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 x1 − 4x3 + x4 − x6 = 6 x2 + 2x3 = 1 x4 + 3x5 − x6 = 8 x5 + 9x6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. Example 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent.    1. 3x − y + z = 3 2x − 4y + 3z = 16 . 2x + 3y − z = 1 10x − z = 2 4x − 9y + 2z = 5    3. 3x1 + x2 + x4 = 6 2x1 + x2 − x3 = 4 x2 − 3x3 − 2x4 = 0 Solution. 1. For deﬁnitiveness, we label the topmost equation in the system E1, the equation beneath that E2, and so forth. We now attempt to put the system in triangular form using an algorithm known as Gaussian Elimination. What this means is that, starting with x, we transform the system so that conditions 2 and 3 in Deﬁnition 8.3 are satisﬁed. Then we move on to the next variable, in this case y, and repeat. Since the variables in all of the equations have a consistent ordering from left to right, our ﬁrst move is to get an x in E1’s spot with a coeﬃcient of 1. While there are many ways to do this, the easiest is to apply the ﬁrst move listed in Theorem 8.1 and interchange E1 and E3.    3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→    x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) To satisfy Deﬁnition 8.3, we need to eliminate the x’s from E2 and E3. We accomplish this by replacing each of them with a sum of themselves and a multiple of E1. To eliminate the x from E2, we need to multiply E1 by −2 then add; to eliminate the x from E3, we need to multiply E1 by −3 then add. Applying the third move listed in Theorem 8.1 twice, we get    x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) Replace E2 with −2E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −3E1 + E3    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 12If letters are used instead of subscripted variables, Deﬁnition 8.3 can be suitably modiﬁed using alphabetical order of the variables instead of numerical order on the subscripts of the variables. 558 Systems of Equations and Matrices Now we enforce the conditions stated in Deﬁnition 8.3 for the variable y. To that end we need to get the coeﬃcient of y in E2 equal to 1. We apply the second move listed in Theorem 8.1 and replace E2 with itself times − 1 2 .    (E1) x − y + z = (E2) −2y + z = (E3) 5 6 2y − 2z = −12 Replace E2 with − 1 2 E2 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 To eliminate the y in E3, we add −2E2 to it.    (E1) x − y + z = (E2) (E3) 5 y − 1 2 z = −3 2y − 2z = −12 Replace E3 with −2E2 + E3 −−−−−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Finally, we apply the second move from Theorem 8.1 one last time and multiply E3 by −1 to satisfy the conditions of Deﬁnition 8.3 for the variable z.    (E1) x − y + z = (E2) (E3) 5 2 z = −3 −z = −6 y − 1 Replace E3 with −1E3 −−−−−−−−−−−−−−→    (E1) x − y + z = (E2) (E3) 5 2 z = −3 6 z = y − 1 Now we proceed to substitute. Plugging in z = 6 into E2 gives y − 3 = −3 so that y = 0. With y = 0 and z = 6, E1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6). We leave it to the reader to check that substituting the respective values for x, y, and z into the original system results in three identities. Since we have found a solution, the system is consistent; since there are no free variables, it is independent. 2. Proceeding as we did in 1, our ﬁrst step is to get an equation with x in the E1 position with 1 as its coeﬃcient. Since there is no easy ﬁx, we multiply E1 by 1 2 .    2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E1 with 1 2 E1 −−−−−−−−−−−−−→    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Now it’s time to take care of the x’s in E2 and E3E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3    2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 3 2 z = 8.1 Systems of Linear Equations: Gaussian Elimination 559 Our next step is to get the coeﬃcient of y in E2 equal to 1. To that end, we have    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) Replace E2 with − 1 15 E2 −−−−−−−−−−−−−−−→    Finally, we rid E3 of y. (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3)    (E1) x + 3 (E2 15 z = 1 5 −15y + 4z = 3 2 (E3) Replace E3 with 15E2 + E3 −−−−−−−−−−−−−−−−−→    (E1) x − y + z = (E2) y − 1 5 2 z = −3 6 0 = (E3) The last equation, 0 = 6, is a contradiction so the system has no solution. According to Theorem 8.1, since this system has no solutions, neither does the original, thus we have an inconsistent system. 3. For our last system, we begin by multiplying E1 by 1 3 to get a coeﬃcient of 1 on x1.    3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Next we eliminate x1 from E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E2 −−−−−−−−−−→ with −2E1 + E2    3 x2 + 1 (E1) x1 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 We switch E2 and E3 to get a coeﬃcient of 1 for x2.    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 3 x2 − x3 − 2 1 3 x4 = 0 (E2) (E3) x2 − 3x3 − 2x4 = 0 Switch E2 and E3 −−−−−−−−−−−→    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Finally, we eliminate x2 in E3. 560 Systems of Equations and Matrices    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 3 x2 − x3 − 2 3 x4 = 0 (E3) 1 Replace E3 −−−−−−−−−−→ with − 1 3 E2 + E3    (E1) x1 + 1 3 x2 + 1 3 x4 = 2 (E2) x2 − 3x3 − 2x4 = 0 0 = 0 (E3) Equation E3 reduces to 0 = 0,which is always true. Since we have no equations with x3 or x4 as leading variables, they are both free, which means we have a consistent dependent system. We parametrize the solution set by letting x3 = s and x4 = t and obtain from E2 that x2 = 3s + 2t. Substituting this and x4 = t into E1, we have x1 + 1 3 t = 2 which gives x1 = 2 − s − t. Our solution is the set {(2 − s − t, 2s + 3t, s, t) \| − ∞ < s, t < ∞}.13 We leave it to the reader to verify that the substitutions x1 = 2 − s − t, x2 = 3s + 2t, x3 = s and x4 = t satisfy the equations in the original system. 3 (3s + 2t) + 1 Like all algorithms, Gaussian Elimination has the advantage of always producing what we need, but it can also be ineﬃcient at times. For example, when solving 2 above, it is clear after we eliminated the x’s in the second step to get the system    (E1) x + 3 (E215y + 4z = −3 −15y + 4z = 3 (E3) that equations E2 and E3 when taken together form a contradiction since we have identical left hand sides and diﬀerent right hand sides. The algorithm takes two more steps to reach this contradiction. We also note that substitution in Gaussian Elimination is delayed until all the elimination is done, thus it gets called back-substitution. This may also be ineﬃcient in many cases. Rest assured, the technique of substitution as you may have learned it in Intermediate Algebra will once again take center stage in Section 8.7. Lastly, we note that the system in 3 above is underdetermined, and as it is consistent, we have free variables in our answer. We close this section with a standard ‘mixture’ type application of systems of linear equations. Example 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has stock solutions of 30% and 90% acid as well as all of the distilled water he wants. Set-up and solve a system of linear equations which determines all of the possible combinations of the stock solutions and water which would produce the required solution. Solution. We are after three unknowns, the amount (in mL) of the 30% stock solution (which we’ll call x), the amount (in mL) of the 90% stock solution (which we’ll call y) and the amount (in mL) of water (which we’ll call w). We now need to determine some relationships between these variables. Our goal is to produce 500 milliliters of a 40% acid solution. This product has two deﬁning characteristics. First, it must be 500 mL; second, it must be 40% acid. We take each 13Here, any choice of s and t will determine a solution which is a point in 4-dimensional space. Yeah, we have trouble visualizing that, too. 8.1 Systems of Linear Equations: Gaussian Elimination 561 of these qualities in turn. First, the total volume of 500 mL must be the sum of the contributed volumes of the two stock solutions and the water. That is amount of 30% stock solution + amount of 90% stock solution + amount of water = 500 mL Using our deﬁned variables, this reduces to x + y + w = 500. Next, we need to make sure the ﬁnal solution is 40% acid. Since water contains no acid, the acid will come from the stock solutions only. We ﬁnd 40% of 500 mL to be 200 mL which means the ﬁnal solution must contain 200 mL of acid. We have amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution is 0.90y. We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes We ﬁrst eliminate the x from the second equation x + y + w = 500 10 y = 200 10 x + 9 3 (E1) x + y + w = 500 10 x + 9 10 y = 200 (E2) 3 Replace E2 with − 3 10 E1 + E2 −−−−−−−−−−−−−−−−−−→ (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Next, we get a coeﬃcient of 1 on the leading variable in E2 (E1) x + y + w = 500 5 y − 3 10 w = 50 (E2) 3 Replace E2 with 5 3 E2 −−−−−−−−−−−−−→ (E1) x + y + w = 500 y − 1 2 w = 250 (E2) 3 3 3 , 1 2 t + 250 2 t + 250 2 t + 250 3 . Substituting into E1 gives x + 1 + t = 500 so that x = − 3 2 t + 1250 Notice that we have no equation to determine w, and as such, w is free. We set w = t and from
E2 get y = 1 2 t + 1250 3 . This system is consistent, dependent and its solution set is {− 3 3 , t \| − ∞ < t < ∞}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The 3 ≥ 0 or t ≥ − 500 constraint x ≥ 0 gives us − 3 3 ≥ 0, or t ≤ 2500 3 . The condition z ≥ 0 yields t ≥ 0, and we see that when we take the set theoretic intersection of 3 , t \| 0 ≤ t ≤ 2500 9 . Our ﬁnal answer is {− 3 these intervals, we get 0 ≤ t ≤ 2500 9 }. Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining and it is due to expire. Can we use all of it to make our required solution? We would have y = 100 so that 1 3 . This means the amount of 30% solution required is x = − 3 3 mL. The reader is invited to check that mixing these three amounts of our constituent solutions produces the required 40% acid mix. 3 = 100, and we get t = 100 + 1250 3 = − 3 3 mL, and for the water, w = t = 100 9 . From y ≥ 0, we get 1 2 t + 250 2 t + 1250 3 = 1100 2 t + 1250 2 t + 1250 2 t + 250 2 t + 250 3 , 1 100 3 2 14We do this only because we believe students can use all of the practice with fractions they can get! 562 Systems of Equations and Matrices 8.1.1 Exercises (Review Exercises) In Exercises 1 - 8, take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically. 1. 3. 5. 7. x + 2y = 5 x = 6 x+2y 4 3x−y 2 = −5 = 1 1 2 x − 1 2y − 3x = 3 y = −1 6 2 x = − 15 3y − . 4. 6. 8. 2y − 3x = 1 y = − + 4y = 6 3 y = 1 12 − 20 3 y = − 7 3 3 y = 10 − 10 In Exercises 9 - 26, put each system of linear equations into triangular form and solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent. −5x + y = 17 x + y = 5 9. 11. 13. 15. 17. 19.    4x − y + z = 5 2y + 6z = 30 17 y − 3z = 0          3x − 2y + z = −5 x + 3y − z = 12 0 x + y + 2z = x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9                   10. 12. 14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 4x − y + z = 5 2y + 6z = 30 x + z = 6 x − 2y + 3z = 7 −3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 8.1 Systems of Linear Equations: Gaussian Elimination 563 21. 23. 25.          x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1 4x − 4y + 4z = −13 6x − 5y + 7z = −25 x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 22. 24. 26.          x − y + z = 8 3x + 3y − 9z = −6 7x − 2y + 5z = 39 2x1 + x2 − 12x3 − x4 = 16 −x1 + x2 + 12x3 − 4x4 = −5 3x1 + 2x2 − 16x3 − 3x4 = 25 x1 + 2x2 − 5x4 = 11 x1 − x2 − 5x3 + 3x4 = −1 x1 + x2 + 5x3 − 3x4 = 0 x2 + 5x3 − 3x4 = 1 x1 − 2x2 − 10x3 + 6x4 = −1 27. Find two other forms of the parametric solution to Exercise 11 above by reorganizing the equations so that x or y can be the free variable. 28. A local buﬀet charges $7.50 per person for the basic buﬀet and $9.25 for the deluxe buﬀet (which includes crab legs.) If 27 diners went out to eat and the total bill was $227.00 before taxes, how many chose the basic buﬀet and how many chose the deluxe buﬀet? 29. At The Old Home Fill’er Up and Keep on a-Truckin’ Cafe, Mavis mixes two diﬀerent types of coﬀee beans to produce a house blend. The ﬁrst type costs $3 per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound? 30. Skippy has a total of $10,000 to split between two investments. One account oﬀers 3% simple interest, and the other account oﬀers 8% simple interest. For tax reasons, he can only earn $500 in interest the entire year. How much money should Skippy invest in each account to earn $500 in interest for the year? 31. A 10% salt solution is to be mixed with pure water to produce 75 gallons of a 3% salt solution. How much of each are needed? 32. At The Crispy Critter’s Head Shop and Patchouli Emporium along with their dried up weeds, sunﬂower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile? 33. Discuss with your classmates how you would approach Exercise 32 above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister. 34. If you were to try to make 100 mL of a 60% acid solution using stock solutions at 20% and 40%, respectively, what would the triangular form of the resulting system look like? Explain. 564 Systems of Equations and Matrices 8.1.2 Answers 1. Consistent independent Solution 6, − 1 2 3. Consistent independent Solution − 16 7 , − 62 7 2. Consistent independent Solution − 7 3 , −3 4. Consistent independent Solution 49 12 , − 25 18 5. Consistent dependent Solution t, 3 for all real numbers t 2 t + 3 7. Inconsistent No solution 6. Consistent dependent Solution (6 − 4t, t) for all real numbers t 8. Inconsistent No solution Because triangular form is not unique, we give only one possible answer to that part of the question. Yours may be diﬀerent and still be correct.          10. 11. 12 + 3z = 15 + 3z = 15 0 = 1 13. x + y + z = −17 y − 3z = 0    14. x − 2y + 3z = y − 11 7 5 z = − 16 5 1 z = Consistent independent Solution (−2, 7) Consistent independent Solution (1, 2, 0) Consistent dependent Solution (−t + 5, −3t + 15, t) for all real numbers t Inconsistent No solution Consistent dependent Solution (−4t − 17, 3t, t) for all real numbers t Consistent independent Solution (2, −1, 1) 8.1 Systems of Linear Equations: Gaussian Elimination 565 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.                                  0 x + y + 2z = 4 y − 7z = 17 z = −2 x − 2y + 2z = − − 3y − 4z = 3 13 z = 4 13 0 = 0 y + 11 − 2z = − = − 11 2 0 0 = Consistent independent Solution (1, 3, −2) Inconsistent no solution Consistent independent Solution (1, 3, −2) Consistent independent Solution −3, 1 2 , 1 Consistent independent Solution 1 3 , 2 3 , 1 Consistent dependent Solution 19 13 , − 11 for all real numbers t 13 t + 51 13 t + 4 13 , t Inconsistent no solution Consistent independent Solution (4, −3, 1) Consistent dependent Solution −2t − 35 for all real numbers t 4 , −t − 11 2 , t x1 + 2 3 x3 − x4 = 25 3 x2 − 16 3 x2 + 4x3 − 3x4 = 2 0 = 0 0 = 0 x1 − x3 = −2 x2 − 1 2 x4 = 0 x3 − 1 2 x4 = x4 = 1 4 Consistent dependent Solution (8s − t + 7, −4s + 3t + 2, s, t) for all real numbers s and t Consistent independent Solution (1, 2, 3, 4) 566 Systems of Equations and Matrices 26.    x1 − x2 − 5x3 + 3x4 = −1 1 x2 + 5x3 − 3x4 = 2 1 0 = 0 0 = Inconsistent No solution 27. If x is the free variable then the solution is (t, 3t, −t + 5) and if y is the free variable then the solution is 1 3 t, t, − 1 3 t + 5. 28. 13 chose the basic buﬀet and 14 chose the deluxe buﬀet. 29. Mavis needs 20 pounds of $3 per pound coﬀee and 30 pounds of $8 per pound coﬀee. 30. Skippy needs to invest $6000 in the 3% account and $4000 in the 8% account. 31. 22.5 gallons of the 10% solution and 52.5 gallons of pure water. 32. 4 3 − 1 2 t pounds of Type I, 2 3 − 1 2 t pounds of Type II and t pounds of Type III where 0 ≤ t ≤ 4 3 . 8.2 Systems of Linear Equations: Augmented Matrices 567 8.2 Systems of Linear Equations: Augmented Matrices In Section 8.1 we introduced Gaussian Elimination as a means of transforming a system of linear equations into triangular form with the ultimate goal of producing an equivalent system of linear equations which is easier to solve. If we take a step back and study the process, we see that all of our moves are determined entirely by the coeﬃcients of the variables involved, and not the variables themselves. Much the same thing happened when we studied long division in Section 3.2. Just as we developed synthetic division to streamline that process, in this section, we introduce a similar bookkeeping device to help us solve systems of linear equations. To that end, we deﬁne a matrix as a rectangular array of real numbers. We typically enclose matrices with square brackets, ‘[ ’ and ‘ ]’, and we size matrices by the number of rows and columns they have. For example, the size (sometimes called the dimension) of 3 0 −1 10 2 −5 is 2 × 3 because it has 2 rows and 3 columns. The individual numbers in a matrix are called its entries and are usually labeled with double subscripts: the ﬁrst tells which row the element is in and the second tells which column it is in. The rows are numbered from top to bottom and the columns are numbered from left to right. Matrices themselves are usually denoted by uppercase letters (A, B, C, etc.) while their entries are usually denoted by the corresponding letter. So, for instance, if we have A = 3 0 −1 10 2 −5 then a11 = 3, a12 = 0, a13 = −1, a21 = 2, a22 = −5, and a23 = 10. We shall explore matrices as mathematical objects with their own algebra in Section 8.3 and introduce them here solely as a bookkeeping device. Consider the system of linear equations from number 2 in Example 8.1.2    2x + 3y − z = 1 (E1) 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 We encode this system into a matrix by assigning each equation to a corresponding row. Within that row, each variable and the constant gets its own column, and to separate the variables on the left hand side of the equation from the constants on the right hand side, we use a vertical bar, \|. Note tha
t in E2, since y is not present, we record its coeﬃcient as 0. The matrix associated with this system is (E1) → (E2) → (E3) →   x 2 10 4 −9 y c z 3 −1 1 0 −1 2 2 5   568 Systems of Equations and Matrices This matrix is called an augmented matrix because the column containing the constants is appended to the matrix containing the coeﬃcients.1 To solve this system, we can use the same kind operations on the rows of the matrix that we performed on the equations of the system. More speciﬁcally, we have the following analog of Theorem 8.1 below. Theorem 8.2. Row Operations: Given an augmented matrix for a system of linear equations, the following row operations produce an augmented matrix which corresponds to an equivalent system of linear equations. Interchange any two rows. Replace a row with a nonzero multiple of itself.a Replace a row with itself plus a nonzero multiple of another row.b aThat is, the row obtained by multiplying each entry in the row by the same nonzero number. bWhere we add entries in corresponding columns. As a demonstration of the moves in Theorem 8.2, we revisit some of the steps that were used in solving the systems of linear equations in Example 8.1.2 of Section 8.1. The reader is encouraged to perform the indicated operations on the rows of the augmented matrix to see that the machinations are identical to what is done to the coeﬃcients of the variables in the equations. We ﬁrst see a demonstration of switching two rows using the ﬁrst step of part 1 in Example 8.1.2.    3x − y + z = 3 (E1) (E2) 2x − 4y + 3z = 16 x − y + z = 5 (E3) Switch E1 and E3 −−−−−−−−−−−→   3 −1 2 −4 1 −1   1 3 3 16 5 1 Switch R1 and R3 −−−−−−−−−−−→      x − y + z = 5 (E1) (E2) 2x − 4y + 3z = 16 3x − y + z = 3 (E3) 1 −1 2 −4 3 −1   5 1 3 16 3 1 Next, we have a demonstration of replacing a row with a nonzero multiple of itself using the ﬁrst step of part 3 in Example 8.1.2.    3x1 + x2 + x4 = 6 (E1) 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 0 Replace E1 with 1 3 E1 −−−−−−−−−−−−−→   3 2 0 1 6 1 0 1 −1 0 4 1 −3 −2 0   Replace R1 with 1 3 R1 −−−−−−−−−−−−−→      3 x2 + 1 (E1) x1 + 1 3 x4 = 2 2x1 + x2 − x3 = 4 (E2) (E3) x2 − 3x3 − 2x4 = 1 0 4 1 −3 −2 0   Finally, we have an example of replacing a row with itself plus a multiple of another row using the second step from part 2 in Example 8.1.2. 1We shall study the coeﬃcient and constant matrices separately in Section 8.3. 8.2 Systems of Linear Equations: Augmented Matrices 569    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3   3 1 10 4 −9 2 − 1 2 0 −1 2   1 2 2 5 Replace R2 with −10R1 + R2 −−−−−−−−−−−−−−−−−−→ Replace R3 with −4R1 + R3      2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y + 4z = −3 −15y + 4z = 15 0 −15   1 2 2 4 −3 3 4 The matrix equivalent of ‘triangular form’ is row echelon form. The reader is encouraged to refer to Deﬁnition 8.3 for comparison. Note that the analog of ‘leading variable’ of an equation is ‘leading entry’ of a row. Speciﬁcally, the ﬁrst nonzero entry (if it exists) in a row is called the leading entry of that row. Deﬁnition 8.4. A matrix is said to be in row echelon form provided all of the following conditions hold: 1. The ﬁrst nonzero entry in each row is 1. 2. The leading 1 of a given row must be to the right of the leading 1 of the row above it. 3. Any row of all zeros cannot be placed above a row with nonzero entries. To solve a system of a linear equations using an augmented matrix, we encode the system into an augmented matrix and apply Gaussian Elimination to the rows to get the matrix into row-echelon form. We then decode the matrix and back substitute. The next example illustrates this nicely. Example 8.2.1. Use an augmented matrix to transform the following system of linear equations into triangular form. Solve the system.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Solution. We ﬁrst encode the system into an augmented matrix.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Encode into the matrix −−−−−−−−−−−−−−→   3 −1 1 2 1 8 2 −1 4 3 −4 10   Thinking back to Gaussian Elimination at an equations level, our ﬁrst order of business is to get x in E1 with a coeﬃcient of 1. At the matrix level, this means getting a leading 1 in R1. This is in accordance with the ﬁrst criteria in Deﬁnition 8.4. To that end, we interchange R1 and R2.   3 −1 1 2 8 1 2 −1 4 3 −4 10   Switch R1 and R2 −−−−−−−−−−−→   1 3 −1 2 2 −1 4 8 1 3 −4 10   570 Systems of Equations and Matrices Our next step is to eliminate the x’s from E2 and E3. From a matrix standpoint, this means we need 0’s below the leading 1 in R1. This guarantees the leading 1 in R2 will be to the right of the leading 1 in R1 in accordance with the second requirement of Deﬁnition 8.4.   1 3 −1 2 2 −1 4 1 8 3 −4 10   Replace R2 with −3R1 + R2 −−−−−−−−−−−−−−−−−→ Replace R3 with −2R1 + R3   2 −1 1 0 −7 0 −1 −2 4 4 −4 2   Now we repeat the above process for the variable y which means we need to get the leading entry in R2 to be 1.   2 −1 1 0 −7 0 −1 −2 4 4 −4 2   Replace R2 with − 1 7 R2 −−−−−−−−−−−−−−→   2 −1 1 1 − 4 0 7 0 −1 −2   4 4 7 2 To guarantee the leading 1 in R3 is to the right of the leading 1 in R2, we get a 0 in the second column of R3.   2 −1 1 1 − 4 0 7 0 −1 −2   4 4 7 2 Replace R3 with R2 + R3 −−−−−−−−−−−−−−−−→    1 0 0 2 −1 1 − 4 7 0 − 18 7    4 4 7 18 7 Finally, we get the leading entry in R3 to be 1.    1 0 0 2 −1 1 − 4 7 0 − 18 7    4 4 7 18 7 Replace R3 with − 7 18 R3 −−−−−−−−−−−−−−−→   1 0 0 2 −1 0   Decoding from the matrix gives a system in triangular form    1 0 0 2 −1 0 We get z = −1 ﬁnal answer of (3, 0, −1). We leave it to the reader to check. Decode from the matrix −−−−−−−−−−−−−−→ 7 (−1) + 4     x + 2y − 1 7 = 0 and x = −2y + z + 4 = −2(0) + (−1) + 4 = 3 for a As part of Gaussian Elimination, we used row operations to obtain 0’s beneath each leading 1 to put the matrix into row echelon form. If we also require that 0’s are the only numbers above a leading 1, we have what is known as the reduced row echelon form of the matrix. Deﬁnition 8.5. A matrix is said to be in reduced row echelon form provided both of the following conditions hold: 1. The matrix is in row echelon form. 2. The leading 1s are the only nonzero entry in their respective columns. 8.2 Systems of Linear Equations: Augmented Matrices 571 Of what signiﬁcance is the reduced row echelon form of a matrix? To illustrate, let’s take the row echelon form from Example 8.2.1 and perform the necessary steps to put into reduced row echelon form. We start by using the leading 1 in R3 to zero out the numbers in the rows above it.   1 0 0 2 −1 0   Replace R1 with R3 + R1 −−−−−−−−−−−−−−−−−→ Replace R2 with 4 7 R3 + R2   1   Finally, we take care of the 2 in R1 above the leading 1 in R21   Replace R1 with −2R2 + R1 −−−−−−−−−−−−−−−−−→   1   To our surprise and delight, when we decode this matrix, we obtain the solution instantly without having to deal with any back-substitution at all1   Decode from the matrix −−−−−−−−−−−−−−→    3 x = 0 y = z = −1 Note that in the previous discussion, we could have started with R2 and used it to get a zero above its leading 1 and then done the same for the leading 1 in R3. By starting with R3, however, we get more zeros ﬁrst, and the more zeros there are, the faster the remaining calculations will be.2 It is also worth noting that while a matrix has several3 row echelon forms, it has only one reduced row echelon form. The process by which we have put a matrix into reduced row echelon form is called Gauss-Jordan Elimination. Example 8.2.2. Solve the following system using an augmented matrix. Use Gauss-Jordan Elimination to put the augmented matrix into reduced row echelon form.    x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Solution. We ﬁrst encode the system into a matrix. (Pay attention to the subscripts!)    x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x2 − x4 = 3 Encode into the matrix −−−−−−−−−−−−−−→   −1 3   Next, we get a leading 1 in the ﬁrst column of R1.   −1 3   Replace R1 with − 1 3 R1 −−−−−−−−−−−−−−→   1 3   2Carl also ﬁnds starting with R3 to be more symmetric, in a purely poetic way. 3inﬁnite, in fact 572 Systems of Equations and Matrices Now we eliminate the nonzero entry below our leading 11 3   Replace R2 with −2R1 + R2 −−−−−−−−−−−−−−−−−→    1 − 1 3 2 0 3 4 0 We proceed to get a leading 1 in R2. 0 − 1 2 4 3 0 −1 3 − 2    3 19 1 3 − 2    Replace R2 with 3 2 R2 −−−−−−−−−−−−−→    1 3 19 2 3    3 19 3 3 We now zero out the entry below the leading 1 in R21 3 19 2 3    Replace R3 with −4R2 + R3 −−−−−−−−−−−−−−−−−→ Next, it’s time for a leading 1 in R3.   1 − 1 3 3 19 1 0 2 0 −24 −5 −35 0 0 − 1 3 − 2 1 6   Replace R3 with − 1 24 R3 −−−−−−−−−−−−−−−→      1 − 1 3 3 19 1 0 2 0 −24 −5 −35 24 3 19 2 35 24    The matrix is now in row echelon form. To get the reduced row echelon form, we start with the last leading 1 we produced and work to get 0’s above it 24 3 19 2 35 24    Replace R2 with −6R3 + R2 −−−−−−−−−−−−−−−−−→ Lastly, we get a 0 above the leading 1 of R2 24 3 − 2 3 3 4 35 24    Replace R1 with 1 3 R2 + R1 −−−−−−−−−−−−−−−−−→    At last, we decode to get 24 12 − 24       3 3 4 35 24 12 3 4 35 24    1 0 0 0 1 0 12 − 24    Decode from the matrix −−−−−−−−−−−−−−→    x1 − 5 x2 − 1 x3 + 5 12 x4 = − 5 4 x4 = 24 x4 = 12 3 4 35 24 12 3 4 35 24 We have that x4 is free and we assign it the parameter t. We obtain x3 = − 5 and x1 = 5 the reader to check. 12 . Our solution is 5 4 t + 3 24 t + 35 4 , 24 , t : −∞ < t < ∞ and leave it to 24 , x2 = 1 24 t + 35 12 t − 5 12 12 , 1 8.2 Systems of Linear Equations: Augmented Matrices 573 Like all good algorithms, putting a matrix in row echelon or reduced row echelon form can easily be programmed into a calculator, and, doubtless, your graphing calculator has such a feature. We use this
