − bi j = di j A + B = B + A (A + B) + C = A + (B + C) Example 11.27 Finding the Sum of Matrices Find the sum of A and B, given Solution Add corresponding entries. A = a b ⎡ ⎣ c d ⎤ ⎦ and ⎤ ⎦ ⎤ ⎦ Example 11.28 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1275 Adding Matrix A and Matrix B Find the sum of A and B. A = ⎡ 4 1 ⎣ 3 2 ⎤ ⎦ and B = ⎡ 5 9 ⎣ 0 7 ⎤ ⎦ Solution Add corresponding entries. Add the entry in row 1, column 1, a11, of matrix A to the entry in row 1, column 1, b11, of B. Continue the pattern until all entries have been added 10 ⎦ ⎣ 3 9 Example 11.29 Finding the Difference of Two Matrices Find the difference of A and B. A = ⎡ −2 3 ⎣ 0 1 ⎤ ⎦ and B = ⎡ 8 1 ⎣ 5 4 ⎤ ⎦ Solution We subtract the corresponding entries of each matrix10 ⎦ ⎣ −5 −3 Example 11.30 Finding the Sum and Difference of Two 3 x 3 Matrices Given A and B : a. Find the sum. b. Find the difference. A = ⎤ ⎡ 2 −10 −2 ⎥ and B = ⎢ 12 10 14 ⎦ ⎣ 2 4 −2 ⎡ ⎢ ⎣ −5 ⎤ 10 −2 6 ⎥ 0 −12 −4 ⎦ 2 −2 1276 Chapter 11 Systems of Equations and Inequalities Solution a. Add the corresponding entries5 10 − 2 2 − 10 − − 12 − 4 10 12 14 ⎦ ⎣ ⎦ − 10 + 10 − 2 − 2 ⎤ ⎡ ⎥ ⎢ 10 − 4 12 − 12 14 + 14 0 ⎣ ⎦ 0 −1 0 b. Subtract the corresponding entries5 ⎤ ⎤ ⎡ 10 −2 6 2 −10 −2 ⎥ ⎥ − ⎢ 0 −12 −4 12 10 14 ⎦ ⎣ ⎦ 2 4 −2 2 −2 2 − 6 −10 − 10 −2 + 2 ⎤ ⎡ ⎥ ⎢ 12 + 12 10 + 4 14 − 0 ⎣ ⎦ −4 −20 0 ⎤ ⎡ ⎥ ⎢ 24 14 14 ⎦ ⎣ −4 4 9 11.20 Add matrix A and matrix B. A = ⎡ ⎤ 6 2 ⎢ ⎥ and B = 0 1 ⎣ ⎦ 1 −3 ⎤ ⎡ 3 −2 ⎥ ⎢ 5 1 ⎦ ⎣ 3 −4 Finding Scalar Multiples of a Matrix Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication. Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in Table 11.2. Lab A Lab B 15 16 16 27 34 34 Computers Computer Tables Chairs Table 11.2 Converting the data to a matrix, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1277 C2013 = ⎡ 15 ⎢ 16 ⎣ 16 ⎤ ⎥ ⎦ 27 34 34 To calculate how much computer equipment will be needed, we multiply all entries in matrix C by 0.15. (0.15)C2013 = ⎡ (0.15)15 ⎢ (0.15)16 ⎢ ⎣ (0.15)16 (0.15)27 (0.15)34 (0.15)34 ⎤ ⎥ ⎥ ⎦ = ⎡ 2.25 ⎢ 2.4 ⎣ 2.4 4.05 5.1 5.1 ⎤ ⎥ ⎦ We must round up to the next integer, so the amount of new equipment needed is Adding the two matrices as shown below, we see the new inventory amounts This means ⎡ 15 ⎢ 16 ⎣ 16 ⎤ ⎥ + ⎦ 27 34 34 ⎡ ⎡ 18 ⎢ 19 ⎣ 19 ⎤ ⎥ ⎦ 32 40 40 C2014 = ⎡ 18 ⎢ 19 ⎣ 19 ⎤ ⎥ ⎦ 32 40 40 Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs. Scalar Multiplication Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given the scalar multiple cA is A = a11 ⎡ ⎣ a21 a12 a22 ⎤ ⎦ a11 a12 ⎤ ⎡ cA = c a21 a22 ⎦ ⎣ ca12 ca11 ⎤ ⎡ ca22 ca21 ⎦ ⎣ = Scalar multiplication is distributive. For the matrices A, B, and C with scalars a and b, a(A + B) = aA + aB (a + b)A = aA + bA Example 11.31 Multiplying the Matrix by a Scalar Multiply matrix A by the scalar 3. A = ⎡ 8 1 ⎣ 5 4 ⎤ ⎦ 1278 Chapter 11 Systems of Equations and Inequalities Solution Multiply each entry in A by the scalar 3. 3A = 24 ⎣ ⎦ 15 12 = = 11.21 Given matrix B, find −2B where B = ⎡ 4 1 ⎣ 3 2 ⎤ ⎦ Example 11.32 Finding the Sum of Scalar Multiples Find the sum 3A + 2B. A = ⎡ 1 −2 ⎢ 0 −1 ⎣ 4 ⎤ 0 ⎥ and B = 2 ⎦ 3 −6 Solution First, find 3A, then 2B. ⎡ −1 ⎢ ⎣ 2 0 −3 0 ⎤ 1 ⎥ 2 ⎦ 1 −4 3A = = 3 ⋅ 1 3(−2(−1) 3 ⋅ 2 ⎥ ⎢ 3(−66 ⎡ ⎤ ⎢ ⎥ 0 −3 6 ⎣ ⎦ 9 −18 12 2B = = 2(−12 ⎢ ⎣ 2 ⋅ 1 ⎤ ⎥ 2(−3) 2 ⋅ 2 ⎥ 2 ⋅ 1 2(−4) ⎦ ⎤ 2 4 ⎥ 0 −6 4 ⎦ 2 −8 0 Now, add 3A + 2B. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1279 3A + 2B = = = ⎡ −2 ⎢ ⎣ ⎤ 4 2 ⎥ 0 −6 4 ⎦ 0 2 −18−8 ⎤ ⎡ 0 3 −6 ⎥ + ⎢ 6 0 −3 ⎣ ⎦ 12 9 −18 3 − 2 −6 + 4 ⎡ ⎢ 0 + 0 −3 − 6 ⎣ 12 + 0 1 −2 ⎡ 2 ⎢ 0 −9 10 ⎣ 12 11 −26 ⎤ ⎥ ⎦ Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If A is an m × r matrix and B is an r × n matrix, then the product matrix AB is an m × n matrix. For example, the product AB is possible because the number of columns in A is the same as the number of rows in B. If the inner dimensions do not match, the product is not defined. We multiply entries of A with entries of B according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers. To obtain the entries in row i of AB, we multiply the entries in row i of A by column j in B and add. For example, the product of AB given matrices A and B, where the dimensions of A are 2 × 3 and the dimensions of B are 3 × 3, will be a 2 × 3 matrix. A = a11 a12 a13 ⎡ a21 a22 a23 ⎣ ⎤ ⎦ and B = b11 b12 b13 ⎡ ⎢ b21 b22 b23 ⎢ b31 b32 b33 ⎣ ⎤ ⎥ ⎥ ⎦ Multiply and add as follows to obtain the first entry of the product matrix AB. 1. To obtain the entry in row 1, column 1 of AB, multiply the first row in A by the first column in B, and add. [a11 a12 a13] ⋅ b11 ⎡ ⎢ b21 ⎢ b31 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b11 + a12 ⋅ b21 + a13 ⋅ b31 2. To obtain the entry in row 1, column 2 of AB, multiply the first row of A by the second column in B, and add. [a11 a12 a13] ⋅ b12 ⎡ ⎢ b22 ⎢ b32 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b12 + a12 ⋅ b22 + a13 ⋅ b32 3. To obtain the entry in row 1, column 3 of AB, multiply the first row of A by the third column in B, and add. [a11 a12 a13] ⋅ b13 ⎡ ⎢ b23 ⎢ b33 ⎣ ⎤ ⎥ ⎥ ⎦ = a11 ⋅ b13 + a12 ⋅ b23 + a13 ⋅ b33 1280 Chapter 11 Systems of Equations and Inequalities We proceed the same way to obtain the second row of AB. In other words, row 2 of A times column 1 of B; row 2 of A times column 2 of B; row 2 of A times column 3 of B. When complete, the product matrix will be AB = a11 ⋅ b11 + a12 ⋅ b21 + a13 ⋅ b31 ⎡ ⎢ a21 ⋅ b11 + a22 ⋅ b21 + a23 ⋅ b31 ⎣ a11 ⋅ b12 + a12 ⋅ b22 + a13 ⋅ b32 a21 ⋅ b12 + a22 ⋅ b22 + a23 ⋅ b32 a11 ⋅ b13 + a12 ⋅ b23 + a13 ⋅ b33 a21 ⋅ b13 + a22 ⋅ b23 + a23 ⋅ b33 ⎤ ⎥ ⎦ Properties of Matrix Multiplication For the matrices A, B, and C the following properties hold. • Matrix multiplication is associative: (AB)C = A(BC). • Matrix multiplication is distributive: C(A + B) = CA + CB, (A + B)C = AC + BC. Note that matrix multiplication is not commutative. Example 11.33 Multiplying Two Matrices Multiply matrix A and matrix B. A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦ and B = ⎡ 5 6 ⎣ 7 8 ⎤ ⎦ Solution First, we check the dimensions of the matrices. Matrix A has dimensions 2 × 2 and matrix B has dimensions 2 × 2. The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions 2 × 2. We perform the operations outlined previously. Example 11.34 Multiplying Two Matrices Given A and B : a. Find AB. b. Find BA. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1281 A = ⎡ −1 2 3 ⎣ 4 0 5 ⎤ ⎦ and B = 5 ⎡ ⎢ −4 ⎣ 2 ⎤ −1 ⎥ 0 ⎦ 3 Solution a. As the dimensions of A are 2×3 and the dimensions of B are 3×2, these matrices can be multiplied together because the number of columns in A matches the number of rows in B. The resulting product will be a 2×2 matrix, the number of rows in A by the number of columns in B. AB = ⎡ −4 ⎣ 2 ⎤ 5 −1 ⎥ 0 ⎦ 3 = = 4(5) + 0(−4) + 5(2) −1(5) + 2(−4) + 3(2) −1(−1) + 2(0) + 3(3) ⎡ ⎤ ⎣ ⎦ 4(−1) + 0(0) + 5(3) ⎡ −7 10 ⎣ 30 11 ⎤ ⎦ b. The dimensions of B are 3×2 and the dimensions of A are 2×3. The inner dimensions match
so the product is defined and will be a 3×3 matrix. BA = = = ⎤ ⎦ ⎡ −1 ⎥ 0 ⎦ 3 ⎡ ⎢ −4 ⎣ 2 5(−1) + −1(4) 5(2) + −1(0) 5(3) + −1(5) ⎡ ⎤ ⎢ ⎥ −4(−1) + 0(4) −4(2) + 0(0) −4(3) + 0(5) ⎢ ⎥ 2(3) + 3(5) 2(2) + 3(0) ⎣ ⎦ ⎡ −9 10 ⎢ ⎣ 2(−1) + 3(4) ⎤ 10 ⎥ 4 −8 −12 ⎦ 21 4 10 Analysis Notice that the products AB and BA are not equal. AB = ⎡ −7 10 ⎣ 30 11 ⎤ ⎦ ≠ ⎤ ⎡ −9 10 10 ⎥ = BA ⎢ 4 −8 −12 ⎦ ⎣ 21 10 4 This illustrates the fact that matrix multiplication is not commutative. Is it possible for AB to be defined but not BA? Yes, consider a matrix A with dimension 3 × 4 and matrix B with dimension 4 × 2. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined. Example 11.35 Using Matrices in Real-World Problems Let’s return to the problem presented at the opening of this section. We have Table 11.3, representing the equipment needs of two soccer teams. 1282 Chapter 11 Systems of Equations and Inequalities Wildcats Mud Cats 6 30 14 Goals Balls Jerseys Table 11.3 10 24 20 We are also given the prices of the equipment, as shown in Table 11.4. Goal $300 Ball $10 Jersey $30 Table 11.4 We will convert the data to matrices. Thus, the equipment need matrix is written as The cost matrix is written as E = ⎡ 6 ⎢ 30 ⎣ 14 ⎤ ⎥ ⎦ 10 24 20 C = [300 10 30] We perform matrix multiplication to obtain costs for the equipment. CE = [300 10 30] ⋅ ⎡ 6 10 ⎢ 30 24 ⎣ 14 20 ⎤ ⎥ ⎦ = ⎡ = ⎡ ⎣300(6) + 10(30) + 30(14) 300(10) + 10(24) + 30(20)⎤ ⎣2,520 3,840⎤ ⎦ ⎦ The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840. Given a matrix operation, evaluate using a calculator. 1. Save each matrix as a matrix variable [A], [B], [C], ... 2. Enter the operation into the calculator, calling up each matrix variable as needed. 3. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1283 Example 11.36 Using a Calculator to Perform Matrix Operations Find AB − C given A = ⎡ −15 25 ⎢ ⎣ ⎤ 32 ⎥, B = 41 −7 −28 ⎦ 10 34 −2 ⎡ 45 ⎢ −24 ⎣ ⎤ 21 −37 ⎥, and C = 52 19 ⎦ 6 −48 −31 ⎤ ⎡ −100 −89 −98 ⎥. ⎢ 74 ⎦ ⎣ 42 −75 25 −56 −67 Solution On the matrix page of the calculator, we enter matrix A above as the matrix variable [A], matrix B above as the matrix variable [B], and matrix C above as the matrix variable [C]. On the home screen of the calculator, we type in the problem and call up each matrix variable as needed. The calculator gives us the following matrix. [A]×[B] − [C] 136 −983 − 462 ⎡ ⎤ 1, 820 1, 897 − 856 ⎢ ⎥ ⎣ ⎦ 413 −311 2, 032 Access these online resources for additional instruction and practice with matrices and matrix operations. • Dimensions of a Matrix (http://openstaxcollege.org/l/matrixdimen) • Matrix Addition and Subtraction (http://openstaxcollege.org/l/matrixaddsub) • Matrix Operations (http://openstaxcollege.org/l/matrixoper) • Matrix Multiplication (http://openstaxcollege.org/l/matrixmult) 1284 Chapter 11 Systems of Equations and Inequalities 11.5 EXERCISES Verbal Can we add any two matrices together? If so, explain 265. why; if not, explain why not and give an example of two matrices that cannot be added together. Can we multiply any column matrix by any row 266. matrix? Explain why or why not. Can both the products AB and BA be defined? If so, 267. explain how; if not, explain why. 268. Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together. Does matrix multiplication commute? That is, does 269. AB = BA ? If so, prove why it does. If not, explain why it does not. Algebraic For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined. A = ⎤ ⎡ 1 3 ⎦, B = ⎣ 0 7 ⎤ ⎡ 2 14 ⎦, C = ⎣ 22 6 ⎤ ⎡ 1 5 ⎥, D = ⎢ 8 92 ⎦ ⎣ 12 6 ⎤ ⎡ 10 14 ⎥, E = ⎢ 7 2 ⎦ ⎣ 5 61 ⎤ ⎡ 6 12 ⎦, F = ⎣ 14 5 ⎡ ⎤ 0 9 ⎢ ⎥ 78 17 ⎣ ⎦ 15 4 270. A + B 271. C + D 272. A + C 273. B − E 274. C + F 275. D − B For the following exercises, use the matrices below to perform scalar multiplication. A = ⎤ ⎡ 4 6 ⎦, B = ⎣ 13 12 ⎡ 3 9 ⎢ 21 12 ⎣ 0 64 ⎤ ⎥, C = ⎦ ⎤ ⎡ 16 3 7 18 ⎦, D = ⎣ 90 5 3 29 ⎡ ⎤ 18 12 13 ⎢ ⎥ 8 14 6 ⎣ ⎦ 7 4 21 276. 