in our next example. Example 8.2.3. Find the quadratic function passing through the points (−1, 3), (2, 4), (5, −2). Solution. According to Deﬁnition 2.5, a quadratic function has the form f (x) = ax2 +bx+c where a = 0. Our goal is to ﬁnd a, b and c so that the three given points are on the graph of f . If (−1, 3) is on the graph of f , then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to a − b + c = 3, an honest-to-goodness linear equation with the variables a, b and c. Since the point (2, 4) is also on the graph of f , then f (2) = 4 which gives us the equation 4a + 2b + c = 4. Lastly, the point (5, −2) is on the graph of f gives us 25a + 5b + c = −2. Putting these together, we obtain a system of three linear equations. Encoding this into an augmented matrix produces    a − b + c = 3 4 4a + 2b + c = 25a + 5b + c = −2 Encode into the matrix −−−−−−−−−−−−−−→   1 −1 2 4 5 25   1 3 1 4 1 −2 18 , b = 13 Using a calculator,4 we ﬁnd a = − 7 9 . Hence, the one and only quadratic which ﬁts the bill is f (x) = − 7 18 x + 37 9 . To verify this analytically, we see that f (−1) = 3, f (2) = 4, and f (5) = −2. We can use the calculator to check our solution as well by plotting the three data points and the function f . 18 and c = 37 18 x2 + 13 The graph of f (x) = − 7 18 x + 37 with the points (−1, 3), (2, 4) and (5, −2) 18 x2 + 13 9 4We’ve tortured you enough already with fractions in this exposition! 574 Systems of Equations and Matrices 8.2.1 Exercises In Exercises 1 - 6, state whether the given matrix is in reduced row echelon form, row echelon form only or in neither of those forms. 1 0 3 0 1 3 1.   4. 5. 3 −1 2 −4 1 −1   1 3 3 16 . 6   In Exercises 7 - 12, the following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that the system is inconsistent. 1 0 −2 7 0 1 7.   10. 11. 1 0 0 −3 20 0 1 0 0 0 1 19   1 0 0 0 0 −8 1 0 0 1 7 4 −. 124 0 0 9 −3 20 0     In Exercises 13 - 26, solve the following systems of linear equations using the techniques discussed in this section. Compare and contrast these techniques with those you used to solve the systems in the Exercises in Section 8.1. 13. −5x + y = 17 x + y = 5          15. 17. 19. 4x − y + z = 5 2y + 6z = 30 x + z = 5 3x − 2y + z = −5 x + 3y − z = 12 0 x + y + 2z = x − y + z = −4 −3x + 2y + 4z = −5 x − 5y + 2z = −18             14. 16. 18. 20. x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 x − 2y + 3z = 7 −3x + y + 2z = −5 3 2x + 2y + z = 2x − y + z = −1 1 4 4x + 3y + 5z = 5y + 3z = 2x − 4y + z = −7 x − 2y + 2z = −2 3 −x + 4y − 2z = 8.2 Systems of Linear Equations: Augmented Matrices 575          21. 23. 25. 2x − y + z = 1 2x + 2y − z = 1 3x + 6y + 4z = 9 x + y + z = 4 2x − 4y − z = −1 x − y = 2 2x − 3y + z = −1 4x − 4y + 4z = −13 6x − 5y + 7z = −25 22. 24. 26.          x − 3y − 4z = 3 3x + 4y − z = 13 2x − 19y − 19z = 2 x − y + z = 8 3x + 3y − 9z = −6 7x − 2y + 5z = 39 x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 27. It’s time for another meal at our local buﬀet. This time, 22 diners (5 of whom were children) feasted for $162.25, before taxes. If the kids buﬀet is $4.50, the basic buﬀet is $7.50, and the deluxe buﬀet (with crab legs) is $9.25, ﬁnd out how many diners chose the deluxe buﬀet. 28. Carl wants to make a party mix consisting of almonds (which cost $7 per pound), cashews (which cost $5 per pound), and peanuts (which cost $2 per pound.) If he wants to make a 10 pound mix with a budget of $35, what are the possible combinations almonds, cashews, and peanuts? (You may ﬁnd it helpful to review Example 8.1.3 in Section 8.1.) 29. Find the quadratic function passing through the points (−2, 1), (1, 4), (3, −2) 30. At 9 PM, the temperature was 60◦F; at midnight, the temperature was 50◦F; and at 6 AM, the temperature was 70◦F . Use the technique in Example 8.2.3 to ﬁt a quadratic function to these data with the temperature, T , measured in degrees Fahrenheit, as the dependent variable, and the number of hours after 9 PM, t, measured in hours, as the independent variable. What was the coldest temperature of the night? When did it occur? 31. The price for admission into the Stitz-Zeager Sasquatch Museum and Research Station is $15 for adults and $8 for kids 13 years old and younger. When the Zahlenreich family visits the museum their bill is $38 and when the Nullsatz family visits their bill is $39. One day both families went together and took an adult babysitter along to watch the kids and the total admission charge was $92. Later that summer, the adults from both families went without the kids and the bill was $45. Is that enough information to determine how many adults and children are in each family? If not, state whether the resulting system is inconsistent or consistent dependent. In the latter case, give at least two plausible solutions. 32. Use the technique in Example 8.2.3 to ﬁnd the line between the points (−3, 4) and (6, 1). How does your answer compare to the slope-intercept form of the line in Equation 2.3? 33. With the help of your classmates, ﬁnd at least two diﬀerent row echelon forms for the matrix 1 2 3 4 12 8 576 Systems of Equations and Matrices 8.2.2 Answers 1. Reduced row echelon form 2. Neither 3. Row echelon form only 4. Reduced row echelon form 5. Reduced row echelon form 6. Row echelon form only 7. (−2, 7) 9. (−3t + 4, −6t − 6, 2, t) for all real numbers t 8. (−3, 20, 19) 10. Inconsistent 11. (8s − t + 7, −4s + 3t + 2, s, t) for all real numbers s and t 12. (−9t − 3, 4t + 20, t) for all real numbers t 13. (−2, 7) 15. (−t + 5, −3t + 15, t) for all real numbers t 17. (1, 3, −2) 19. (1, 3, −2) 14. (1, 2, 0) 16. (2, −1, 1) 18. Inconsistent 20. −3, 1 2 , 1 21. 1 3 , 2 3 , 1 22. 19 13 , t 13 , − 11 13 t + 51 for all real numbers t 13 t + 4 23. Inconsistent 25. −2t − 35 2 , t for all real numbers t 4 , −t − 11 24. (4, −3, 1) 26. (1, 2, 3, 4) 27. This time, 7 diners chose the deluxe buﬀet. 28. If t represents the amount (in pounds) of peanuts, then we need 1.5t − 7.5 pounds of almonds and 17.5 − 2.5t pounds of cashews. Since we can’t have a negative amount of nuts, 5 ≤ t ≤ 7. 29. f (x) = − 4 30. T (t) = 20 5 x2 + 1 27 t2 − 50 5 x + 23 9 t + 60. Lowest temperature of the evening 595 5 12 ≈ 49.58◦F at 12:45 AM. 8.2 Systems of Linear Equations: Augmented Matrices 577 31. Let x1 and x2 be the numbers of adults and children, respectively, in the Zahlenreich family and let x3 and x4 be the numbers of adults and children, respectively, in the Nullsatz family. The system of equations determined by the given information is   15x1 + 8x2 = 38 15x3 + 8x4 = 39 15x1 + 8x2 + 15x3 + 8x4 = 77 15x1 + 15x3 = 45  We subtracted the cost of the babysitter in E3 so the constant is 77, not 92. This system is 5 , t. Our variables repreconsistent dependent and its solution is 8 15 t + 2 sent numbers of adults and children so they must be whole numbers. Running through the values t = 0, 1, 2, 3, 4 yields only one solution where all four variables are whole numbers; t = 3 gives us (2, 1, 1, 3). Thus there are 2 adults and 1 child in the Zahlenreichs and 1 adult and 3 kids in the Nullsatzs. 5 , −t + 4, − 8 15 t + 13 578 Systems of Equations and Matrices 8.3 Matrix Arithmetic In Section 8.2, we used a special class of matrices, the augmented matrices, to assist us in solving systems of linear equations. In this section, we study matrices as mathematical objects of their own accord, temporarily divorced from systems of linear equations. To do so conveniently requires some more notation. When we write A = [aij]m×n, we mean A is an m by n matrix1 and aij is the entry found in the ith row and jth column. Schematically, we have j counts columns from left to right −−−−−−−−−−−−−−−→ a1n a11 a12 a2n a21 a22 ... ... ... · · · amn am1 am2 · · · · · ·      counts rows from top to bottom With this new notation we can deﬁne what it means for two matrices to be equal. Deﬁnition 8.6. Matrix Equality: Two matrices are said to be equal if they are the same size and their corresponding entries are equal. More speciﬁcally, if A = [aij]m×n and B = [bij]p×r, we write A = B provided 1. m = p and n = r 2. aij = bij for all 1 ≤ i ≤ m and all 1 ≤ j ≤ n. Essentially, two matrices are equal if they are the same size and they have the same numbers in the same spots.2 For example, the two 2 × 3 matrices below are, despite appearances, equal. 0 −2 9 25 117 −3 = √ 3 ln(1) e2 ln(3) −8 1252/3 32 · 13 log(0.001) Now that we have an agreed upon understanding of what it means for two matrices to equal each other, we may begin deﬁning arithmetic operations on matrices. Our ﬁrst operation is addition. Deﬁnition 8.7. Matrix Addition: Given two matrices of the same size, the matrix obtained by adding the corresponding entries of the two matrices is called the sum of the two matrices. More speciﬁcally, if A = [aij]m×n and B = [bij]m×n, we deﬁne A + B = [aij]m×n + [bij]m×n = [aij + bij]m×n As an example, consider the sum below. 1Recall that means A has m rows and n columns. 2Critics may well ask: Why not leave it at that? Why the need for all the notation in Deﬁnition 8.6? It is the authors’ attempt to expose you to the wonderful world of mathematical precision. 8.3 Matrix Arithmetic 579   3 2 4 −1 0 −7    +  −1 4 −5 −3 1 8    =  2 + (−1) 4 + (−5) 0 + 8 3 + 4 (−1) + (−3) (−71 −4 8 −6 It is worth the reader’s time to think what would have happened had we reversed the order of the summands above. As we would expect, we arrive at the same answer. In general, A + B = B + A for matrices A and B, provided they are the same size so that the sum is deﬁned in the ﬁrst place. This is the commutative property of matrix addition. To see why this is true in general, we appeal to the deﬁnition of matrix addition. Given A = [aij]m×n and B = [bij]m×n, A + B = [aij]m×n + [bij]m×n = [aij + bij]m×n = [bij + aij]m×n = [bij]m×n + [aij]m×n = B + A where the second equali
ty is the deﬁnition of A + B, the third equality holds by the commutative law of real number addition, and the fourth equality is the deﬁnition of B + A. In other words, matrix addition is commutative because real number addition is. A similar argument shows the associative property of matrix addition also holds, inherited in turn from the associative law of real number addition. Speciﬁcally, for matrices A, B, and C of the same size, (A + B) + C = A + (B + C). In other words, when adding more than two matrices, it doesn’t matter how they are grouped. This means that we can write A + B + C without parentheses and there is no ambiguity as to what this means.3 These properties and more are summarized in the following theorem. Theorem 8.3. Properties of Matrix Addition Commutative Property: For all m × n matrices, A + B = B + A Associative Property: For all m × n matrices, (A + B) + C = A + (B + C) Identity Property: If 0m×n is the m × n matrix whose entries are all 0, then 0m×n is called the m × n additive identity and for all m × n matrices A A + 0m×n = 0m×n + A = A Inverse Property: For every given m × n matrix A, there is a unique matrix denoted −A called the additive inverse of A such that A + (−A) = (−A) + A = 0m×n The identity property is easily veriﬁed by resorting to the deﬁnition of matrix addition; just as the number 0 is the additive identity for real numbers, the matrix comprised of all 0’s does the same job for matrices. To establish the inverse property, given a matrix A = [aij]m×n, we are looking for a matrix B = [bij]m×n so that A + B = 0m×n. By the deﬁnition of matrix addition, we must have that aij + bij = 0 for all i and j. Solving, we get bij = −aij. Hence, given a matrix A, its additive inverse, which we call −A, does exist and is unique and, moreover, is given by the formula: −A = [−aij]m×n. The long and short of this is: to get the additive inverse of a matrix, 3A technical detail which is sadly lost on most readers. 580 Systems of Equations and Matrices take additive inverses of each of its entries. With the concept of additive inverse well in hand, we may now discuss what is meant by subtracting matrices. You may remember from arithmetic that a − b = a + (−b); that is, subtraction is deﬁned as ‘adding the opposite (inverse).’ We extend this concept to matrices. For two matrices A and B of the same size, we deﬁne A − B = A + (−B). At the level of entries, this amounts to A − B = A + (−B) = [aij]m×n + [−bij]m×n = [aij + (−bij)]m×n = [aij − bij]m×n Thus to subtract two matrices of equal size, we subtract their corresponding entries. Surprised? Our next task is to deﬁne what it means to multiply a matrix by a real number. Thinking back to arithmetic, you may recall that multiplication, at least by a natural number, can be thought of as ‘rapid addition.’ For example. We know from algebra4 that 3x = x + x + x, so it seems natural that given a matrix A, we deﬁne 3A = A + A + A. If A = [aij]m×n, we have 3A = A + A + A = [aij]m×n + [aij]m×n + [aij]m×n = [aij + aij + aij]m×n = [3aij]m×n In other words, multiplying the matrix in this fashion by 3 is the same as multiplying each entry by 3. This leads us to the following deﬁnition. Deﬁnition 8.8. Scalara Multiplication: We deﬁne the product of a real number and a matrix to be the matrix obtained by multiplying each of its entries by said real number. More speciﬁcally, if k is a real number and A = [aij]m×n, we deﬁne kA = k [aij]m×n = [kaij]m×n aThe word ‘scalar’ here refers to real numbers. ‘Scalar multiplication’ in this context means we are multiplying a matrix by a real number (a scalar). One may well wonder why the word ‘scalar’ is used for ‘real number.’ It has everything to do with ‘scaling’ factors.5 A point P (x, y) in the plane can be represented by its position matrix, P : (x, y) ↔ P = x y Suppose we take the point (−2, 1) and multiply its position matrix by 3. We have 3P = 3 −2 1 = 3(−2) 3(1) = −6 3 which corresponds to the point (−6, 3). We can imagine taking (−2, 1) to (−6, 3) in this fashion as a dilation by a factor of 3 in both the horizontal and vertical directions. Doing this to all points (x, y) in the plane, therefore, has the eﬀect of magnifying (scaling) the plane by a factor of 3. 4The Distributive Property, in particular. 5See Section 1.7. 8.3 Matrix Arithmetic 581 As did matrix addition, scalar multiplication inherits many properties from real number arithmetic. Below we summarize these properties. Theorem 8.4. Properties of Scalar Multiplication Associative Property: For every m × n matrix A and scalars k and r, (kr)A = k(rA). Identity Property: For all m × n matrices A, 1A = A. Additive Inverse Property: For all m × n matrices A, −A = (−1)A. Distributive Property of Scalar Multiplication over Scalar Addition: For every m × n matrix A and scalars k and r, (k + r)A = kA + rA Distributive Property of Scalar Multiplication over Matrix Addition: For all m × n matrices A and B scalars k, k(A + B) = kA + kB Zero Product Property: If A is an m × n matrix and k is a scalar, then kA = 0m×n if and only if k = 0 or A = 0m×n As with the other results in this section, Theorem 8.4 can be proved using the deﬁnitions of scalar multiplication and matrix addition. For example, to prove that k(A + B) = kA + kB for a scalar k and m × n matrices A and B, we start by adding A and B, then multiplying by k and seeing how that compares with the sum of kA and kB. k(A + B) = k [aij]m×n + [bij]m×n = k [aij + bij]m×n = [k (aij + bij)]m×n = [kaij + kbij]m×n As for kA + kB, we have kA + kB = k [aij]m×n + k [bij]m×n = [kaij]m×n + [kbij]m×n = [kaij + kbij]m×n which establishes the property. The remaining properties are left to the reader. The properties in Theorems 8.3 and 8.4 establish an algebraic system that lets us treat matrices and scalars more or less as we would real numbers and variables, as the next example illustrates. Example 8.3.1. Solve for the matrix A: 3A − 2 −1 5 3 using the deﬁnitions and properties of matrix arithmetic. + 5A = −4 2 6 −2 + 1 3 9 12 −3 39 582 Solution. Systems of Equations and Matrices 3A − 3A + − 3A + (−1) 2 −1 3 5 2 −1 5 3 2 −1 3 5 + 5A = + 5A + 5A = = = 2 −1 3 5 2 −1 5 3 (−1)(−1) (−1)(5) −2 3A + 1 −3 −5 + (−1)(5A) + (−1)(5A) = + ((−1)(5))A = + (−5)A = 3A + (−1) 3A + (−1) 3A + (−1)(2) (−1)(3) 3A + (−5)A + −2 1 −3 −5 (3 + (−5))A + −2 1 −3 −5 + − −2 1 −3 −5 = = (−2)A + 02×2 = (−2)A = (−2)A = −4 2 6 −2 −4 2 6 −2 −4 2 6 −2 1 3 + + + 9 12 −3 39 1 (9) 1 3 3 (−3) 1 1 3 3 4 3 −1 13 (12) (39) −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 −1 6 5 11 + − −2 1 −3 −5 −1 6 5 11 −1 − (−2) − −2 1 −3 −5 6 − 1 5 − (−3) 11 − (−5) − 1 2 ((−2)A) = − 1 2 − 1 2 (−2) A = 1 5 8 16 1 5 8 16 (1) − 1 (5) 2 (16) (8 1A = A = − 1 2 − 5 2 −4 − 16 2 − 1 2 − 5 2 −4 −8 The reader is encouraged to check our answer in the original equation. 8.3 Matrix Arithmetic 583 While the solution to the previous example is written in excruciating detail, in practice many of the steps above are omitted. We have spelled out each step in this example to encourage the reader to justify each step using the deﬁnitions and properties we have established thus far for matrix arithmetic. The reader is encouraged to solve the equation in Example 8.3.1 as they would any other linear equation, for example: 3a − (2 + 5a) = −4 + 1 We now turn our attention to matrix multiplication - that is, multiplying a matrix by another matrix. Based on the ‘no surprises’ trend so far in the section, you may expect that in order to multiply two matrices, they must be of the same size and you ﬁnd the product by multiplying the corresponding entries. While this kind of product is used in other areas of mathematics,6 we deﬁne matrix multiplication to serve us in solving systems of linear equations. To that end, we begin by deﬁning the product of a row and a column. We motivate the general deﬁnition with an example. Consider the two matrices A and B below. 3 (9). A = 2 −10 0 −8 1 8 −5 9 0 −2 −12   Let R1 denote the ﬁrst row of A and C1 denote the ﬁrst column of B. To ﬁnd the ‘product’ of R1 with C1, denoted R1 · C1, we ﬁrst ﬁnd the product of the ﬁrst entry in R1 and the ﬁrst entry in C1. Next, we add to that the product of the second entry in R1 and the second entry in C1. Finally, we take that sum and we add to that the product of the last entry in R1 and the last entry in C1. Using entry notation, R1·C1 = a11b11 +a12b21 +a13b31 = (2)(3)+(0)(4)+(−1)(5) = 6+0+(−5) = 1. We can visualize this schematically as follows 2 −10 0 −8 1 8 −5 9 0 −2 −12   −−−−−−−−−→ 0 −1 2 a11b11 (2)(3) 3 4 5      −−−−−−−−−→ 0 −1 2 + + a12b21 (0)(4) 3 4 5      −−−−−−−−−→ 0 −1 2 + + a13b31 (−1)(5) 3 4 5      To ﬁnd R2 · C3 where R2 denotes the second row of A and C3 denotes the third column of B, we proceed similarly. We start with ﬁnding the product of the ﬁrst entry of R2 with the ﬁrst entry in C3 then add to it the product of the second entry in R2 with the second entry in C3, and so forth. Using entry notation, we have R2·C3 = a21b13+a22b23+a23b33 = (−10)(2)+(3)(−5)+(5)(−2) = −45. Schematically, 2 −10 0 −8 1 8 −5 9 0 −2 −12   6See this article on the Hadamard Product. 584 Systems of Equations and Matrices −−−−−−−−−→ −10 3 5 2 −5 −2      a21b13 = (−10)(2) = −20 + −−−−−−−−−→ −10 5 3 2 −5 −2      −−−−−−−−−→ −10 3 5 2 −5 −2      a22b23 = (3)(−5) = −15 a23b33 = (5)(−2) = −10 + Generalizing this process, we have the following deﬁnition. Deﬁnition 8.9. Product of a Row and a Column: Suppose A = [aij]m×n and B = [bij]n×r. Let Ri denote the ith row of A and let Cj denote the jth column of B. The product of Ri and Cj, denoted Ri · Cj is the real number deﬁned by Ri · Cj = ai1b1j + ai2b2j + . . . ainbnj Note that in order to multiply a row by a column, the number of entries in the row must match the number of entries in the column. We are now in the position to deﬁne matrix multiplication. Deﬁnition 8.10. Matrix Multiplication: Suppose A = [aij]m×n and B = [bij]n×r. Let Ri denote the ith row of
A and let Cj denote the jth column of B. The product of A and B, denoted AB, is the matrix deﬁned by that is AB = [Ri · Cj]m×r      AB = R1 · C1 R1 · C2 R2 · C1 R2 · C2 . . . R1 · Cr . . . R2 · Cr ... ... ... Rm · C1 Rm · C2 . . . Rm · Cr      There are a number of subtleties in Deﬁnition 8.10 which warrant closer inspection. First and foremost, Deﬁnition 8.10 tells us that the ij-entry of a matrix product AB is the ith row of A times the jth column of B. In order for this to be deﬁned, the number of entries in the rows of A must match the number of entries in the columns of B. This means that the number of columns of A must match7 the number of rows of B. In other words, to multiply A times B, the second dimension of A must match the ﬁrst dimension of B, which is why in Deﬁnition 8.10, Am×n is being multiplied by a matrix Bn×r. Furthermore, the product matrix AB has as many rows as A and as many columns of B. As a result, when multiplying a matrix Am×n by a matrix Bn×r, the result is the matrix ABm×r. Returning to our example matrices below, we see that A is a 2 × 3 matrix and B is a 3 × 4 matrix. This means that the product matrix AB is deﬁned and will be a 2 × 4 matrix. A = 2 −10 0 −8 1 8 −5 9 0 −2 −12   7The reader is encouraged to think this through carefully. 8.3 Matrix Arithmetic 585 Using Ri to denote the ith row of A and Cj to denote the jth column of B, we form AB according to Deﬁnition 8.10. AB = R1 · C1 R1 · C2 R1 · C3 R1 · C4 R2 · C1 R2 · C2 R2 · C3 R2 · C4 = 1 7 2 14 −45 6 −4 47 Note that the product BA is not deﬁned, since B is a 3 × 4 matrix while A is a 2 × 3 matrix; B has more columns than A has rows, and so it is not possible to multiply a row of B by a column of A. Even when the dimensions of A and B are compatible such that AB and BA are both deﬁned, the product AB and BA aren’t necessarily equal.8 In other words, AB may not equal BA. Although there is no commutative property of matrix multiplication in general, several other real number properties are inherited by matrix multiplication, as illustrated in our next theorem. Theorem 8.5. Properties of Matrix Multiplication Let A, B and C be matrices such that all of the matrix products below are deﬁned and let k be a real number. Associative Property of Matrix Multiplication: (AB)C = A(BC) Associative Property with Scalar Multiplication: k(AB) = (kA)B = A(kB) Identity Property: For a natural number k, the k × k identity matrix, denoted Ik, is deﬁned by Ik = [dij]k×k where dij = 1, if i = j 0, otherwise For all m × n matrices, ImA = AIn = A. Distributive Property of Matrix Multiplication over Matrix Addition: A(B ± C) = AB ± AC and (A ± B)C = AC ± BC The one property in Theorem 8.5 which begs further investigation is, without doubt, the multiplicative identity. The entries in a matrix where i = j comprise what is called the main diagonal of the matrix. The identity matrix has 1’s along its main diagonal and 0’s everywhere else. A few examples of the matrix Ik mentioned in Theorem 8.5 are given below. The reader is encouraged to see how they match the deﬁnition of the identity matrix presented there   I2 I3 [1] I1         I4 8And may not even have the same dimensions. For example, if A is a 2 × 3 matrix and B is a 3 × 2 matrix, then AB is deﬁned and is a 2 × 2 matrix while BA is also deﬁned... but is a 3 × 3 matrix! 586 Systems of Equations and Matrices The identity matrix is an example of what is called a square matrix as it has the same number of rows as columns. Note that to in order to verify that the identity matrix acts as a multiplicative identity, some care must be taken depending on the order of the multiplication. For example, take the matrix 2 × 3 matrix A from earlier A = 2 −10 0 −1 5 3 In order for the product IkA to be deﬁned, k = 2; similarly, for AIk to be deﬁned, k = 3. We leave it to the reader to show I2A = A and AI3 = A. In other words, and 1 0 0 1 2 −10 0 −1 5 3 = 2 −10 0 −1 5 3 2 −10 0 −10 0 −1 5 3 While the proofs of the properties in Theorem 8.5 are computational in nature, the notation becomes quite involved very quickly, so they are left to a course in Linear Algebra. The following example provides some practice with matrix multiplication and its properties. As usual, some valuable lessons are to be learned. Example 8.3.2. 1. Find AB for A = −23 −1 46 17 2 −34 and B =   −3 2 1 5 −4 3   2. Find C2 − 5C + 10I2 for C = 1 −2 4 3 3. Suppose M is a 4 × 4 matrix. Use Theorem 8.5 to expand (M − 2I4) (M + 3I4). Solution. 1. We have AB = −23 −1 46 17 2 −34   −3 2 1 5 −. Just as x2 means x times itself, C2 denotes the matrix C times itself. We get 8.3 Matrix Arithmetic 587 C2 − 5C + 10I2 = = = = 1 −2 4 3 1 −2 4 3 2 − 5 1 −2 4 3 1 −2 4 3 + + 10 1 0 0 1 −5 10 −15 −20 10 0 0 10 + + 5 10 −15 −10 −5 −10 10 15 0 0 0 0 3. We expand (M − 2I4) (M + 3I4) with the same pedantic zeal we showed in Example 8.3.1. The reader is encouraged to determine which property of matrix arithmetic is used as we proceed from one step to the next. (M − 2I4) (M + 3I4) = (M − 2I4) M + (M − 2I4) (3I4) = M M − (2I4) M + M (3I4) − (2I4) (3I4) = M 2 − 2 (I4M ) + 3 (M I4) − 2 (I4 (3I4)) = M 2 − 2M + 3M − 2 (3 (I4I4)) = M 2 + M − 6I4 Example 8.3.2 illustrates some interesting features of matrix multiplication. First note that in part 1, neither A nor B is the zero matrix, yet the product AB is the zero matrix. Hence, the the zero product property enjoyed by real numbers and scalar multiplication does not hold for matrix multiplication. Parts 2 and 3 introduce us to polynomials involving matrices. The reader is encouraged to step back and compare our expansion of the matrix product (M − 2I4) (M + 3I4) in part 3 with the product (x − 2)(x + 3) from real number algebra. The exercises explore this kind of parallel further. As we mentioned earlier, a point P (x, y) in the xy-plane can be represented as a 2 × 1 position matrix. We now show that matrix multiplication can be used to rotate these points, and hence graphs of equations. Example 8.3.3. Let . Plot P (2, −2), Q(4, 0), S(0, 3), and T (−3, −3) in the plane as well as the points RP , RQ, RS, and RT . Plot the lines y = x and y = −x as guides. What does R appear to be doing to these points? 2. If a point P is on the hyperbola x2 − y2 = 4, show that the point RP is on the curve y = 2 x . 588 Systems of Equations and Matrices Solution. For P (2, −2), the position matrix is P = 2 −2 , and RP = = √ 2 √ √ √ We have that R takes (2, −2) to (2 2), (0, 3) √ 2 2 , 3 is moved to 2). Plotting these in the coordinate plane along with the lines y = x and y = −x, we see that the matrix R is rotating these points counterclockwise by 45◦. 2, 0). Similarly, we ﬁnd (4, 0) is moved to (2 √ , and (−3, −3) is moved to (0, −3 − 3 2 RQ RP RS −4 −3 −2 −1 −1 1 2 3 Q x T −2 −3 −4 P RT For a generic point P (x, y) on the hyperbola x2 − y2 = 4, we have RP = = √ which means R takes (x, y) to y = 2 x , we replace x with √ , √ 2 2 y and y with √ √ + . To show that this point is on the curve √ 2 2 y and simplify. 8.3 Matrix Arithmetic 589 √ √ = 2 2 x − y2 x2 x2 − y2 Since (x, y) is on the hyperbola x2 − y2 = 4, we know that this last equation is true. Since all of our steps are reversible, this last equation is equivalent to our original equation, which establishes the point is, indeed, on the graph of y = 2 x is a hyperbola, and it is none other than the hyperbola x2 − y2 = 4 rotated counterclockwise by 45◦.9 Below we have the graph of x2 − y2 = 4 (solid line) and y = 2 x . This means the graph of y = 2 x (dashed line) for comparison3 −1 −1 −2 −3 When we started this section, we mentioned that we would temporarily consider matrices as their own entities, but that the algebra developed here would ultimately allow us to solve systems of linear equations. To that end, consider the system    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 In Section 8.2, we encoded this system into the augmented matrix   3 −1 1 2 8 1 2 −1 4 3 −4 10   9See Section 7.5 for more details. 590 Systems of Equations and Matrices Recall that the entries to the left of the vertical line come from the coeﬃcients of the variables in the system, while those on the right comprise the associated constants. For that reason, we may form the coeﬃcient matrix A, the unknowns matrix X and the constant matrix B as below   A = 3 −1 1 2 1 2 −1 3 − 10 x y z We now consider the matrix equation AX = B. AX = B   3 −1 3 −4 x y z 3x − y + z x + 2y − z 2x + 3y − 4z   =   =         8 4 10 8 4 10   We see that ﬁnding a solution (x, y, z) to the original system corresponds to ﬁnding a solution X for the matrix equation AX = B. If we think about solving the real number equation ax = b, we would simply ‘divide’ both sides by a. Is it possible to ‘divide’ both sides of the matrix equation AX = B by the matrix A? This is the central topic of Section 8.4. 8.3 Matrix Arithmetic 8.3.1 Exercises For each pair of matrices A and B in Exercises 1 - 7, ﬁnd the following, if deﬁned 3A A − 2B −B AB 1. A = 3. A = 2 −3 1 4 −1 3 5 2 , B = 5 −3 1 4 A2 BA 2. A = −1 5 −3 6 , B = 2 10 1 −7 41 3 −5 11 7 −9 591     5. A = 7. A =  , B = 1 2 3 7 8 9 6. A =   −3 1 −2 4 5 −6  , B = −5 1 8 2 −3 3 −7 5 1 −2 1 −1   , B =  1 2 1 17 33 19 10 19 11   In Exercises 8 - 21, use the matrices 3 2 −5 C = 10 − 11 13 9 0 −5 to compute the following or state that the indicated operation is undeﬁned. 8. 7B − 4A 11. E + D 14. A − 4I2 9. AB 12. ED 15. A2 − B2 10. BA 13. CD + 2I2A 16. (A + B)(A − B) 17. A2 − 5A − 2I2 18. E2 + 5E − 36I3 19. EDC 20. CDE 22. Let A = a b c d e f E1 = 21. ABCEDI2 0 1 1 0 E2 = 5 0 0 1 E3 = 1 −2 1 0 Compute E1A, E2A and E3A. What eﬀect did each of the Ei matrices have on the rows of A? Create E4 so that its eﬀect on A is to multiply the bottom row by −6. How would you extend this idea to matrices with more than two rows? 592 Systems of Equations and Matrices In Exercises 23 - 29, consider the following scenario. In the small village of Pedimaxus i
n the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, 90% of those who subscribe to the Pedimaxus Tribune want to keep getting it, but 10% want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, 80% want to continue with it and 20% want switch to the Tribune. We can express this situation using matrices. Speciﬁcally, let X be the ‘state matrix’ given by X = T P where T is the number of people who get the Tribune and P is the number of people who get the Picayune in a given week. Let Q be the ‘transition matrix’ given by Q = 0.90 0.20 0.10 0.80 such that QX will be the state matrix for the next week. 23. Let’s assume that when Pedimaxus was founded, all 150 residents got the Tribune. (Let’s call this Week 0.) This would mean X = 150 0 Since 10% of that 150 want to switch to the Picayune, we should have that for Week 1, 135 people get the Tribune and 15 people get the Picayune. Show that QX in this situation is indeed QX = 135 15 24. Assuming that the percentages stay the same, we can get to the subscription numbers for Week 2 by computing Q2X. How many people get each paper in Week 2? 25. Explain why the transition matrix does what we want it to do. 26. If the conditions do not change from week to week, then Q remains the same and we have what’s known as a Stochastic Process10 because Week n’s numbers are found by computing QnX. Choose a few values of n and, with the help of your classmates and calculator, ﬁnd out how many people get each paper for that week. You should start to see a pattern as n → ∞. 27. If you didn’t see the pattern, we’ll help you out. Let Xs = 100 50 . Show that QXs = Xs This is called the steady state because the number of people who get each paper didn’t change for the next week. Show that QnX → Xs as n → ∞. 10More speciﬁcally, we have a Markov Chain, which is a special type of stochastic process. 8.3 Matrix Arithmetic 593 28. Now let Show that Qn → S as n → ∞. 29. Show that SY = Xs for any matrix Y of the form Y = y 150 − y This means that no matter how the distribution starts in Pedimaxus, if Q is applied often enough, we always end up with 100 people getting the Tribune and 50 people getting the Picayune. 30. Let z = a + bi and w = c + di be arbitrary complex numbers. Associate z and w with the matrices Z = a b −b a and W = c d −d c Show that complex number addition, subtraction and multiplication are mirrored by the associated matrix arithmetic. That is, show that Z + W , Z − W and ZW produce matrices which can be associated with the complex numbers z + w, z − w and zw, respectively. 31. Let A = 1 2 3 4 and B = 0 −3 2 −5 Compare (A + B)2 to A2 + 2AB + B2. Discuss with your classmates what constraints must be placed on two arbitrary matrices A and B so that both (A + B)2 and A2 + 2AB + B2 exist. When will (A + B)2 = A2 + 2AB + B2? In general, what is the correct formula for (A + B)2? In Exercises 32 - 36, consider the following deﬁnitions. A square matrix is said to be an upper triangular matrix if all of its entries below the main diagonal are zero and it is said to be a lower triangular matrix if all of its entries above the main diagonal are zero. For example9 0 −5 from Exercises 8 - 21 above is an upper triangular matrix whereas F = 1 0 3 0 is a lower triangular matrix. questions with your classmates. (Zeros are allowed on the main diagonal.) Discuss the following 594 Systems of Equations and Matrices 32. Give an example of a matrix which is neither upper triangular nor lower triangular. 33. Is the product of two n × n upper triangular matrices always upper triangular? 34. Is the product of two n × n lower triangular matrices always lower triangular? 35. Given the matrix A = 1 2 3 4 write A as LU where L is a lower triangular matrix and U is an upper triangular matrix? 36. Are there any matrices which are simultaneously upper and lower triangular? 8.3 Matrix Arithmetic 8.3.2 Answers 1. For A = 2 −3 4 1 and B = 5 −2 8 4 595 3A = 6 −9 12 3 −B = −5 2 −4 −8 A2 = 1 −18 13 6 A − 2B = −8 1 −7 −12 AB = −2 −28 30 21 BA = 8 −23 20 16 2. For A = −1 5 −3 6 and B = 2 10 1 −7 3A = −3 15 −9 18 −B = −2 −10 7 −1 A2 = −14 25 −15 21 A − 2B = −5 −15 4 11 AB = −37 −5 −48 −24 BA = −32 70 4 −29 3. For A = −1 3 5 2 and B = 7 0 8 −3 1 4 3A = −3 9 15 6 −B = −7 0 −8 3 −1 −4 A2 = 16 3 5 19 A − 2B is not deﬁned AB = −16 3 4 29 2 48 BA is not deﬁned 4. For A = 2 4 6 8 and B = −1 3 −5 11 7 −9 3A = 6 12 18 24 −B = 1 −3 5 9 −11 −7 A2 = 28 40 60 88 A − 2B is not deﬁned AB = 26 −30 34 50 −54 58 BA is not deﬁned 596 5. For A =   3A = 7 8 9     and B = 1 2 3   21 24 27 A2 is not deﬁned AB =   7 14 21 8 16 24 9 18 27   6. For A =   −3 1 −2 4 5 −6   and B = −5 1 8 3A =   3 −6 −9 12 15 −18   A2 is not deﬁned AB is not deﬁned Systems of Equations and Matrices −B = −1 −2 −3 A − 2B is not deﬁned BA = [50] −B = 5 −1 −8 A − 2B is not deﬁned BA = 32 −34 7. For A =   2 −3 3 −7 5 1 −2 1 −1    and B =  1 1 2 17 33 19 10 19 11   3A = A2 =     6 −9 9 −21 15 3 −6 3 −3   −40 −4 23 −10 −4 11 15 21 −36   AB = 8. 7B − 4A =     4 −29 −47 −2 −B =   −1 −2 −1 −17 −33 −19 −10 −19 −11   A − 2B =   0 −7 3 −31 −65 −40 −27 −37 −23   BA =     . AB = −10 1 −20 −1 8.3 Matrix Arithmetic 597 10. BA = 12. ED = −9 −12 1 −2    67 11 3 − 178 3 −72 −30 −40    14. A − 4I2 = −3 2 3 0 16. (A + B)(A − B) = −7 3 46 2 11. E + D is undeﬁned 13. CD + 2I2A = 238 3 −126 361 863 5 15 15. A2 − B2 = 17. A2 − 5A − 2I2 = −8 16 3 25 0 0 0 0 18. E2 + 5E − 36I3 =   −30 20 −15 0 −36 0 −36 0 0   19. EDC =    3449 15 − 407 15 − 101 99 6 − 9548 3 −648 −324 −35 −360    20. CDE is undeﬁned 21. ABCEDI2 = − 90749 15 − 156601 15 − 28867 5 − 47033 5 d e f c a b E1 interchanged R1 and R2 of A. 22. E1A = E2A = E3A = d 5a 5b 5c f a − 2d b − 2e c − 2f f e e d E4 = 1 0 0 −6 E2 multiplied R1 of A by 5. E3 replaced R1 in A with R1 − 2R2. 598 Systems of Equations and Matrices 8.4 Systems of Linear Equations: Matrix Inverses We concluded Section 8.3 by showing how we can rewrite a system of linear equations as the matrix equation AX = B where A and B are known matrices and the solution matrix X of the equation corresponds to the solution of the system. In this section, we develop the method for solving such an equation. To that end, consider the system 2x − 3y = 16 3x + 4y = 7 To write this as a matrix equation, we follow the procedure outlined on page 590. We ﬁnd the coeﬃcient matrix A, the unknowns matrix X and constant matrix B to be A = 2 −3 4 3 X = x y B = 16 7 In order to motivate how we solve a matrix equation like AX = B, we revisit solving a similar equation involving real numbers. Consider the equation 3x = 5. To solve, we simply divide both sides by 3 and obtain x = 5 3 . How can we go about deﬁning an analogous process for matrices? To answer this question, we solve 3x = 5 again, but this time, we pay attention to the properties of real numbers being used at each step. Recall that dividing by 3 is the same as multiplying by 1 3 = 3−1, the so-called multiplicative inverse 1 of 3. 3x = 5 3−1(3x) = 3−1(5) Multiply by the (multiplicative) inverse of 3 Associative property of multiplication Inverse property Multiplicative Identity 3−1 · 3 x = 3−1(5) 1 · x = 3−1(5) x = 3−1(5) If we wish to check our answer, we substitute x = 3−1(5) into the original equation 3x 3 3−1(5) 3 · 3−1 (5 Associative property of multiplication ? = 5 = 5 Multiplicative Identity Inverse property Thinking back to Theorem 8.5, we know that matrix multiplication enjoys both an associative property and a multiplicative identity. What’s missing from the mix is a multiplicative inverse for the coeﬃcient matrix A. Assuming we can ﬁnd such a beast, we can mimic our solution (and check) to 3x = 5 as follows 1Every nonzero real number a has a multiplicative inverse, denoted a−1, such that a−1 · a = a · a−1 = 1. 8.4 Systems of Linear Equations: Matrix Inverses 599 Solving AX = B Checking our answer AX = B A−1(AX) = A−1B A−1A X = A−1B I2X = A−1B X = A−1B AX A A−1B AA−1 B I2B The matrix A−1 is read ‘A-inverse’ and we will deﬁne it formally later in the section. At this stage, we have no idea if such a matrix A−1 exists, but that won’t deter us from trying to ﬁnd it.2 We want A−1 to satisfy two equations, A−1A = I2 and AA−1 = I2, making A−1 necessarily a 2 × 2 matrix.3 Hence, we assume A−1 has the form A−1 = x1 x2 x3 x4 for real numbers x1, x2, x3 and x4. For reasons which will become clear later, we focus our attention on the equation AA−1 = I2. We have AA−1 = I2 x1 x2 2 −3 x3 x4 4 3 2x1 − 3x3 2x2 − 3x4 3x1 + 4x3 3x2 + 4x4 = = 1 0 0 1 1 0 0 1 This gives rise to two more systems of equations 2x1 − 3x3 = 1 3x1 + 4x3 = 0 2x2 − 3x4 = 0 3x2 + 4x4 = 1 At this point, it may seem absurd to continue with this venture. After all, the intent was to solve one system of equations, and in doing so, we have produced two more to solve. Remember, the objective of this discussion is to develop a general method which, when used in the correct scenarios, allows us to do far more than just solve a system of equations. If we set about to solve these systems using augmented matrices using the techniques in Section 8.2, we see that not only do both systems have the same coeﬃcient matrix, this coeﬃcient matrix is none other than the matrix A itself. (We will come back to this observation in a moment.) 2Much like Carl’s quest to ﬁnd Sasquatch. 3Since matrix multiplication isn’t necessarily commutative, at this stage, these are two diﬀerent equations. 600 Systems of Equations and Matrices 2x1 − 3x3 = 1 3x1 + 4x3 = 0 2x2 − 3x4 = 0 3x2 + 4x4 = 1 Encode into a matrix −−−−−−−−−−−−−→ Encode into a matrix −−−−−−−−−−−−−→ 3 2 −3 1 4 0 2 −3 0 4 1 3 To solve these two systems, we use Gauss-Jordan Elimination to put the augmented matrices into reduced row echelon form. (We leave the details to the reader.) For the ﬁrst system, we get 2 −3 1 4 0 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 4
1 0 17 0 1 − 3 17 which gives x1 = 4 17 . To solve the second system, we use the exact same row operations, in the same order, to put its augmented matrix into reduced row echelon form (Think about why that works.) and we obtain 17 and x3 = − 3 2 −3 0 4 1 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 1 0 0 1 3 17 2 17 which means x2 = 3 17 and x4 = 2 17 . Hence, A−1 = x1 x2 x3 x4 = 4 17 − 3 17 3 17 2 17 We can check to see that A−1 behaves as it should by computing AA−1 As an added bonus, AA−1 = 2 −3 4 3 4 17 − 3 17 3 17 2 17 1 0 0 1 = = I2 A−1A = 4 17 − 3 17 3 17 2 17 2 −3 4 3 = 1 0 0 1 = I2 We can now return to the problem at hand. From our discussion at the beginning of the section on page 599, we know X = A−1B = 4 17 − 3 17 3 17 2 17 16 7 = 5 −2 so that our ﬁnal solution to the system is (x, y) = (5, −2). As we mentioned, the point of this exercise was not just to solve the system of linear equations, but to develop a general method for ﬁnding A−1. We now take a step back and analyze the foregoing discussion in a more general context. In solving for A−1, we used two augmented matrices, both of which contained the same entries as A 8.4 Systems of Linear Equations: Matrix Inverses 601 3 2 −3 1 4 0 2 − We also note that the reduced row echelon forms of these augmented matrices can be written as 4 1 0 17 0 1 − 3 17 3 17 2 17 1 0 0 1 I2 I2 = = x1 x3 x2 x4 where we have identiﬁed the entries to the left of the vertical bar as the identity I2 and the entries to the right of the vertical bar as the solutions to our systems. The long and short of the solution process can be summarized as A 1 0 A 0 1 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 I2 x1 x3 x2 x4 Since the row operations for both processes are the same, all of the arithmetic on the left hand side of the vertical bar is identical in both problems. The only diﬀerence between the two processes is what happens to the constants to the right of the vertical bar. As long as we keep these separated into columns, we can combine our eﬀorts into one ‘super-sized’ augmented matrix and describe the above process as A 1 0 0 1 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 x1 x2 x3 x4 We have the identity matrix I2 appearing as the right hand side of the ﬁrst super-sized augmented matrix and the left hand side of the second super-sized augmented matrix. To our surprise and delight, the elements on the right hand side of the second super-sized augmented matrix are none other than those which comprise A−1. Hence, we have A I2 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ I2 A−1 In other words, the process of ﬁnding A−1 for a matrix A can be viewed as performing a series of row operations which transform A into the identity matrix of the same dimension. We can view this process as follows. In trying to ﬁnd A−1, we are trying to ‘undo’ multiplication by the matrix A. The identity matrix in the super-sized augmented matrix [A\|I] keeps a running memory of all of the moves required to ‘undo’ A. This results in exactly what we want, A−1. We are now ready 602 Systems of Equations and Matrices to formalize and generalize the foregoing discussion. We begin with the formal deﬁnition of an invertible matrix. Deﬁnition 8.11. An n × n matrix A is said to be invertible if there exists a matrix A−1, read ‘A inverse’, such that A−1A = AA−1 = In. Note that, as a consequence of our deﬁnition, invertible matrices are square, and as such, the conditions in Deﬁnition 8.11 force the matrix A−1 to be same dimensions as A, that is, n × n. Since not all matrices are square, not all matrices are invertible. However, just because a matrix is square doesn’t guarantee it is invertible. (See the exercises.) Our ﬁrst result summarizes some of the important characteristics of invertible matrices and their inverses. Theorem 8.6. Suppose A is an n × n matrix. 1. If A is invertible then A−1 is unique. 2. A is invertible if and only if AX = B has a unique solution for every n × r matrix B. The proofs of the properties in Theorem 8.6 rely on a healthy mix of deﬁnition and matrix arithmetic. To establish the ﬁrst property, we assume that A is invertible and suppose the matrices B and C act as inverses for A. That is, BA = AB = In and CA = AC = In. We need to show that B and C are, in fact, the same matrix. To see this, we note that B = InB = (CA)B = C(AB) = CIn = C. Hence, any two matrices that act like A−1 are, in fact, the same matrix.4 To prove the second property of Theorem 8.6, we note that if A is invertible then the discussion on page 599 shows the solution to AX = B to be X = A−1B, and since A−1 is unique, so is A−1B. Conversely, if AX = B has a unique solution for every n × r matrix B, then, in particular, there is a unique solution X0 to the equation AX = In. The solution matrix X0 is our candidate for A−1. We have AX0 = In by deﬁnition, but we need to also show X0A = In. To that end, we note that A (X0A) = (AX0) A = InA = A. In other words, the matrix X0A is a solution to the equation AX = A. Clearly, X = In is also a solution to the equation AX = A, and since we are assuming every such equation as a unique solution, we must have X0A = In. Hence, we have X0A = AX0 = In, so that X0 = A−1 and A is invertible. The foregoing discussion justiﬁes our quest to ﬁnd A−1 using our super-sized augmented matrix approach A In Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ In A−1 We are, in essence, trying to ﬁnd the unique solution to the equation AX = In using row operations. What does all of this mean for a system of linear equations? Theorem 8.6 tells us that if we write the system in the form AX = B, then if the coeﬃcient matrix A is invertible, there is only one solution to the system − that is, if A is invertible, the system is consistent and independent.5 We also know that the process by which we ﬁnd A−1 is determined completely by A, and not by the 4If this proof sounds familiar, it should. See the discussion following Theorem 5.2 on page 380. 5It can be shown that a matrix is invertible if and only if when it serves as a coeﬃcient matrix for a system of equations, the system is always consistent independent. It amounts to the second property in Theorem 8.6 where the matrices B are restricted to being n × 1 matrices. We note that, owing to how matrix multiplication is deﬁned, being able to ﬁnd unique solutions to AX = B for n × 1 matrices B gives you the same statement about solving such equations for n × r matrices − since we can ﬁnd a unique solution to them one column at a time. 8.4 Systems of Linear Equations: Matrix Inverses 603 constants in B. This answers the question as to why we would bother doing row operations on a super-sized augmented matrix to ﬁnd A−1 instead of an ordinary augmented matrix to solve a system; by ﬁnding A−1 we have done all of the row operations we ever need to do, once and for all, since we can quickly solve any equation AX = B using one multiplication, A−1B. Example 8.4.1. Let A =   1 3 0 −1 1 2   2 5 4 1. Use row operations to ﬁnd A−1. Check your answer by ﬁnding A−1A and AA−1. 2. Use A−1 to solve the following systems of equations    (a) 3x + y + 2z = 26 −y + 5z = 39 2x + y + 4z = 117    (b) 3x + y + 2z = 4 −y + 5z = 2 2x + y + 4z = 5    (c) 3x + y + 2z = 1 −y + 5z = 0 2x + y + 4z = 0 Solution. 1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination.   Replace R1 −−−−−−−→ with 1 3 R1 Replace R3 with −−−−−−−−−−→ −2R1 + R3  1 1 3 0 −1 1 2  1 1 3 0 −1 1 0 3  1 0 0  Replace R2 −−−−−−−−→ with (−1)R2  1 0 0 Replace R3 with −−−−−−−−−−→ 3 R2 + R3 − 1  Replace R3 −−−−−−−→ with 3 13 R3   1 1 2 1 1 3 0 −5 13 1 0 3 1 0 3 0 −5 0 1 0 3 0 −1 1 13 1 − 2 13       0 0 1 0 0 1 0 0 3 13 5 1 8 3 1 0 3 0 −5 13 5 0 1 − 2 13 1 0 3 0 −1 1 3 3 1 0 3 0 −1 1 13   13 Replace R1 with − 2 3 R3 + R1 −−−−−−−−−−−−→ Replace R2 with 5R3 + R2    1 0 0 39 − 2 39 − 2 13 − 8 17 1 0 3 1 0 − 10 0 1 − 2 13 13 1 13    13 15 13 3 13 604 Systems of Equations and Matrices    1 0 0 39 − 2 39 − 2 13 − 8 17 1 0 3 1 0 − 10 0 1 − 2 13 13 1 13    Replace R1 with −−−−−−−−−−→ 3 R2 + R1 − 1    2 13 − 7 9 1 0 0 13 0 1 0 − 10 0 0 1 − 2 13 13 − 8 13 1 13    13 15 13 3 13 13 15 13 3 13 We ﬁnd A−1 =    2 13 − 7 13 15 13 3 13 9 13 − 10 − 2 13  13 − 8 13 1 13 A−1A =   and   . To check our answer, we compute 2 13 − 7 9 13 − 10 − 2 13 13 − 8 13 1 13 13 15 13 3 13       1 3 0 −   = I3  AA−1 =    1 3 0 − 13 − 10 − 2 13 13 − 8 13 1 13 2 13 −   = I3  13 15 13 3 13 2. Each of the systems in this part has A as its coeﬃcient matrix. The only diﬀerence between the systems is the constants which is the matrix B in the associated matrix equation AX = B. We solve each of them using the formula X = A−1B. (a) X = A−1B = (b) X = A−1B = (c) X = A−1B =          2 13 − 7 13 − 8 13 − 7 13 − 8 13 − 7 13 − 8 9 13 − 10 − 2 13 9 13 − 10 − 2 13 9 13 − 10 − 2 13 13 1 13 2 13 1 13 2 13 1 13                   26 39 117  13 15 13 3 13 13 15 13 3 13 13 15 13 3 13     =   −39 91   . Our solution is (−39, 91, 26). 26  . We get 5  13 , 19 13 , 9 13 . 5 13 19 13 9 13  . We ﬁnd 9  13 , − 10 13 , − 2 13 .6 9 13 − 10 13 − 2 13 In Example 8.4.1, we see that ﬁnding one inverse matrix can enable us to solve an entire family of systems of linear equations. There are many examples of where this comes in handy ‘in the wild’, and we chose our example for this section from the ﬁeld of electronics. We also take this opportunity to introduce the student to how we can compute inverse matrices using the calculator. 6Note that the solution is the ﬁrst column of the A−1. The reader is encouraged to meditate on this ‘coincidence’. 8.4 Systems of Linear Equations: Matrix Inverses 605 Example 8.4.2. Consider the circuit diagram below.7 We have two batteries with source voltages VB1 and VB2, measured in volts V , along with six resistors with resistances R1 through R6, measured in kiloohms, kΩ. Using
Ohm’s Law and Kirchhoﬀ’s Voltage Law, we can relate the voltage supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to ﬁnd the four ‘mesh’ currents: i1, i2, i3 and i4, measured in milliamps, mA. If we think of electrons ﬂowing through the circuit, we can think of the voltage sources as providing the ‘push’ which makes the electrons move, the resistors as obstacles for the electrons to overcome, and the mesh current as a net rate of ﬂow of electrons around the indicated loops. The system of linear equations associated with this circuit is    (R1 + R3) i1 − R3i2 − R1i4 = VB1 −R3i1 + (R2 + R3 + R4) i2 − R4i3 − R2i4 = 0 −R4i2 + (R4 + R6) i3 − R6i4 = −VB2 −R1i1 − R2i2 − R6i3 + (R1 + R2 + R5 + R6) i4 = 0 1. Assuming the resistances are all 1kΩ, ﬁnd the mesh currents if the battery voltages are (a) VB1 = 10V and VB2 = 5V (b) VB1 = 10V and VB2 = 0V (c) VB1 = 0V and VB2 = 10V (d) VB1 = 10V and VB2 = 10V 2. Assuming VB1 = 10V and VB2 = 5V , ﬁnd the possible combinations of resistances which would yield the mesh currents you found in 1(a). 7The authors wish to thank Don Anthan of Lakeland Community College for the design of this example. VB1R5R1R2R6VB2R3R4i1i2i3i41 606 Solution. Systems of Equations and Matrices 1. Substituting the resistance values into our system of equations, we get    2i1 − i2 − i4 = VB1 −i1 + 3i2 − i3 − i4 = 0 −i2 + 2i3 − i4 = −VB2 −i1 − i2 − i3 + 4i4 = 0 This corresponds to the matrix equation AX = B where     A = 2 −1 −1 0 −1 0 −1 3 −1 −1 2 −1 4 −1 −1 −1     X =         i1 i2 i3 i4 B =     VB1 0 −VB2 0     When we input the matrix A into the calculator, we ﬁnd from which we have A−1 =     1.625 1.25 1.125 1 1.25 1.125 1.5 1.25 1.25 1.625 To solve the four systems given to us, we ﬁnd X = A−1B where the value of B is determined by the given values of VB1 and VB2 1 (a) B =         10 0 −5 0 , 1 (b) B =         10 0 0 0 , 1 (c10 0 , 1 (d) B =         10 0 10 0 (a) For VB1 = 10V and VB2 = 5V , the calculator gives i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA, and i4 = 5 mA. We include a calculator screenshot below for this part (and this part only!) for reference. 8.4 Systems of Linear Equations: Matrix Inverses 607 (b) By keeping VB1 = 10V and setting VB2 = 0V , we are removing the eﬀect of the second battery. We get i1 = 16.25 mA, i2 = 12.5 mA, i3 = 11.25 mA, and i4 = 10 mA. (c) Part (c) is a symmetric situation to part (b) in so much as we are zeroing out VB1 and making VB2 = 10. We ﬁnd i1 = −11.25 mA, i2 = −12.5 mA, i3 = −16.25 mA, and i4 = −10 mA, where the negatives indicate that the current is ﬂowing in the opposite direction as is indicated on the diagram. The reader is encouraged to study the symmetry here, and if need be, hold up a mirror to the diagram to literally ‘see’ what is happening. (d) For VB1 = 10V and VB2 = 10V , we get i1 = 5 mA, i2 = 0 mA, i3 = −5 mA, and i4 = 0 mA. The mesh currents i2 and i4 being zero is a consequence of both batteries ‘pushing’ in equal but opposite directions, causing the net ﬂow of electrons in these two regions to cancel out. 2. We now turn the tables and are given VB1 = 10V , VB2 = 5V , i1 = 10.625 mA, i2 = 6.25 mA, i3 = 3.125 mA and i4 = 5 mA and our unknowns are the resistance values. Rewriting our system of equations, we get    1.25R2 − 4.375R3 + 3.125R4 = 5.625R1 + 4.375R3 = 10 0 −3.125R4 − 1.875R6 = −5 0 −5.625R1 − 1.25R2 + 5R5 + 1.875R6 = The coeﬃcient matrix for this system is 4 × 6 (4 equations with 6 unknowns) and is therefore not invertible. We do know, however, this system is consistent, since setting all the resistance values equal to 1 corresponds to our situation in problem 1a. This means we have an underdetermined consistent system which is necessarily dependent. To solve this system, we encode it into an augmented matrix 0 4.375 1.25 −4.375     5.25 0 0 0 −5.625 −1.25 0 3.125 0 −3.125 0 0 0 0 0 10 0 0 0 −1.875 −5 0 1.875 5     and use the calculator to write in reduced row echelon form 608 Systems of Equations and Matrices     1 0 0 0 0.7 0 1 −3..7 0 0 −1.5 −4 0.6 1.6 0 1 1 0     Decoding this system from the matrix, we get    R1 + 0.7R3 = 1.7 R2 − 3.5R3 − 1.5R6 = −4 R4 + 0.6R6 = 1.6 1 R5 = We can solve for R1, R2, R4 and R5 leaving R3 and R6 as free variables. Labeling R3 = s and R6 = t, we have R1 = −0.7s + 1.7, R2 = 3.5s + 1.5t − 4, R4 = −0.6t + 1.6 and R5 = 1. Since resistance values are always positive, we need to restrict our values of s and t. We know R3 = s > 0 and when we combine that with R1 = −0.7s + 1.7 > 0, we get 0 < s < 16 7 . Similarly, R6 = t > 0 and with R4 = −0.6t + 1.6 > 0, we ﬁnd 0 < t < 8 In order visualize the inequality R2 = 3.5s + 1.5t − 4 > 0, we graph the 3 . line 3.5s + 1.5t − 4 = 0 on the st-plane and shade accordingly.8 Imposing the additional conditions 0 < s < 16 3 , we ﬁnd our values of s and t restricted to the region depicted on the right. Using the roster method, the values of s and t are pulled from the region (s, t) : 0 < s < 16 3 , 3.5s + 1.5t − 4 > 0. The reader is encouraged to check that the solution presented in 1(a), namely all resistance values equal to 1, corresponds to a pair (s, t) in the region. 7 and = 16 7 −2 −1 1 2 4 t −2 −1 1 2 4 t −1 −1 The region where 3.5s + 1.5t − 4 > 0 The region for our parameters s and t. t = 8 3 8See Section 2.4 for a review of this procedure. 8.4 Systems of Linear Equations: Matrix Inverses 609 8.4.1 Exercises In Exercises 1 - 8, ﬁnd the inverse of the matrix or state that the matrix is not invertible. 1. A = 3. C = 1 2 3 4 6 15 14 35 2. B = 4. D = 12 −7 −5 3 2 −1 16 −9     5. E = 7. G = 0 3 2 −1 4 3 2 −5   3 2 3 11 4 19 −3 1 2 3   6. F = 83 4 −3 6 2 1 2 −2 0 0 0 −3 8 16 4 − In Exercises 9 - 11, use one matrix inverse to solve the following systems of linear equations. 3x + 7y = 26 5x + 12y = 39 9. 10. 3x + 7y = 0 5x + 12y = −1 11. 3x + 7y = −7 5 5x + 12y = In Exercises 12 - 14, use the inverse of E from Exercise 5 above to solve the following systems of linear equations.    12. 3x + 4z = 1 2x − y + 3z = 0 −3x + 2y − 5z = 0    13. 3x + 4z = 0 2x − y + 3z = 1 −3x + 2y − 5z = 0    14. 3x + 4z = 0 2x − y + 3z = 0 −3x + 2y − 5z = 1 15. This exercise is a continuation of Example 8.3.3 in Section 8.3 and gives another application of matrix inverses. Recall that given the position matrix P for a point in the plane, the matrix RP corresponds to a point rotated 45◦ counterclockwise from P where a) Find R−1. (b) If RP rotates a point counterclockwise 45◦, what should R−1P do? Check your answer by ﬁnding R−1P for various points on the coordinate axes and the lines y = ±x. (c) Find R−1P where P corresponds to a generic point P (x, y). Verify that this takes points on the curve y = 2 x to points on the curve x2 − y2 = 4. 610 Systems of Equations and Matrices 16. A Sasquatch’s diet consists of three primary foods: Ippizuti Fish, Misty Mushrooms, and Sun Berries. Each serving of Ippizuti Fish is 500 calories, contains 40 grams of protein, and has no Vitamin X. Each serving of Misty Mushrooms is 50 calories, contains 1 gram of protein, and 5 milligrams of Vitamin X. Finally, each serving of Sun Berries is 80 calories, contains no protein, but has 15 milligrams of Vitamin X.9 (a) If an adult male Sasquatch requires 3200 calories, 130 grams of protein, and 275 milligrams of Vitamin X daily, use a matrix inverse to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries he needs to eat each day. (b) An adult female Sasquatch requires 3100 calories, 120 grams of protein, and 300 milligrams of Vitamin X daily. Use the matrix inverse you found in part (a) to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day. (c) An adolescent Sasquatch requires 5000 calories, 400 grams of protein daily, but no Vitamin X daily.10 Use the matrix inverse you found in part (a) to ﬁnd how many servings each of Ippizuti Fish, Misty Mushrooms, and Sun Berries she needs to eat each day. 17. Matrices can be used in cryptography. Suppose we wish to encode the message ‘BIGFOOT LIVES’. We start by assigning a number to each letter of the alphabet, say A = 1, B = 2 and so on. We reserve 0 to act as a space. Hence, our message ‘BIGFOOT LIVES’ corresponds to the string of numbers ‘2, 9, 7, 6, 15, 15, 20, 0, 12, 9, 22, 5, 19.’ To encode this message, we use an invertible matrix. Any invertible matrix will do, but for this exercise, we choose   A = 2 −3 3 −7 5 1 −2 1 −1   Since A is 3 × 3 matrix, we encode our message string into a matrix M with 3 rows. To do this, we take the ﬁrst three numbers, 2 9 7, and make them our ﬁrst column, the next three numbers, 6 15 15, and make them our second column, and so on. We put 0’s to round out the matrix.   M = 6 20 2 9 15 7 15 12 9 19 0 0 0 22 5   To encode the message, we ﬁnd the product AM AM =   2 −3 3 −7 5 1 −2 1 −1     6 20 2 9 15 7 15 12 9 19 0 0 0 22 5    =  12 1 42 3 100 −23 38 57 39 36 −12 −42 −152 −46 −133   9Misty Mushrooms and Sun Berries are the only known ﬁctional sources of Vitamin X. 10Vitamin X is needed to sustain Sasquatch longevity only. 8.4 Systems of Linear Equations: Matrix Inverses 611 So our coded message is ‘12, 1, −12, 42, 3, −42, 100, 36, −152, −23, 39, −46, 38, 57, −133.’ To decode this message, we start with this string of numbers, construct a message matrix as we did earlier (we should get the matrix AM again) and then multiply by A−1. (a) Find A−1. (b) Use A−1 to decode the message and check this method actually works. (c) Decode the message ‘14, 37, −76, 128, 21, −151, 31, 65, −140’ (d) Choose another invertible matrix and encode and decode your own messages. 18. Using the matrices A from Exercise 1, B from Exercise 2 and D from Exercise 4, show AB = D and D−1 = B−1A−1. That is, show that (AB)−1 = B−1A−1. 19. Let M and N be invertible n × n matrices. Show that (M N )−1 = N −1M −1 and compare