5A 277. 3B 278. −2B 279. −4C 280. C 1 2 This content is available for free at https://cnx.org/content/col11758/1.5 281. 100D For the following exercises, use the matrices below to perform matrix multiplication. A = ⎤ ⎡ −1 5 ⎦, 8 0 12 ⎤ ⎦, C = ⎡ 4 10 ⎢ −2 6 ⎣ 9 5 ⎤ ⎥, D = ⎦ ⎤ ⎡ 12 2 −10 0 282. AB 283. BC 284. CA 285. BD 286. DC 287. CB For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. A = ⎤ ⎡ 2 −5 ⎦, B = ⎣ 7 6 ⎡ ⎤ −9 6 ⎦, C = ⎣ −4 2 ⎡ ⎤ 0 9 ⎦, D = ⎣ 7 1 ⎤ ⎡ −8 7 −5 ⎥, 6 −5 ⎦ ⎣ 9 0 1 288. A + B − C 289. 4A + 5D 290. 2C + B 291. 3D + 4E 292. C−0.5D 293. 100D−10E For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A2 = A ⋅ A ) A = ⎡ −10 20 ⎣ 5 25 ⎤ ⎦, B = ⎤ ⎡ 40 10 ⎦, C = ⎣ −20 30 ⎡ −1 ⎢ ⎣ ⎤ 0 ⎥ 0 −1 ⎦ 0 1 294. AB 295. BA 296. CA 297. BC Chapter 11 Systems of Equations and Inequalities 1285 298. A2 299. B2 300. C 2 301. B2 A2 302. A2 B2 303. (AB)2 304. (BA)2 For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A2 = A ⋅ A ) 317. BC 318. ABC Extensions For the following exercises, use the matrix below to perform the indicated operation on the given matrix ⎤ ⎥ ⎦ 319. 320. 321. 322. B2 B3 B4 B5 A = ⎤ ⎡ 1 0 ⎦, B = ⎣ 2 3 ⎤ ⎡ 4 −2 3 ⎦, C = ⎣ −1 1 −5 ⎤ ⎡ 0.5 0.1 ⎥, D = ⎢ 1 0.2 ⎦ ⎣ −0.5 0.3 ⎡ ⎢ −6 7 ⎣ 4 2 ⎤ 1 0 −1 ⎥ 5 ⎦ 1 Using the above questions, find a formula for Bn 323. Test the formula for B201 and B202, using a calculator. . 305. AB 306. BA 307. BD 308. DC 309. D2 310. A2 311. D3 312. (AB)C 313. A(BC) Technology For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution. A = −2 0 ⎡ ⎢ ⎣ 0.5 4 ⎤ 9 ⎥, B = 1 8 −3 ⎦ 5 ⎡ 0.5 3 0 ⎢ −4 1 6 ⎣ 8 7 2 ⎤ ⎥, ⎤ ⎥ ⎦ 314. AB 315. BA 316. CA 1286 Chapter 11 Systems of Equations and Inequalities 11.6 \| Solving Systems with Gaussian Elimination Learning Objectives In this section, you will: 11.6.1 Write the augmented matrix of a system of equations. 11.6.2 Write the system of equations from an augmented matrix. 11.6.3 Perform row operations on a matrix. 11.6.4 Solve a system of linear equations using matrices. Figure 11.30 German mathematician Carl Friedrich Gauss (1777–1855). Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries. We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables. In this section, we will revisit this technique for solving systems, this time using matrices. Writing the Augmented Matrix of a System of Equations A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. For example, consider the following 2 × 2 system of equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1287 We can write this system as an augmented matrix: 3x + 4y = 7 4x−2y = 5 We can also write a matrix containing just the coefficients. This is called the coefficient matrix. ⎡ 4 3 ⎣ 4 −2 ⎤ ⎦ 7 5 \| A three-by-three system of equations such as has a coefficient matrix and is represented by the augmented matrix ⎤ ⎡ 3 4 ⎦ ⎣ 4 −2 3x − 2x−3z = 2 ⎡ ⎤ 3 −1 −3 2 Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form ax + by + cz = d so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0. ⎡ 3 −1 −1 ⎢ 1 0 1 ⎣ 0 −3 2 ⎤ ⎥ ⎦ 0 5 2 \| Given a system of equations, write an augmented matrix. 1. Write the coefficients of the x-terms as the numbers down the first column. 2. Write the coefficients of the y-terms as the numbers down the second column. 3. If there are z-terms, write the coefficients as the numbers down the third column. 4. Draw a vertical line and write the constants to the right of the line. Example 11.37 Writing the Augmented Matrix for a System of Equations Write the augmented matrix for the given system of equations. x + 2y − z = 3 2x − y + 2z = 6 x − 3y + 3z = 4 Solution The augmented matrix displays the coefficients of the variables, and an additional column for the constants. ⎡ 1 ⎢ 2 −1 ⎣ 1 −3 2 − 1288 Chapter 11 Systems of Equations and Inequalities 11.22 Write the augmented matrix of the given system of equations. 4x−3y = 11 3x + 2y = 4 Writing a System of Equations from an Augmented Matrix We can use augmented matrices to help us sol
ve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form. Example 11.38 Writing a System of Equations from an Augmented Matrix Form Find the system of equations from the augmented matrix. ⎡ ⎢ ⎣ −3 1 −3 −5 2 −5 −4 4 5 ⎤ ⎥ ⎦ \| −2 5 6 Solution When the columns represent the variables x, y, and z, ⎡ ⎢ ⎣ −3 1 −3 −5 2 −5 −4 4 5 \| −2 5 6 ⎤ ⎥ → ⎦ x − 3y − 5z = − 2 2x − 5y − 4z = 5 −3x + 5y + 4z = 6 11.23 Write the system of equations from the augmented matrix. 1 −1 1 ⎡ ⎢ 2 −9 \| Performing Row Operations on a Matrix Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows. Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown. Row-echelon form ⎤ ⎥ ⎦ We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form. 1. In any nonzero row, the first nonzero number is a 1. It is called a leading 1. 2. Any all-zero rows are placed at the bottom on the matrix. 3. Any leading 1 is below and to the right of a previous leading 1. 4. Any column containing a leading 1 has zeros in all other positions in the column. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1289 To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution. 1. Interchange rows. (Notation: Ri ↔ R j ) 2. Multiply a row by a constant. (Notation: cRi ) 3. Add the product of a row multiplied by a constant to another row. (Notation: Ri + cR j) Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in rowechelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows. Gaussian Elimination The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix A with the number 1 as the entry down the main diagonal and have all zeros below. A = a11 a12 a13 ⎡ a21 a22 a23 ⎢ ⎣ a31 a32 a33 ⎤ ⎥ ⎦ After Gaussian elimination A = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → 1 b12 b13 ⎡ ⎢ 1 b23 0 ⎢ ⎣ 1 0 0 ⎤ ⎥ ⎥ ⎦ The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below. Given an augmented matrix, perform row operations to achieve row-echelon form. 1. The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary. 2. Use row operations to obtain zeros down the first column below the first entry of 1. 3. Use row operations to obtain a 1 in row 2, column 2. 4. Use row operations to obtain zeros down column 2, below the entry of 1. 5. Use row operations to obtain a 1 in row 3, column 3. 6. Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below. 7. If any rows contain all zeros, place them at the bottom. Example 11.39 Solving a 2×2 System by Gaussian Elimination Solve the given system by Gaussian elimination. 2x + 3y = 6 x − y = 1 2 Solution First, we write this as an augmented matrix. 1290 Chapter 11 Systems of Equations and Inequalities We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2. ⎡ 3 ⎢2 1 −1 ⎣ ⎤ ⎥ ⎦ \| 6 1 2 We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by −2, and then adding the result to row 2. R1 ↔ R2 → ⎡ ⎢1 −2R1 + R2 = R2 → We only have one more step, to multiply row 2 by 1 5 . ⎤ ⎥ ⎦ ⎡ ⎢1 − R2 = R2 → 1 5 ⎡ ⎢1 −1 ⎣ 0 Use back-substitution. The second row of the matrix represents y = 1. Back-substitute y = 1 into the first equation. x − (1) = 1 2 x = 3 2 The solution is the point ⎛ ⎝ ⎞ , 1 ⎠. 3 2 11.24 Solve the given system by Gaussian elimination. 4x + 3y = 11 x−3y = −1 Example 11.40 Using Gaussian Elimination to Solve a System of Equations Use Gaussian elimination to solve the given 2 × 2 system of equations. Solution Write the system as an augmented matrix. 2x + y = 1 4x + 2y = This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1291 Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by 1 2 . Next, we want a 0 in row 2, column 1. Multiply row 1 by −4 and add row 1 to row 2. 1 2 R1 = R1 → ⎡ ⎢ \| The second row represents the equation 0 = 4. Therefore, the system is inconsistent and has no solution. −4R1 + R2 = R2 → ⎡ ⎢ \| Example 11.41 Solving a Dependent System Solve the system of equations. 3x + 4y = 12 6x + 8y = 24 Solution Perform row operations on the augmented matrix to try and achieve row-echelon form \| 12 24 − 1 2 R2 + R1 = R1 → ⎡ 6 ⎣ 0 R1 ↔ R2 → 8 ⎤ ⎦ 0 \| 24 0 ⎡ 0 ⎣ 6 0 24 ⎤ ⎦ 0 8 \| The matrix ends up with all zeros in the last row: 0y = 0. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for y. 3x + 4y = 12 4y = 12−3x y = 3 − 3 4 x So the solution to this system is ⎛ ⎝x, 3 − 3 4 x⎞ ⎠. Example 11.42 1292 Chapter 11 Systems of Equations and Inequalities Performing Row Operations on a 3×3 Augmented Matrix to Obtain Row-Echelon Form Perform row operations on the given matrix to obtain row-echelon form. ⎡ ⎢ ⎣ −3 1 −3 4 2 − Solution The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by −2 and add it to row 2. Then replace row 2 with the result. −2R1 + R2 = R2 → Next, obtain a zero in row 3, column 1. 3R1 + R3 = R3 → Next, obtain a zero in row 3, column 2. 6R2 + R3 = R3 → The last step is to obtain a 1 in row 3, column 3. ⎡ 1 −3 ⎢ 0 ⎣ −3 4 1 −2 ⎡ 1 −3 ⎢ 0 ⎣ 0 −6 ⎡ 1 −3 ⎢ 0 ⎣ 0 4 1 − 15 4 1 −2 3 4 \| 3 16 \| 3 4 \| 3 \| ⎤ 3 ⎥ −6 ⎥ 21 ⎦ 2 0 15 ⎤ ⎥ ⎦ 1 2 R3 = R3 → ⎡ ⎢1 −3 0 ⎢ 0 ⎣ 4 1 −2 1 0 11.25 Write the system of equations in row-echelon form. x − 2y + 3z = 9 − x + 3y = − 4 2x − 5y + 5z = 17 Solving a System of Linear Equations Using Matrices We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and backsubstitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables. Example 11.43 Solving a System of Linear Equations Using Matrices Solve the system of linear equations using matrices. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1293 x − y + z = 8 2x + 3y − z = −2 3x − 2y − 9z = 9 Solution First, we write the augmented matrix. Next, we perform row operations to obtain row-echelon form. ⎡ 1 1 −1 ⎢ 3 −1 2 ⎣ 3 −2 −9 ⎤ 8 ⎥ −2 ⎦ 9 \| −2R1 + R2 = R2 → 1 5 −3 ⎡ 1 −1 ⎢ 0 ⎣ 3 −2 −9 \| 8 −18 9 ⎤ ⎥ ⎦ −3R1 + R3 = R3 → The easiest way to obtain a 1 in row 2 of column 1 is to interchange R2 and R3. Interchange R2 and R3 → ⎡ 1 −12 −15 ⎦ −3 −18 5 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 8 −18 −15 ⎡ ⎢1 −3 1 −12\| 8 ⎤ ⎥ ⎥ ⎥ −15 ⎥ ⎥ ⎦ 1 1 1\| Then −5R2 + R3 = R3 → ⎡ 1 −1 ⎢ 0 ⎣ 0 1 1 −12 0 57 \| ⎤ 8 ⎥ −15 ⎦ 57 − 1 57 R3 = R3 → ⎡ ⎢1 −12 The last matrix represents the equivalent system. x − y + z = 8 y − 12z = −15 z = 1 Using back-substitution, we obtain the solution as (4, −3, 1). Example 11.44 Solving a Dependent System of Linear Equations Using Matrices Solve the following system of linear equations using matrices. −x−2y + z = −1 2x + 3y = 2 y−2z = 0 1294 Chapter 11 Systems of Equations and Inequalities Solution Write the augmented matrix. First, multiply row 1 by −1 to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form. ⎡ 1 −1 −2 ⎢ 0 3 ⎣ 1 −2 2 0 ⎤ ⎥ ⎦ \| −1 2 0 −R1 → R2 ↔ R3 → ⎡ 2R1 + R3 = R3 → R2 + R3 = R3 → ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ 2 −1 1 3 2 0 1 −2 0 2 −1 1 −2 0 0 3 2 ⎡ 2 −1 1 ⎢ 1 −2 0 ⎣ 0 −1 ⎡ 1 ⎢ 0 ⎣ 0 2 −1 1 − ⎤ ⎥ ⎦ The last matrix represents the following system. x + 2y − z = 1 y − 2z = 0 0 = 0 We see by the identity 0 = 0 that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for y and substituting it into the first equation we can solve for z in terms of x. x + 2y − z = 1 y = 2z x + 2(2z) − z = 1 x + 3z = 1 z = 1 − x 3 Now we substitute the expression for z into the second equation to solve for y in terms of x. y − 2z = ⎝ The generic solution is ⎛ ⎝x, 2−2x 3 , 1 − x 3 ⎞ ⎠. − 2x 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1295 11.26 Solve the system using matrices. x + 4y − z = 4 2x + 5y + 8z = 15 x + 3y−3z = 1 Can any system of linear equations be solved by Gaussian elimination? Yes, a system of linear equations of any size can be solved by Gaussian elimination. Given a system of equations,