your work to Exercise 31 in Section 5.2. 612 Systems of Equations and Matrices 8.4.2 Answers 1. A−1 = −2 3 1 2 − 1 2 3. C is not invertible 5. E−1 =   −1 8 4 1 −3 −1 1 −6 −3   2. B−1 = 3 7 5 12 4. D−1 = 9 2 − 1 2 8 −1 6. F −    16 3 0 2 − 35 −90 − 1 2 0 1 5 0 −7 −36      0 7 2 0 1 7. G is not invertible 8. H −1 = The coeﬃcient matrix is B−1 from Exercise 2 above so the inverse we need is (B−1)−1 = B. 12 −7 −5 3 26 39 9. = 39 −13 10. 11. 12 −7 −5 3 0 −1 12 −7 −5 3 −7 5 = = 7 −3 So x = 39 and y = −13. So x = 7 and y = −3. −119 50 So x = −119 and y = 50. The coeﬃcient matrix is E =   3 0 2 −1 4 3 2 −5   from Exercise 5, so E−1 =   −1 8 4 1 −3 −1 1 −6 −3   12. 13. 14.       −1 8 4 1 −3 −1 1 −6 −3 −1 8 4 1 −3 −1 1 −6 −3 −1 8 4 1 −3 −1 1 −6 −1 1 1 8 −3 −6 4 −1 −3   So x = −1, y = 1 and z = 1.   So x = 8, y = −3 and z = −6.   So x = 4, y = −1 and z = −3. 8.4 Systems of Linear Equations: Matrix Inverses 613 16. (a) The adult male Sasquatch needs: 3 servings of Ippizuti Fish, 10 servings of Misty Mush- rooms, and 15 servings of Sun Berries daily. (b) The adult female Sasquatch needs: 3 servings of Ippizuti Fish and 20 servings of Sun Berries daily. (No Misty Mushrooms are needed!) (c) The adolescent Sasquatch requires 10 servings of Ippizuti Fish daily. (No Misty Mush- rooms or Sun Berries are needed!) 17. (a) A−1 =     1 2 1 17 33 19 10 19 11     (b) 1 1 2 17 33 19 10 19 11   42 3 12 1 100 −23 38 57 39 36 −12 −42 −152 −46 −133    =  6 20 2 9 15 7 15 12 0 22 5 9 19 0 0   (c) ‘LOGS RULE’ 614 Systems of Equations and Matrices 8.5 Determinants and Cramer’s Rule 8.5.1 Definition and Properties of the Determinant In this section we assign to each square matrix A a real number, called the determinant of A, which will eventually lead us to yet another technique for solving consistent independent systems of linear equations. The determinant is deﬁned recursively, that is, we deﬁne it for 1 × 1 matrices and give a rule by which we can reduce determinants of n × n matrices to a sum of determinants of (n − 1) × (n − 1) matrices.1 This means we will be able to evaluate the determinant of a 2 × 2 matrix as a sum of the determinants of 1 × 1 matrices; the determinant of a 3 × 3 matrix as a sum of the determinants of 2 × 2 matrices, and so forth. To explain how we will take an n × n matrix and distill from it an (n − 1) × (n − 1), we use the following notation. Deﬁnition 8.12. Given an n × n matrix A where n > 1, the matrix Aij is the (n − 1) × (n − 1) matrix formed by deleting the ith row of A and the jth column of A. For example, using the matrix A below, we ﬁnd the matrix A23 by deleting the second row and third column of A1 5 1 4 2   Delete R2 and C3 −−−−−−−−−−−→ A23 = 3 1 2 1 We are now in the position to deﬁne the determinant of a matrix. Deﬁnition 8.13. Given an n × n matrix A the determinant of A, denoted det(A), is deﬁned as follows If n = 1, then A = [a11] and det(A) = det ([a11]) = a11. If n > 1, then A = [aij]n×n and det(A) = det [aij]n×n = a11 det (A11) − a12 det (A12) + − . . . + (−1)1+na1n det (A1n) ‘det(A)’ and ‘\|A\|’ There are two commonly used notations for the determinant of a matrix A: We have chosen to use the notation det(A) as opposed to \|A\| because we ﬁnd that the latter is often confused with absolute value, especially in the context of a 1 × 1 matrix. In the expansion a11 det (A11)−a12 det (A12)+− . . .+(−1)1+na1n det (A1n), the notation ‘+−. . .+(−1)1+na1n’ means that the signs alternate and the ﬁnal sign is dictated by the sign of the quantity (−1)1+n. Since the entries a11, a12 and so forth up through a1n comprise the ﬁrst row of A, we say we are ﬁnding the determinant of A by ‘expanding along the ﬁrst row’. Later in the section, we will develop a formula for det(A) which allows us to ﬁnd it by expanding along any row. Applying Deﬁnition 8.13 to the matrix A = 4 −3 1 2 we get 1We will talk more about the term ‘recursively’ in Section 9.1. 8.5 Determinants and Cramer’s Rule 615 det(A) = det 4 −3 1 2 = 4 det (A11) − (−3) det (A12) = 4 det([1]) + 3 det([2]) = 4(1) + 3(2) = 10 For a generic 2 × 2 matrix A = a b c d we get det(A) = det a b c d = a det (A11) − b det (A12) = a det ([d]) − b det ([c]) = ad − bc This formula is worth remembering Equation 8.1. For a 2 × 2 matrix, det a b c d = ad − bc Applying Deﬁnition 8.13 to the 3 × 3 matrix A =   3 1 0 −1 1 2 2 5 4   we obtain det(A) = det     3 1 0 − det (A11) − 1 det (A12) + 2 det (A13) = 3 det −1 5 1 4 − det 0 5 2 4 + 2 det 0 −1 1 2 = 3((−1)(4) − (5)(1)) − ((0)(4) − (5)(2)) + 2((0)(1) − (−1)(2)) = 3(−9) − (−10) + 2(2) = −13 To evaluate the determinant of a 4 × 4 matrix, we would have to evaluate the determinants of four 3 × 3 matrices, each of which involves the ﬁnding the determinants of three 2 × 2 matrices. As you can see, our method of evaluating determinants quickly gets out of hand and many of you may be reaching for the calculator. There is some mathematical machinery which can assist us in calculating determinants and we present that here. Before we state the theorem, we need some more terminology. 616 Systems of Equations and Matrices Deﬁnition 8.14. Let A be an n × n matrix and Aij be deﬁned as in Deﬁnition 8.12. The ij minor of A, denoted Mij is deﬁned by Mij = det (Aij). The ij cofactor of A, denoted Cij is deﬁned by Cij = (−1)i+jMij = (−1)i+j det (Aij). We note that in Deﬁnition 8.13, the sum a11 det (A11) − a12 det (A12) + − . . . + (−1)1+na1n det (A1n) can be rewritten as a11(−1)1+1 det (A11) + a12(−1)1+2 det (A12) + . . . + a1n(−1)1+n det (A1n) which, in the language of cofactors is a11C11 + a12C12 + . . . + a1nC1n We are now ready to state our main theorem concerning determinants. Theorem 8.7. Properties of the Determinant: Let A = [aij]n×n. We may ﬁnd the determinant by expanding along any row. That is, for any 1 ≤ k ≤ n, det(A) = ak1Ck1 + ak2Ck2 + . . . + aknCkn If A is the matrix obtained from A by: – interchanging any two rows, then det(A) = − det(A). – replacing a row with a nonzero multiple (say c) of itself, then det(A) = c det(A) – replacing a row with itself plus a multiple of another row, then det(A) = det(A) If A has two identical rows, or a row consisting of all 0’s, then det(A) = 0. If A is upper or lower triangular,a then det(A) is the product of the entries on the main diagonal.b If B is an n × n matrix, then det(AB) = det(A) det(B). det (An) = det(A)n for all natural numbers n. A is invertible if and only if det(A) = 0. In this case, det A−1 = 1 det(A) . aSee Exercise 8.3.1 in 8.3. bSee page 585 in Section 8.3. Unfortunately, while we can easily demonstrate the results in Theorem 8.7, the proofs of most of these properties are beyond the scope of this text. We could prove these properties for generic 2 × 2 8.5 Determinants and Cramer’s Rule 617 or even 3 × 3 matrices by brute force computation, but this manner of proof belies the elegance and symmetry of the determinant. We will prove what few properties we can after we have developed some more tools such as the Principle of Mathematical Induction in Section 9.3.2 For the moment, let us demonstrate some of the properties listed in Theorem 8.7 on the matrix A below. (Others will be discussed in the Exercises.) A =   3 1 0 −1 1 2   2 5 4 We found det(A) = −13 by expanding along the ﬁrst row. To take advantage of the 0 in the second row, we use Theorem 8.7 to ﬁnd det(A) = −13 by expanding along that row.   det   1 3 0 −1 1 2 2 5 4     = 0C21 + (−1)C22 + 5C23 = (−1)(−1)2+2 det (A22) + 5(−1)2+3 det (A23) = − det 3 2 2 4 − 5 det 3 1 2 1 = −((3)(4) − (2)(2)) − 5((3)(1) − (2)(1)) = −8 − 5 = −13 In general, the sign of (−1)i+j in front of the minor in the expansion of the determinant follows an alternating pattern. Below is the pattern for 2 × 2, 3 × 3 and 4 × 4 matrices, and it extends naturally to higher dimensions. + − − +   + − + − + − + − +       + − + − − + − + + − + − − + − +     The reader is cautioned, however, against reading too much into these sign patterns. In the example above, we expanded the 3 × 3 matrix A by its second row and the term which corresponds to the second entry ended up being negative even though the sign attached to the minor is (+). These signs represent only the signs of the (−1)i+j in the formula; the sign of the corresponding entry as well as the minor itself determine the ultimate sign of the term in the expansion of the determinant. To illustrate some of the other properties in Theorem 8.7, we use row operations to transform our 3 × 3 matrix A into an upper triangular matrix, keeping track of the row operations, and labeling 2For a very elegant treatment, take a course in Linear Algebra. There, you will most likely see the treatment of determinants logically reversed than what is presented here. Speciﬁcally, the determinant is deﬁned as a function which takes a square matrix to a real number and satisﬁes some of the properties in Theorem 8.7. From that function, a formula for the determinant is developed. 618 Systems of Equations and Matrices each successive matrix.3   3 1 0 −1 1 2 A   2 5 4 Replace R3 −−−−−−−−−−→ with − 2 3 R1 + R3   3 1 0 − Replace R3 with −−−−−−−−−−→ 1 3 R2 + R3   3 1 0 −1 0 0 C   2 5 13 3 Theorem 8.7 guarantees us that det(A) = det(B) = det(C) since we are replacing a row with itself plus a multiple of another row moving from one matrix to the next. Furthermore, since C is upper triangular, det(C) is the product of the entries on the main diagonal, in this case det(C) = (3)(−1) 13 = −13. This demonstrates the utility of using row operations to assist in 3 calculating determinants. This also sheds some light on the connection between a determinant and invertibility. Recall from Section 8.4 that in order to ﬁnd A−1, we attempt to transform A to In using row operations A In Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ In A−1 As we apply our allowable row operations on A to put it into reduced row echelon form, the determinant of the intermedia
te matrices can vary from the determinant of A by at most a nonzero multiple. This means that if det(A) = 0, then the determinant of A’s reduced row echelon form must also be nonzero, which, according to Deﬁnition 8.4 means that all the main diagonal entries on A’s reduced row echelon form must be 1. That is, A’s reduced row echelon form is In, and A is invertible. Conversely, if A is invertible, then A can be transformed into In using row operations. Since det (In) = 1 = 0, our same logic implies det(A) = 0. Basically, we have established that the determinant determines whether or not the matrix A is invertible.4 It is worth noting that when we ﬁrst introduced the notion of a matrix inverse, it was in the context of solving a linear matrix equation. In eﬀect, we were trying to ‘divide’ both sides of the matrix equation AX = B by the matrix A. Just like we cannot divide a real number by 0, Theorem 8.7 tells us we cannot ‘divide’ by a matrix whose determinant is 0. We also know that if the coeﬃcient matrix of a system of linear equations is invertible, then system is consistent and independent. It follows, then, that if the determinant of said coeﬃcient is not zero, the system is consistent and independent. 8.5.2 Cramer’s Rule and Matrix Adjoints In this section, we introduce a theorem which enables us to solve a system of linear equations by means of determinants only. As usual, the theorem is stated in full generality, using numbered unknowns x1, x2, etc., instead of the more familiar letters x, y, z, etc. The proof of the general case is best left to a course in Linear Algebra. 3Essentially, we follow the Gauss Jordan algorithm but we don’t care about getting leading 1’s. 4In Section 8.5.2, we learn determinants (speciﬁcally cofactors) are deeply connected with the inverse of a matrix. 8.5 Determinants and Cramer’s Rule 619 Theorem 8.8. Cramer’s Rule: Suppose AX = B is the matrix form of a system of n linear equations in n unknowns where A is the coeﬃcient matrix, X is the unknowns matrix, and B is the constant matrix. If det(A) = 0, then the corresponding system is consistent and independent and the solution for unknowns x1, x2, . . . xn is given by: xj = det (Aj) det(A) , where Aj is the matrix A whose jth column has been replaced by the constants in B. In words, Cramer’s Rule tells us we can solve for each unknown, one at a time, by ﬁnding the ratio of the determinant of Aj to that of the determinant of the coeﬃcient matrix. The matrix Aj is found by replacing the column in the coeﬃcient matrix which holds the coeﬃcients of xj with the constants of the system. The following example ﬂeshes out this method. Example 8.5.1. Use Cramer’s Rule to solve for the indicated unknowns. 1. Solve 2x1 − 3x2 = 4 5x1 + x2 = −2 for x1 and x2 2. Solve    2x − 3y + z = −1 1 0 x − y + z = 3x − 4z = for z. Solution. 1. Writing this system in matrix form, we ﬁnd A = 2 −3 1 5 X = x1 x2 B = 4 −2 To ﬁnd the matrix A1, we remove the column of the coeﬃcient matrix A which holds the coeﬃcients of x1 and replace it with the corresponding entries in B. Likewise, we replace the column of A which corresponds to the coeﬃcients of x2 with the constants to form the matrix A2. This yields A1 = 4 −3 1 −2 A2 = 2 4 5 −2 Computing determinants, we get det(A) = 17, det (A1) = −2 and det (A2) = −24, so that x1 = det (A1) det(A) = − 2 17 x2 = det (A2) det(A) = − 24 17 The reader can check that the solution to the system is − 2 17 , − 24 17 . 620 Systems of Equations and Matrices 2. To use Cramer’s Rule to ﬁnd z, we identify x3 as z. We have   A = 2 −3 1 −1 3 1 1 0 −1 1 0    A3 = Az =  2 −3 −1 1 −1 1 0 0 3   Expanding both det(A) and det (Az) along the third rows (to take advantage of the 0’s) gives z = det (Az) det(A) = −12 −10 = 6 5 The reader is encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alternative method for ﬁnding the inverse of a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 A =   3 1 0 −1 1 2   2 5 4 We found through a variety of methods that det(A) = −13. To our surprise and delight, its inverse below has a remarkable number of 13’s in the denominators of its entries. This is no coincidence.   2   A−1 = 13 − 7 9 13 − 10 − 2 13 Recall that to ﬁnd A−1, we are essentially solving the matrix equation AX = I3, where X = [xij]3×3 is a 3 × 3 matrix. Because of how matrix multiplication is deﬁned, the ﬁrst column of I3 is the product of A with the ﬁrst column of X, the second column of I3 is the product of A with the second column of X and the third column of I3 is the product of A with the third column of X. In other words, we are solving three equations6 13 − 8 13 15 13 3 13 13 1 13    A  x11 x21 x31    =    1 0 0  A  x12 x22 x32    =    0 1 0  A  x13 x23 x33    =    0 0 1 We can solve each of these systems using Cramer’s Rule. Focusing on the ﬁrst system, we have A1 =   1 1 0 −1 1 0 2 5 4    A2 =  3 1 2 0 0 5 2 0 4    A3 =  3 1 0 −1 1 2   1 0 0 5We are developing a method in the forthcoming discussion. As with the discussion in Section 8.4 when we developed the ﬁrst algorithm to ﬁnd matrix inverses, we ask that you indulge us. 6The reader is encouraged to stop and think this through. 8.5 Determinants and Cramer’s Rule 621 If we expand det (A1) along the ﬁrst row, we get det (A1) = det = det −1 5 1 4 −1 5 1 4 − det 0 5 0 4 + 2 det 0 −1 1 0 Amazingly, this is none other than the C11 cofactor of A. The reader is invited to check this, as well as the claims that det (A2) = C12 and det (A3) = C13.7 (To see this, though it seems unnatural to do so, expand along the ﬁrst row.) Cramer’s Rule tells us x11 = det (A1) det(A) = C11 det(A) , x21 = det (A2) det(A) = C12 det(A) , x31 = det (A3) det(A) = C13 det(A) So the ﬁrst column of the inverse matrix X is:   =   x11 x21 x31                   C11 det(A) C12 det(A) C13 det(A) = 1 det(A)     C11 C12 C13 Notice the reversal of the subscripts going from the unknown to the corresponding cofactor of A. This trend continues and we get   =   x12 x22 x32 1 det(A)     C21 C22 C23   =   x13 x23 x33 1 det(A)     C31 C32 C33 Putting all of these together, we have obtained a new and surprising formula for A−1, namely A−1 = 1 det(A)   C11 C21 C31 C12 C22 C32 C13 C23 C33   To see that this does indeed yield A−1, we ﬁnd all of the cofactors of A C11 = −9, C21 = −2, C31 = C12 = 10, C22 = C13 = 7 8, C32 = −15 2, C23 = −1, C33 = −3 And, as promised, 7In a solid Linear Algebra course you will learn that the properties in Theorem 8.7 hold equally well if the word ‘row’ is replaced by the word ‘column’. We’re not going to get into column operations in this text, but they do make some of what we’re trying to say easier to follow. 622 Systems of Equations and Matrices A−1 = 1 det(A)   C11 C21 C31 C12 C22 C32 C13 C23 C33   = −   1 13 −9 −2 7 8 −15 10 2 −1 −3   =    2 13 − 7 9 13 − 10 − 2 13 13 − 8 13 1 13    13 15 13 3 13 To generalize this to invertible n × n matrices, we need another deﬁnition and a theorem. Our deﬁnition gives a special name to the cofactor matrix, and the theorem tells us how to use it along with det(A) to ﬁnd the inverse of a matrix. Deﬁnition 8.15. Let A be an n × n matrix, and Cij denote the ij cofactor of A. The adjoint of A, denoted adj(A) is the matrix whose ij-entry is the ji cofactor of A, Cji. That is adj(A) =      C11 C21 C12 C22 ... ... C1n C2n . . . Cn1 . . . Cn2 ... . . . Cnn      This new notation greatly shortens the statement of the formula for the inverse of a matrix. Theorem 8.9. Let A be an invertible n × n matrix. Then A−1 = 1 det(A) adj(A) For 2 × 2 matrices, Theorem 8.9 reduces to a fairly simple formula. Equation 8.2. For an invertible 2 × 2 matrix, a b c d −1 = 1 ad − bc d −b a −c The proof of Theorem 8.9 is, like so many of the results in this section, best left to a course in Linear Algebra. In such a course, not only do you gain some more sophisticated proof techniques, you also gain a larger perspective. The authors assure you that persistence pays oﬀ. If you stick around a few semesters and take a course in Linear Algebra, you’ll see just how pretty all things matrix really are - in spite of the tedious notation and sea of subscripts. Within the scope of this text, we will prove a few results involving determinants in Section 9.3 once we have the Principle of Mathematical Induction well in hand. Until then, make sure you have a handle on the mechanics of matrices and the theory will come eventually. 8.5 Determinants and Cramer’s Rule 623 8.5.3 Exercises In Exercises 1 - 8, compute the determinant of the given matrix. (Some of these matrices appeared in Exercises 1 - 8 in Section 8.4.) 1. B = 12 −7 −5 3 3. Q = x x2 1 2x     5. F = 73 4 −3 6 2 i −1 k j 5 0 9 −4 −2   2. C = 6 15 14 35           4. L = 6. G = 8. H = 1 x3 3 x4     ln(x) x3 1 − 3 ln(x) x4   2 3 3 11 4 19 − 1 2 3 1 2 −2 0 0 0 −3 0 8 7 16 0 4 1 −5 1     In Exercises 9 - 14, use Cramer’s Rule to solve the system of linear equations. 3x + 7y = 26 5x + 12y = 39 9. 11. 13.    x + y = 8000 0.03x + 0.05y = 250 x + y + z = 3 2x − y + z = 0 −3x + 5y + 7z = 7 In Exercises 15 - 16, use Cramer’s Rule to solve for x4. 15.    x1 − x3 = −2 2x2 − x4 = 0 x1 − 2x2 + x3 = 0 −x3 + x4 = 1 16. 10. 2x − 4y = 5 10x + 13y = −6 12 6x + 7y = 3 14.       3x + y − 2z = 10 4x − y + z = 5 x − 3y − 4z = −1 4x1 + x2 = x2 − 3x3 = 10x1 + x3 + x4 = 4 1 0 −x2 + x3 = −3 624 Systems of Equations and Matrices In Exercises 17 - 18, ﬁnd the inverse of the given matrix using their determinants and adjoints. 17. B = 12 −7 −5 3 183 4 −3 6 2 19. Carl’s Sasquatch Attack! Game Card Collection is a mixture of common and rare cards. Each common card is worth $0.25 while each rare card is worth $0.75. If his entire 117 card collection is worth $48.75, how many of each kin
d of card does he own? 20. How much of a 5 gallon 40% salt solution should be replaced with pure water to obtain 5 gallons of a 15% solution? 21. How much of a 10 liter 30% acid solution must be replaced with pure acid to obtain 10 liters of a 50% solution? 22. Daniel’s Exotic Animal Rescue houses snakes, tarantulas and scorpions. When asked how many animals of each kind he boards, Daniel answered: ‘We board 49 total animals, and I am responsible for each of their 272 legs and 28 tails.’ How many of each animal does the Rescue board? (Recall: tarantulas have 8 legs and no tails, scorpions have 8 legs and one tail, and snakes have no legs and one tail.) 23. This exercise is a continuation of Exercise 16 in Section 8.4. Just because a system is consistent independent doesn’t mean it will admit a solution that makes sense in an applied setting. Using the nutrient values given for Ippizuti Fish, Misty Mushrooms, and Sun Berries, use Cramer’s Rule to determine the number of servings of Ippizuti Fish needed to meet the needs of a daily diet which requires 2500 calories, 1000 grams of protein, and 400 milligrams of Vitamin X. Now use Cramer’s Rule to ﬁnd the number of servings of Misty Mushrooms required. Does a solution to this diet problem exist? 11 −7 −3 15 9 6 1 −5 9 6 −7 11 24. Let R = , and a) Show that det(RS) = det(R) det(S) (b) Show that det(T ) = − det(R) (c) Show that det(U ) = −3 det(S) 25. For M , N , and P below, show that det(M ) = 0, det(N ) = 0 and det(P ) = 02 −4 −6 9 8 7   8.5 Determinants and Cramer’s Rule 625 26. Let A be an arbitrary invertible 3 × 3 matrix. (a) Show that det(I3) = 1. (See footnote8 below.) (b) Using the facts that AA−1 = I3 and det(AA−1) = det(A) det(A−1), show that det(A−1) = 1 det(A) The purpose of Exercises 27 - 30 is to introduce you to the eigenvalues and eigenvectors of a matrix.9 We begin with an example using a 2 × 2 matrix and then guide you through some exercises using a 3 × 3 matrix. Consider the matrix C = 6 15 14 35 from Exercise 2. We know that det(C) = 0 which means that CX = 02×2 does not have a unique solution. So there is a nonzero matrix Y with CY = 02×2. In fact, every matrix of the form Y = − 5 2 t t is a solution to CX = 02×2, so there are inﬁnitely many matrices such that CX = 02×2. But consider the matrix X41 = 3 7 It is NOT a solution to CX = 02×2, but rather, 3 7 6 15 14 35 CX41 = = 123 287 = 41 3 7 In fact, if Z is of the form Z = 3 7 t t then CZ = 6 15 14 35 3 7 t t = 123 7 t 41t = 41 3 7 t t = 41Z for all t. The big question is “How did we know to use 41?” We need a number λ such that CX = λX has nonzero solutions. We have demonstrated that λ = 0 and λ = 41 both worked. Are there others? If we look at the matrix equation more closely, what 8If you think about it for just a moment, you’ll see that det(In) = 1 for any natural number n. The formal proof of this fact requires the Principle of Mathematical Induction (Section 9.3) so we’ll stick with n = 3 for the time being. 9This material is usually given its own chapter in a Linear Algebra book so clearly we’re not able to tell you everything you need to know about eigenvalues and eigenvectors. They are a nice application of determinants, though, so we’re going to give you enough background so that you can start playing around with them. 626 Systems of Equations and Matrices we really wanted was a nonzero solution to (C − λI2)X = 02×2 which we know exists if and only if the determinant of C − λI2 is zero.10 So we computed det(C − λI2) = det 6 − λ 15 14 35 − λ = (6 − λ)(35 − λ) − 14 · 15 = λ2 − 41λ This is called the characteristic polynomial of the matrix C and it has two zeros: λ = 0 and λ = 41. That’s how we knew to use 41 in our work above. The fact that λ = 0 showed up as one of the zeros of the characteristic polynomial just means that C itself had determinant zero which we already knew. Those two numbers are called the eigenvalues of C. The corresponding matrix solutions to CX = λX are called the eigenvectors of C and the ‘vector’ portion of the name will make more sense after you’ve studied vectors. Now it’s your turn. In the following exercises, you’ll be using the matrix G from Exercise 6 11 4 19 27. Show that the characteristic polynomial of G is p(λ) = −λ(λ − 1)(λ − 22). That is, compute det (G − λI3). 28. Let G0 = G. Find the parametric description of the solution to the system of linear equations given by GX = 03×3. 29. Let G1 = G − I3. Find the parametric description of the solution to the system of linear equations given by G1X = 03×3. Show that any solution to G1X = 03×3 also has the property that GX = 1X. 30. Let G22 = G − 22I3. Find the parametric description of the solution to the system of linear equations given by G22X = 03×3. Show that any solution to G22X = 03×3 also has the property that GX = 22X. 10Think about this. 8.5 Determinants and Cramer’s Rule 627 8.5.4 Answers 1. det(B) = 1 3. det(Q) = x2 5. det(F ) = −12 2. det(C) = 0 4. det(L) = 1 x7 6. det(G) = 0 7. det(V ) = 20i + 43j + 4k 8. det(H) = −2 9. x = 39, y = −13 11. x = 7500, y = 500 13. x = 1, y = 2, z = 0 15. x4 = 4 17. B−1 = 3 7 5 12 18. F − 10. x = 41 66 , y = − 31 33 12. x = 76 47 , y = − 45 47 14. x = 121 60 , y = 131 60 , z = − 53 60 16. x4 = −1    19. Carl owns 78 common cards and 39 rare cards. 20. 3.125 gallons. 21. 20 7 ≈ 2.85 liters. 22. The rescue houses 15 snakes, 21 tarantulas and 13 scorpions. 23. Using Cramer’s Rule, we ﬁnd we need 53 servings of Ippizuti Fish to satisfy the dietary requirements. The number of servings of Misty Mushrooms required, however, is −1120. Since it’s impossible to have a negative number of servings, there is no solution to the applied problem, despite there being a solution to the mathematical problem. A cautionary tale about using Cramer’s Rule: just because you are guaranteed a mathematical answer for each variable doesn’t mean the solution will make sense in the ‘real’ world. 628 Systems of Equations and Matrices 8.6 Partial Fraction Decomposition This section uses systems of linear equations to rewrite rational functions in a form more palatable to Calculus students. In College Algebra, the function f (x) = x2 − x − 6 x4 + x2 (1) is written in the best form possible to construct a sign diagram and to ﬁnd zeros and asymptotes, but certain applications in Calculus require us to rewrite f (x) as f (x) = x + 7 x2 + 1 − 1 x − 6 x2 (2) If we are given the form of f (x) in (2), it is a matter of Intermediate Algebra to determine a common denominator to obtain the form of f (x) given in (1). The focus of this section is to develop a method by which we start with f (x) in the form of (1) and ‘resolve it into partial fractions’ to obtain the form in (2). Essentially, we need to reverse the least common denominator process. Starting with the form of f (x) in (1), we begin by factoring the denominator x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) We now think about which individual denominators could contribute to obtain x2 x2 + 1 as the least common denominator. Certainly x2 and x2 + 1, but are there any other factors? Since x2 + 1 is an irreducible quadratic1 there are no factors of it that have real coeﬃcients which can contribute to the denominator. The factor x2, however, is not irreducible, since we can think of it as x2 = xx = (x − 0)(x − 0), a so-called ‘repeated’ linear factor.2 This means it’s possible that a term with a denominator of just x contributed to the expression as well. What about something like x x2 + 1? This, too, could contribute, but we would then wish to break down that denominator into x and x2 + 1, so we leave out a term of that form. At this stage, we have guessed x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = ? x + ? x2 + ? x2 + 1 Our next task is to determine what form the unknown numerators take. It stands to reason that since the expression x2−x−6 is ‘proper’ in the sense that the degree of the numerator is less than x4+x2 the degree of the denominator, we are safe to make the ansatz that all of the partial fraction resolvents are also. This means that the numerator of the fraction with x as its denominator is just a constant and the numerators on the terms involving the denominators x2 and x2 + 1 are at most linear polynomials. That is, we guess that there are real numbers A, B, C, D and E so that x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = A x + Bx + C x2 + Dx + E x2 + 1 1Recall this means it has no real zeros; see Section 3.4. 2Recall this means x = 0 is a zero of multiplicity 2. 8.6 Partial Fraction Decomposition 629 However, if we look more closely at the term Bx+C term B Hence, we drop it and, after re-labeling, we ﬁnd ourselves with our new guess: x2 . The x which means it contributes nothing new to our expansion. x has the same form as the term A , we see that Bx+C x2 = Bx x2 + C x2 = B x + C x2 x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = A x + B x2 + Cx + D x2 + 1 Our next task is to determine the values of our unknowns. Clearing denominators gives x2 − x − 6 = Ax x2 + 1 + B x2 + 1 + (Cx + D)x2 Gathering the like powers of x we have x2 − x − 6 = (A + C)x3 + (B + D)x2 + Ax + B In order for this to hold for all values of x in the domain of f , we equate the coeﬃcients of corresponding powers of x on each side of the equation3 and obtain the system of linear equations    (E1) A + C = (E2) B + D = (E3) (E4) 0 From equating coeﬃcients of x3 1 From equating coeﬃcients of x2 A = −1 From equating coeﬃcients of x B = −6 From equating the constant terms To solve this system of equations, we could use any of the methods presented in Sections 8.1 through 8.5, but none of these methods are as eﬃcient as the good old-fashioned substitution you learned in Intermediate Algebra. From E3, we have A = −1 and we substitute this into E1 to get C = 1. Similarly, since E4 gives us B = −6, we have from E2 that D = 7. We get x2 − x − 6 x4 + x2 = x2 − x − 6 x2 (x2 + 1) = − 1 x − 6 x2 + x + 7 x2 + 1 which matches the formula given in (2). As we have seen in
this opening example, resolving a rational function into partial fractions takes two steps: ﬁrst, we need to determine the form of the decomposition, and then we need to determine the unknown coeﬃcients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coeﬃcients can be factored over the real numbers as a product of linear factors and irreducible quadratic factors. Once we have this factorization of the denominator of a rational function, the next theorem tells us the form the decomposition takes. The reader is encouraged to review the Factor Theorem (Theorem 3.6) and its connection to the role of multiplicity to fully appreciate the statement of the following theorem. 3We will justify this shortly. 630 Systems of Equations and Matrices Theorem 8.10. Suppose R(x) = than the degree of D(x) and N (x) and D(x) have no common factors.a is a rational function where the degree of N (x) less N (x) D(x) If α is a real zero of D of multiplicity m which corresponds to the linear factor ax + b, the partial fraction decomposition includes A1 ax + b + A2 (ax + b)2 + . . . + Am (ax + b)m for real numbers A1, A2, . . . Am. If α is a non-real zero of D of multiplicity m which corresponds to the irreducible quadratic ax2 + bx + c, the partial fraction decomposition includes B1x + C1 ax2 + bx + c + B2x + C2 (ax2 + bx + c)2 + . . . + Bmx + Cm (ax2 + bx + c)m for real numbers B1, B2, . . . Bm and C1, C2, . . . Cm. aIn other words, R(x) is a proper rational function which has been fully reduced. The proof of Theorem 8.10 is best left to a course in Abstract Algebra. Notice that the theorem provides for the general case, so we need to use subscripts, A1, A2, etc., to denote diﬀerent unknown coeﬃcients as opposed to the usual convention of A, B, etc.. The stress on multiplicities is to help us correctly group factors in the denominator. For example, consider the rational function 3x − 1 (x2 − 1) (2 − x − x2) Factoring the denominator to ﬁnd the zeros, we get (x + 1)(x − 1)(1 − x)(2 + x). We ﬁnd x = −1 and x = −2 are zeros of multiplicity one but that x = 1 is a zero of multiplicity two due to the two diﬀerent factors (x − 1) and (1 − x). One way to handle this is to note that (1 − x) = −(x − 1) so 3x − 1 (x + 1)(x − 1)(1 − x)(2 + x) = 3x − 1 −(x − 1)2(x + 1)(x + 2) = 1 − 3x (x − 1)2(x + 1)(x + 2) from which we proceed with the partial fraction decomposition 1 − 3x (x − 1)2(x + 1)(x + 2) = A x − 1 + B (x − 1) Turning our attention to non-real zeros, we note that the tool of choice to determine the irreducibility of a quadratic ax2 + bx + c is the discriminant, b2 − 4ac. If b2 − 4ac < 0, the quadratic admits a pair of non-real complex conjugate zeros. Even though one irreducible quadratic gives two distinct non-real zeros, we list the terms with denominators involving a given irreducible quadratic only once to avoid duplication in the form of the decomposition. The trick, of course, is factoring the 8.6 Partial Fraction Decomposition 631 denominator or otherwise ﬁnding the zeros and their multiplicities in order to apply Theorem 8.10. We recommend that the reader review the techniques set forth in Sections 3.3 and 3.4. Next, we state a theorem that if two polynomials are equal, the corresponding coeﬃcients of the like powers of x are equal. This is the principal by which we shall determine the unknown coeﬃcients in our partial fraction decomposition. Theorem 8.11. Suppose anxn + an−1xn−1 + · · · + a2x2 + a1x + a0 = bmxm + mm−1xm−1 + · · · + b2x2 + b1x + b0 for all x in an open interval I. Then n = m and ai = bi for all i = 1 . . . n. Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Deﬁne p(x) to be the diﬀerence of the left hand side of the equation in Theorem 8.11 and the right hand side. Then p(x) = 0 for all x in the open interval I. If p(x) were a nonzero polynomial of degree k, then, by Theorem 3.14, p could have at most k zeros in I, and k is a ﬁnite number. Since p(x) = 0 for all the x in I, p has inﬁnitely many zeros, and hence, p is the zero polynomial. This means there can be no nonzero terms in p(x) and the theorem follows. Arguably, the best way to make sense of either of the two preceding theorems is to work some examples. Example 8.6.1. Resolve the following rational functions into partial fractions. 1. R(x) = x + 5 2x2 − x − 1 4. R(x) = 4x3 x2 − 2 Solution. 2. R(x) = 5. R(x) = 3 x3 − 2x2 + x x3 + 5x − 1 x4 + 6x2 + 9 3. R(x) = 6. R(x) = 3 x3 − x2 + x 8x2 x4 + 16 1. We begin by factoring the denominator to ﬁnd 2x2 − x − 1 = (2x + 1)(x − 1). We get x = − 1 2 and x = 1 are both zeros of multiplicity one and thus we know x + 5 2x2 − x − 1 = x + 5 (2x + 1)(x − 1) = A 2x + 1 + B x − 1 Clearing denominators, we get x+5 = A(x−1)+B(2x+1) so that x+5 = (A+2B)x+B −A. Equating coeﬃcients, we get the system A + 2B = 1 −A + B = 5 This system is readily handled using the Addition Method from Section 8.1, and after adding both equations, we get 3B = 6 so B = 2. Using back substitution, we ﬁnd A = −3. Our answer is easily checked by getting a common denominator and adding the fractions. x + 5 2x2 − 2x + 1 632 Systems of Equations and Matrices 2. Factoring the denominator gives x3 − 2x2 + x = x x2 − 2x + 1 = x(x − 1)2 which gives x = 0 as a zero of multiplicity one and x = 1 as a zero of multiplicity two. We have 3 x3 − 2x2 + x = 3 x(x − 1)x − 1)2 Clearing denominators, we get 3 = A(x − 1)2 + Bx(x − 1) + Cx, which, after gathering up the like terms becomes 3 = (A + B)x2 + (−2A − B + C)x + A. Our system is    A + B = 0 −2A − B + C = 0 A = 3 Substituting A = 3 into A + B = 0 gives B = −3, and substituting both for A and B in −2A − B + C = 0 gives C = 3. Our ﬁnal answer is 3 x3 − 2x2 + x − 1)2 3. The denominator factors as x x2 − x + 1. We see immediately that x = 0 is a zero of multiplicity one, but the zeros of x2 − x + 1 aren’t as easy to discern. The quadratic doesn’t factor easily, so we check the discriminant and ﬁnd it to be (−1)2 − 4(1)(1) = −3 < 0. We ﬁnd its zeros are not real so it is an irreducible quadratic. The form of the partial fraction decomposition is then 3 x3 − x2 + x = 3 x (x2 − x + 1) = A x + Bx + C x2 − x + 1 Proceeding as usual, we clear denominators and get 3 = A x2 − x + 1 + (Bx + C)x or 3 = (A + B)x2 + (−A + C)x + A. We get    A + B = 0 −A + C = 0 A = 3 From A = 3 and A + B = 0, we get B = −3. From −A + C = 0, we get C = A = 3. We get 3 x3 − x2 + x = 3 x + 3 − 3x x2 − x + 1 4. Since 4x3 x2−2 isn’t proper, we use long division and we get a quotient of 4x with a remainder of 8x. That is, 4x3 x2−2 so we now work on resolving 8x x2−2 into partial fractions. The quadratic x2 −2, though it doesn’t factor nicely, is, nevertheless, reducible. Solving x2 −2 = 0 x2−2 = 4x + 8x 8.6 Partial Fraction Decomposition 633 √ gives us x = ± enables us to now factor x2 − 2 = x − 2, and each of these zeros must be of multiplicity one since Theorem 3.14 2. Hence, 2 x + √ √ 8x x2 − 2 = x − √ √ 8x or 8x = (A + B)x + (A − B) √ 2. Clearing fractions, we get 8x = A x + We get the system √ A + B = 8 2 = 0 (A − B) √ From (A − B) Hence, A = B = 4 and we get 2 = 0, we get A = B, which, when substituted into A + B = 8 gives B = 4. 4x3 x2 − 2 = 4x + 8x x2 − 2 = 4x + . At ﬁrst glance, the denominator D(x) = x4 + 6x2 + 9 appears irreducible. However, D(x) has three terms, and the exponent on the ﬁrst term is exactly twice that of the second. Rewriting + 6x2 + 9, we see it is a quadratic in disguise and factor D(x) = x2 + 32 D(x) = x22 . Since x2 + 3 clearly has no real zeros, it is irreducible and the form of the decomposition is x3 + 5x − 1 x4 + 6x2 + 9 x3 + 5x − 1 (x2 + 3)2 = When we clear denominators, we ﬁnd x3 + 5x − 1 = (Ax + B) x2 + 3 + Cx + D which yields x3 + 5x − 1 = Ax3 + Bx2 + (3A + C)x + 3B + D. Our system is Cx + D (x2 + 3)2 Ax + B x2 + 3 = +    A = 1 B = 0 5 3A + C = 3B + D = −1 We have A = 1 and B = 0 from which we get C = 2 and D = −1. Our ﬁnal answer is x3 + 5x − 1 x4 + 6x2 + 9 = x x2 + 3 + 2x − 1 (x2 + 3)2 6. Once again, the diﬃculty in our last example is factoring the denominator. In an attempt to get a quadratic in disguise, we write x4 + 16 = x22 + 42 = x22 + 8x2 + 42 − 8x2 = x2 + 42 − 8x2 634 Systems of Equations and Matrices and obtain a diﬀerence of two squares: x2 + 42 and 8x2 = 2x √ 22 . Hence, x4 + 16 = x2 + 4 − 2x √ 2 x2 + 4 + 2x √ 2 = x2 − 2x √ 2 + 4 x2 + 2x √ 2 + 4 The discrimant of both of these quadratics works out to be −8 < 0, which means they are irreducible. We leave it to the reader to verify that, despite having the same discriminant, these quadratics have diﬀerent zeros. The partial fraction decomposition takes the form 8x2 x4 + 16 = √ x2 − 2x √ 8x2 2 + 4 x2 + 2x √ 2 + 4 = 2 + 4 + (Cx + D) x2 − 2x x2 − 2x Ax + B √ √ 2 + 4 or + Cx + D √ x2 + 2x 2 + 4 2 + 4 We get 8x2 = (Ax + B) x2 + 2x 8x2 = (A + C)x3 + (2A √ 2 + B − 2C √ 2 + D)x2 + (4A + 2B √ 2 + 4C − 2D √ 2)x + 4B + 4D which gives the system    √ 2A 4A + 2B 2 + B − 2C √ 2 + 4C − 2D √ 4B + 4D = 0 √ We choose substitution as the weapon of choice to solve this system. From A + C = 0, we get A = −C; from 4B + 4D = 0, we get B = −D. Substituting these into the remaining two equations, we get −2C √ √ 2 − D − 2C √ 2 + 4C − 2D √ 2 + D = 8 2 = 0 or −4C − 2D √ √ −4C −4D 2 = 8 2 = 0 We get C = − √ 2 so that A = −C = √ 2 and D = 0 which means B = −D = 0. We get 8x2 x4 + 16 = √ x x2 − 2x 2 √ − 2 + 4 √ x x2 + 2x 2 √ 2 + 4 8.6 Partial Fraction Decomposition 635 8.6.1 Exercises In Exercises 1 - 6, ﬁnd only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to ﬁnd the actual decomposition. 1. 3. 5. 7 (x − 3)(x + 5) m (7x − 6)(x2 + 9) A polynomial of degree < 9 (x + 4)5(x2 + 1)2 2. 4. 6. 5x + 4 x(x − 2)(2 − x) ax2 + bx + c x3(5x + 9)(3x2 + 7x + 9) A polynomial of degree < 7 x(4x − 1)2(x2 + 5)(9x2 + 16) In Exercises 7 - 18, ﬁnd the partial fraction decomposition of the following rational expressions. 7. 9. 11. 13. 15. 1
7. 2x x2 − 1 11x2 − 5x − 10 5x3 − 5x2 −x2 + 15 4x4 + 40x2 + 36 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 8. 10. 12. 14. 16. 18. −7x + 43 3x2 + 19x − 14 −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 −21x2 + x − 16 3x3 + 4x2 − 3x + 2 x6 + 5x5 + 16x4 + 80x3 − 2x2 + 6x − 43 x3 + 5x2 + 16x + 80 −10x4 + x3 − 19x2 + x − 10 x5 + 2x3 + x 2x2 + 3x + 14 (x2 + 2x + 9)(x2 + x + 5) 19. As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors ﬁnd students make is something along the lines of Think about why if the above were true, this section would have no need to exist. 8 x2 − 9 = 8 x2 − 8 9 636 Systems of Equations and Matrices 8.6.2 Answers 1. 3. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18 7x − 6 A x + 4 A x + 2x x2 − 1 + + Bx + C x2 + 9 C B (x + 4)3 + (x + 4)2 + C B (4x − 1)2 + 4x − x + 4)4 + E (x + 4)5 + Dx + E x2 + 5 + F x + G 9x2 + 16 2. 4x − 2)2 + B x2 + F x + G x2 + 1 C x3 + + D 5x + 9 Hx + I (x2 + 1)2 + Ex + F 3x2 + 7x + 9 −7x + 43 3x2 + 19x − 14 11x2 − 5x − 10 5x3 − 5x2 = = 5 3x − x2 − 4 5(x − 1) −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 = − 9 x + 4 + 7x − 8 x2 + 4 −x2 + 15 4x4 + 40x2 + 36 = 1 2(x2 + 1) − 3 4(x2 + 9) = − −21x2 + x − 16 6 3x3 + 4x2 − 3x + 2 x + 2 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 − 3x + 5 3x2 − 2x + 1 = 5x2 − 4x + 1 + 9 x − 3 − 1 (x − 3)2 x6 + 5x5 + 16x4 + 80x3 − 2x2 + 6x − 43 x3 + 5x2 + 16x + 80 = x3 + x + 1 x2 + 16 − 3 x + 5 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 = − 7 x + 6 + 8 (x + 6)2 − 4 (x + 6)3 −10x4 + x3 − 19x2 + x − 10 x5 + 2x3 + x = − 10 x + 1 x2 + 1 + x (x2 + 1)2 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 = 1 x − 2 + 4 (x − 2)2 + 3x + 1 x2 + 4 2x2 + 3x + 14 (x2 + 2x + 9)(x2 + x + 5) = 1 x2 + 2x + 9 + 1 x2 + x + 5 8.7 Systems of Non-Linear Equations and Inequalities 637 8.7 Systems of Non-Linear Equations and Inequalities In this section, we study systems of non-linear equations and inequalities. Unlike the systems of linear equations for which we have developed several algorithmic solution techniques, there is no general algorithm to solve systems of non-linear equations. Moreover, all of the usual hazards of non-linear equations like extraneous solutions and unusual function domains are once again present. Along with the tried and true techniques of substitution and elimination, we shall often need equal parts tenacity and ingenuity to see a problem through to the end. You may ﬁnd it necessary to review topics throughout the text which pertain to solving equations involving the various functions we have studied thus far. To get the section rolling we begin with a fairly routine example. Example 8.7.1. Solve the following systems of equations. Verify your answers algebraically and graphically. x2 + y2 = 4 4x2 + 9y2 = 36 x2 + y2 = 4 4x2 − 9y2 = 36 1. 2. Solution: x2 + y2 = 4 y − 2x = 0 x2 + y2 = 4 y − x2 = 0 3. 4. 1. Since both equations contain x2 and y2 only, we can eliminate one of the variables as we did in Section 8.1. (E1) x2 + y2 = 4 (E2) 4x2 + 9y2 = 36 Replace E2 with −−−−−−−−−−→ −4E1 + E2 (E1) x2 + y2 = 4 5y2 = 20 (E2) From 5y2 = 20, we get y2 = 4 or y = ±2. To ﬁnd the associated x values, we substitute each value of y into one of the equations to ﬁnd the resulting value of x. Choosing x2 + y2 = 4, we ﬁnd that for both y = −2 and y = 2, we get x = 0. Our solution is thus {(0, 2), (0, −2)}. To check this algebraically, we need to show that both points satisfy both of the original equations. We leave it to the reader to verify this. To check our answer graphically, we sketch both equations and look for their points of intersection. The graph of x2 + y2 = 4 is a circle centered at (0, 0) with a radius of 2, whereas the graph of 4x2 + 9y2 = 36, when written in the standard form x2 4 = 1 is easily recognized as an ellipse centered at (0, 0) with a major axis along the x-axis of length 6 and a minor axis along the y-axis of length 4. We see from the graph that the two curves intersect at their y-intercepts only, (0, ±2). 9 + y2 2. We proceed as before to eliminate one of the variables (E1) x2 + y2 = 4 (E2) 4x2 − 9y2 = 36 Replace E2 with −−−−−−−−−−→ −4E1 + E2 (E1) x2 + y2 = 4 (E2) −13y2 = 20 638 Systems of Equations and Matrices Since the equation −13y2 = 20 admits no real solution, the system is inconsistent. To verify this graphically, we note that x2 + y2 = 4 is the same circle as before, but when writing the second equation in standard form, x2 4 = 1, we ﬁnd a hyperbola centered at (0, 0) opening to the left and right with a transverse axis of length 6 and a conjugate axis of length 4. We see that the circle and the hyperbola have no points in common. 9 − y2 y 1 −1 −3 −2 −1 1 2 3 x −3 −2 −1 y 1 −1 1 2 3 x Graphs for x2 + y2 = 4 4x2 + 9y2 = 36 Graphs for x2 + y2 = 4 4x2 − 9y2 = 36 √ 3. Since there are no like terms among the two equations, elimination won’t do us any good. We turn to substitution and from the equation y − 2x = 0, we get y = 2x. Substituting this √ into x2 + y2 = 4 gives x2 + (2x)2 = 4. Solving, we ﬁnd 5x2 = 4 or x = ± 2 5 5 . Returning √ 5 when x = 2 5 to the equation we used for the substitution, y = 2x, we ﬁnd y = 4 5 , so one solution is . We . Similarly, we ﬁnd the other solution to be leave it to the reader that both points satisfy both equations, so that our ﬁnal answer is 2 . The graph of x2 + y2 = 4 is our circle from before and the graph of y − 2x = 0 is a line through the origin with slope 2. Though we cannot verify the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required. While it may be tempting to solve y − x2 = 0 as y = x2 and substitute, we note that this system is set up for elimination.1 (E1) x2 + y2 = 4 y − x2 = 0 (E2) Replace E2 with −−−−−−−−−−→ E1 + E2 (E1) x2 + y2 = 4 y2 + y = 4 (E2) From y2 + y = 4 we get y2 + y − 4 = 0 which gives y = −1± . Due to the complicated 2 nature of these answers, it is worth our time to make a quick sketch of both equations to head oﬀ any extraneous solutions we may encounter. We see that the circle x2 + y2 = 4 intersects the parabola y = x2 exactly twice, and both of these points have a positive y value. Of the two solutions for y, only y = −1+ is positive, so to get our solution, we substitute this 2 17 17 √ √ 1We encourage the reader to solve the system using substitution to see that you get the same solution. 8.7 Systems of Non-Linear Equations and Inequalities 639 into y − x2 = 0 and solve for x. We get 2+2 2 17 , −1+ 2 17 , − −2+2 2 17 , −1+ 2 17 √ −1+ 2 17 = ± √ √ −2+2 2 17 . Our solution is , which we leave to the reader to verify. y 1 −3 −2 −1 1 2 3 x −3 −2 −1 y 1 −1 1 2 3 x Graphs for x2 + y2 = 4 y − 2x = 0 Graphs for x2 + y2 = 4 y − x2 = 36 A couple of remarks about Example 8.7.1 are in order. First note that, unlike systems of linear equations, it is possible for a system of non-linear equations to have more than one solution without having inﬁnitely many solutions. In fact, while we characterize systems of nonlinear equations as being ‘consistent’ or ‘inconsistent,’ we generally don’t use the labels ‘dependent’ or ‘independent’. Secondly, as we saw with number 4, sometimes making a quick sketch of the problem situation can save a lot of time and eﬀort. While in general the curves in a system of non-linear equations may not be easily visualized, it sometimes pays to take advantage when they are. Our next example provides some considerable review of many of the topics introduced in this text. Example 8.7.2. Solve the following systems of equations. Verify your answers algebraically and graphically, as appropriate. x2 + 2xy − 16 = 0 y2 + 2xy − 16 = 0 1. y + 4e2x = 1 y2 + 2ex = 1 2.    3. z(x − 2) = x yz = y (x − 2)2 + y2 = 1 Solution. 1. At ﬁrst glance, it doesn’t appear as though elimination will do us any good since it’s clear that we cannot completely eliminate one of the variables. The alternative, solving one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it is possible to eliminate the troublesome xy term, and the constant term as well, by elimination and doing so we get a more tractable relationship between x and y (E1) x2 + 2xy − 16 = 0 (E2) y2 + 2xy − 16 = 0 Replace E2 with −−−−−−−−−−→ −E1 + E2 (E1) x2 + 2xy − 16 = 0 y2 − x2 = 0 (E2) 640 Systems of Equations and Matrices √ √ 4 3 or x = ± 4 We get y2 − x2 = 0 or y = ±x. Substituting y = x into E1 we get x2 + 2x2 − 16 = 0 so √ 3 that x2 = 16 3 . On the other hand, when we substitute y = −x into E1, we get √ 3 x2 − 2x2 − 16 = 0 or x2 = −16 which gives no real solutions. Substituting each of x = ± 4 3 into the substitution equation y = x yields the solution . We leave it to the reader to show that both points satisfy both equations and now turn to verifying our solution graphically. We begin by solving x2+2xy−16 = 0 for y to obtain y = 16−x2 2x . This function is easily graphed using the techniques of Section 4.2. Solving the second equation, y2 + 2xy − 16 = 0, for y, however, is more complicated. We use the quadratic formula to x2 + 16 which would require the use of Calculus or a calculator to graph. obtain y = −x ± Believe it or not, we don’t need either because the equation y2 + 2xy − 16 = 0 can be obtained from the equation x2 + 2xy − 16 = 0 by interchanging y and x. Thinking back to Section 5.2, this means we can obtain the graph of y2 + 2xy − 16 = 0 by reﬂecting the graph of x2 + 2xy − 16 = 0 across the line y = x. Doing so conﬁrms that the two graphs intersect twice: once in Quadrant I, and once in Quadrant III as required4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 The graphs of x2 + 2xy − 16 = 0 and y2 + 2xy − 16 = 0 2. Unlik
e the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving y + 4e2x = 1 for y, we get y = 1 − 4e2x which, when substituted into the second equation, yields 1 − 4e2x2 + 2ex = 1. After expanding and gathering like terms, we get 16e4x − 8e2x + 2ex = 0. Factoring gives us 2ex 8e3x − 4ex + 1 = 0, and since 2ex = 0 for any real x, we are left with solving 8e3x − 4ex + 1 = 0. We have three terms, and even though this is not a ‘quadratic in disguise’, we can beneﬁt from the substitution u = ex. The equation becomes 8u3−4u+1 = 0. Using the techniques set forth in Section 3.3, we ﬁnd u = 1 2 8u2 + 4u − 2. We is a zero and use synthetic division to factor the left hand side as u − 1 2 √ use the quadratic formula to solve 8u2 + 4u − 2 = 0 and ﬁnd u = −1± 5 . Since u = ex, we 4 = − ln(2). As now must solve ex = 1 for ex = −1± has no real solutions. We are 4 2 and ex = −1± 4 , we ﬁrst note that −1− . From ex = 1 4 < 0, so ex = −1− 2 , we get x = ln .7 Systems of Non-Linear Equations and Inequalities 641 left with ex = −1+ 4 accompanying y values for each of our solutions for x. For x = − ln(2), we get , so that x = ln . We now return to y = 1 − 4e2x to ﬁnd the 5 5 √ −1+ 4 √ y = 1 − 4e2x = 1 − 4e−2 ln(2) = 1 − 4eln For x = ln √ −1+ 4 5 , we have y = 1 − 4e2x 2 ln = 1 − 4e √ 5 −1+ 4 √ 2 5 2 5 ln −1+ 4 √ −1+ 4 √ 3− 8 5 = 1 − 4e = 1+ 2 5 √ 5 −1+ 4 , −1+ 2 √ 5 (0, − ln(2)), We get two solutions, . It is a good review of the properties of logarithms to verify both solutions, so we leave that to the reader. We are able to sketch y = 1 − 4e2x using transformations, but the second equation is more diﬃcult and we resort to the calculator. We note that to graph y2 + 2ex = 1, we need to graph both the positive and negative roots, y = ± 1 − 2ex. After some careful zooming,2 we get ln √ The graphs of y = 1 − 4e2x and y = ± √ 1 − 2ex. 3. Our last system involves three variables and gives some insight on how to keep such systems organized. Labeling the equations as before, we have 2The calculator has trouble conﬁrming the solution (− ln(2), 0) due to its issues in graphing square root functions. If we mentally connect the two branches of the thicker curve, we see the intersection. 642 Systems of Equations and Matrices    z(x − 2) = x E1 E2 yz = y E3 (x − 2)2 + y2 = 1 The easiest equation to start with appears to be E2. While it may be tempting to divide both sides of E2 by y, we caution against this practice because it presupposes y = 0. Instead, we take E2 and rewrite it as yz − y = 0 so y(z − 1) = 0. From this, we get two cases: y = 0 or z = 1. We take each case in turn. Case 1: y = 0. Substituting y = 0 into E1 and E3, we get E1 z(x − 2) = x E3 (x − 2)2 = 1 Solving E3 for x gives x = 1 or x = 3. Substituting these values into E1 gives z = −1 when x = 1 and z = 3 when x = 3. We obtain two solutions, (1, 0, −1) and (3, 0, 3). Case 2: z = 1. Substituting z = 1 into E1 and E3 gives us E1 (1)(x − 2) = x E3 (1 − 2)2 + y2 = 1 Equation E1 gives us x − 2 = x or −2 = 0, which is a contradiction. This means we have no solution to the system in this case, even though E3 is solvable and gives y = 0. Hence, our ﬁnal answer is {(1, 0, −1), (3, 0, 3)}. These points are easy enough to check algebraically in our three original equations, so that is left to the reader. As for verifying these solutions graphically, they require plotting surfaces in three dimensions and looking for intersection points. While this is beyond the scope of this book, we provide a snapshot of the graphs of our three equations near one of the solution points, (1, 0, −1). Example 8.7.2 showcases some of the ingenuity and tenacity mentioned at the beginning of the section. Sometimes you just have to look at a system the right way to ﬁnd the most eﬃcient method to solve it. Sometimes you just have to try something. 8.7 Systems of Non-Linear Equations and Inequalities 643 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we ﬁrst introduced in Section 2.4. Before we embark on some examples, a little motivation is in order. Suppose we wish to solve x2 < 4 − y2. If we mimic the algorithms for solving nonlinear inequalities in one variable, we would gather all of the terms on one side and leave a 0 on the other to obtain x2 + y2 − 4 < 0. Then we would ﬁnd the zeros of the left hand side, that is, where is x2 + y2 − 4 = 0, or x2 + y2 = 4. Instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve, in this case, a circle, which divides the plane into two regions - the ‘inside’ and ‘outside’ of the circle - with the circle itself as the boundary between the two. Just like we used test values to determine whether or not an interval belongs to the solution of the inequality, we use test points in the each of the regions to see which of these belong to our solution set.3 We choose (0, 0) to represent the region inside the circle and (0, 3) to represent the points outside of the circle. When we substitute (0, 0) into x2 + y2 − 4 < 0, we get −4 < 4 which is true. This means (0, 0) and all the other points inside the circle are part of the solution. On the other hand, when we substitute (0, 3) into the same inequality, we get 5 < 0 which is false. This means (0, 3) along with all other points outside the circle are not part of the solution. What about points on the circle itself? Choosing a point on the circle, say (0, 2), we get 0 < 0, which means the circle itself does not satisfy the inequality.4 As a result, we leave the circle dashed in the ﬁnal diagram. y 2 −2 2 x −2 The solution to x2 < 4 − y2 We put this technique to good use in the following example. Example 8.7.3. Sketch the solution to the following nonlinear inequalities in the plane. 1. y2 − 4 ≤ x < y + 2 Solution. 2. x2 + y2 ≥ 4 x2 − 2x + y2 − 2y ≤ 0 1. The inequality y2 − 4 ≤ x < y + 2 is a compound inequality. It translates as y2 − 4 ≤ x and x < y + 2. As usual, we solve each inequality and take the set theoretic intersection to determine the region which satisﬁes both inequalities. To solve y2 − 4 ≤ x, we write 3The theory behind why all this works is, surprisingly, the same theory which guarantees that sign diagrams work the way they do - continuity and the Intermediate Value Theorem - but in this case, applied to functions of more than one variable. 4Another way to see this is that points on the circle satisfy x2 + y2 − 4 = 0, so they do not satisfy x2 + y2 − 4 < 0. 644 Systems of Equations and Matrices y2 − x − 4 ≤ 0. The curve y2 − x − 4 = 0 describes a parabola since exactly one of the variables is squared. Rewriting this in standard form, we get y2 = x + 4 and we see that the vertex is (−4, 0) and the parabola opens to the right. Using the test points (−5, 0) and (0, 0), we ﬁnd that the solution to the inequality includes the region to the right of, or ‘inside’, the parabola. The points on the parabola itself are also part of the solution, since the vertex (−4, 0) satisﬁes the inequality. We now turn our attention to x < y + 2. Proceeding as before, we write x − y − 2 < 0 and focus our attention on x − y − 2 = 0, which is the line y = x − 2. Using the test points (0, 0) and (0, −4), we ﬁnd points in the region above the line y = x − 2 satisfy the inequality. The points on the line y = x − 2 do not satisfy the inequality, since the y-intercept (0, −2) does not. We see that these two regions do overlap, and to make the graph more precise, we seek the intersection of these two curves. That is, we need to solve the system of nonlinear equations (E1) y2 = x + 4 y = x − 2 (E2) Solving E1 for x, we get x = y2 − 4. Substituting this into E2 gives y = y2 − 4 − 2, or y2 − y − 6 = 0. We ﬁnd y = −2 and y = 3 and since x = y2 − 4, we get that the graphs intersect at (0, −2) and (5, 3). Putting all of this together, we get our ﬁnal answer below. y 3 y y −5−4 x 2 3 4 5 x −5−4 2 3 4 5 x −3 −3 −3 y2 − 4 ≤ x x < y + 2 y2 − 4 ≤ x < y + 2 2. To solve this system of inequalities, we need to ﬁnd all of the points (x, y) which satisfy both inequalities. To do this, we solve each inequality separately and take the set theoretic intersection of the solution sets. We begin with the inequality x2 + y2 ≥ 4 which we rewrite as x2 + y2 − 4 ≥ 0. The points which satisfy x2 + y2 − 4 = 0 form our friendly circle x2 + y2 = 4. Using test points (0, 0) and (0, 3) we ﬁnd that our solution comprises the region outside the circle. As far as the circle itself, the point (0, 2) satisﬁes the inequality, so the circle itself is part of the solution set. Moving to the inequality x2 − 2x + y2 − 2y ≤ 0, we start with x2 − 2x + y2 − 2y = 0. Completing the squares, we obtain (x − 1)2 + (y − 1)2 = 2, which is 2. Choosing (1, 1) to represent the inside of the a circle centered at (1, 1) with a radius of circle, (1, 3) as a point outside of the circle and (0, 0) as a point on the circle, we ﬁnd that the solution to the inequality is the inside of the circle, including the circle itself. Our ﬁnal answer, then, consists of the points on or outside of the circle x2 + y2 = 4 which lie on or √ 8.7 Systems of Non-Linear Equations and Inequalities 645 inside the circle (x − 1)2 + (y − 1)2 = 2. To produce the most accurate graph, we need to ﬁnd where these circles intersect. To that end, we solve the system (E1) x2 + y2 = 4 (E2) x2 − 2x + y2 − 2y = 0 We can eliminate both the x2 and y2 by replacing E2 with −E1 + E2. Doing so produces −2x − 2y = −4. Solving this for y, we get y = 2 − x. Substituting this into E1 gives x2 + (2 − x)2 = 4 which simpliﬁes to x2 + 4 − 4x + x2 = 4 or 2x2 − 4x = 0. Factoring yields 2x(x − 2) which gives x = 0 or x = 2. Substituting these values into y = 2 − x gives the points (0, 2) and (2, 0). The intermediate graphs and ﬁnal solution are below. y 1 −3 −2 −1 2 x −3 −2 −1 2 x −1 −2 −3 −1 −2 −3 x2 + y2 ≥ 4 x2 − 2x + y2 − 2y ≤ 0 Solution to the system. 646 Systems of Equations and Mat