solve with matrices using a calculator. 1. Save the augmented matrix as a matrix variable [A], [B], [C], … . 2. Use the ref( function in the calculator, calling up each matrix variable as needed. Example 11.45 Solving Systems of Equations with Matrices Using a Calculator Solve the system of equations. 5x + 3y + 9z = −1 −2x + 3y − z = −2 −x−4y + 5z = 1 Solution Write the augmented matrix for the system of equations. On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [A]. ⎡ 5 ⎢ −2 ⎣ −1 −4 3 9 3 −1 5 ⎤ 5 ⎥ −2 ⎦ −1 \| [A] = ⎡ 5 ⎢ −2 ⎣ −1 −4 ⎤ 9 −1 3 ⎥ 3 −1 −2 ⎦ 1 5 Use the ref( function in the calculator, calling up the matrix variable [A]. ref([A]) Evaluate 13 − 4 ⎢ 7 21 ⎢ 1 − 24 0 0 ⎣ 187 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ → + 13 21 z = − 24 187 Using back-substitution, the solution is ⎛ ⎝ 61 187 , − 92 187 , − 24 187 ⎞ ⎠. 1296 Chapter 11 Systems of Equations and Inequalities Example 11.46 Applying 2 × 2 Matrices to Finance Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate? Solution We have a system of two equations in two variables. Let x = the amount invested at 10.5% interest, and y = the amount invested at 12% interest. x + y = 12,000 0.105x + 0.12y = 1,335 As a matrix, we have Multiply row 1 by −0.105 and add the result to row 2. ⎡ 1 1 ⎣ 0.105 0.12 12,000 1,335 ⎤ ⎦ \| Then, ⎡ 1 1 ⎣ 0 0.015 12,000 75 ⎤ ⎦ \| 0.015y = 75 y = 5,000 So 12,000−5,000 = 7,000. Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest. Example 11.47 Applying 3 × 3 Matrices to Finance Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate? Solution We have a system of three equations in three variables. Let x be the amount invested at 5% interest, let y be the amount invested at 8% interest, and let z be the amount invested at 9% interest. Thus, x + y + z = 10, 000 0.05x + 0.08y + 0.09z = 770 2x − z = 0 As a matrix, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1297 Now, we perform Gaussian elimination to achieve row-echelon form. 1 1 ⎡ 1 ⎢ 0.05 0.08 0.09 ⎣ 0 −1 2 ⎤ 10, 000 ⎥ 770 ⎦ 0 \| −0.05R1 + R2 = R2 → −2R1 + R3 = R3 → 1 0.03 R2 = R2 → 2R2 + R3 = R3 → 1 0.04 1 0.04 ⎡ ⎣ ⎤ ⎥ 270 ⎦ 0 ⎤ 10,000 ⎥ 270 ⎦ −20,000 1 0.03 0 1 0.03 −2 1 −1 \| 10,000 −3 \| 0 −2 −3\| 3\| 10,000 ⎤ ⎥ ⎥ 9,000 ⎥ −2,000 ⎦ ⎤ 10,000 ⎥ 9,000 ⎥ −20,000 ⎦ The third row tells us − 1 3 z = −2,000; thus z = 6,000. The second row tells us y + 4 3 z = 9,000. Substituting z = 6,000, we get (6,000) = 9,000 y + 4 3 y + 8,000 = 9,000 y = 1,000 The first row tells us x + y + z = 10, 000. Substituting y = 1, 000 and z = 6, 000, we get x + 1, 000 + 6, 000 = 10,000 x = 3,000 The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest. 11.27 A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate. Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination. • Solve a System of Two Equations Using an Augmented Matrix (http://openstaxcollege.org/l/system2augmat) • Solve a System of Three Equations Using an Augmented Matrix (http://openstaxcollege.org/l/system3augmat) • Augmented Matrices on the Calculator (http://openstaxcollege.org/l/augmatcalc) 1298 Chapter 11 Systems of Equations and Inequalities For the following exercises, solve the system by Gaussian elimination. 11.6 EXERCISES Verbal 324. Can any system of linear equations be written as an augmented matrix? Explain why or why not. Explain how to write that augmented matrix. 325. Can any matrix be written as a system of linear equations? Explain why or why not. Explain how to write that system of equations. 336. 337. 338. 326. Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix ⎡ 3 9 ⎣ 1 −2 ⎤ ⎦. 6 \| 0 Can a matrix whose entry is 0 on the diagonal be 327. solved? Explain why or why not. What would you do to remedy the situation? Can a matrix that has 0 entries for an entire row have 328. one solution? Explain why or why not. Algebraic For the following exercises, write the augmented matrix for the linear system. 329. 8x−37y = 8 2x + 12y = 3 330. 16y = 4 9x − y = 2 331. 332. 333. 3x + 2y + 10z = 3 −6x + 2y + 5z = 13 4x + z = 18 x + 5y + 8z = 19 12x + 3y = 4 3x + 4y + 9z = −7 6x + 12y + 16z = 4 19x−5y + 3z = −9 x + 2y = −8 For the following exercises, write the linear system from the augmented matrix. 334. ⎡ −2 ⎣ 5 6 −18 335. ⎡ 4 3 ⎣ 10 17 ⎤ ⎦ 5 26 \| \| 10 439 ⎤ ⎦ This content is available for free at https://cnx.org/content/col11758/1.5 3 ⎡ 2 0 ⎢ −1 −9 4 ⎣ 5 7 8 ⎡ 8 29 1 ⎢ −2 ⎢ 0 1 58 ⎣ 8 7 −3 ⎤ 3 ⎥ −1 ⎦ 8 ⎤ ⎥ ⎦ 43 38 10 \| \| \| 12 ⎤ ⎥ 2 ⎦ −5 339. 340. 341. 342. 343. 3441 ⎣ 2 4 −5 ⎤ −3 ⎦ 6 ⎡ −2 0 ⎣ 0 2 ⎤ 1 ⎦ −1 2x − 3y = − 9 5x + 4y = 58 345. 6x + 2y = −4 3x + 4y = −17 346. 2x + 3y = 12 4x + y = 14 347. −4x−3y = −2 3x−5y = −13 348. −5x + 8y = 3 10x + 6y = 5 349. 3x + 4y = 12 −6x−8y = −24 350. −60x + 45y = 12 20x−15y = −4 351. 11x + 10y = 43 15x + 20y = 65 Chapter 11 Systems of Equations and Inequalities 1299 352. 2x − y = 2 3x + 2y = 17 y − z = −3 ⎡ −0.1 0.3 −0.1 ⎢ 0.1 −0.4 0.2 ⎣ 0.7 0.6 0.1 ⎤ 0.2 ⎥ 0.8 ⎦ −0.8 Extensions For the following exercises, use Gaussian elimination to solve the system. −1.06x−2.25y = 5.51 −5.03x−1.08y = 5.40 1 y = 3 ⎤ ⎥ ⎦ 31 45 87 ⎤ 50 ⎥ 20 ⎦ −90 ⎤ ⎥ ⎦ 2x + 3y − 2z = 3 4x + 2y − z = 9 4x − 8y + 2z = −6 x + y − 4z = −4 5x − 3y − 2z = 0 2x + 6y + 7z = 30 2x + 3y + 2z = 1 −4x − 6y − 4z = −2 10x + 15y + 10z = 5 x + 2y − z = 1 −x − 2y + 2z = −2 3x + 6y − 3z = 5 x + 2y − z = 1 −x−2y + 2z = −2 3x + 6y−3z = 3 353. 354. 355. 356. 357. 358. 359. 360. 361. 362. 363. 364. 365. 366. 367. 368. 369. x + y + z = 100 x + 2z = 125 −y + 2z = 25 = − 53 14 z = 3 z = 23 15 − 10 z = − 29 6 z = 431 210 z = − 49 45 370. x−1 7 + y−2 8 + z− + 2y + z−3 3 = 5 371. x−1 4 − y + 1 4 + 3z = −−2 2 − z = 4 = 1 372. 373. x−3 4 x + 5 2 + − y−1 3 y + 5 2 + + 2z = −−3 10 x + 5 4 x−1 4 + y + 3 2 y−1 8 y + 4 2 − −2z = 3 + z = 3 2 + 3z = 3 2 Chapter 11 Systems of Equations and Inequalities ice cream, find out the percentage of ice cream sales each individual ice cream made last year. A bag of mixed nuts contains cashews, pistachios, 383. and almonds. There are 1,000 total nuts in the bag, and there are 100 less almonds than pistachios. The cashews weigh 3 g, pistachios weigh 4 g, and almonds weigh 5 g. If the bag weighs 3.7 kg, find out how many of each type of nut is in the bag. A bag of mixed nuts contains cashews, pistachios, 384. and almonds. Originally there were 900 nuts in the bag. 30% of the almonds, 20% of the cashews, and 10% of the pistachios were eaten, and now there are 770 nuts left in the bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with. 1300 374. x−3 4 x + 5 2 + − y−1 3 y + 5 2 + 2z = − Real-World Applications For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. Every day, a cupcake store sells 5,000 cupcakes in 375. chocolate and vanilla flavors. If the chocolate flavor is 3 times as popular as the vanilla flavor, how many of each cupcake sell per day? At a competing cupcake store, $4,520 worth of 376. cupcakes are sold daily. The chocolate cupcakes cost $2.25 and the red velvet cupcakes cost $1.75. If the total number of cupcakes sold per day is 2,200, how many of each flavor are sold each day? You invested $10,000 into two accounts: one that has 377. simple 3% interest, the other with 2.5% interest. If your total interest payment after one year was $283.50, how much was in each account after the year passed? You invested $2,300 into account 1, and $2,700 into 378. account 2. If the total amount of interest after one year is $254, and account 2 has 1.5 times the interest rate of account 1, what are the interest rates? Assume simple interest rates. Bikes’R’Us manufactures bikes, which sell for $250. 379. It costs the manufacturer $180 per bike, plus a startup fee of $3,500. After how many bikes sold will the manufacturer break even? A major appliance store is considering purchasing 380. vacuums from a small manufacturer. The store would be able to purchase the vacuums for $86 each, with a delivery fee of $9,200, regardless of how many vacuums are sold. If the store needs to start seeing a profit after 230 units are sold, how much should they charge for the vacuums? The three most popular ice cream flavors are 381. chocolate, strawberry, and vanilla, comprising 83% of the flavors sold at an ice cream shop. If vanilla sells 1% more than twice strawberry, and chocolate sells 11% more than vanilla, how much of the total ice cream consumption are the vanilla, chocolate, and strawberry flavors? At an ice cream shop, three flavors are increasing in 382. demand. Last year, banana, pumpkin, and rocky road ice cream made up 12% of total ice cream sales. This year, the same three ice creams made up 16.9% of ice cream sales. The rocky road sales doubled, the banana sales increased by 50%, and the pumpkin sales increased by 20%. If the rocky road ice cream had one less percent of sales than the banana This content is available for free at https://cnx.org/content