rices 8.7.1 Exercises In Exercises 1 - 6, solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. x2 − y = 4 x2 + y2 = 4 x2 + y2 = 16 9x2 − 16y2 = 144 1. 4. 2. 5. x2 + y2 = 4 x2 − y = 5 x2 + y2 = 16 16 x2 = 1 9 y2 − 1 1 x2 + y2 = 16 16x2 + 4y2 = 64 x2 + y2 = 16 x − y = 2 3. 6. In Exercises 9 - 15, solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. 7. 10. x2 − y2 = 1 x2 + 4y2 = 4 (x − 2)2 + y2 = 1 x2 + 4y2 = 4 √ x + 1 − y = 0 x2 + 4y2 = 4 x2 + y2 = 25 y − x = 1 8. 11. 13. y = x3 + 8 y = 10x − x2 14. x2 − xy = 8 y2 − xy = 8 x + 2y2 = 2 x2 + 4y2 = 4 9. 12. 15.    x2 + y2 = 25 x2 + (y − 3)2 = 10 x2 + y2 = 25 4x2 − 9y = 0 3y2 − 16x = 0 16. A certain bacteria culture follows the Law of Uninbited Growth, Equation 6.4. After 10 minutes, there are 10,000 bacteria. Five minutes later, there are 14,000 bacteria. How many bacteria were present initially? How long before there are 50,000 bacteria? Consider the system of nonlinear equations below    1 If we let u = 1 x and v = 1 y then the system becomes 4u + 3v = 1 3u + 2v = −1 This associated system of linear equations can then be solved using any of the techniques presented earlier in the chapter to ﬁnd that u = −5 and v = 7. Thus x = 1 We say that the original system is linear in form because its equations are not linear but a few substitutions reveal a structure that we can treat like a system of linear equations. Each system in Exercises 17 - 19 is linear in form. Make the appropriate substitutions and solve for x and y. 5 and .7 Systems of Non-Linear Equations and Inequalities 647 4x3 + 3 3x3 + 2 √ √ y = 1 y = −1 17. 18. 4ex + 3e−y = 1 3ex + 2e−y = −1 19. 4 ln(x) + 3y2 = 1 3 ln(x) + 2y2 = −1 20. Solve the following system    √ x2 + √ 3x2 − 2 −5x2 + 3 √ y + log2(z) = 6 y + 2 log2(z) = 5 y + 4 log2(z) = 13 In Exercises 21 - 26, sketch the solution to each system of nonlinear inequalities in the plane. 21. 23. 25. x2 − y2 ≤ 1 x2 + 4y2 ≥ 4 (x − 2)2 + y2 < 1 x2 + 4y2 < 4 x + 2y2 > 2 x2 + 4y2 ≤ 4 22. 24. 26. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y < y > 10x − x2 x3 + 8 x2 + y2 ≥ 25 y − x ≤ 1 27. Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to ﬁnd the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables x, y and z are the dimensions of the box and λ is called a Lagrange multiplier. With the help of your classmates, solve the system.5    2y + 2z = λyz 2x + 2z = λxz 2y + 2x = λxy xyz = 1000 28. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors. In this exercise, we present a method for obtaining that factorization. (a) Show that p has no real zeros. (b) Because p has no real zeros, its factorization must be of the form (x2 +ax+b)(x2 +cx+d) where each factor is an irreducible quadratic. Expand this quantity and gather like terms together. (c) Create and solve the system of nonlinear equations which results from equating the coeﬃcients of the expansion found above with those of x4 + 4. You should get four equations in the four unknowns a, b, c and d. Write p(x) in factored form. 29. Factor q(x) = x4 + 6x2 − 5x + 6. 5If using λ bothers you, change it to w when you solve the system. 648 Systems of Equations and Matrices 8.7.2 Answers √ 1. (±2, 0), ± y 3, −1 2 1 −2 −1 1 2 x −1 −2 −3 −4 3. (0, ±4) y 4 3 2 1 2. No solution y 2 1 −2 −1 1 2 x −1 −2 −3 −4 4. (±4, 0) y 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −2 −3 −4 √ ± 4 5. 7 5 , ± 12 √ 5 y 2 √ 6. 1 + 7, −1 + y √ 7, 1 − √ 7, −1 − √ 7 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −2 −3 −4 √ 15 5 , ± √ 7. ± 2 10 5 √ 104−3−2−1 −1 1 2 3 4 x −2 −3 −4 8. (0, 1) 9. (0, ±1), (2, 0) 11. (3, 4), (−4, −3) 12. (±3, 4) 13. (−4, −56), (1, 9), (2, 16) 14. (−2, 2), (2, −2) 15. (3, 4) 16. Initially, there are 250000 49 ≈ 5102 bacteria. It will take 5 ln(49/5) ln(7/5) ≈ 33.92 minutes for the colony to grow to 50,000 bacteria. 8.7 Systems of Non-Linear Equations and Inequalities 649 √ 17. − 3 5, 49 20. (1, 4, 8), (−1, 4, 8) 21. x2 − y2 ≤ 1 x2 + 4y2 ≥ 4 y 2 1 −2 −1 1 2 x −1 −2 23. (x − 2)2 + y2 < 1 x2 + 4y2 < 4 y 1 −1 1 2 x 25. x + 2y2 > 2 x2 + 4y2 ≤ 4 y 1 −1 18. No solution 19. e−5, ± √ 7 22. x2 + y2 < 25 x2 + (y − 3)2 ≥ 10 y 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 24. y > 10x − x2 x3 + 8 y < y 16 9 −4−3−2−1 1 2 x −56 26. x2 + y2 ≥ 25 5 −3 −1 1 3 5 x −3 −5 650 Systems of Equations and Matrices 27. x = 10, y = 10, z = 10, λ = 2 5 28. (c) x4 + 4 = (x2 − 2x + 2)(x2 + 2x + 2) 29. x4 + 6x2 − 5x + 6 = (x2 − x + 1)(x2 + x + 6) Chapter 9 Sequences and the Binomial Theorem 9.1 Sequences When we ﬁrst introduced a function as a special type of relation in Section 1.3, we did not put any restrictions on the domain of the function. All we said was that the set of x-coordinates of the points in the function F is called the domain, and it turns out that any subset of the real numbers, regardless of how weird that subset may be, can be the domain of a function. As our exploration of functions continued beyond Section 1.3, we saw fewer and fewer functions with ‘weird’ domains. It is worth your time to go back through the text to see that the domains of the polynomial, rational, exponential, logarithmic and algebraic functions discussed thus far have fairly predictable domains which almost always consist of just a collection of intervals on the real line. This may lead some readers to believe that the only important functions in a College Algebra text have domains which consist of intervals and everything else was just introductory nonsense. In this section, we introduce sequences which are an important class of functions whose domains are the set of natural numbers.1 Before we get to far ahead of ourselves, let’s look at what the term ‘sequence’ means mathematically. Informally, we can think of a sequence as an inﬁnite list of numbers. For example, consider the sequence 1 2 , − 3 4 , 9 8 , − 27 16 , . . . (1) As usual, the periods of ellipsis, . . ., indicate that the proposed pattern continues forever. Each of the numbers in the list is called a term, and we call 1 8 the ‘third term’ and so forth. In numbering them this way, we are setting up a function, which we’ll call a per tradition, between the natural numbers and the terms in the sequence. 4 the ‘second term’, 9 2 the ‘ﬁrst term’, − 3 1Recall that this is the set {1, 2, 3, . . .}. 652 Sequences and the Binomial Theorem 1 2 n a(n − 27 16 ... ... In other words, a(n) is the nth term in the sequence. We formalize these ideas in our deﬁnition of a sequence and introduce some accompanying notation. Deﬁnition 9.1. A sequence is a function a whose domain is the natural numbers. The value a(n) is often written as an and is called the nth term of the sequence. The sequence itself is usually denoted using the notation: an, n ≥ 1 or the notation: {an}∞ n=1. 2 , a2 = − 3 Applying the notation provided in Deﬁnition 9.1 to the sequence given (1), we have a1 = 1 4 , a3 = 9 8 and so forth. Now suppose we wanted to know a117, that is, the 117th term in the sequence. While the pattern of the sequence is apparent, it would beneﬁt us greatly to have an explicit formula for an. Unfortunately, there is no general algorithm that will produce a formula for every sequence, so any formulas we do develop will come from that greatest of teachers, experience. In other words, it is time for an example. Example 9.1.1. Write the ﬁrst four terms of the following sequences. 1. an = 5n−1 3n , n ≥ 1 3. {2n − 1}∞ n=1 , k ≥ 0 2. bk = (−1)k 2k + 1 1 + (−1)i i 4. ∞ i=2 5. a1 = 7, an+1 = 2 − an, n ≥ 1 6. f0 = 1, fn = n · fn−1, n ≥ 1 Solution. 32 = 5 1. Since we are given n ≥ 1, the ﬁrst four terms of the sequence are a1, a2, a3 and a4. Since the notation a1 means the same thing as a(1), we obtain our ﬁrst term by replacing every occurrence of n in the formula for an with n = 1 to get a1 = 51−1 3 . Proceeding similarly, we get a2 = 52−1 33 = 25 27 and a4 = 54−1 2. For this sequence we have k ≥ 0, so the ﬁrst four terms are b0, b1, b2 and b3. Proceeding as before, replacing in this case the variable k with the appropriate whole number, beginning with 0, we get b0 = (−1)0 2(3)+1 = − 1 7 . (This sequence is called an alternating sequence since the signs alternate between + and −. The reader is encouraged to think what component of the formula is producing this eﬀect.) 2(0)+1 = 1, b1 = (−1)1 5 and b3 = (−1)3 3 , b2 = (−1)2 9 , a3 = 53−1 2(1)+1 = − 1 2(2)+1 = 1 34 = 125 81 . 31 = 1 9.1 Sequences 653 3. From {2n − 1}∞ n=1, we have that an = 2n − 1, n ≥ 1. We get a1 = 1, a2 = 3, a3 = 5 and a4 = 7. (The ﬁrst four terms are the ﬁrst four odd natural numbers. The reader is encouraged to examine whether or not this pattern continues indeﬁnitely.) 4. Here, we are using the letter i as a counter, not as the imaginary unit we saw in Section 3.4. 2 and a5 = 0. Proceeding as before, we set ai = 1+(−1)i , i ≥ 2. We ﬁnd a2 = 1, a3 = 0, a4 = 1 i 5. To obtain the terms of this sequence, we start with a1 = 7 and use the equation an+1 = 2−an for n ≥ 1 to generate successive terms. When n = 1, this equation becomes a1 + 1 = 2 − a1 which simpliﬁes to a2 = 2−a1 = 2−7 = −5. When n = 2, the equation becomes a2 + 1 = 2−a2 so we get a3 = 2 − a2 = 2 − (−5) = 7. Finally, when n = 3, we get a3 + 1 = 2 − a3 so a4 = 2 − a3 = 2 − 7 = −5. 6. As with the problem above, we are given a place to start with f0 = 1 and given a formula to build other terms of the sequence. Substituting n = 1 into the equation fn = n · fn−1, we get f1 = 1 · f0 = 1 · 1 = 1. Advancing to n = 2, we get f2 = 2 · f1 = 2 · 1 = 2. Finally, f3 = 3 · f2 = 3 · 2 = 6. Some remarks about Example 9.1.1 are in order. We ﬁrst note that since sequences are functions, we can graph them
in the same way we graph functions. For example, if we wish to graph the sequence {bk}∞ k=0 from Example 9.1.1, we graph the equation y = b(k) for the values k ≥ 0. That is, we plot the points (k, b(k)) for the values of k in the domain, k = 0, 1, 2, . . .. The resulting collection of points is the graph of the sequence. Note that we do not connect the dots in a pleasing fashion as we are used to doing, because the domain is just the whole numbers in this case, not a collection of intervals of real numbers. If you feel a sense of nostalgia, you should see Section 1.21 − 3 2 1 2 3 x Graphing y = bk = (−1)k 2k + 1 , k ≥ 0 Speaking of {bk}∞ k=0, the astute and mathematically minded reader will correctly note that this technically isn’t a sequence, since according to Deﬁnition 9.1, sequences are functions whose domains are the natural numbers, not the whole numbers, as is the case with {bk}∞ k=0. In other words, to satisfy Deﬁnition 9.1, we need to shift the variable k so it starts at k = 1 instead of k = 0. To see how we can do this, it helps to think of the problem graphically. What we want is to shift the graph of y = b(k) to the right one unit, and thinking back to Section 1.7, we can accomplish this by replacing k with k − 1 in the deﬁnition of {bk}∞ k=0. Speciﬁcally, let ck = bk−1 where k − 1 ≥ 0. We get ck = (−1)k−1 2k−1 , where now k ≥ 1. We leave to the reader to verify that {ck}∞ k=0, but the former satisﬁes Deﬁnition 2(k−1)+1 = (−1)k−1 k=1 generates the same list of numbers as does {bk}∞ 654 Sequences and the Binomial Theorem 9.1, while the latter does not. Like so many things in this text, we acknowledge that this point is pedantic and join the vast majority of authors who adopt a more relaxed view of Deﬁnition 9.1 to include any function which generates a list of numbers which can then be matched up with the natural numbers.2 Finally, we wish to note the sequences in parts 5 and 6 are examples of sequences described recursively. In each instance, an initial value of the sequence is given which is then followed by a recursion equation − a formula which enables us to use known terms of the sequence to determine other terms. The terms of the sequence in part 6 are given a special name: fn = n! is called n-factorial. Using the ‘!’ notation, we can describe the factorial sequence as: 0! = 1 and n! = n(n − 1)! for n ≥ 1. After 0! = 1 the next four terms, written out in detail, are 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 · 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24. From this, we see a more informal way of computing n!, which is n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as a special case. (We will study factorials in greater detail in Section 9.4.) The world famous Fibonacci Numbers are deﬁned recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight into what kinds of patterns to look for. Two patterns in particular are given in the next deﬁnition. Deﬁnition 9.2. Arithmetic and Geometric Sequences: Suppose {an}∞ n=k is a sequencea If there is a number d so that an+1 = an + d for all n ≥ k, then {an}∞ arithmetic sequence. The number d is called the common diﬀerence. n=k is called an If there is a number r so that an+1 = ran for all n ≥ k, then {an}∞ n=k is called a geometric sequence. The number r is called the common ratio. aNote that we have adjusted for the fact that not all ‘sequences’ begin at n = 1. Both arithmetic and geometric sequences are deﬁned in terms of recursion equations. In English, an arithmetic sequence is one in which we proceed from one term to the next by always adding the ﬁxed number d. The name ‘common diﬀerence’ comes from a slight rewrite of the recursion equation from an+1 = an + d to an+1 − an = d. Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same ﬁxed number r. If r = 0, we can rearrange the recursion equation to get an+1 = r, hence the name ‘common ratio.’ Some an sequences are arithmetic, some are geometric and some are neither as the next example illustrates.3 Example 9.1.2. Determine if the following sequences are arithmetic, geometric or neither. arithmetic, ﬁnd the common diﬀerence d; if geometric, ﬁnd the common ratio r. If 1. an = 5n−1 3n , n ≥ 1 3. {2n − 1}∞ n=1 2. bk = (−1)k 2k + 1 , k ≥ 0 4 , − 27 16 , . . . 2We’re basically talking about the ‘countably inﬁnite’ subsets of the real number line when we do this. 3Sequences which are both arithmetic and geometric are discussed in the Exercises. 9.1 Sequences 655 Solution. A good rule of thumb to keep in mind when working with sequences is “When in doubt, write it out!” Writing out the ﬁrst several terms can help you identify the pattern of the sequence should one exist. 1. From Example 9.1.1, we know that the ﬁrst four terms of this sequence are 1 27 and 125 81 . To see if this is an arithmetic sequence, we look at the successive diﬀerences of terms. We ﬁnd that a2 − a1 = 5 27 . Since we get diﬀerent numbers, there is no ‘common diﬀerence’ and we have established that the sequence is not arithmetic. To investigate whether or not it is geometric, we compute the ratios of successive terms. The ﬁrst three ratios 9 and a3 − a2 = 25 9 = 10 27 − 5 3 = 2 9 − 1 9 , 25 3 , 5 a2 a1 = 5 9 1 3 = 5 3 , a3 a2 = = 5 3 and a4 a3 = 25 27 5 9 125 81 25 27 = 5 3 suggest that the sequence is geometric. To prove it, we must show that an+1 an = r for all n. an+1 an = 5(n+1)−1 3n+1 5n−1 3n = 5n 3n+1 · 3n 5n−1 = 5 3 This sequence is geometric with common ratio r = 5 3 . 3 , 1 2. Again, we have Example 9.1.1 to thank for providing the ﬁrst four terms of this sequence: 15 . Hence, the sequence is not 5 . Since there is no 1, − 1 3 and b2 − b1 = 8 3 and b2 arithmetic. To see if it is geometric, we compute b1 b1 b0 ‘common ratio,’ we conclude the sequence is not geometric, either. 7 . We ﬁnd b1 − b0 = − 4 5 and − 1 = − 1 = − 3 3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ n=1 generates the odd numbers: 1, 3, 5, 7, . . .. Computing the ﬁrst few diﬀerences, we ﬁnd a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This suggests that the sequence is arithmetic. To verify this, we ﬁnd an+1 − an = (2(n + 1) − 1) − (2n − 1) = 2n + 2 − 1 − 2n + 1 = 2 This establishes that the sequence is arithmetic with common diﬀerence d = 2. To see if it is geometric, we compute a2 3 . Since these ratios are diﬀerent, we conclude the a1 sequence is not geometric. = 3 and a3 a2 = 5 4. We met our last sequence at the beginning of the section. Given that a2 − a1 = − 5 4 and a3 − a2 = 15 8 , the sequence is not arithmetic. Computing the ﬁrst few ratios, however, gives us 2 , a3 a2 = − 3 2 . Since these are the only terms given to us, we assume that a2 a1 the pattern of ratios continue in this fashion and conclude that the sequence is geometric. 2 and a4 a3 = − 3 = − 3 We are now one step away from determining an explicit formula for the sequence given in (1). We know that it is a geometric sequence and our next result gives us the explicit formula we require. 656 Sequences and the Binomial Theorem Equation 9.1. Formulas for Arithmetic and Geometric Sequences: An arithmetic sequence with ﬁrst term a and common diﬀerence d is given by an = a + (n − 1)d, n ≥ 1 A geometric sequence with ﬁrst term a and common ratio r = 0 is given by an = arn−1, n ≥ 1 While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in Section 9.3, we attempt to motivate them here. According to Deﬁnition 9.2, given an arithmetic sequence with ﬁrst term a and common diﬀerence d, the way we get from one term to the next is by adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d, . . . . We see that to reach the nth term, we add d to a exactly (n − 1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, we start with a and go from one term to the next by multiplying by r. We get a, ar, ar2, ar3 and so forth. The nth term results from multiplying a by r exactly (n − 1) times. We note here that the reason r = 0 is excluded from Equation 9.1 is to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we ﬁnally have the tools required to ﬁnd an explicit formula for the nth term of the sequence given in (1). We know from Example 9.1.2 that it is geometric with common ratio r = − 3 2 . The ﬁrst term is a = 1 for n ≥ 1. After a touch of simplifying, we get an = (−3)n−1 for n ≥ 1. Note that we can easily check our answer by substituting in values of n and seeing that the formula generates the sequence given in (1). We leave this to the reader. Our next example gives us more practice ﬁnding patterns. 2 so by Equation 9.1 we get an = arn−1 = 1 − 3 2 n−1 2n 2 Example 9.1.3. Find an explicit formula for the nth term of the following sequences. 1. 0.9, 0.09, 0.009, 0.0009, . . . 2. 2 5 , 2. 1, − 2 7 , 4 13 , − 8 19 , . . . Solution. 1. Although this sequence may seem strange, the reader can verify it is actually a geometric for 10n , n ≥ 1. There is more to this sequence than meets the sequence with common ratio r = 0.1 = 1 n ≥ 0. Simplifying, we get an = 9 eye and we shall return to this example in the next section. 10 . With a = 0.9 = 9 10 , we get an = 9 1 10 n−1 10 2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to ﬁnd a pattern, we rewrite the second term with a denominator to make all the terms appear as fractions. We have 2 7 , . . .. If we associate the negative ‘−’ of the last two −3 , 2 1 , 2 terms with the denominators we get 2 −7 , . . .. This tells us that we can tentatively sketch out the formula for the sequence as an = 2 where dn is the sequence of denominators. dn 4See the footnotes on page 237 in Section 3.1 and page 418 of Section 6.1. 9.1 Sequences 657 Looking at the denominators 5, 1, −3, −7, . . ., we ﬁnd that they go f
rom one term to the next by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 2 9−4n , n ≥ 1. 3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem to try to get patterns individually for the numerator and denominator. Letting cn and dn denote the sequence of numerators and denominators, respectively, we have an = cn . After dn some experimentation,5 we choose to write the ﬁrst term as a fraction and associate the negatives ‘−’ with the numerators. This yields 1 19 , . . .. The numerators form the sequence 1, −2, 4, −8, . . . which is geometric with a = 1 and r = −2, so we get cn = (−2)n−1, for n ≥ 1. The denominators 1, 7, 13, 19, . . . form an arithmetic sequence with a = 1 and d = 6. Hence, we get dn = 1 + 6(n − 1) = 6n − 5, for n ≥ 1. We obtain our formula for an = cn dn 6n−5 , for n ≥ 1. We leave it to the reader to show that this checks out. = (−2)n−1 13 , −8 1 , −2 7 , 4 While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17, . . ., we would be hard-pressed to ﬁnd a formula for an if we restrict our attention to these two archetypes. We said before that there is no general algorithm for ﬁnding the explicit formula for the nth term of a given sequence, and it is only through experience gained from evaluating sequences from explicit formulas that we learn to begin to recognize number patterns. The pattern 1, 4, 9, 16, . . . is rather recognizable as the squares, so the formula an = n2, n ≥ 1 may not be too hard to determine. With this in mind, it’s possible to see 2, 5, 10, 17, . . . as the sequence 1 + 1, 4 + 1, 9 + 1, 16 + 1, . . ., so that an = n2 + 1, n ≥ 1. Of course, since we are given only a small sample of the sequence, we shouldn’t be too disappointed to ﬁnd out this isn’t the only formula which generates this sequence. For example, consider the sequence deﬁned by bn = − 1 2 n − 5, n ≥ 1. The reader is encouraged to verify that it also produces the terms 2, 5, 10, 17. In fact, it can be shown that given any ﬁnite sample of a sequence, there are inﬁnitely many explicit formulas all of which generate those same ﬁnite points. This means that there will be inﬁnitely many correct answers to some of the exercises in this section.6 Just because your answer doesn’t match ours doesn’t mean it’s wrong. As always, when in doubt, write your answer out. As long as it produces the same terms in the same order as what the problem wants, your answer is correct. Sequences play a major role in the Mathematics of Finance, as we have already seen with Equation 6.2 in Section 6.5. Recall that if we invest P dollars at an annual percentage rate r and compound the interest n times per year, the formula for Ak, the amount in the account after k compounding periods, is Ak = P 1 + r , k ≥ 1. We now spot this as a geometric sequence with ﬁrst term P 1 + r . In retirement planning, it is seldom n the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in the next section once we have developed more mathematical machinery. 1 + r n and common ratio n2 + 25 2 n3 − 31 4 n4 + 5 k−1 k n 5Here we take ‘experimentation’ to mean a frustrating guess-and-check session. 6For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test. 658 Sequences and the Binomial Theorem 9.1.1 Exercises In Exercises 1 - 13, write out the ﬁrst four terms of the given sequence. 1. an = 2n − 1, n ≥ 0 3. {5k − 2}∞ k=1 5. xn n2 ∞ n=1 2. dj = (−1) j(j+1) 2 , j ≥ 1 4. 6. ∞ n=0 ∞ n2 + 1 n + 1 ln(n) n n=1 7. a1 = 3, an+1 = an − 1, n ≥ 1 8. d0 = 12, dm = dm-1 100 , m ≥ 1 9. b1 = 2, bk+1 = 3bk + 1, k ≥ 1 11. a1 = 117, an+1 = 1 an , n ≥ 1 10. c0 = −2, cj = cj-1 (j + 1)(j + 2) , j ≥ 1 12. s0 = 1, sn+1 = xn+1 + sn, n ≥ 0 13. F0 = 1, F1 = 1, Fn = Fn-1 + Fn-2, n ≥ 2 (This is the famous Fibonacci Sequence ) In Exercises 14 - 21 determine if the given sequence is arithmetic, geometric or neither. If it is arithmetic, ﬁnd the common diﬀerence d; if it is geometric, ﬁnd the common ratio r. 14. {3n − 5}∞ n=1 16. 1 3 , 1 6 , 1 12 , 1 24 , . . . 15. an = n2 + 3n + 2, n ≥ 1 3 1 5 17. n−1∞ n=1 18. 17, 5, −7, −19, . . . 19. 2, 22, 222, 2222, . . . 20. 0.9, 9, 90, 900, . . . 21. an = n! 2 , n ≥ 0. In Exercises 22 - 30, ﬁnd an explicit formula for the nth term of the given sequence. Use the formulas in Equation 9.1 as needed. 22. 3, 5, 7, 9, . . . 25. 1, 2 3 , 1 3 , 4 27 , . . . 23. 1 , . . . 26. 1, 1 4 , 1 9 , 1 16 , . . . , . . . 24. 1, 2 3 , 27. x, − 4 5 x3 3 , , 8 7 x5 5 , − x7 7 , . . . 9.1 Sequences 659 28. 0.9, 0.99, 0.999, 0.9999, . . . 29. 27, 64, 125, 216, . . . 30. 1, 0, 1, 0, . . . 31. Find a sequence which is both arithmetic and geometric. (Hint: Start with an = c for all n.) 32. Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms. 33. Thomas Robert Malthus is credited with saying, “The power of population is indeﬁnitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the ﬁrst power in comparison with the second.” (See this webpage for more information.) Discuss this quote with your classmates from a sequences point of view. 34. This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for 30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30th day? What is the total dollar value of the gift you have received? 35. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates, show that a given term of a arithmetic sequence ak, k ≥ 2 is the arithmetic mean of the term immediately preceding, ak−1 it and immediately following it, ak+1. State and prove an analogous result for geometric sequences. 36. Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the nth term of the sequence i, −1, −i, 1, i, . . .. List out the ﬁrst four terms of the matrix sequences we discussed in Exercise 8.3.1 in Section 8.3. 660 Sequences and the Binomial Theorem 9.1.2 Answers 1. 0, 1, 3, 7 3. 3, 8, 13, 18 5. x, x2 4 , x3 9 , x4 16 7. 3, 2, 1, 0 9. 2, 7, 22, 67 11. 117, 1 117 , 117, 1 117 13. 1, 1, 2, 3 14. arithmetic, d = 3 16. geometric, r = 1 2 18. arithmetic, d = −12 20. geometric, r = 10 2. −1, −1, 1, 1 4. 1, 1, 5 3 , 5 2 6. 0, ln(2) 2 , ln(3) 3 , ln(4) 4 8. 12, 0.12, 0.0012, 0.000012 10. −2, − 1 3 , − 1 36 , − 1 720 12. 1, x + 1, x2 + x + 1, x3 + x2 + x + 1 15. neither 17. geometric, r = 1 5 19. neither 21. neither 22. an = 1 + 2n, n ≥ 1 23. an = − 1 2 n−1 , n ≥ 1 24. an = 2n−1 2n−1 , n ≥ 1 25. an = n 3n−1 , n ≥ 1 26. an = 1 n2 , n ≥ 1 27. (−1)n−1x2n−1 2n−1 , n ≥ 1 28. an = 10n−1 10n , n ≥ 1 29. an = (n + 2)3, n ≥ 1 30. an = 1+(−1)n−1 2 , n ≥ 1 9.2 Summation Notation 661 9.2 Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a deﬁnition, which, while intimidating, is meant to make our lives easier. Deﬁnition 9.3. Summation Notation: Given a sequence {an}∞ satisfying k ≤ m ≤ p, the summation from m to p of the sequence {an} is written n=k and numbers m and p p n=m an = am + am+1 + . . . + ap The variable n is called the index of summation. The number m is called the lower limit of summation while the number p is called the upper limit of summation. In English, Deﬁnition 9.3 is simply deﬁning a short-hand notation for adding up the terms of the sequence {an}∞ n=k from am through ap. The symbol Σ is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence an = 2n − 1 for n ≥ 1, we can write the sum a3 + a4 + a5 + a6 as 6 (2n − 1) = (2(3) − 1) + (2(4) − 1) + (2(5) − 1) + (2(6) − 1) n=3 = 5 + 7 + 9 + 11 = 32 The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without aﬀecting the value of the summation. For instance, 6 (2n − 1) = n=3 6 k=3 (2k − 1) = 6 j=3 (2j − 1) One place you may encounter summation notation is in mathematical deﬁnitions. For example, summation notation allows us to deﬁne polynomials as functions of the form f (x) = n k=0 akxk for real numbers ak, k = 0, 1, . . . n. The reader is invited to compare this with what is given in Deﬁnition 3.1. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the ith row Ri of a matrix A = [aij]m×n and the jth column Cj of a matrix B = [bij]n×r as Ri · Cj = n k=1 aikbkj 662 Sequences and the Binomial Theorem Again, the reader is encouraged to write out the sum and compare it to Deﬁnition 8.9. Our next example gives us practice with this new notation. Example 9.2.1. 1. Find the following sums. (a) 4 k=1 13 100k (b) 4 n=0 n! 2 (c) 5 n=1 (−1)n+1 n (x − 1)n 2. Write the following sums using summation notation. (b) 1 − (a) 1 + 3 + 5 + . . . + 117 1 3 1 4 (c) 0.9 + 0.09 + 0.
009 + . . . 0 zeros + − . . . + 1 117 1 2 + − Solution. 1. (a) We substitute k = 1 into the formula 13 100k and add successive terms until we reach k = 4. 4 k=1 13 100k = 13 1002 + 13 1001 + 13 1003 + = 0.13 + 0.0013 + 0.000013 + 0.00000013 = 0.13131313 13 1004 (b) Proceeding as in (a), we replace every occurrence of n with the values 0 through 4. We recall the factorials, n! as deﬁned in number Example 9.1.1, number 6 and get: 4 n=0 n! 2 + + + 2! 2 2 · 1 2 = 4! 3 + 12 1! 2 1 2 1 2 = = 0! 2 1 2 1 2 = 17 = + 4 · 3 · 2 · 1 2 (c) We proceed as before, replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 9.2 Summation Notation 663 5 n=1 (−1)n+1 n (x − 1)n = (−1)1+1 1 (x − 1)1 + (−1)2+1 2 (x − 1)2 + (−1)3+1 3 (x − 1)3 + (−1)1+4 4 = (x − 1) − (x − 1)4 + (x − 1)2 2 + (−1)1+5 5 (x − 1)3 3 (x − 1)5 − (x − 1)4 4 + (x − 1)5 5 2. The key to writing these sums with summation notation is to ﬁnd the pattern of the terms. To that end, we make good use of the techniques presented in Section 9.1. (a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 2. We get a formula for the nth term of the sequence using Equation 9.1 to get an = 1 + (n − 1)2 = 2n − 1, n ≥ 1. At this stage, we have the formula for the terms, namely 2n − 1, and the lower limit of the summation, n = 1. To ﬁnish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of n produces the term 117. Setting an = 117, we get 2n − 1 = 117 or n = 59. Our ﬁnal answer is 1 + 3 + 5 + . . . + 117 = 59 (2n − 1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘−’ with the numerators to get 1 1 + −1 2 + 1 3 + −1 4 + . . . + 1 117 The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1 for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for n ≥ 1. Hence, we get the formula an = (−1)n−1 for our terms, and we ﬁnd the lower and upper limits of summation to be n = 1 and n = 117, respectively. Thus 117 = 117 n=1 (−1)n−1 n (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 9 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n − 1 zeros to the right of the decimal point before the 9, we need a denominator of 10n. Hence, n is 1This is indeed a geometric sequence with ﬁrst term a = 1 and common ratio r = −1. 2It is an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 1. 664 Sequences and the Binomial Theorem the upper limit of summation. Since n is used in the limits of the summation, we need to choose a diﬀerent letter for the index of summation.3 We choose k and get 0.9 + 0.09 + 0.009 + . . . 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, “When in doubt, write it out!” Theorem 9.1. Properties of Summation Notation: Suppose {an} and {bn} are sequences so that the following sums are deﬁned. p n=m p n=m p (an ± bn) = p p bn an ± n=m n=m c an = c p n=m an, for any real number c. j an + p an = an, for any natural number m ≤ j < j + 1 ≤ p. n=m n=m n=j+1 p an = p+r n=m n=m+r an−r, for any whole number r. We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the ﬁrst n terms. To derive a formula for S, we write it out in two diﬀerent ways S = S = (a + (n − 1)d) + (a + (n − 2)d) + . . . + (a + d) + a + . . . + (a + (n − 2)d) + (a + (n − 1)d) (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get 2S = (2a + (n − 1)d) + (2a + (n − 1)d) + . . . + (2a + (n − 1)d) + (2a + (n − 1)d) The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d) so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula 3To see why, try writing the summation using ‘n’ as the index. 9.2 Summation Notation 665 S = n 2 (2a + (n − 1)d) If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a1 + an, we get the formula S = n a1 + an 2 A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n and the average of the ﬁrst and nth terms. To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−1, k ≥ 1, and let S once again denote the sum of the ﬁrst n terms. Comparing S and rS, we get S = a + ar + ar2 + . . . + arn−2 + arn−1 rS = ar + ar2 + . . . + arn−2 + arn−1 + arn Subtracting the second equation from the ﬁrst forces all of the terms except a and arn to cancel out and we get S − rS = a − arn. Factoring, we get S(1 − r) = a (1 − rn). Assuming r = 1, we can divide both sides by the quantity (1 − r) to obtain S = a 1 − rn 1 − r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula S = a1 − an+1 1 − r In the case when r = 1, we get the formula times = n a Our results are summarized below. 666 Sequences and the Binomial Theorem Equation 9.2. Sums of Arithmetic and Geometric Sequences: The sum S of the ﬁrst n terms of an arithmetic sequence ak = a + (k − 1)d for k ≥ 1 is S = n k=1 ak = n a1 + an 2 = n 2 (2a + (n − 1)d) The sum S of the ﬁrst n terms of a geometric sequence ak = ark−1 for k ≥ 1 is 1. S = 2. S = n k=1 n k=1 ak = a1 − an+1 1 − r = a 1 − rn 1 − r , if r = 1. ak = n k=1 a = na, if r = 1. While we have made an honest eﬀort to derive the formulas in Equation 9.2, formal proofs require the machinery in Section 9.3. An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the ﬁrst n natural numbers. The natural numbers themselves are a sequence4 1, 2, 3, . . . which is arithmetic with a = d = 1. Applying Equation 9.2(n + 1) 2 2 = 5050. So, for example, the sum of the ﬁrst 100 natural numbers5 is 100(101) An important application of the geometric sum formula is the investment plan called an annuity. Annuities diﬀer from the kind of investments we studied in Section 6.5 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.6 Suppose you have an account with annual interest rate r which is compounded n times per year. We let i = r n denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the amount in the account after k compounding periods. Then A1 = P , because we have made our ﬁrst deposit at the end of the ﬁrst compounding period and no interest has been earned. During the second compounding period, we earn interest on A1 so that our initial investment has grown to A1(1 + i) = P (1 + i) in accordance with Equation 6.1. When we add our second payment at the end of the second period, we get A2 = A1(1 + i) + P = P (1 + i) + P = P (1 + i) 1 + 1 1 + i The reason for factoring out the P (1 + i) will become apparent in short order. During the third compounding period, we earn interest on A2 which then grows to A2(1 + i). We add our third 4This is the identity function on the natural numbers! 5There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this problem in primary school and devised a very clever solution. 6The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding. 9.2 Summation Notation 667 payment at the end of the third compounding period to obtain A3 = A2(1 + i) + P = P (1 + i) 1 + 1 1 + i (1 + i) + P = P (1 + i)1 + i)2 During the fourth compounding period, A3 grows to A3(1+i), and when we add the fourth payment, we factor out P (1 + i)3 to get A4 = P (1 + i)1 + i)2 + 1 (1 + i)3 This pattern continues so that at the end of the kth compounding, we get Ak = P (1 + i)k−1 + i)2 + . . . + 1 (1 + i)k−1 The sum in the parentheses above is the sum of the ﬁrst k terms of a geometric sequence with a = 1 and r = 1 1+i . Using Equation 9.2, we get 1 + 1 1 + i + 1 (1 + i)2 + . . . + 1 (1 + i)k−1 = 1     Hence, we get 1 − 1 (1 + i)1 + i) 1 − (1 + i)−k i Ak = P (1 + i)k−1 (1 + i) 1 − (1 + i)−k i P (1 + i)k − 1 i = If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity. Equation 9.3. Future Value of an Ordinary Annuity: Suppose an annuity oﬀers an annual interest rate r compounded n times per year. Let i = r n be the interest rate per compounding period. If a deposit P is made at the end of each compounding period, the amount A in the account after t years is given by A = P (1 + i)nt − 1 i The reader is encouraged to substitute i = r n into Equation 9.3 and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit P is made a the beginning of the compounding period instead of at the end, the annuity is called an annuitydue. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader. 668 Sequences and the Binomial Theorem Example 9.2.2. An ordinary annuity oﬀers a 6% annual interest rate, compounded monthly. 1. If monthly payments of $50 are made, ﬁnd the value of the annuity in 30 years. 2. How many years will it take for the annuity to grow to $100,000? Solution. 1. We have r = 0.06 and n = 12 so that i = r n = 0.06 12 = 0.005. With P = 50 and t = 30, A = 50 (1 + 0.005)(12)(30) − 1 0.005 ≈ 5022
5.75 Our ﬁnal answer is $50,225.75. 2. To ﬁnd how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take natural logs of both sides of the equation. 100000 = 50 (1 + 0.005)12t − 1 0.005 10 = (1.005)12t − 1 (1.005)12t = 11 ln (1.005)12t = ln(11) 12t ln(1.005) = ln(11) ln(11) t = 12 ln(1.005) ≈ 40.06 This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see that in just 10 additional years, the value of the annuity nearly doubles. This is a lesson worth remembering. We close this section with a peek into Calculus by considering inﬁnite sums, called series. Consider the number 0.9. We can write this number as 0.9 = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + . . . From Example 9.2.1, we know we can write the sum of the ﬁrst n of these terms as 0. 9 · · · 9 n nines = .9 + 0.09 + 0.009 + . . . 0. 0 · · · 0 n − 1 zeros 9 = n k=1 9 10k Using Equation 9.2, we have 9.2 Summation Notation 669   k=1   1 − n 9 10 9 10k = 1 10n+1 1 10 It stands to reason that 0.9 is the same value of 1 − 1 10n+1 as n → ∞. Our knowledge of exponential 10n+1 → 1. We have expressions from Section 6.1 tells us that just argued that 0.9 = 1, which may cause some distress for some readers.7 Any non-terminating decimal can be thought of as an inﬁnite sum whose denominators are the powers of 10, so the phenomenon of adding up inﬁnitely many terms and arriving at a ﬁnite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series. 10n+1 → 0 as n → ∞, so 1 − 1   = 1 − 1 10n+1 1 − 1 Theorem 9.2. Geometric Series: Given the sequence ak = ark−1 for k ≥ 1, where \|r\| < 1, a + ar + ar2 + . . . = ∞ ark−1 = k=1 If \|r\| ≥ 1, the sum a + ar + ar2 + . . . is not deﬁned. a 1 − r The justiﬁcation of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the ﬁrst n terms of a geometric sequence and examining the formula as n → ∞. Assuming \|r\| < 1 means −1 < r < 1, so rn → 0 as n → ∞. Hence as n → ∞, n k=1 ark−1 = a 1 − rn 1 − r → a 1 − r As to what goes wrong when \|r\| ≥ 1, we leave that to Calculus as well, but will explore some cases in the exercises. 7To make this more palatable, it is usually accepted that 0.3 = 1 3 so that 0.9 = 3 0.3 = 3 1 3 = 1. Feel better? 670 Sequences and the Binomial Theorem 9.2.1 Exercises In Exercises 1 - 8, ﬁnd the value of each sum using Deﬁnition 9.3. 1. 5. 9 (5g + 3) g=4 4 i=1 1 4 (i2 + 1) 2. 6. 8 k=3 1 k 100 (−1)n n=1 3. 7. 5 j=0 5 n=1 2j (n + 1)! n! 4. 8. 2 (3k − 5)xk k=0 3 j=1 5! j! (5 − j)! In Exercises 9 - 16, rewrite the sum using summation notation. 9. 8 + 11 + 14 + 17 + 20 10 11. x − x3 3 + x5 5 − x7 7 13 15 16 + 1 25 − 1 36 12. 1 + 2 + 4 + · · · + 229 14. − ln(3) + ln(4) − ln(5) + · · · + ln(20) 16. 1 2 (x − 5) + 1 4 (x − 5)2 + 1 6 (x − 5)3 + 1 8 (x − 5)4 In Exercises 17 - 28, use the formulas in Equation 9.2 to ﬁnd the sum. 17. 20. 10 n=1 5n + 3 n 10 n=1 1 2 18. 21. 20 n=1 2n − 1 n 5 n=1 3 2 19. 22. 15 k=0 3 − k k 5 k=0 1 4 2 23. 1 + 4 + 7 + . . . + 295 24. 4 + 2 + 0 − 2 − . . . − 146 25. 1 + 3 + 9 + . . . + 2187 26 256 27 256 28. 10 n=1 −2n + n 5 3 In Exercises 29 - 32, use Theorem 9.2 to express each repeating decimal as a fraction of integers. 29. 0.7 30. 0.13 31. 10.159 32. −5.867 9.2 Summation Notation 671 In Exercises 33 - 38, use Equation 9.3 to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly. 33. payments are $300, interest rate is 2.5%, term is 17 years. 34. payments are $50, interest rate is 1.0%, term is 30 years. 35. payments are $100, interest rate is 2.0%, term is 20 years 36. payments are $100, interest rate is 2.0%, term is 25 years 37. payments are $100, interest rate is 2.0%, term is 30 years 38. payments are $100, interest rate is 2.0%, term is 35 years 39. Suppose an ordinary annuity oﬀers an annual interest rate of 2%, compounded monthly, for 30 years. What should the monthly payment be to have $100,000 at the end of the term? 40. Prove the properties listed in Theorem 9.1. 41. Show that the formula for the future value of an annuity due is A = P (1 + i) (1 + i)nt − 1 i 42. Discuss with your classmates what goes wrong when trying to ﬁnd the following sums.8 (a) ∞ k=1 2k−1 ∞ (1.0001)k−1 (b) k=1 ∞ (−1)k−1 (c) k=1 8When in doubt, write them out! 672 Sequences and the Binomial Theorem 9.2.2 Answers 1. 213 5. 17 2 9. 13. 5 (3k + 5) k=1 5 k=1 k + 1 k 17. 305 21. 633 32 25. 3280 29. 7 9 2. 341 280 6. 0 10. 14. 8 k=1 20 k=3 (−1)k−1k (−1)k ln(k) 18. 400 22. 26. 30. 1365 512 255 256 13 99 3. 63 7. 20 11. 15. 4 k=1 6 k=1 (−1)k−1 x2k−1 2k − 1 (−1)k−1 k2 19. −72 4. −5 − 2x + x2 8. 25 12. 16. 20. 30 k=1 4 k=1 2k−1 1 2k (x − 5)k 1023 1024 23. 14652 24. −5396 27. 31. 513 256 3383 333 28. 17771050 59049 32. − 5809 990 33. $76,163.67 34. $20,981.40 35. $29,479.69 36. $38,882.12 37. 49,272.55 38. 60,754.80 39. For $100,000, the monthly payment is ≈ $202.95. 9.3 Mathematical Induction 673 9.3 Mathematical Induction The Chinese philosopher Confucius is credited with the saying, “A journey of a thousand miles begins with a single step.” In many ways, this is the central theme of this section. Here we introduce a method of proof, Mathematical Induction, which allows us to prove many of the formulas we have merely motivated in Sections 9.1 and 9.2 by starting with just a single step. A good example is the formula for arithmetic sequences we touted in Equation 9.1. Arithmetic sequences are deﬁned recursively, starting with a1 = a and then an+1 = an + d for n ≥ 1. This tells us that we start the sequence with a and we go from one term to the next by successively adding d. In symbols, a, a + d, a + 2d, a + 3d, a + 4d + . . . The pattern suggested here is that to reach the nth term, we start with a and add d to it exactly n − 1 times, which lead us to our formula an = a + (n − 1)d for n ≥ 1. But how do we prove this to be the case? We have the following. The Principle of Mathematical Induction (PMI): Suppose P (n) is a sentence involving the natural number n. IF 1. P (1) is true and 2. whenever P (k) is true, it follows that P (k + 1) is also true THEN the sentence P (n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.1 It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for all natural numbers n, we start by showing it is true for n = 1 (the ‘base step’) and then show that if it is true for a generic natural number k, it must be true for the next natural number, k + 1 (the ‘inductive step’). The notation P (n) acts just like function notation. For example, if P (n) is the sentence (formula) ‘n2 + 1 = 3’, then P (1) would be ‘12 + 1 = 3’, which is false. The construction P (k + 1) would be ‘(k + 1)2 + 1 = 3’. As usual, this new concept is best illustrated with an example. Returning to our quest to prove the formula for an arithmetic sequence, we ﬁrst identify P (n) as the formula an = a + (n − 1)d. To prove this formula is valid for all natural numbers n, we need to do two things. First, we need to establish that P (1) is true. In other words, is it true that a1 = a + (1 − 1)d? The answer is yes, since this simpliﬁes to a1 = a, which is part of the deﬁnition of the arithmetic sequence. The second thing we need to show is that whenever P (k) is true, it follows that P (k + 1) is true. In other words, we assume P (k) is true (this is called the ‘induction hypothesis’) and deduce that P (k + 1) is also true. Assuming P (k) to be true seems to invite disaster - after all, isn’t this essentially what we’re trying to prove in the ﬁrst place? To help explain this step a little better, we show how this works for speciﬁc values of n. We’ve already established P (1) is true, and we now want to show that P (2) 1Another word for this you may have seen is ‘axiom.’ 674 Sequences and the Binomial Theorem is true. Thus we need to show that a2 = a + (2 − 1)d. Since P (1) is true, we have a1 = a, and by the deﬁnition of an arithmetic sequence, a2 = a1 +d = a+d = a+(2−1)d. So P (2) is true. We now use the fact that P (2) is true to show that P (3) is true. Using the fact that a2 = a + (2 − 1)d, we show a3 = a +(3 − 1)d. Since a3 = a2 + d, we get a3 = (a +(2 − 1)d)+ d = a +2d = a +(3 − 1)d, so we have shown P (3) is true. Similarly, we can use the fact that P (3) is true to show that P (4) is true, and so forth. In general, if P (k) is true (i.e., ak = a+(k −1)d) we set out to show that P (k +1) is true (i.e., ak+1 = a + ((k + 1) − 1)d). Assuming ak = a + (k − 1)d, we have by the deﬁnition of an arithmetic sequence that ak+1 = ak + d so we get ak+1 = (a + (k − 1)d) + d = a + kd = a + ((k + 1) − 1)d. Hence, P (k + 1) is true. In essence, by showing that P (k + 1) must always be true when P (k) is true, we are showing that the formula P (1) can be used to get the formula P (2), which in turn can be used to derive the formula P (3), which in turn can be used to establish the formula P (4), and so on. Thus as long as P (k) is true for some natural number k, P (n) is true for all of the natural numbers n which follow k. Coupling this with the fact P (1) is true, we have established P (k) is true for all natural numbers which follow n = 1, in other words, all natural numbers n. One might liken Mathematical Induction to a repetitive process like climbing stairs.2 If you are sure that (1) you can get on the stairs (the base case) and (2) you can climb from any one step to the next step (the inductive step), then presumably you can climb the entire staircase.3 We get some more practice with induction in the following example. Example 9.3.1. Prove the following assertions using the Principle of Mathematical Inducti
on. 1. The sum formula for arithmetic sequences: n (a + (j − 1)d) = j=1 2. For a complex number z, (z)n = zn for n ≥ 1. n 2 (2a + (n − 1)d). 3. 3n > 100n for n > 5. 4. Let A be an n × n matrix and let A be the matrix obtained by replacing a row R of A with cR for some real number c. Use the deﬁnition of determinant to show det(A) = c det(A). Solution. 1. We set P (n) to be the equation we are asked to prove. For n = 1, we compare both sides of the equation given in P (n) 1 j=1 (a + (j − 1)d) a + (1 − 1)2a + (1 − 1)d) (2a) a = a 2Falling dominoes is the most widely used metaphor in the mainstream College Algebra books. 3This is how Carl climbed the stairs in the Cologne Cathedral. Well, that, and encouragement from Kai. 9.3 Mathematical Induction 675 This shows the base case P (1) is true. Next we assume P (k) is true, that is, we assume k (a + (j − 1)d) = j=1 k 2 (2a + (k − 1)d) and attempt to use this to show P (k + 1) is true. Namely, we must show k+1 (a + (j − 1)d) = j=1 k + 1 2 (2a + (k + 1 − 1)d) To see how we can use P (k) in this case to prove P (k + 1), we note that the sum in P (k + 1) is the sum of the ﬁrst k + 1 terms of the sequence ak = a + (k − 1)d for k ≥ 1 while the sum in P (k) is the sum of the ﬁrst k terms. We compare both side of the equation in P (k + 1). k+1 (a + (j − 1)d) j=1 summing the ﬁrst k + 1 terms k (a + (j − 1)d) + (a + (k + 1 − 1)d) j=1 summing the ﬁrst k terms adding the (k + 1)st term ? = k + 1 2 (2a + (k + 1 − 1)d) ? = k + 1 2 (2a + kd) k 2 (2a + (k − 1)d) Using P (k) +(a + kd) k(2a + (k − 1)d) + 2(a + kd) 2 2ka + 2a + k2d + kd 2 ? = ? = = (k + 1)(2a + kd) 2 2ka + k2d + 2a + kd 2 2ka + 2a + k2d + kd 2 Since all of our steps on both sides of the string of equations are reversible, we conclude that the two sides of the equation are equivalent and hence, P (k + 1) is true. By the Principle of Mathematical Induction, we have that P (n) is true for all natural numbers n. 2. We let P (n) be the formula (z)n = zn. The base case P (1) is (z)1 = z1, which reduces to z = z which is true. We now assume P (k) is true, that is, we assume (z)k = zk and attempt to show that P (k + 1) is true. Since (z)k+1 = (z)k z, we can use the induction hypothesis and 676 Sequences and the Binomial Theorem write (z)k = zk. Hence, (z)k+1 = (z)k z = zk z. We now use the product rule for conjugates4 to write zk z = zkz = zk+1. This establishes (z)k+1 = zk+1, so that P (k + 1) is true. Hence, by the Principle of Mathematical Induction, (z)n = zn for all n ≥ 1. 3. The ﬁrst wrinkle we encounter in this problem is that we are asked to prove this formula for n > 5 instead of n ≥ 1. Since n is a natural number, this means our base step occurs at n = 6. We can still use the PMI in this case, but our conclusion will be that the formula is valid for all n ≥ 6. We let P (n) be the inequality 3n > 100n, and check that P (6) is true. Comparing 36 = 729 and 100(6) = 600, we see 36 > 100(6) as required. Next, we assume that P (k) is true, that is we assume 3k > 100k. We need to show that P (k + 1) is true, that is, we need to show 3k+1 > 100(k + 1). Since 3k+1 = 3 · 3k, the induction hypothesis gives 3k+1 = 3 · 3k > 3(100k) = 300k. We are done if we can show 300k > 100(k + 1) for k ≥ 6. Solving 300k > 100(k + 1) we get k > 1 2 . Since k ≥ 6, we know this is true. Putting all of this together, we have 3k+1 = 3 · 3k > 3(100k) = 300k > 100(k + 1), and hence P (k + 1) is true. By induction, 3n > 100n for all n ≥ 6. 4. To prove this determinant property, we use induction on n, where we take P (n) to be that the property we wish to prove is true for all n × n matrices. For the base case, we note that if A is a 1 × 1 matrix, then A = [a] so A = [ca]. By deﬁnition, det(A) = a and det(A) = ca so we have det(A) = c det(A) as required. Now suppose that the property we wish to prove is true for all k × k matrices. Let A be a (k + 1) × (k + 1) matrix. We have two cases, depending on whether or not the row R being replaced is the ﬁrst row of A. Case 1: The row R being replaced is the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p 1p = (−1)(1+p) det A where the 1p cofactor of A is C 1p is the k × k matrix obtained by deleting the 1st row and pth column of A.5 Since the ﬁrst row of A is c times the ﬁrst 1p = c a1p. In addition, since the remaining rows of A are identical to row of A, we have a 1p = A1p. (To obtain these matrices, the ﬁrst row of A is removed.) Hence those of A, A det A 1p = det (A1p), so that C 1p = C1p. As a result, we get and A 1p det(A) = n p=1 1pC a 1p = n p=1 n c a1pC1p = c p=1 a1pC1p = c det(A), as required. Hence, P (k + 1) is true in this case, which means the result is true in this case for all natural numbers n ≥ 1. (You’ll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where 4See Exercise 54 in Section 3.4. 5See Section 8.5 for a review of this notation. 9.3 Mathematical Induction 677 it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R being replaced is the not the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p, where in this case, a 1p = a1p, since the ﬁrst rows of A and A are the same. The matrices 1p and A1p, on the other hand, are diﬀerent but in a very predictable way − the row in A A 1p which corresponds to the row cR in A is exactly c times the row in A1p which corresponds to the row R in A. In other words, A 1p and A1p are k × k matrices which satisfy the induction hypothesis. Hence, we know det A 1p = c det (A1p) and C 1p = c C1p. We get det(A) = n p=1 1pC a 1p = n p=1 n a1pc C1p = c p=1 a1pC1p = c det(A), which establishes P (k + 1) to be true. Hence by induction, we have shown that the result holds in this case for n ≥ 1 and we are done. While we have used the Principle of Mathematical Induction to prove some of the formulas we have merely motivated in the text, our main use of this result comes in Section 9.4 to prove the celebrated Binomial Theorem. The ardent Mathematics student will no doubt see the PMI in many courses yet to come. Sometimes it is explicitly stated and sometimes it remains hidden in the background. If ever you see a property stated as being true ‘for all natural numbers n’, it’s a solid bet that the formal proof requires the Principle of Mathematical Induction. 678 Sequences and the Binomial Theorem 9.3.1 Exercises In Exercises 1 - 7, prove each assertion using the Principle of Mathematical Induction. 1. 2. n j=1 n j=1 j2 = n(n + 1)(2n + 1) 6 j3 = n2(n + 1)2 4 3. 2n > 500n for n > 12 4. 3n ≥ n3 for n ≥ 4 5. Use the Product Rule for Absolute Value to show \|xn\| = \|x\|n for all real numbers x and all natural numbers n ≥ 1 6. Use the Product Rule for Logarithms to show log (xn) = n log(x) for all real numbers x > 0 and all natural numbers n ≥ 1. an a 0 0 0 b 0 bn n = for n ≥ 1. 7. 8. Prove Equations 9.1 and 9.2 for the case of geometric sequences. That is: (a) For the sequence a1 = a, an+1 = ran, n ≥ 1, prove an = arn−1, n ≥ 1. (b) n j=1 arn−1 = a 1 − rn 1 − r , if r = 1, n j=1 arn−1 = na, if r = 1. 9. Prove that the determinant of a lower triangular matrix is the product of the entries on the main diagonal. (See Exercise 8.3.1 in Section 8.3.) Use this result to then show det (In) = 1 where In is the n × n identity matrix. 10. Discuss the classic ‘paradox’ All Horses are the Same Color problem with your classmates. 9.3 Mathematical Induction 679 9.3.2 Selected Answers 1. Let P (n) be the sentence n j=1 j2 = n(n + 1)(2n + 1) 6 . For the base case, n = 1, we get 1 j=1 j2 ? = (1)(1 + 1)(2(1) + 1) 6 12 = 1 We now assume P (k) is true and use it to show P (k + 1) is true. We have k+1 j=1 j2 ? = (k + 1)((k + 1) + 1)(2(k + 1) + 1) 6 k j=1 j2 + (k + 1)2 k(k + 1)(2k + 1) 6 Using P (k) +(k + 1)2 k(k + 1)(2k + 1) 6 + 6(k + 1)2 6 k(k + 1)(2k + 1) + 6(k + 1)2 6 (k + 1)(k(2k + 1) + 6(k + 1)) 6 (k + 1) 2k2 + 7k + 6 6 (k + 1)(k + 2)(2k + 3k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 By induction, n j=1 j2 = n(n + 1)(2n + 1) 6 is true for all natural numbers n ≥ 1. 4. Let P (n) be the sentence 3n > n3. Our base case is n = 4 and we check 34 = 81 and 43 = 64 so that 34 > 43 as required. We now assume P (k) is true, that is 3k > k3, and try to show P (k + 1) is true. We note that 3k+1 = 3 · 3k > 3k3 and so we are done if we can show 3k3 > (k + 1)3 for k ≥ 4. We can solve the inequality 3x3 > (x + 1)3 using the techniques of Section 5.3, and doing so gives us x > 1 ≈ 2.26. Hence, for k ≥ 4, 3√ 3k+1 = 3 · 3k > 3k3 > (k + 1)3 so that 3k+1 > (k + 1)3. By induction, 3n > n3 is true for all natural numbers n ≥ 4. 3−1 680 Sequences and the Binomial Theorem 6. Let P (n) be the sentence log (xn) = n log(x). For the duration of this argument, we assume x > 0. The base case P (1) amounts checking that log x1 = 1 log(x) which is clearly true. Next we assume P (k) is true, that is log xk = k log(x) and try to show P (k + 1) is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get xk+1 log = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log xk+1 = (k + 1) log(x). By induction log (xn) = n log(x) is true for all x > 0 and all natural numbers n ≥ 1. 9. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a, so the result is (trivially) true. Next suppose the result is true for k × k lower triangular matrices. Let A be a (k + 1) × (k + 1) lower triangular matrix. Expanding det(A) along the ﬁrst row, we have det(A) = n p=1 a1pC1p Since a1p = 0 for 2 ≤ p ≤ k + 1, this simpliﬁes det(A) = a11C11. By deﬁnition, we know that C11 = (−1)1+1 det (A11) = det (A11) where A11 is k × k matrix obtained by deleting t