/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1301 11.7 \| Solving Systems with Inverses Learning Objectives In this section, you will: 11.7.1 Find the inverse of a matrix. 11.7.2 Solve a system of linear equations using an inverse matrix. Nancy plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix. Finding the Inverse of a Matrix We know that the multiplicative inverse of a real number a is a−1, and aa−1 = a−1 a = ⎛ ⎝ 1 a ⎞ ⎠a = 1. For example, 1 2 and ⎛ ⎝ 2−1 = 1 2 ⎞ ⎠2 = 1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A−1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by In where n represents the dimension of the matrix. Equation 11.1 and Equation 11.2 are the identity matrices for a 2×2 matrix and a 3×3 matrix, respectively. I2 = ⎡ 1 ⎣ 0 ⎤ ⎦ 0 1 I3 = ⎡ 11.1) (11.2) The identity matrix acts as a 1 in matrix algebra. For example, AI = IA = A. A matrix that has a multiplicative inverse has the properties AA−1 = I A−1 A = I A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, AA−1 = A−1 A = I, is a requirement. Not all square matrices have an inverse, but if A is invertible, then A−1 is unique. We will look at two methods for finding the inverse of a 2×2 matrix and a third method that can be used on both 2×2 and 3×3 matrices. The Identity Matrix and Multiplicative Inverse The identity matrix, In, is a square matrix containing ones down the main diagonal and zeros everywhere else. ⎤ ⎦ I2 = ⎡ I3 = ⎤ ⎥ ⎦ If A is an n × n matrix and B is an n × n matrix such that AB = BA = In, then B = A−1, the multiplicative inverse of a matrix A. 1302 Chapter 11 Systems of Equations and Inequalities Example 11.48 Showing That the Identity Matrix Acts as a 1 Given matrix A, show that AI = IA = A2 5 Solution Use matrix multiplication to show that the product of A and the identity is equal to the product of the identity and A. AI = AI = ⎤ ⎡ 1 ⎣ ⎦ 0 ⎡ 4 3 ⎣ −2 5 ⎡ ⎤ 4 3 ⎦ ⎣ − ⋅ (−2) ⎣ 0 ⋅ 3 + 1 ⋅ (−22 5 ⎡ 4 3 ⎣ −2 5 ⎤ ⎦ ⎤ ⎦ Given two matrices, show that one is the multiplicative inverse of the other. 1. Given matrix A of order n × n and matrix B of order n × n multiply AB. 2. If AB = I, then find the product BA. If BA = I, then B = A−1 and A = B−1. Example 11.49 Showing That Matrix A Is the Multiplicative Inverse of Matrix B Show that the given matrices are multiplicative inverses of each other. A = 1 ⎤ ⎡ 5 ⎦, B = ⎣ −2 −9 ⎤ ⎡ −9 −5 ⎦ ⎣ 1 2 Solution Multiply AB and BA. If both products equal the identity, then the two matrices are inverses of each other. AB = = = 1 ⎤ ⎦ ⎡ ⎤ −9 −2 −9 2 ⎤ ⎡ 1(−5) + 5(1) 1(−9) + 5(2) ⎦ ⎣ −2(−9)−9(2) −2(−5)−9(19 −5 ⎦ · ⎣ 1 ⎤ ⎡ −9(5)−5(−9) −9(1)−5(−2) ⎦ ⎣ 2(1) + 1(−2) 2(−5) + 1(−92 −9 0 1 ⎤ ⎦ 1 2 BA = = = A and B are inverses of each other. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1303 11.28 Show that the following two matrices are inverses of each other. A = 1 ⎤ ⎡ 4 ⎦, B = ⎣ −1 −3 ⎤ ⎡ −3 −4 ⎦ ⎣ 1 1 Finding the Multiplicative Inverse Using Matrix Multiplication We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. Example 11.50 Finding the Multiplicative Inverse Using Matrix Multiplication Use matrix multiplication to find the inverse of the given matrix. A = ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 Solution For this method, we multiply A by a matrix containing unknown constants and set it equal to the identity. ⎤ ⎡ 1 −2 ⎦ ⎣ 2 − ⎤ ⎦ Find the product of the two matrices on the left side of the equal sign. ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 a b ⎡ ⎣ c d ⎤ ⎦ = 1a−2c 1b−2d ⎡ ⎣ 2a−3c 2b−3d ⎤ ⎦ Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0. 1a−2c = 1 R1 2a−3c = 0 R2 Using row operations, multiply and add as follows: (−2)R1 + R2 → R2. Add the equations, and solve for c. 1a − 2c = 1 0 + 1c = − 2 c = − 2 Back-substitute to solve for a. a−2(−23 Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity. Using row operations, multiply and add as follows: (−2)R1 + R2 = R2. Add the two equations and solve for d. 1b−2d = 0 R1 2b−3d = 1 R2 1304 Chapter 11 Systems of Equations and Inequalities Once more, back-substitute and solve for b. 1b−2d = 0 0 + 1d = 1 d = 1 b−2(1) = 0 b−2 = 0 b = 2 ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 A−1 = Finding the Multiplicative Inverse by Augmenting with the Identity Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I, the augmented matrix I transforms into A−1. For example, given augment A with the identity Perform row operations with the goal of turning A into the identity. Switch row 1 and row 2. 2. Multiply row 2 by −2 and add to row 1. 3. Multiply row 1 by −2 and add to row 2. 4. Add row 2 to row 1. 5. Multiply row 2 by −1. The matrix we have found is A−11 \| −2 ⎤ 1 ⎦ 5 −2 ⎡ 0 1 ⎣ 0 −1 \| ⎤ 3 −1 ⎦ 5 −1 ⎦ 2 −5 A−1 = ⎡ ⎣ −5 ⎤ 3 −1 ⎦ 2 Finding the Multiplicative Inverse of 2×2 Matrices Using a Formula When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1305 If A is a 2×2 matrix, such as the multiplicative inverse of A is given by the formula −1 = 1 ad − bc ⎡ ⎣ −c ⎤ d −b ⎦ a (11.3) where ad − bc ≠ 0. If ad − bc = 0, then A has no inverse. Example 11.51 Using the Formula to Find the Multiplicative Inverse of Matrix A Use the formula to find the multiplicative inverse of A = ⎤ ⎡ 1 −2 ⎦ ⎣ 2 −3 Solution Using the formula, we have ⎡ −3 2 ⎣ −2 1 ⎤ ⎦ A−1 = = = 1 (1)(−3) − (−2)(2) ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 1 −3 + 4 ⎤ ⎡ −3 2 ⎦ ⎣ −2 1 Analysis We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment A with the identity. Perform row operations with the goal of turning A into the identity. ⎡ 1 −2 ⎣ 2 −3 ⎤ ⎦ \| 1 0 0 1 1. Multiply row 1 by −2 and add to row 2. 2. Multiply row 1 by 2 and add to row 1. So, we have verified our original solution2 ⎣ 0 1 ⎤ ⎦ −2 1 \| 1 0 \| −3 2 ⎤ ⎦ −2 1 A−1 = ⎡ −3 2 ⎣ −2 1 ⎤ ⎦ Use the formula to find the inverse of matrix A. Verify your answer by augmenting with the identity 11.29 matrix. A = ⎤ ⎡ 1 −1 ⎦ ⎣ 2 3 1306 Chapter 11 Systems of Equations and Inequalities Example 11.52 Finding the Inverse of the Matrix, If It Exists Find the inverse, if it exists, of the given matrix. A = ⎡ 3 6 ⎣ 1 2 ⎤ ⎦ Solution We will use the method of augmenting with the identity. 1. Switch row 1 and row 2. Multiply row 1 by −3 and add it to row 23 1 ⎤ ⎦ 3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse. Finding the Multiplicative Inverse of 3×3 Matrices Unfortunately, we do not have a formula similar to the one for a 2×2 matrix to find the inverse of a 3×3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a 3×3 matrix augment A with the identity matrix \| ⎤ ⎥ ⎦ \| To begin, we write the augmented matrix with the identity on the right and A on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example. Given a 3 × 3 matrix, find the inverse 1. Write the original matrix augmented with the identity matrix on the right. 2. Use elementary row operations so that the identity appears on the left. 3. What is obtained on the right is the inverse of the original matrix. 4. Use matrix multiplication to show that AA−1 = I and A−1 A = I. Example 11.53 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1307 Finding the Inverse of a 3 × 3 Matrix Given the 3 × 3 matrix A, find the inverse ⎤ ⎥ ⎦ Solution Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A. The matrix on the right will be the inverse of A. Interchange R2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ and R1 → ⎡ R2 + R1 = R1 → −R2 + R3 = R3 → ⎡ R3 ↔ R2 → ⎤ ⎥ ⎦ \| 1 1 0 ⎥ 1 0 0 ⎦ −1 1 0 −1 0 1 1 0 0 −1 −2 −3 −1 −1 −2R1 + R3 = R3 → −3R2 + R3 = R3 → Thus, A−1 = B = 0 −1 1 ⎡ ⎢ −1 0 1 ⎣ 6 −2 −3 ⎤ ⎥ ⎦ Analysis To prove that B = A−1, let’s multiply the two matrices together to see if the product equals the identity, if AA−1 = I and A−1 A = I. AA−1 = = = ⎤ ⎥ ⎦ ⎡ −1 ⎢ −2 −(−1) + 3(−1) + 1(6) 2(1) + 3(0) + 1(−2) 2(0) + 3(1) + 1(−3) ⎤ ⎡ ⎥ ⎢ 3(−1) + 3(−1) + 1(6) 3(1) + 3(0) + 1(−2) 3(0) + 3(1) + 1(−3) ⎥ ⎢ 2(−1) + 4(−1) + 1(6) 2(1) + 4(0) + 1(−2) 2(0) + 4(1) + 1(−3 ⎤ ⎥ ⎦ 1308 Chapter 11 Systems of Equations and Inequalitie
s A−2 −3 2 4 1 −1(2) + 1(3) + 0(2) −1(2) + 0(3) + 1(2) ⎡ −1 ⎢ −1 ⎣ −1(1) + 1(1) + 0(1) −1(3) + 1(3) + 0(4) ⎤ ⎡ ⎥ ⎢ −1(1) + 0(1) + 1(1) −1(3) + 0(3) + 1(4) ⎥ ⎢ 6(2) + −2(3) + −3(2) 6(3) + −2(3) + −3(4) 6(1) + −2(1) + −3(1 ⎤ ⎥ ⎦ 11.30 Find the inverse of the 3×3 matrix. A = 2 −17 11 ⎡ ⎤ ⎢ ⎥ −1 11 −7 ⎣ ⎦ 0 3 −2 Solving a System of Linear Equations Using the Inverse of a Matrix Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system AX = B. For example, look at the following system of equations. AX = B From this system, the coefficient matrix is The variable matrix is And the constant matrix is Then AX = B looks like a1 x + b1 y = c1 a2 x + b2 y = c2 A = a1 b1 ⎡ ⎣ a2 b2 ⎤ ⎦ X = x ⎡ y ⎣ ⎤ ⎦ B = c1 ⎡ c2 ⎣ ⎤ ⎦ a1 b1 ⎡ ⎣ a2 b2 ⎤ ⎦ x ⎡ y ⎣ ⎤ ⎦ = c1 ⎡ c2 ⎣ ⎤ ⎦ Recall the discussion earlier in this section regarding multiplying a real number by its inverse, (2−1) 2 = ⎞ ⎠ 2 = 1. To solve a single linear equation ax = b for x, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a. Thus, 1 2 ⎛ ⎝ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1309 ⎛ ⎝ ⎞ ⎠b ax = b ⎛ ⎞ 1 1 ⎠ax = a a ⎝ (a−1 )ax = (a−1)b [(a−1)a]x = (a−1)b 1x = (a−1)b x = (a−1)b The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system. Solving a System of Equations Using the Inverse of a Matrix Given a system of equations, write the coefficient matrix A, the variable matrix X, and the constant matrix B. Then Multiply both sides by the inverse of A to obtain the solution. AX = B ⎛ ⎝A−1⎞ ⎠AX = ⎡ ⎠A⎤ ⎛ ⎝A−1⎞ ⎣ ⎦X = IX = X = ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. Example 11.54 Solving a 2 × 2 System Using the Inverse of a Matrix Solve the given system of equations using the inverse of a matrix. 3x + 8y = 5 4x + 11y = 7 Solution Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. Then A = ⎡ 3 8 ⎣ 4 11 ⎤ ⎦, X = x ⎡ y ⎣ ⎤ ⎦, 11 ⎤ ⎦ First, we need to calculate A−1. Using the formula to calculate the inverse of a 2 by 2 matrix, we have: 1310 Chapter 11 Systems of Equations and Inequalities ⎤ ⎦ d −b ⎡ ⎣ −c a ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 A−1 = = 1 ad − bc 1 3(11)−8(4) ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 = 1 1 So, Now we are ready to solve. Multiply both sides of the equation by A−1. A−1 = ⎤ ⎡ 11 −8 ⎦ ⎣ −4 3 ⎤ ⎡ 11 −8 ⎦ ⎣ 3 −4 The solution is (−1, 1). ⎤ ⎦ ⎤ ⎦ x ⎡ y ⎣ ⎡ 5 ⎣ 7 (A−1)AX = (A−1)B ⎤ ⎡ ⎡ 11 −4 3 4 11 ⎡ ⎤ 11(5) + (−8)4(5) + 3(7) ⎡ ⎤ − ⎣ Can we solve for X by finding the product BA−1? No, recall that matrix multiplication is not commutative, so A−1 B ≠ BA−1. Consider our steps for solving the matrix equation. ⎛ ⎝A−1⎞ ⎠AX = ⎡ ⎠A⎤ ⎛ ⎝A−1⎞ ⎣ ⎦X = IX = ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B ⎛ ⎝A−1⎞ ⎠B X = Notice in the first step we multiplied both sides of the equation by A−1, but the A−1 was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters. Example 11.55 Solving a 3 × 3 System Using the Inverse of a Matrix Solve the following system using the inverse of a matrix. 5x + 15y + 56z = 35 −4x−11y−41z = −26 −x−3y−11z = −7 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1311 Write the equation AX = B. ⎤ ⎡ 56 15 5 ⎥ ⎢ −4 −11 −41 ⎦ ⎣ −1 −3 −11 x ⎡ y ⎢ ⎣ z ⎤ ⎥ = ⎦ ⎤ ⎡ 35 ⎥ ⎢ −26 ⎦ ⎣ −7 First, we will find the inverse of A by augmenting with the identity. Multiply row 1 by 1 5 . Multiply row 1 by 4 and add to row 2. Add row 1 to row 3. Multiply row 2 by −3 and add to row 1. Multiply row 3 by 5. Multiply row 3 by 1 5 and add to row 1 19 ⎢ 19 ⎢ 5 ⎣ 0 0 1 5 ⎡ 56 15 ⎢ −4 −11 −41 ⎣ −1 −3 −11 56 5 ⎡ ⎢ ⎢ −4 −11 −41 ⎣ −1 −3 −11 1 −3 −11 56 5 19 56 ⎢ 5 ⎢ 0 1 19 ⎢ − 11 5 4 5 1 5 − 11 19 ⎢ 5 ⎣ 0 0 1 −2 −3 1 4 5 1 1 0 0 5 ⎤ ⎥ ⎥ ⎦ 1312 Chapter 11 Systems of Equations and Inequalities Multiply row 3 by − 19 5 and add to row 2. So−1 = 1 ⎤ −2 −3 ⎥ −3 1 −19 ⎦ 0 1 5 \| 1 ⎤ ⎡ −2 −3 ⎥ ⎢ −3 1 −19 ⎦ ⎣ 0 1 5 Multiply both sides of the equation by A−1. We want A−1 AX = A−1 B : ⎡ −2 −3 ⎢ −3 ⎣ 1 ⎤ 1 ⎥ 1 −19 ⎦ 5 0 5 ⎤ ⎡ 56 15 ⎥ ⎢ −4 −11 −41 ⎦ ⎣ −1 −3 −11 2 −3 ⎢ −3 ⎣ 1 ⎤ 1 ⎥ 1 −19 ⎦ 5 0 ⎤ ⎡ 35 ⎥ ⎢ −26 ⎦ ⎣ −7 Thus, The solution is (1, 2, 0). A−1 B = ⎡ ⎤ −70 + 78−7 ⎢ ⎥ = −105−26 + 133 ⎣ ⎦ 35 + 0−35 ⎤ ⎥ ⎦ ⎡ 1 ⎢ 2 ⎣ 0 11.31 Solve the system using the inverse of the coefficient matrix. 2x − 17y + 11z = 0 − x + 11y − 7z = 8 3y − 2z = −2 Given a system of equations, solve with matrix inverses using a calculator. 1. Save the coefficient matrix and the constant matrix as matrix variables [A] and [B]. 2. Enter the multiplication into the calculator, calling up each matrix variable as needed. 3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. Example 11.56 Using a Calculator to Solve a System of Equations with Matrix Inverses Solve the system of equations with matrix inverses using a calculator 2x + 3y + z = 32 3x + 3y + z = −27 2x + 4y + z = −2 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1313 On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A], and enter the constant matrix as the matrix variable [B]. [A ⎤ ⎥, [B] = ⎦ ⎤ ⎡ 32 ⎥ ⎢ −27 ⎦ ⎣ −2 On the home screen of the calculator, type in the multiplication to solve for X, calling up each matrix variable as needed. Evaluate the expression. [A]−1×[B] ⎤ ⎡ −59 ⎥ ⎢ −34 ⎦ ⎣ 252 Access these online resources for additional instruction and practice with solving systems with inverses. • The Identity Matrix (http://openstaxcollege.org/l/identmatrix) • Determining Inverse Matrices (http://openstaxcollege.org/l/inversematrix) • Using a Matrix Equation to Solve a System of Equations (http://openstaxcollege.org/l/ matrixsystem) 1314 Chapter 11 Systems of Equations and Inequalities 11.7 EXERCISES Verbal 385. In a previous section, we showed that matrix multiplication is not commutative, that is, AB ≠ BA in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, A−1 A = AA−1 ? Does every 2×2 matrix have an inverse? Explain 386. why or why not. Explain what condition is necessary for an inverse to exist. Can you explain whether a 2×2 matrix with an entire 387. row of zeros can have an inverse? Can a matrix with an entire column of zeros have an 388. inverse? Explain why or why not. Can a matrix with zeros on the diagonal have an 389. inverse? If so, find an example. If not, prove why not. For simplicity, assume a 2×2 matrix. Algebraic In the following exercises, show that matrix A is the inverse of matrix B. 390. 391. 392. 393. 394. 395. 396. A = ⎡ 1 0 ⎣ −1 1 ⎤ ⎦, ⎤ ⎦, ⎤ ⎦, B = ⎡ −0 ⎢ 1 ⎣ 5 1 7 − 4 35 ⎤ ⎥ ⎥ ⎦ A = ⎡ ⎤ ⎢−2 1 ⎥, B = 2 ⎣ ⎦ 3 −1 ⎤ ⎡ −2 −1 ⎦ ⎣ −6 −4 A = ⎡ ⎤ 1 0 1 ⎥, B = 1 ⎢ 0 1 − ⎤ ⎥, B = 1 4 ⎦ ⎤ ⎡ 0 −2 6 ⎥ ⎢ 17 −3 −5 ⎦ ⎣ 4 −12 12 ⎤ ⎥, B = 1 36 ⎦ ⎤ ⎡ −6 84 −6 ⎥ ⎢ 7 −26 1 ⎦ ⎣ −1 −22 5 For the following exercises, find the multiplicative inverse of each matrix, if it exists. This content is available for free at https://cnx.org/content/col11758/1.5 397. 398. 399. 400. 401. 402. 403. 404. 405. 406. 407. 408. 409. 410. ⎡ ⎤ 3 −2 ⎣ ⎦ 1 9 ⎡ −2 2 ⎣ 3 1 ⎤ ⎦ ⎡ −4 −3 ⎣ ⎦ −.5 1.5 ⎦ ⎣ 1 −0.5 ⎡ 1 0 6 ⎢ −1 1 ⎥ ⎢ 1 −3 4 ⎦ ⎣ −2 −4 −5 ⎤ ⎡ 1 9 −2 7 3 ⎤ ⎡ 1 −2 ⎥ ⎢ −4 8 −12 ⎦ ⎣ ⎤ ⎥ ⎦ For the following exercises, solve the system using the inverse of a 2 × 2 matrix. 411. Chapter 11 Systems of Equations and Inequalities 1315 5x − 6y = − 61 4x + 3y = − 2 412. 8x + 4y = −100 3x−4y = 1 413. 3x−2y = 6 −x + 5y = −2 414. 5x−4y = −5 4x + y = 2.3 415. −3x−4y = 9 12x + 4y = −6 416. 417. 418. −2x + 3y = 3 10 − x + 5y = 10 For the following exercises, solve a system using the inverse of a 3×3 matrix. 3x−2y + 5z = 21 5x + 4y = 37 x−2y−5z = 5 4x + 4y + 4z = 40 2x − 3y + 4z = −12 − x + 3y + 4z = 9 6x − 5y − z = 31 − x + 2y + z = −6 3x + 3y + 2z = 13 6x−5y + 2z = −4 2x + 5y − z = 12 2x + 5y + z = 12 4x−2y + 3z = −12 2x + 2y−9z = 33 6y−4z = 1 419. 420. 421. 422. 423. 424. y + 4z = −41 2 z = −101 x − 1 1 10 5 x−20y + 2 1 5 5 3 10 x + 4y − 3 10 z = 23 425. z = 31 100 40 426. 0.1x + 0.2y + 0.3z = −1.4 0.1x−0.2y + 0.3z = 0.6 0.4y + 0.9z = −2 Technology For the following exercises, use a calculator to solve the system of equations with matrix inverses. 427. 2x − y = −3 −x + 2y = 2.3 428. 429. 430 + 11 2 5 y = − 43 20 y = 31 4 12.3x−2y−2.5z = 2 36.9x + 7y−7.5z = −7 8y−5z = −10 0.5x−3y + 6z = −0.8 0.7x−2y = −0.06 0.5x + 4y + 5z = 0 Extensions For the following exercises, find the inverse of the given matrix. 431. 432. 4331 ⎢ 0 ⎢ 0 ⎣ 1 −2 ⎢ 1 0 0 2 ⎢ 4 −2 3 1 ⎣ 1 1 0 −5 ⎤ ⎥ ⎥ ⎦ Chapter 11 Systems of Equations and Inequalities Three roommates shared a package of 12 ice cream 443. bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate
three less than Tom, how many ice cream bars did each roommate eat? A farmer constructed a chicken coop out of chicken 444. wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood $10 per square foot, and the plywood $5 per square foot. The farmer spent a total of $51, and the total amount of materials used was 14 ft2. He used 3 ft2 more chicken wire than plywood. How much of each material in did the farmer use? Jay has lemon, orange, and pomegranate trees in his 445. backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick? 1316 434. 435 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Real-World Applications For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. 2,400 tickets were sold for a basketball game. If the 436. prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket? In the previous exercise, if you were told there were 437. 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket? A food drive collected two different types of canned 438. goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated? Students were asked to bring their favorite fruit to 439. class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit? A sorority held a bake sale to raise money and sold 440. brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold? A clothing store needs to order new inventory. It has 441. three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at $7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold? Anna, Ashley, and Andrea weigh a combined 370 lb. 442. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1317 11.8 \| Solving Systems with Cramer's Rule Learning Objectives In this section, you will: 11.8.1 Evaluate 2 × 2 determinants. 11.8.2 Use Cramer’s Rule to solve a system of equations in two variables. 11.8.3 Evaluate 3 × 3 determinants. 11.8.4 Use Cramer’s Rule to solve a system of three equations in three variables. 11.8.5 Know the properties of determinants. We have learned how to solve systems of equations in two variables and three variables, and by multiple methods: substitution, addition, Gaussian elimination, using the inverse of a matrix, and graphing. Some of these methods are easier to apply than others and are more appropriate in certain situations. In this section, we will study two more strategies for solving systems of equations. Evaluating the Determinant of a 2×2 Matrix A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an invertible matrix and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section. Find the Determinant of a 2 × 2 Matrix The determinant of a 2 × 2 matrix, given is defined as A = a b ⎡ ⎣ c d ⎤ ⎦ Notice the change in notation. There are several ways to indicate the determinant, including det(A) and replacing the brackets in a matrix with straight lines, \|A\|. Example 11.57 Finding the Determinant of a 2 × 2 Matrix Find the determinant of the given matrix. A = ⎡ 5 2 ⎣ −6 3 ⎤ ⎦ Solution 1318 Chapter 11 Systems of Equations and Inequalities det(A) = \| 5 2 −6 3\| = 5(3) − (−6)(2) = 27 Using Cramer’s Rule to Solve a System of Two Equations in Two Variables We will now introduce a final method for solving systems of equations that uses determinants. Known as Cramer’s Rule, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704-1752), who introduced it in 1750 in Introduction à l'Analyse des lignes Courbes algébriques. Cramer’s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns. Cramer’s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used. To understand Cramer’s Rule, let’s look closely at how we solve systems of linear equations using basic row operations. Consider a system of two equations in two variables. a1 x + b1 y = c1 (1) a2 x + b2 y = c2 (2) We eliminate one variable using row operations and solve for the other. Say that we wish to solve for x. If equation (2) is multiplied by the opposite of the coefficient of y in equation (1), equation (1) is multiplied by the coefficient of y in equation (2), and we add the two equations, the variable y will be eliminated. Now, solve for x. b2 a1 x + b2 b1 y = b2 c1 −b1 a2 x − b1 b2 y = − b1 c2 Multiply R2 by − b1 ________________________________________________________ Multiply R1 by b2 b2 a1 x − b1 a2 x = b2 c1 − b1 c2 b2 a1 x − b1 a2 x = b2 c1 − b1 c2 x(b2 a1 − b1 a2) = b2 c1 − b1 c2 x = b2 c1 − b1 c2 b2 a1 − b1 a2 = c1 b1 ⎤ ⎡ ⎦ ⎣ c2 b2 a1 b1 ⎤ ⎡ ⎦ ⎣ a2 b2 Similarly, to solve for y, we will eliminate x. a2 a1 x + a2 b1 y = a2 c1 −a1 a2 x − a1 b2 y = − a1 c2 Multiply R2 by − a1 ________________________________________________________ a2 b1 y − a1 b2 y = a2 c1 − a1 c2 Multiply R1 by a2 Solving for y gives (11.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1319 a2 b1 y − a1 b2 y = a2 c1 − a1 c2 y(a2 b1 − a1 b2) = a2 c1 − a1 c2 y = a2 c1 − a1 c2 a2 b1 − a1 b2 = a1 c2 − a2 c1 a1 b2 − a2 b1 a1 c1 a2 c2\| = \| \|a1 b1 a2 b2\| Notice that the denominator for both x and y is the determinant of the coefficient matrix. We can use these formulas to solve for x and y, but Cramer’s Rule also introduces new notation: • D : determinant of the coefficient matrix • D x : determinant of the numerator in the solution of x • D y : determinant of the numerator in the solution of The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express x and y as a quotient of two determinants. Cramer’s Rule for 2×2 Systems Cramer’s Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. Consider a system of two linear equations in two variables. The solution using Cramer’s Rule is given as a1 x + b1 y = c1 a2 x + b2 y = c2 x = , D ≠ 0; y = D x D = \|c1 b1 c2 b2\| \|a1 b1 a2 b2\| D y D = \| a1 c1 a2 c2\| \|a1 b1 a2 b2\| , D ≠ 0. (11.5) If we are solving for x, the x column is replaced with the constant column. If we are solving for y, the y column is replaced with the constant column. Example 11.58 Using Cramer’s Rule to Solve a 2 × 2 System Solve the following 2 × 2 system using Cramer’s Rule. 12x + 3y = 15 2x − 3y = 13 Solution 1320 Chapter 11 Systems of Equations and Inequalities Solve for x. Solve for y. x = D x 3 D = \|15 13 −3\| \|12 2 −3\| 3 = −45 − 39 −36 − 6 = −84 −42 = 2 y = The solution is (2, −3). D y D = \|12 15 2 13\| \|12 2 −3\| 3 = 156 − 30 −36 − 6 = − 126 42 = −3 11.32 Use Cramer’s Rule to solve the 2 × 2 system of equations. x + 2y = −11 −2x + y = −13 Evaluating the Determinant of a 3 × 3 Matrix Finding the determinant of a 2×2 matrix is straightforward, but finding the determinant of a 3×3 matrix is more complicated. One method is to augment the 3×3 matrix with a repetition of the first two columns, giving a 3×5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example. Find the determinant of the 3×3 matrix. 1. Augment A with the first two columns. A = a1 b1 c1 ⎡ ⎢ a2 b2 c2 ⎢ a3 b3 c3 ⎣ ⎤ ⎥ ⎥ ⎦ 2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal. a1 b1 c1 a2 b2 c2 a3 b3 c3 a1 a2 a3 \| det(A) =\| b1 b2 b3\| 3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product
of entries up the third diagonal. The algebra is as follows: \|A\| = a1 b2 c3 + b1 c2 a3 + c1 a2 b3 − a3 b2 c1 − b3 c2 a1 − c3 a2 b1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1321 Example 11.59 Finding the Determinant of a 3 × 3 Matrix Find the determinant of the 3 × 3 matrix given A = ⎡ 0 2 1 ⎢ 3 −1 1 ⎣ 4 0 1 ⎤ ⎥ ⎦ Solution Augment the matrix with the first two columns and then follow the formula. Thus, \|A\| =\|0 2 1 3 −1 0\| = 0(−1)(1) + 2(1)(4) + 1(3)(0) − 4(−1)(1) − 0(1)(0) − 1(3)(2 11.33 Find the determinant of the 3 × 3 matrix. 1 1 1 det(A) =\|1 −3 7 1 −2 3\| Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2 × 2 and 3 × 3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramer’s Rule to Solve a System of Three Equations in Three Variables Now that we can find the determinant of a 3 × 3 matrix, we can apply Cramer’s Rule to solve a system of three equations in three variables. Cramer’s Rule is straightforward, following a pattern consistent with Cramer’s Rule for 2 × 2 matrices. As the order of the matrix increases to 3 × 3, however, there are many more calculations required. When we calculate the determinant to be zero, Cramer’s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system. Consider a 3 × 3 system of equations. where = Dz D , D ≠ 0 1322 Chapter 11 Systems of Equations and Inequalities If we are writing the determinant D x, we replace the x column with the constant column. If we are writing the determinant D y, we replace the y column with the constant column. If we are writing the determinant Dz, we replace the z column with the constant column. Always check the answer. Example 11.60 Solving a 3 × 3 System Using Cramer’s Rule Find the solution to the given 3 × 3 system using Cramer’s Rule. x + y − z = 6 3x − 2y + z = −5 x + 3y − 2z = 14 Solution Use Cramer’s Rule. 3 −2 1 D =\|1 1 −1 1 3 −2\|, D x =\| 6 1 −1 −5 −2 1 14 3 −2\|, D y =\|1 6 −1 1 14 −2\|, Dz =\|1 1 1 3 14\| 6 3 −2 −5 3 −5 1 Then3 −3 D y D = −9 −3 Dz D = 6 −3 = 1 = 3 = − 2 The solution is (1, 3, −2). 11.34 Use Cramer’s Rule to solve the 3 × 3 matrix. x − 3y + 7z = 13 x + y + z = 1 x − 2y + 3z = 4 (11.6) Example 11.61 Using Cramer’s Rule to Solve an Inconsistent System Solve the system of equations using Cramer’s Rule. 3x − 2y = 4 (1) 6x − 4y = 0 (2) Solution We begin by finding the determinants D, D x, and D y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1323 We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables. D = \|3 −2 6 −4\| = 3(−4) − 6(−2) = 0 1. Multiply equation (1) by −2. 2. Add the result to equation (2). −6x + 4y = −8 6x − 4y = 0 _______________ 0 = −8 (11.7) We obtain the equation 0 = −8, which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 11.31. Figure 11.31 Example 11.62 Use Cramer’s Rule to Solve a Dependent System Solve the system with an infinite number of solutions. x − 2y + 3z = 0 (1) 3x + y − 2z = 0 (2) 2x − 4y + 6z = 0 (3) Solution Let’s find the determinant first. Set up a matrix augmented by the first two columns. 1324 Chapter 11 Systems of Equations and Inequalities Then, 3 1 −2 6 3 2 −4 \|1 −2 \| 1 −2 1 3 2 −4\| 1(1)(6) + (−2)(−2)(2) + 3(3)(−4) − 2(1)(3) − (−4)(−2)(1) − 6(3)(−2) = 0 As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (1) by −2 and add the result to equation (3): −2x + 4y − 6x = 0 2x − 4y + 6z = 0 0 = 0 2. Obtaining an answer of 0 = 0, a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See Figure 11.32. Figure 11.32 Understanding Properties of Determinants There are many properties of determinants. Listed here are some properties that may be helpful in calculating the determinant of a matrix. Properties of Determinants 1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. 2. When two rows are interchanged, the determinant changes sign. 3. 4. If either two rows or two columns are identical, the determinant equals zero. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. 5. The determinant of an inverse matrix A−1 is the reciprocal of the determinant of the matrix A. 6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. Example 11.63 Illustrating Properties of Determinants Illustrate each of the properties of determinants. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1325 Solution Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal1 ⎤ ⎥ ⎦ Augment A with the first two columns. A = Then ⎡ 1 det(A) = 1(2)(−1) + 2(1)(0) + 3(0)(0) − 0(2)(3) − 0(1)(1) + 1(0)(2) = −2 Property 2 states that interchanging rows changes the sign. Given A = ⎤ ⎡ −1 5 ⎦, det(A) = (−1)(−3) − (4)(5) = 3 − 20 = −17 ⎣ 4 −3 B = ⎡ ⎤ 4 −3 ⎦, det(B) = (4)(5) − (−1)(−3) = 20 − 3 = 17 ⎣ −1 5 Property 3 states that if two rows or two columns are identical, the determinant equals zero1 \| det(A) = 1(2)(2) + 2(2)(−1) + 2(2)(2) + 1(2)(2) − 2(2)(1) − 2(2)(2 Property 4 states that if a row or column equals zero, the determinant equals zero. Thus, A = ⎡ 1 2 ⎣ 0 0 ⎤ ⎦, det(A) = 1(0) − 2(0) = 0 Property 5 states that the determinant of an inverse matrix A−1 is the reciprocal of the determinant A. Thus, A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, det(A) = 1(4) − 3(2) = −2 A−1 = ⎡ −2 1 ⎢ 3 − 1 ⎣ 2 2 ⎤ ⎥, det ⎦ ⎝A−1⎞ ⎛ ⎛ ⎝− ⎞ ⎠(1) = − 1 2 Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus, A = ⎡ 1 2 ⎣ 3 4 ⎤ ⎦, det(A) = 1(4) − 2(3) = −2 B = ⎤ ⎡ 2(1) 2(2) ⎦, det(B) = 2(4) − 3(4) = −4 ⎣ 3 4 1326 Chapter 11 Systems of Equations and Inequalities Example 11.64 Using Cramer’s Rule and Determinant Properties to Solve a System Find the solution to the given 3 × 3 system. 2x + 4y + 4z = 2 (1) 3x + 7y + 7z = −5 (2) x + 2y + 2z = 4 (3) Solution Using Cramer’s Rule, we have Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 3 7 7 D =\|2 4 4 1 2 2\| 1. Multiply equation (3) by –2 and add the result to equation (1). −2x − 4y − 4x = − 8 2x + 4y + 4z = 2 0 = − 6 Obtaining a statement that is a contradiction means that the system has no solution. Access these online resources for additional instruction and practice with Cramer’s Rule. • Solve a System of Two Equations Using Cramer's Rule (http://openstaxcollege.org/l/ system2cramer) • Solve a Systems of Three Equations using Cramer's Rule (http://openstaxcollege.org/l/ system3cramer) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1327 11.8 EXERCISES Verbal Explain why we can always evaluate the determinant 446. of a square matrix. 447. Examining Cramer’s Rule, explain why there is no unique solution to the system when the determinant of your matrix is 0. For simplicity, use a 2 × 2 matrix. Explain what it means in terms of an inverse for a 448. matrix to have a 0 determinant. The determinant of 2 × 2 matrix A is 3. 449. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer. Algebraic For the following exercises, find the determinant. 450. 451. 452. 453. 454. 455. 456. 457. 458. 459. 460. 461. 0 2 20 −1 \|1 2 3 4\| \|−1 3 −4\| \| 2 −5 6\| \|−8 4 −1 5\| \|1 3 −4\| \|10 0 −10\| \|10 0.2 5 0.1\| \|6 −3 4\| \|−2 3.1 4, 000\| \|−1.1 7.2 −0.5\| \|−1 0 0 0 −3\| 0 1 0.6 −3 0 0 8 463. 462. 464. 465. 0 3 2 5 −5 −4 0 2 0 1 0 3 −4 1 4 1 2 −8 −4 −3 1 \|−1 4 0 0 −3\| \|1 0 1 1 0 0\| \| 2 −3 1 6 1\| \|−2 2 −8 −3\| \| 6 −1 9 −1\| \|5 3 −6 −3\| \|1.1 4.1 −0.4 2.5\| \| \| 81 0 0 1 −1 1 3 1.1 −9.3 1 4 1 7 1 −4 0 0 2 469. 466. 467. 468. 2 −1.6 3.1 3 −8 0 2\| For the following exercises, solve the system of linear equations using Cramer’s Rule. 470. 2x − 3y = −1 4x + 5y = 9 471. 5x − 4y = 2 −4x + 7y = 6 472. 6x − 3y = 2 −8x + 9y = −1 473. 2x + 6y = 12 5x − 2y = 13 474. 1328 Chapter 11 Systems of Equations and Inequalities 4x + 3y = 23 2x − y = −1 475. 10x − 6y = 2 −5x + 8y = −1 476. 4x − 3y = −3 2x + 6y = −4 477. 4x − 5y = 7 −3x + 9y = 0 478. 4x + 10y = 180 −3x − 5y = −105 479. 8x − 2y = −3 −4x + 6y = 4 For the following exercises, solve the system of linear equations using Cramer’s Rule. x + 2y − 4z = − 1 7x + 3y + 5z = 26 −2x − 6y + 7z = − 6 −5x + 2y − 4z = − 47 4x − 3y − z = − 94 3x − 3y + 2z = 94 4x + 5y − z = −7 −2x − 9y + 2z = 8 5y + 7z = 21 4x − 3y + 4z = 10 5x − 2z = − 2 3x + 2y − 5z = − 9 4x − 2y + 3z = 6 − 6x + y = − 2 2x + 7y + 8z = 24 5x + 2y − z = 1 −7x − 8y + 3z = 1.5 6x − 12y + z = 7 13x − 17y + 16z = 73 −11x + 15y + 17z = 61 46x + 10y − 30z = − 18 480. 481. 482. 483. 484. 485. 486. 487. This content is available for free at https://cnx.org/content/col11758/1.5 −4x − 3y − 8z = − 7 2x − 9y + 5z = 0.5 5x − 6y − 5z = − 2 488. 489. 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 x + y + z = 1 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 12x + 18y − 24z = − 30 Technology For the following exercises, use the determinant function on a graphing utility. 4
90. 491. 492. 4939 3 0 1 2 1 3 0 −2 −1 1 0 2 4 3\| \| 1 −2\| \| \| 7 8 9 0 100 2 2,000 0 4 5 2\| Real-World Applications For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution. Two numbers add up to 56. One number is 20 less 494. than the other. Two numbers add up to 104. If you add two times the 495. first number plus two times the second number, your total is 208 Three numbers add up to 106. The first number is 3 496. less than the second number. The third number is 4 more than the first number. 497. Chapter 11 Systems of Equations and Inequalities 1329 Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined. For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. You invest $10,000 into two accounts, which receive 498. 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account? You invest $80,000 into two accounts, $22,000 in one 499. account, and $58,000 in the other account. At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts? A movie theater needs to know how many adult 500. tickets and children tickets were sold out of the 1,200 total tickets. If children’s tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many children’s tickets and adult tickets were sold? A concert venue sells single tickets for $40 each and 501. couple’s tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many couple’s tickets were sold? You decide to paint your kitchen green. You create the 502. color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix? You sold two types of scarves at a farmers’ market 503. and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold? Your garden produced two types of tomatoes, one 504. green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have? At a market, the three most popular vegetables make 505. up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share? At the same market, the three most popular fruits 506. make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold. Three bands performed at a concert venue. The first 507. band charged $15 per ticket, the second band charged $45 per ticket, and the final band charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band? A movie theatre sold tickets to three movies. The 508. tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie were sold? Men aged 20–29, 30–39, and 40–49 made up 78% of 509. the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20–29 age group increased by 20%, the 30–39 age group increased by 2%, and the 40–49 age group decreased to 3 4 of their previous population. Originally, the 30–39 age group had 2% more prisoners than the 20–29 age group. Determine the prison population percentage for each age group last year. 510. At a women’s prison down the road, the total number of inmates aged 20–49 totaled 5,525. This year, the 20–29 age group increased by 10%, the 30–39 age group decreased by 20%, and the 40–49 age group doubled. There are now 6,040 prisoners. Originally, there were 500 more in the 30–39 age group than the 20–29 age group. Determine the prison population for each age group last year. For the following exercises, use this scenario: A healthconscious company decides to make a trail mix out of almonds, chocolate-covered cashews. The nutritional information for these items is shown in Table 11.5. cranberries, dried and Fat (g) 6 Almonds (10) Cranberries (10) 0.02 Protein (g) Carbohydrates (g) 2 0 3 8 7 3.5 5.5 Cashews (10) Table 11.5 511. 1330 Chapter 11 Systems of Equations and Inequalities For the special “low-carb”trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix? For the “hiking” mix, there are 1,000 pieces in the 512. mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix? For the “energy-booster” mix, there are 1,000 pieces 513. in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together amount of equivalent cranberries, how many of each item is in the trail mix? to the is This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1331 CHAPTER 11 REVIEW KEY TERMS addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable augmented matrix brackets a coefficient matrix adjoined with the constant column separated by a vertical line within the matrix break-even point the point at which a cost function intersects a revenue function; where profit is zero coefficient matrix a matrix that contains only the coefficients from a system of equations column a set of numbers aligned vertically in a matrix consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system cost function costs the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable Cramer’s Rule a method for solving systems of equations that have the same number of equations as variables using determinants dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system determinant a number calculated using the entries of a square matrix that determines such information as whether there is a solution to a system of equations entry an element, coefficient, or constant in a matrix feasible region the solution to a system of nonlinear inequalities that is the region of the graph where the shaded regions of each inequality intersect Gaussian elimination using elementary row operations to obtain a matrix in row-echelon form identity matrix algebra a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common independent system a system of linear equations with exactly one solution pair (x, y) main diagonal entries from the upper left corner diagonally to the lower right corner of a square matrix matrix a rectangular array of numbers multiplicative inverse of a matrix a matrix that, when multiplied by the original, equals the identity matrix nonlinear inequality an inequality containing a nonlinear expression partial fraction decomposition the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions partial fractions the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression profit function the profit function is written as P(x) = R(x) − C(x), revenue minus cost revenue function the function that is used to calculate revenue, simply written as R = xp, where x = quantity and p = price row a set of numbers aligned horizontally in a matrix 1332 Chapter 11 Systems of Equations and Inequalities row operations adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the goal of achieving row-echelon form row-echelon form after performing row operations, the matrix form that contains ones down the main diagonal and zeros at every space below the diagonal row-equivalent two matrices A and B are row-equivalent if one can be obtained from the other by performing basic row operations scalar multiple an entry of a matrix that has been multiplied by a scalar solution set the set of all ordered pairs or triples that satisfy all equations in a system of equations substitution method an algebraic technique used to solve systems of linear equations in which one of the
two equations is solved for one variable and then substituted into the second equation to solve for the second variable system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously. system of nonlinear equations a system of equations containing at least one equation that is of degree larger than one system of nonlinear inequalities inequality that is not linear KEY EQUATIONS Identity matrix for a 2×2 matrix Identity matrix for a 3×3 matrix a system of two or more inequalities in two or more variables containing at least one I2 = ⎡ 1 0 ⎣ 0 1 ⎤ ⎦ I3 = ⎡ ⎤ ⎥ ⎦ Multiplicative inverse of a 2×2 matrix A−1 = 1 ad − bc d −b ⎡ ⎣ −c a ⎤ ⎦, where ad − bc ≠ 0 KEY CONCEPTS 11.1 Systems of Linear Equations: Two Variables • A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. • The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 11.1. • Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. • One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 11.2. • Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 11.3. • A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 11.4. • It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 11.5, Example 11.6, and Example 11.7. • Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 11.8. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1333 • The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 11.9. • Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 11.10 and Example 11.11. 11.2 Systems of Linear Equations: Three Variables • A solution set is an ordered triple ⎧ ⎨(x, y, z)⎫ ⎭ ⎩ ⎬ that represents the intersection of three planes in space. See Example 11.12. • A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example 11.13. • Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example 11.14. • A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example 11.15. • Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. • A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example 11.16. • Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables • There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 11.17. • There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See Example 11.18. • There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle: (1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example 11.19. • An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example 11.20. • Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example 11.21. 11.4 Partial Fractions • Decompose P(x) Q(x) by writing the partial fractions as A a1 x + b1 + B a2 x + b2 . Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See Example 11.22. • The decomposition of P(x) Q(x) with repeated linear factors must account for the factors of the denominator in increasing powers. See Example 11.23. 1334 Chapter 11 Systems of Equations and Inequalities • The decomposition of P(x) Q(x) with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in A x + Bx + C ⎝ax2 + bx + c⎞ ⎛ ⎠ . See Example 11.24. • In the decomposition of P(x) Q(x) , where Q(x) has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as Ax + B ⎝ax2 + bx + c⎞ ⎛ ⎠ + A2 x + B2 ⎝ax2 + bx + c⎞ ⎛ ⎠ 2 + ⋯ + An x + Bn ⎝ax2 + bx + c⎞ ⎛ ⎠ n. See Example 11.25. 11.5 Matrices and Matrix Operations • A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. • The dimensions of a matrix refer to the number of rows and the number of columns. A 3×2 matrix has three rows and two columns. See Example 11.26. • We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See Example 11.27, Example 11.28, Example 11.29, and Example 11.30. • Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example 11.31. • Scalar multiplication is often required before addition or subtraction can occur. See Example 11.32. • Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second. • The product of two matrices, A and B, is obtained by multiplying each entry in row 1 of A by each entry in column 1 of B; then multiply each entry of row 1 of A by each entry in columns 2 of B, and so on. See Example 11.33 and Example 11.34. • Many real-world problems can often be solved using matrices. See Example 11.35. • We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example 11.36. 11.6 Solving Systems with Gaussian Elimination • An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example 11.37. • A matrix augmented with the constant column can be represented as the original system of equations. See Example 11.38. • Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows. • We can use Gaussian elimination to solve a system of equations. See Example 11.39, Example 11.40, and Example 11.41. • Row operations are performed on matrices to obtain row-echelon form. See Example 11.42. • To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions. See Example 11.43 and Example 11.44. • A calculator can be used to solve systems of equations using matrices. See Example 11.45. • Many real-world problems can be solved using augmented matrices. See Example 11.46 and Example 11.47. 11.7 Solving Systems with Inverses • An identity matrix has the property AI = IA = A. See Example 11.48. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1335 • An invertible matrix has the property AA−1 = A−1 A = I. See Example 11.49. • Use matrix multiplication and the identity to find the inverse of a 2×2 matrix. See Example 11.50. • The multiplicative inverse can be found using a formula. See Example 11.51. • Another method of finding the inverse is by augmenting with the identity. See Example 11.52. • We can augment a 3×3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example 11.53. • Write the system of equations as AX = B, and multiply both sides by the inverse of A : A−1 AX = A−1 B. See Example 11.54 and Example 11.55. • We can also use a calculator to solve a system of equations with matrix inverses. See Example 11.56. 11.8 Solving Systems with Cramer's Rule a b • The determinant for ⎡
⎣ c d ⎤ is ad − bc. See Example 11.57. ⎦ • Cramer’s Rule replaces a variable column with the constant column. Solutions are . See Example 11.58. • To find the determinant of a 3×3 matrix, augment with the first two columns. Add the three diagonal entries (upper left to lower right) and subtract the three diagonal entries (lower left to upper right). See Example 11.59. • To solve a system of three equations in three variables using Cramer’s Rule, replace a variable column with the constant column for each desired solution = Dz D . See Example 11.60. • Cramer’s Rule is also useful for finding the solution of a system of equations with no solution or infinite solutions. See Example 11.61 and Example 11.62. • Certain properties of determinants are useful for solving problems. For example: ◦ If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. ◦ When two rows are interchanged, the determinant changes sign. ◦ ◦ If either two rows or two columns are identical, the determinant equals zero. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. ◦ The determinant of an inverse matrix A−1 is the reciprocal of the determinant of the matrix A. ◦ If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. See Example 11.63 and Example 11.64. CHAPTER 11 REVIEW EXERCISES Systems of Linear Equations: Two Variables For the following exercises, determine whether the ordered pair is a solution to the system of equations. 514. 3x − y = 4 x + 4y = − 3 and ( − 1, 1) 515. 6x − 2y = 24 −3x + 3y = 18 and (9, 15) For the following exercises, use substitution to solve the system of equations. 516. 10x + 5y = −5 3x − 2y = −12 517 = 43 70 y = − 2 3 1336 Chapter 11 Systems of Equations and Inequalities 518. 5x + 6y = 14 4x + 8y = 8 For the following exercises, use addition to solve the system of equations. 519. 3x + 2y = −7 2x + 4y = 6 520. 3x + 4y = 2 9x + 12y = 3 521. 8x + 4y = 2 6x − 5y = 0.7 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 522. has A factory C(x) = 150x + 15,000 and revenue R(x) = 200x. What is the break-even point? cost of a a production function 523. A performer charges C(x) = 50x + 10,000, where x is the total number of attendees at a show. The venue charges $75 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? Systems of Linear Equations: Three Variables 529. 530. 531. 3x + 4z = −11 x − 2y = 5 4y − z = −10 2x − 3y + z = 0 2x + 4y − 3z = 0 6x − 2y − z = 0 6x − 4y − 2z = 2 3x + 2y − 5z = 4 6y − 7z = 5 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 532. Three odd numbers sum up to 61. The smaller is onethird the larger and the middle number is 16 less than the larger. What are the three numbers? 533. A local theatre sells out for their show. They sell all 500 tickets for a total purse of $8,070.00. The tickets were priced at $15 for students, $12 for children, and $18 for adults. If the band sold three times as many adult tickets as children’s tickets, how many of each type was sold? Systems of Nonlinear Equations and Inequalities: Two Variables For the following exercises, solve the system of nonlinear equations. For the following exercises, solve the system of three equations using substitution or addition. 534. y = x2 − 7 y = 5x − 13 524. 525. 526. 527. 528. 0.5x − 0.5y = 10 − 0.2y + 0.2x = 4 0.1x + 0.1z = 2 5x + 3y − z = 5 3x − 2y + 4z = 13 4x + 3y + 5z = 22 x + y + z = 1 2x + 2y + 2z = 1 3x + 3y = 2 2x − 3y + z = −1 x + y + z = −4 4x + 2y − 3z = 33 3x + 2y − z = −10 x − y + 2z = 7 −x + 3y + z = −2 This content is available for free at https://cnx.org/content/col11758/1.5 535. y = x2 − 4 y = 5x + 10 536. x2 + y2 = 16 y = x − 8 537. x2 + y2 = 25 y = x2 + 5 538. x2 + y2 = 4 y − x2 = 3 For the following exercises, graph the inequality. 539. y > x2 − 1 Chapter 11 Systems of Equations and Inequalities 1337 540. 1 4 x2 + y2 < 4 the following exercises, graph the system of For inequalities. 541. x2 + y2 + 2x < 3 y > − x2 − 3 542. x2 − 2x + y2 − 4x < 4 y < − x + 4 543. x2 + y2 < 1 y2 < x Partial Fractions the following exercises, decompose into partial For fractions. 544. −2x + 6 x2 + 3x + 2 545. 10x + 2 4x2 + 4x + 1 546. 7x + 20 x2 + 10x + 25 547. x − 18 x2 − 12x + 36 548. −x2 + 36x + 70 x3 − 125 549. −5x2 + 6x − 2 x3 + 27 553. 10D − 6E 554. B + C 555. AB 556. BA 557. BC 558. CB 559. DE 560. ED 561. EC 562. CE 563. A3 Solving Systems with Gaussian Elimination For the following exercises, write the system of linear equations from the augmented matrix. Indicate whether there will be a unique solution. 564. 565. ⎡ 1 0 −2 ⎣ 0 0 0 ⎤ 7 ⎥ −5 ⎦ 0 ⎤ −9 ⎥ 4 ⎦ 3 \| \| For the following exercises, write the augmented matrix from the system of linear equations. 550. x3 − 4x2 + 3x + 11 (x2 − 2)2 551. 4x4 − 2x3 + 22x2 − 6x + 48 x(x2 + 4)2 Matrices and Matrix Operations the following exercises, perform the requested For operations on the given matrices. A = ⎡ 4 −2 ⎣ 3 1 ⎤ ⎦, B = ⎡ 6 ⎣ 11 −2 ⎤ 7 −3 ⎦, C = 4 ⎡ ⎤ 7 6 ⎢ ⎥, D = 11 −2 ⎣ ⎦ 14 0 ⎡ 1 −4 ⎢ 10 ⎣ 2 9 5 −7 5 8 ⎤ ⎥, E = ⎦ ⎡ ⎤ 7 −14 3 ⎢ ⎥ 2 −1 3 ⎣ ⎦ 1 9 0 552. −4A 566. 567. 568. −2x + 2y + z = 7 2x − 8y + 5z = 0 19x − 10y + 22z = 3 4x + 2y − 3z = 14 −12x + 3y + z = 100 9x − 6y + 2z = 31 x + 3z = 12 −x + 4y = 0 y + 2z = − 7 1338 Chapter 11 Systems of Equations and Inequalities For the following exercises, solve the system of linear equations using Gaussian elimination. 569. 3x − 4y = − 7 −6x + 8y = 14 570. 3x − 4y = 1 −6x + 8y = 6 571. −1.1x − 2.3y = 6.2 −5.2x − 4.1y = 4.3 572. 573. 2x + 3y + 2z = 1 −4x − 6y − 4z = − 2 10x + 15y + 10z = 0 −x + 2y − 4z = 8 3y + 8z = − 4 −7x + y + 2z = 1 Solving Systems with Inverses For the following exercises, find the inverse of the matrix. 574. ⎡ −0.2 ⎣ 1.4 1.2 −0.4 ⎤ ⎦ 575. 576. 577 ⎡ 12 9 −6 ⎢ 2 −1 3 ⎣ −4 − ⎤ ⎥ ⎦ For the following exercises, computing the inverse of the matrix. find the solutions by 578. 0.3x − 0.1y = − 10 −0.1x + 0.3y = 14 579. 0.4x − 0.2y = − 0.6 −0.1x + 0.05y = 0.3 580. 4x + 3y − 3z = − 4.3 5x − 4y − z = − 6.1 x + z = − 0.7 This content is available for free at https://cnx.org/content/col11758/1.5 581. −2x − 3y + 2z = 3 −x + 2y + 4z = − 5 −2y + 5z = − 3 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 582. Students were asked to bring their favorite fruit to class. 90% of the fruits consisted of banana, apple, and oranges. If oranges were half as popular as bananas and apples were 5% more popular than bananas, what are the percentages of each individual fruit? 583. A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $2 and the chocolate chip cookies at $1. They raised $250 and sold 175 items. How many brownies and how many cookies were sold? Solving Systems with Cramer's Rule For the following exercises, find the determinant. 584. 585. 586. 587. 0 \|100 0 0\| \|0.2 −0.6 0.7 −1.1\| \|−1 4 3 0 0 −3\| \| 2 0 2 For the following exercises, use Cramer’s Rule to solve the linear systems of equations. 588. 4x − 2y = 23 −5x − 10y = − 35 589. 0.2x − 0.1y = 0 −0.3x + 0.3y = 2.5 590. −0.5x + 0.1y = 0.3 −0.25x + 0.05y = 0.15 591. x + 6y + 3z = 4 2x + y + 2z = 3 3x − 2y + z = 0 Chapter 11 Systems of Equations and Inequalities 1339 592. 4x − 3y + 5z = − 5 2 7x − 9y − 3z = 3 2 x − 5y − 5z = 5 2 3 10 1 10 593. x − 1 5 x − 1 10 x − 1 2 y − 3 10 50 z = − 9 50 z = − 1 5 2 5 CHAPTER 11 PRACTICE TEST Is the following ordered pair a solution to the system of equations? 602. y2 + x2 = 25 y2 − 2x2 = 1 594. −5x − y = 12 x + 4y = 9 with ( − 3, 3) For the following exercises, solve the systems of linear and nonlinear equations using substitution or elimination. Indicate if no solution exists. 595. 596 − 4y = 4 2x + 16y = 2 597. 5x − y = 1 −10x + 2y = − 2 4x − 6y − 2z = 1 10 x − 7y + 5z = − 1 4 3x + 6y − 9z = 6 5 x + z = 20 x + y + z = 20 x + 2y + z = 10 5x − 4y − 3z = 0 2x + y + 2z = 0 x − 6y − 7z = 0 598. 599. 600. 601. y = x2 + 2x − 3 y = x − 1 the For inequalities. following exercises, graph the following 603. y < x2 + 9 604. x2 + y2 > 4 y < x2 + 1 For the following exercises, write the partial fraction decomposition. 605. −8x − 30 x2 + 10x + 25 606. 13x + 2 (3x + 1)2 607. x4 − x3 + 2x − 1 x(x2 + 1)2 For the following exercises, perform the given matrix operations. 6082 3 ⎡ ⎤ −6 12 ⎣ ⎦ 4 −8 609. ⎡ 1 4 −7 ⎢ 5 −2 9 ⎣ 12 0 −4 ⎤ ⎥ ⎦ ⎡ ⎤ 3 −4 ⎢ ⎥ 1 3 ⎣ ⎦ 5 10 610. − 1340 Chapter 11 Systems of Equations and Inequalities 611. det\| 0 400 4,000\| 0 612. det\| 613. If det(A) = −6, what would be the determinant if you switched rows 1 and 3, multiplied the second row by 12, and took the inverse? 621. 0.1x + 0.1y − 0.1z = − 1.2 0.1x − 0.2y + 0.4z = − 1.2 0.5x − 0.3y + 0.8z = − 5.9 For the following exercises, solve using a system of linear equations. 622. A factory producing cell phones has the following cost and revenue functions: C(x) = x2 + 75x + 2,688 and R(x) = x2 + 160x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. 614. Rewrite the system of linear equations as an augmented matrix. 14x − 2y + 13z = 140 −2x + 3y − 6z = − 1 x − 5y + 12z = 11 623. A small fair charges $1.50 for students, $1 for children, and $2 for adults. In one day, three times as many children as adults attended. A total of 800 tickets were sold for a total revenue of $1,050. How many of each type of ticket was sold? 615. Rewrite the augmented matrix as a system of linear equations. ⎤ ⎥ −5 ⎦ 8 ⎡ 1 0 3 ⎢ −2 4 9 ⎣ −6 1 2\| 12 For the following exercises, use Gaussian elimination to solve the systems of equations. x − 6y = 4 2x − 12y = 0 616. 617. 2x + y + z = − 3 x − 2y + 3z = 6 x − y − z = 6 For the following exercises, use the inverse of a matrix t

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