he ﬁrst row and ﬁrst column of A. Since A is lower triangular, so is A11 and, as such, the induction hypothesis applies to A11. In other words, det (A11) is the product of the entries along A11’s main diagonal. Now, the entries on the main diagonal of A11 are the entries a22, a33, . . . , a(k+1)(k+1) from the main diagonal of A. Hence, det(A) = a11 det (A11) = a11 a22a33 · · · a(k+1)(k+1) = a11a22a33 · · · a(k+1)(k+1) We have det(A) is the product of the entries along its main diagonal. This shows P (k + 1) is true, and, hence, by induction, the result holds for all n × n upper triangular matrices. The n × n identity matrix In is a lower triangular matrix whose main diagonal consists of all 1’s. Hence, det (In) = 1, as required. 9.4 The Binomial Theorem 681 9.4 The Binomial Theorem In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities (a + b)n for natural numbers n. In Elementary and Intermediate Algebra, you should have seen speciﬁc instances of the formula, namely (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3 If we wanted the expansion for (a + b)4 we would write (a + b)4 = (a + b)(a + b)3 and use the formula that we have for (a+b)3 to get (a+b)4 = (a+b) a3 + 3a2b + 3ab2 + b3 = a4+4a3b+6a2b2+4ab3+b4. Generalizing this a bit, we see that if we have a formula for (a + b)k, we can obtain a formula for (a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k. Clearly this means Mathematical Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the theorem we need to revisit the sequence of factorials which were introduced in Example 9.1.1 number 6 in Section 9.1. Deﬁnition 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of the sequence f0 = 1, fn = n · fn−1, n ≥ 1. Recall this means 0! = 1 and n! = n(n − 1)! for n ≥ 1. Using the recursive deﬁnition, we get: 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 · 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24. Informally, n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as our ‘base case.’ Our ﬁrst example familiarizes us with some of the basic computations involving factorials. Example 9.4.1. 1. Simplify the following expressions. (a) 3! 2! 0! (b) 7! 5! (c) 1000! 998! 2! (d) (k + 2)! (k − 1)! , k ≥ 1 2. Prove n! > 3n for all n ≥ 7. Solution. 1. We keep in mind the mantra, “When in doubt, write it out!” as we simplify the following. (a) We have been programmed to react with alarm to the presence of a 0 in the denominator, but in this case 0! = 1, so the fraction is deﬁned after all. As for the numerator, 3! = 3 · 2 · 1 = 6 and 2! = 2 · 1 = 2, so we have 3! 2! 0! = (6)(2) 1 = 12. 1It’s pretty much the reason Section 9.3 is in the book. 682 Sequences and the Binomial Theorem (b) We have 7 = 5040 while 5 = 120. Dividing, we get 120 = 42. While this is correct, we note that we could have saved ourselves some 7! 5! = 5040 of time had we proceeded as follows 7! 5 = 42 In fact, should we want to fully exploit the recursive nature of the factorial, we can write 7! 5! = 7 · 6 · 5! 5! = 7 · 6 · 5! 5! = 42 (c) Keeping in mind the lesson we learned from the previous problem, we have 1000! 998! 2! = 1000 · 999 · 998! 998! · 2! = 1000 · 999 · 998! 998! · 2! = 999000 2 = 499500 (d) This problem continues the theme which we have seen in the previous two problems. We ﬁrst note that since k + 2 is larger than k − 1, (k + 2)! contains all of the factors of (k − 1)! and as a result we can get the (k − 1)! to cancel from the denominator. To see this, we begin by writing out (k + 2)! starting with (k + 2) and multiplying it by the numbers which precede it until we reach (k − 1): (k + 2)! = (k + 2)(k + 1)(k)(k − 1)!. As a result, we have (k + 2)! (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! (k − 1)! = (k + 2)(k + 1)(k) (k − 1)! (k − 1)! = k(k + 1)(k + 2) The stipulation k ≥ 1 is there to ensure that all of the factorials involved are deﬁned. 2. We proceed by induction and let P (n) be the inequality n! > 3n. The base case here is n = 7 and we see that 7! = 5040 is larger than 37 = 2187, so P (7) is true. Next, we assume that P (k) is true, that is, we assume k! > 3k and attempt to show P (k + 1) follows. Using the properties of the factorial, we have (k + 1)! = (k + 1)k! and since k! > 3k, we have (k + 1)! > (k + 1)3k. Since k ≥ 7, k + 1 ≥ 8, so (k + 1)3k ≥ 8 · 3k > 3 · 3k = 3k+1. Putting all of this together, we have (k + 1)! = (k + 1)k! > (k + 1)3k > 3k+1 which shows P (k + 1) is true. By the Principle of Mathematical Induction, we have n! > 3n for all n ≥ 7. Of all of the mathematical animals we have discussed in the text, factorials grow most quickly. In problem 2 of Example 9.4.1, we proved that n! overtakes 3n at n = 7. ‘Overtakes’ may be too polite a word, since n! thoroughly trounces 3n for n ≥ 7, as any reasonable set of data will show. It can be shown that for any real number x > 0, not only does n! eventually overtake xn, but the ratio xn Applications of factorials in the wild often involve counting arrangements. For example, if you have ﬁfty songs on your mp3 player and wish arrange these songs in a playlist in which the order of the n! → 0 as n → ∞.2 2This fact is far more important than you could ever possibly imagine. 9.4 The Binomial Theorem 683 songs matters, it turns out that there are 50! diﬀerent possible playlists. If you wish to select only ten of the songs to create a playlist, then there are 50! 40! such playlists. If, on the other hand, you just want to select ten song ﬁles out of the ﬁfty to put on a ﬂash memory card so that now the order 50! 40!10! ways to achieve this.3 While some of these ideas are explored no longer matters, there are in the Exercises, the authors encourage you to take courses such as Finite Mathematics, Discrete Mathematics and Statistics. We introduce these concepts here because this is how the factorials make their way into the Binomial Theorem, as our next deﬁnition indicates. Deﬁnition 9.5. Binomial Coeﬃcients: Given two whole numbers n and j with n ≥ j, the binomial coeﬃcient n j (read, n choose j) is the whole number given by n j = n! j!(n − j)! The name ‘binomial coeﬃcient’ will be justiﬁed shortly. For now, we can physically interpret as the number of ways to select j items from n items where the order of the items selected is n j unimportant. For example, suppose you won two free tickets to a special screening of the latest Hollywood blockbuster and have ﬁve good friends each of whom would love to accompany you to ways to choose who goes with you. Applying Deﬁnition 9.5, we get the movies. There are 5 2 5 2 = 5! 2!(5 − 2)! = 5! 2!3! = 5 · 4 2 = 10 So there are 10 diﬀerent ways to distribute those two tickets among ﬁve friends. (Some will see it as 10 ways to decide which three friends have to stay home.) The reader is encouraged to verify this by actually taking the time to list all of the possibilities. We now state anf prove a theorem which is crucial to the proof of the Binomial Theorem. Theorem 9.3. For natural numbers n and j with n ≥ j = The proof of Theorem 9.3 is purely computational and uses the deﬁnition of binomial coeﬃcients, the recursive property of factorials and common denominators. 3For reference, 50! 50! 40! 50! 40!10! = 30414093201713378043612608166064768844377641568960512000000000000, = 37276043023296000, and = 10272278170 684 Sequences and the Binomial Theorem ! (j − 1)!(n − (j − 1))! + n! j!(n − j)! n! (j − 1)!(n − j + 1)! + n! j!(n − j)! n! (j − 1)!(n − j + 1)(n − j)! + n! j(j − 1)!(n − j)! n! j j(j − 1)!(n − j + 1)(n − j)! + n!(n − j + 1) j(j − 1)!(n − j + 1)(n − j)! n! j j!(n − j + 1)! + n!(n − j + 1) j!(n − j + 1)! n! j + n!(n − j + 1) j!(n − j + 1)! n! (j + (n − j + 1)) j!(n − j + 1)! (n + 1)n! j!(n + 1 − j))! (n + 1)! j!((n + 1) − j))! n + 1 j We are now in position to state and prove the Binomial Theorem where we see that binomial coeﬃcients are just that - coeﬃcients in the binomial expansion. Theorem 9.4. Binomial Theorem: For nonzero real numbers a and b, (a + b)n = an−jbj n j=0 n j for all natural numbers n. To get a feel of what this theorem is saying and how it really isn’t as hard to remember as it may ﬁrst appear, let’s consider the speciﬁc case of n = 4. According to the theorem, we have 9.4 The Binomial Theorem 685 (a + b)4 = 4 a4−jbj 4 j j=0 4 0 4 0 = = a4−0b0 + a4−1b1 + 4 1 a4−2b2 + 4 2 a4−3b3 + a4−4b4 4 4 a4 + a3b + 4 1 4 2 a2b2 + 4 3 ab3 + 4 3 4 4 b4 We forgo the simpliﬁcation of the coeﬃcients in order to note the pattern in the expansion. First note that in each term, the total of the exponents is 4 which matched the exponent of the binomial (a + b)4. The exponent on a begins at 4 and decreases by one as we move from one term to the next while the exponent on b starts at 0 and increases by one each time. Also note that the binomial coeﬃcients themselves have a pattern. The upper number, 4, matches the exponent on the binomial (a + b)4 whereas the lower number changes from term to term and matches the exponent of b in that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If we think of obtaining (a + b)4 by multiplying (a + b)(a + b)(a + b)(a + b), our answer is the sum of all possible products with exactly four factors - some a, some b. If we wish to count, for instance, the number of ways we obtain 1 factor of b out of a total of 4 possible factors, thereby forcing the remaining 3 factors to be a, the answer is 4 a3b is in the expansion. The . Hence, the term 4 1 1 other terms which appear cover the remaining cases. While this discussion gives an indication as to why the theorem is true, a formal proof requires Mathematical Induction.4 To prove the Binomial Theorem, we let P (n) be the expansion formula given in the statement of the theorem and we note that P (1) is true since (a + b)1 a + b ? = ? = 1 a1−jbj 1 j j=0 1 0 a1−0b0 + a1−1b1 1 1 a +
b = a + b Now we assume that P (k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. 4and a fair amount of tenacity and attention to detail. 686 Sequences and the Binomial Theorem (a + b)k+1 = (a + b)(a + b)k k = (a + b) k j ak−jbj j=0 = a k j=0 k j ak−jbj + b = k j=0 k j ak+1−jbj + k j=0 k j=0 k j ak−jbj ak−jbj+1 k j Our goal is to combine as many of the terms as possible within the two summations. As the counter j in the ﬁrst summation runs from 0 through k, we get terms involving ak+1, akb, ak−1b2, . . . , abk. In the second summation, we get terms involving akb, ak−1b2, . . . , abk, bk+1. In other words, apart from the ﬁrst term in the ﬁrst summation and the last term in the second summation, we have terms common to both summations. Our next move is to ‘kick out’ the terms which we cannot combine and rewrite the summations so that we can combine them. To that end, we note k j=0 k j ak+1−jbj = ak+1 + k j=1 k j ak+1−jbj ak−jbj+1 = k j=0 k j k−1 j=0 k j ak−jbj+1 + bk+1 and so that (a + b)k+1 = ak+1 + ak+1−jbj + k j=1 k j k−1 j=0 k j ak−jbj+1 + bk+1 We now wish to write ak+1−jbj + k j=1 k j k−1 j=0 k j ak−jbj+1 as a single summation. The wrinkle is that the ﬁrst summation starts with j = 1, while the second starts with j = 0. Even though the sums produce terms with the same powers of a and b, they do so for diﬀerent values of j. To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. 9.4 The Binomial Theorem 687 k−1 j=0 k j ak−jbj+1 = k−1+1 k j − 1 j=0+1 ak−(j−1)b(j−1)+1 = k k j − 1 j=1 ak+1−jbj We can now combine our two sums using Theorem 9.1 and simplify using Theorem 9.3 k j=1 k j ak+1−jbj + ak−jbj+1 = k−1 j=0 k j = = k j=1 k j=1 k j=1 ak+1−jbj + k k ak+1−jbj k j j=1 j − 1 ak+1−jbj ak+1−jbj Using this and the fact that k+1 0 = 1 and k+1 k+1 = 1, we get (a + b)k+1 = ak+1 + k k + 1 j ak+1−jbj + bk+1 j=1 k + 1 0 ak+1b0 + k+1 j=0 k + 1 j a(k+1)−jbj = = k j=1 k + 1 j ak+1−jbj + k + 1 k + 1 a0bk+1 which shows that P (k + 1) is true. Hence, by induction, we have established that the Binomial Theorem holds for all natural numbers n. Example 9.4.2. Use the Binomial Theorem to ﬁnd the following. 1. (x − 2)4 2. 2.13 3. The term containing x3 in the expansion (2x + y)5 Solution. 1. Since (x − 2)4 = (x + (−2))4, we identify a = x, b = −2 and n = 4 and obtain 688 Sequences and the Binomial Theorem (x − 2)4 = 4 x4−j(−2)j 4 j j=0 4 0 = x4−0(−2)0 + x4−1(−2)1 + 4 1 4 2 x4−2(−2)2 + 4 3 x4−3(−2)3 + x4−4(−2)4 4 4 = x4 − 8x3 + 24x2 − 32x + 16 2. At ﬁrst this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3. Identifying a = 2, b = 0.1 = 1 10 and n = 3, we get 3 23−j 3 j j 1 10 2 + 3 1 10 = = j=0 3 0 23−0 0 1 10 3 1 + 23−1 1 1 10 3 2 + 23−2 2 1 10 3 3 + 23−3 3 1 10 = 8 + 12 10 + 6 100 + 1 1000 = 8 + 1.2 + 0.06 + 0.001 = 9.261 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives (2x + y)5 = (2x)5−jyj 5 j=0 5 j Since we are concerned with only the term containing x3, there is no need to expand the entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the exponent on y must be 2. Plucking out the term j = 2, we get 5 2 (2x)5−2y2 = 10(2x)3y2 = 80x3y2 We close this section with Pascal’s Triangle, named in honor of the mathematician Blaise Pascal. Pascal’s Triangle is obtained by arranging the binomial coeﬃcients in the triangular fashion below. 9.4 The Binomial Theorem 689 ... for all whole numbers n, we get that each row of Pascal’s Triangle = 1 and n Since n n 0 begins and ends with 1. To generate the numbers in the middle of the rows (from the third row onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance, and so forth. This relationship is indicated by the arrows in the + array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start with the ﬁrst two rows, 1 and 1 1. From that point on, each successive row begins and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the ﬁrst ﬁve rows of Pascal’s Triangle −−−−−−→ −−−−−−→ −−−−−−→ 690 Sequences and the Binomial Theorem To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppose we wish to for j = 0, 1, 2, 3, 4 and are the numbers which expand (3x − y)4. The coeﬃcients we need are 4 j form the ﬁfth row of Pascal’s Triangle. Since we know that the exponent of 3x in the ﬁrst term is 4 and then decreases by one as we go from left to right while the exponent of −y starts at 0 in the ﬁrst term and then increases by one as we move from left to right, we quickly obtain (3x − y)4 = (1)(3x)4 + (4)(3x)3(−y) + (6)(3x)2(−y)2 + 4(3x)(−y)3 + 1(−y)4 = 81x4 − 108x3y + 54x2y2 − 12xy3 + y4 We would like to stress that Pascal’s Triangle is a very quick method to expand an entire binomial. If only a term (or two or three) is required, then the Binomial Theorem is deﬁnitely the way to go. 9.4 The Binomial Theorem 691 9.4.1 Exercises In Exercises 1 - 9, simplify the given expression. 1. (3!)2 4. 7. 9! 4!3!2! 8 3 2. 5. 8. 10! 7! (n + 1)! n! 117 0 , n ≥ 0. 3. 6. 9. 7! 233! (k − 1)! (k + 2)! n n − 2 , k ≥ 1. , n ≥ 2 In Exercises 10 - 13, use Pascal’s Triangle to expand the given binomial. 10. (x + 2)5 11. (2x − 1)4 12. 1 3 x + y23 13. x − x−14 In Exercises 14 - 17, use Pascal’s Triangle to simplify the given power of a complex number. 14. (1 + 2i)4 √ 16. 3 i 3 2 + 1 2 15. −1 + i √ √ 33 √ 4 i 2 2 − 2 2 17. In Exercises 18 - 22, use the Binomial Theorem to ﬁnd the indicated term. 18. The term containing x3 in the expansion (2x − y)5 19. The term containing x117 in the expansion (x + 2)118 √ 7 2 in the expansion ( 20. The term containing x 21. The term containing x−7 in the expansion 2x − x−35 22. The constant term in the expansion x + x−18 23. Use the Prinicple of Mathematical Induction to prove n! > 2n for n ≥ 4. x − 3)8 24. Prove n j=0 n j = 2n for all natural numbers n. (HINT: Use the Binomial Theorem!) 25. With the help of your classmates, research Patterns and Properties of Pascal’s Triangle. 26. You’ve just won three tickets to see the new ﬁlm, ‘8.9.’ Five of your friends, Albert, Beth, Chuck, Dan, and Eugene, are interested in seeing it with you. With the help of your classmates, list all the possible ways to distribute your two extra tickets among your ﬁve friends. Now suppose you’ve come down with the ﬂu. List all the diﬀerent ways you can distribute the three tickets among these ﬁve friends. How does this compare with the ﬁrst list you made? ? = 5 What does this have to do with the fact that 5 3 2 692 Sequences and the Binomial Theorem 9.4.2 Answers 1. 36 4. 1260 7. 56 2. 720 5. n + 1 8. 1 3. 105 6. 1 k(k+1)(k+2) 9. n(n−1) 2 10. (x + 2)5 = x5 + 10x4 + 40x3 + 80x2 + 80x + 32 11. (2x − 1)4 = 16x4 − 32x3 + 24x2 − 8x + 1 12. 1 3 x + y23 13. x − x−14 = 1 27 x3 + 1 3 x2y2 + xy4 + y6 = x4 − 4x2 + 6 − 4x−2 + x−4 14. −7 − 24i 15. 8 16. i 17. −1 18. 80x3y2 19. 236x117 20. −24x 7 2 21. −40x−7 22. 70 Index nth root of a complex number, 1000, 1001 principal, 397 nth Roots of Unity, 1006 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value deﬁnition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix, 622 alkalinity of a solution pH, 432 amplitude, 794, 881 angle acute, 694 between two vectors, 1035, 1036 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 deﬁnition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle side opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary deﬁnition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 1069 Index 1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arcsine deﬁnition of, 820 graph of, 820 properties of, 820 arctangent deﬁnition of, 824 graph of, 823 properties of, 824 argument of a complex number deﬁnition of, 991 properties of, 995 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1015 scalar multiplication, 1018 asymptote horizontal formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 of a hyperbola, 531 slant determination of, 312 formal deﬁnition of, 311 slant (oblique), 311 vertical formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 905 binomial coeﬃcient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buﬀer solution, 478 cardioid, 951 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1071 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 deﬁnition of, 498 fr
om slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1015 dot product, 1034 complementary angles, 696 Complex Factorization Theorem, 290 complex number nth root, 1000, 1001 nth Roots of Unity, 1006 argument deﬁnition of, 991 properties of, 995 conjugate deﬁnition of, 288 properties of, 289 deﬁnition of, 2, 287, 991 imaginary part, 991 imaginary unit, i, 287 modulus deﬁnition of, 991 properties of, 993 polar form cis-notation, 995 principal argument, 991 real part, 991 rectangular form, 991 set of, 2 complex plane, 991 component form of a vector, 1013 composite function deﬁnition of, 360 properties of, 367 compound interest, 470 conic sections deﬁnition, 495 conjugate axis of a hyperbola, 532 conjugate of a complex number deﬁnition of, 288 properties of, 289 Conjugate Pairs Theorem, 291 consistent system, 553 constant function as a horizontal line, 156 formal deﬁnition of, 101 intuitive deﬁnition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 919 rectangular, 919 correlation coeﬃcient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1072 cost average, 82, 346 ﬁxed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 Cramer’s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix deﬁnition of, 614 properties of, 616 Diﬀerence Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 diﬀerence quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 981 of a parabola, 505 discriminant Index of a conic, 979 of a quadratic equation, 195 trichotomy, 195 distance deﬁnition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1034 scalar multiplication, 1018 DMS, 695 domain applied, 60 deﬁnition of, 45 implied, 58 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to orthogonality, 1037 relation to vector magnitude, 1034 work, 1042 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 981 eigenvalue, 626 eigenvector, 626 ellipse center, 516 deﬁnition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1073 reﬂective property, 523 standard equation, 519 vertices, 516 ellipsis (. . . ), 31, 651 empty set, 2 end behavior of f (x) = axn, n even, 240 of f (x) = axn, n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 ﬁxed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 981 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 881 of a sinusoid, 795 ordinary, 708, 881 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite deﬁnition of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 diﬀerence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1074 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-deﬁned, 62 polynomial, 235 price-demand, 82 product, 76 proﬁt, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 938 Fundamental Theorem of Algebra, 290 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of, 2 greatest integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regression, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 deﬁnition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60
modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior, 239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 1018 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of, 795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁnition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zero
ace the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we ﬁnd 5π 8 ≈ 1.96 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 706 Foundations of Trigonometry 4. Since 117 is positive, the arc corresponding to t = 117 begins at (1, 0) and proceeds counterclockwise. As 117 is much greater than 2π, we wrap around the Unit Circle several times before ﬁnally reaching our endpoint. We approximate 117 2π as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of 5 8 of a revolution to spare. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III2 t = 117 10.1.1 Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure, a whole world of applications awaits us. Our ﬁrst excursion into this realm comes by way of circular motion. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. Q θ r s P Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction. Note that with this convention the formula we used to deﬁne radian measure, namely θ = , still holds since a negative value of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. In Physics, the average velocity of the object, denoted v and read as ‘v-bar’, is deﬁned as the average rate of change of the position of the object with respect to time.16 As a result, we s r 16See Deﬁnition 2.3 in Section 2.1 for a review of this concept. 10.1 Angles and their Measure 707 s t = time . The quantity v has units of length have v = displacement time and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of clockwise motion), so that the quantity \|v\| quantiﬁes how fast the object is moving - it is the speed of the object. Measuring θ in radians we have θ = thus s = rθ and s r v = s t = rθ t = r · θ t θ t is called the average angular velocity of the object. It is denoted by ω and is The quantity read ‘omega-bar’. The quantity ω is the average rate of change of the angle θ with respect to time and thus has units radians time . If ω is constant throughout the duration of the motion, then it can be shown17 that the average velocities involved, namely v and ω, are the same as their instantaneous counterparts, v and ω, respectively. In this case, v is simply called the ‘velocity’ of the object and is the instantaneous rate of change of the position of the object with respect to time.18 Similarly, ω is called the ‘angular velocity’ and is the instantaneous rate of change of the angle with respect to time. If the path of the object were ‘uncurled’ from a circle to form a line segment, then the velocity of the object on that line segment would be the same as the velocity on the circle. For this reason, the quantity v is often called the linear velocity of the object in order to distinguish it from the angular velocity, ω. Putting together the ideas of the previous paragraph, we get the following. Equation 10.2. Velocity for Circular Motion: For an object moving on a circular path of radius r with constant angular velocity ω, the (linear) velocity of the object is given by v = rω. time . Thus the left hand side of the equation v = rω has units length We need to talk about units here. The units of v are length time , the units of r are length only, and the units of ω are radians time , whereas time = length·radians the right hand side has units length · radians . The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identiﬁed with real numbers so that the units length·radians time . We are long overdue for an example. reduce to the units length time time Example 10.1.5. Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at 41.628◦ north latitude, and it can be shown19 that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns. Solution. To use the formula v = rω, we ﬁrst need to compute the angular velocity ω. The earth π makes one revolution in 24 hours, and one revolution is 2π radians, so ω = 2π radians 12 hours , 24 hours = 17You guessed it, using Calculus . . . 18See the discussion on Page 161 for more details on the idea of an ‘instantaneous’ rate of change. 19We will discuss how we arrived at this approximation in Example 10.2.6. 708 Foundations of Trigonometry where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity’s sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0.) Hence, the linear velocity is v = 2960 miles · π 12 hours ≈ 775 miles hour It is worth noting that the quantity 1 revolution in Example 10.1.5 is called the ordinary frequency 24 hours of the motion and is usually denoted by the variable f . The ordinary frequency is a measure of how often an object makes a complete cycle of the motion. The fact that ω = 2πf suggests that ω is also a frequency. Indeed, it is called the angular frequency of the motion. On a related note, the quantity T = is called the period of the motion and is the amount of time it takes for the 1 f object to complete one cycle of the motion. In the scenario of Example 10.1.5, the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is ﬁxed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period’s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the ‘magniﬁcation factor’ which relates ω and v. We will have more to say about frequencies and periods in Section 11.1. While we have exhaustively discussed velocities associated with circular motion, we have yet to discuss a more natural question: if an object is moving on a circular path of radius r with a ﬁxed angular velocity (frequency) ω, what is the position of the object at time t? The answer to this question is the very heart of Trigonometry and is answered in the next section. 10.1 Angles and their Measure 709 10.1.2 Exercises In Exercises 1 - 4, convert the angles into the DMS system. Round each of your answers to the nearest second. 1. 63.75◦ 2. 200.325◦ 3. −317.06◦ 4. 179.999◦ In Exercises 5 - 8, convert the angles into decimal degrees. Round each of your answers to three decimal places. 5. 125◦50 6. −32◦1012 7. 502◦35 8. 237◦5843 In Exercises 9 - 28, graph the oriented angle in standard position. Classify each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 330◦ 10. −135◦ 13. −270◦ 17. 3π 4 21. − π 2 25. −2π 14. 5π 6 18. − π 3 22. 7π 6 26. − π 4 11. 120◦ 15. − 11π 3 19. 7π 2 23. − 5π 3 27. 15π 4 12. 405◦ 16. 20. 5π 4 π 4 24. 3π 28. − 13π 6 In Exercises 29 - 36, convert the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0◦ 33. −315◦ 30. 240◦ 34. 150◦ 31. 135◦ 35. 45◦ 32. −270◦ 36. −225◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 41. π 3 38. − 2π 3 42. 5π 3 39. 7π 6 43. − π 6 40. 44. 11π 6 π 2 710 Foundations of Trigonometry In Exercises 45 - 49, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 45. t = 5π 6 46. t = −π 47. t = 6 48. t = −2 49. t = 12 50. A yo-yo which is 2.25 inches in diameter spins at a rate of 4500 revolutions per minute. How fast is the edge of the yo-yo spinning in miles per hour? Round your answer to two decimal places. 51. How many revolutions per minute would the yo-yo in exercise 50 have to complete if the edge of the yo-yo is to be spinning at a rate of 42 miles per hour? Round your answer to two decimal places. 52. In the yo-yo trick ‘Around the World,’ the performer throws the yo-yo so it sweeps out a vertical circle whose radius is the yo-yo string. If the yo-yo string is 28 inches long and the yo-yo takes 3 seconds to complete one revolution of the circle, compute the speed of the yo-yo in miles per hour. Round your answer to two decimal places. 53. A computer hard drive contains a circular disk with diameter 2.5 inches and spins at a rate of 7200 RPM (revolutions per minute). Find the linear speed of a point on the edge of the disk in miles per hour. 54. A rock got stuck in the tread of my tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? (The tire has a diameter of 23 inches.) 55. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds.20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measure
d in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 (Hint: Use the proportion s circumference of the circle .) A area of the circle = 2 r2θ. r s θ r 20Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = π 6 , r = 12 60. θ = π, r = 1 58. θ = 5π 4 , r = 100 61. θ = 240◦, r = 5 59. θ = 330◦, r = 9.3 62. θ = 1◦, r = 117 63. Imagine a rope tied around the Earth at the equator. Show that you need to add only 2π feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.) 64. With the help of your classmates, look for a proof that π is indeed a constant. 712 Foundations of Trigonometry 10.1.3 Answers 1. 63◦45 5. 125.833◦ 2. 200◦1930 3. −317◦336 4. 179◦5956 6. −32.17◦ 7. 502.583◦ 8. 237.979◦ 9. 330◦ is a Quadrant IV angle coterminal with 690◦ and −30◦ 10. −135◦ is a Quadrant III angle coterminal with 225◦ and −4954 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 11. 120◦ is a Quadrant II angle coterminal with 480◦ and −240◦ 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −3154 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 13. −270◦ lies on the positive y-axis 14. coterminal with 90◦ and −630◦ is a Quadrant II angle 5π 6 coterminal with 17π 6 and − 7π 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 10.1 Angles and their Measure 713 15. − 11π 3 is a Quadrant I angle 5π π 3 3 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 17. 19. is a Quadrant II angle 3π 4 coterminal with 11π 4 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 7π 2 coterminal with 3π 2 and − π 2 y 4 3 2 1 16. is a Quadrant III angle 5π 4 coterminal with 13π 4 and − 3π 4 −3 −2 −1 −1 −2 −3 −4 18. − π 3 is a Quadrant IV angle 7π 3 and − 5π 3 coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x π 4 coterminal with 9π 4 and − 7π 4 y 4 3 2 1 lies on the negative y-axis 20. is a Quadrant I angle −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 714 21. − π 2 lies on the negative y-axis 5π 3π 2 2 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x Foundations of Trigonometry 22. is a Quadrant III angle 7π 6 coterminal with 19π 6 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 23. − 5π 3 is a Quadrant I angle 24. 3π lies on the negative x-axis coterminal with y 4 3 2 1 π 3 and − 11π 3 coterminal with π and −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 25. −2π lies on the positive x-axis 26. − π 4 is a Quadrant IV angle coterminal with 2π and −4π coterminal with 7π 4 and − 9π 4 −4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 10.1 Angles and their Measure 715 27. 15π 4 is a Quadrant IV angle π 4 and − 7π 4 coterminal with y 28. − 13π 6 is a Quadrant IV angle π 11π 6 6 and − coterminal with y 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x 29. 0 33. − 7π 4 37. 180◦ 41. 60◦ 45. t = 5π 6 47. t = 6 30. 34. 4π 3 5π 6 38. −120◦ 42. 300◦ 1 x y 1 y 1 31. 35. 3π 4 π 4 39. 210◦ 43. −30◦ 46. t = −π 48. t = −2 32. − 36. − 3π 2 5π 4 40. 330◦ 44. 90 716 Foundations of Trigonometry 49. t = 12 (between 1 and 2 revolutions) y 1 1 x 50. About 30.12 miles per hour 51. About 6274.52 revolutions per minute 52. About 3.33 miles per hour 53. About 53.55 miles per hour 54. 70 miles per hour 55. About 4.32 miles per hour 57. 12π square units 59. 79.2825π ≈ 249.07 square units 61. 50π 3 square units 58. 6250π square units 60. π 2 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosine and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ).1 The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the deﬁnition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). y 1 θ 1 x y 1 P (cos(θ), sin(θ)) θ 1 x Example 10.2.1. Find the cosine and sine of the following angles. 1. θ = 270◦ 2. θ = −π 3. θ = 45◦ 4. θ = π 6 5. θ = 60◦ Solution. 1. To ﬁnd cos (270◦) and sin (270◦), we plot the angle θ = 270◦ in standard position and ﬁnd the point on the terminal side of θ which lies on the Unit Circle. Since 270◦ represents 3 4 of a counter-clockwise revolution, the terminal side of θ lies along the negative y-axis. Hence, the point we seek is (0, −1) so that cos (270◦) = 0 and sin (270◦) = −1. 2. The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry y 1 θ = 270◦ P (0, −1) y 1 P (−1, 0) 1 x 1 x θ = −π Finding cos (270◦) and sin (270◦) Finding cos (−π) and sin (−π) 3. When we sketch θ = 45◦ in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of ﬁnding the cosine and sine values a bit more diﬃcult. Let P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle. By deﬁnition, x = cos (45◦) and y = sin (45◦). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45◦ − 45◦ − 90◦ right triangle whose legs have lengths x and y units. From Geometry,2 we get y = x. Since P (x, y) lies on the Unit Circle, we have √ 2 x2 + y2 = 1. Substituting y = x into this equation yields 2x2 = 1, or x = ± 2 . √ 2 Since P (x, y) lies in the ﬁrst quadrant, x > 0, so x = cos (45◦) = 2 and with y = x we have y = sin (45◦) = = 45◦ P (x, y) x 1 P (x, y) 45◦ y θ = 45◦ x 2Can you show this? 10.2 The Unit Circle: Cosine and Sine 719 4. As before, the terminal side of θ = π 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P (x, y) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form = 1 a 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we ﬁnd y = 1 2 . Since P (x, y) lies on the Unit Circle, we substitute y = 1 4 , or x = ± 2 into x2 + y2 = 1 to get x2 = 3 2 . Here, x > 0 so x = cos π 2 so sin x, y) x 1 P (x, y) 60◦ y θ = π 6 = 30◦ x 5. Plotting θ = 60◦ in standard position, we ﬁnd it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 30◦ − 60◦ − 90◦ right triangle and, after the usual computations, ﬁnd x = cos (60◦) = 1 2 and y = sin (60◦) = √ 3 2 . y 1 P (x, y) θ = 60◦ x 1 P (x, y) 30◦ y θ = 60◦ x 3Again, can you show this? 720 Foundations of Trigonometry In Example 10.2.1, it was quite easy to ﬁnd the cosine and sine of the quadrantal angles, but for In these latter cases, we made good non-quadrantal angles, the task was much more involved. use of the fact that the point P (x, y) = (cos(θ), sin(θ)) lies on the Unit Circle, x2 + y2 = 1. If we substitute x = cos(θ) and y = sin(θ) into x2 + y2 = 1, we get (cos(θ))2 + (sin(θ))2 = 1. An unfortunate4 convention, which the authors are compelled to perpetuate, is to write (cos(θ))2 as cos2(θ) and (sin(θ))2 as sin2(θ). Rewriting the identity using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometry. Theorem 10.1. The Pythagorean Identity: For any angle θ, cos2(θ) + sin2(θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known, Theorem 10.1 can be used to determine the other, up to a (±) sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguity of the (±) and completely determine the missing value as the next example illustrates. Example 10.2.2. Using the given information about θ, ﬁnd the indicated value. 1. If θ is a Quadrant II angle with sin(θ) = 3 5 , ﬁnd cos(θ). 2. If π < θ < 3π 2 with cos(θ) = − √ 5 5 , ﬁnd sin(θ). 3. If sin(θ) = 1, ﬁnd cos(θ). Solution. 1. When we substitute sin(θ) = 3 5 into The Pythagorean Identity, cos2(θ) + sin2(θ) = 1, we obtain cos2(θ) + 9 5 . Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the x-coordinates are negative in Quadrant II, cos(θ) is too. Hence, cos(θ) = − 4 5 . 25 = 1. Solving, we ﬁnd cos(θ) = ± 4 √ √ 5 5 2. Substituting cos(θ) = − 5 . Since we 5 are given that π < θ < 3π 2 , we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sin(θ) = − 2 into cos2(θ) + sin2(θ) = 1 gives sin(θ) = ± 2. When we substitute sin(θ) = 1 into cos2(θ) + sin2(θ) = 1, we ﬁnd cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6 . We plot θ in standard position below and, as usual, let P (x, y) denote the poin
t on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos π 2 . This means 6 2 and sin π = 1 = √ 3 6 4This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5See Sections 1.1 and 7.2 for details. 10.2 The Unit Circle: Cosine and Sine 721 that the point on the terminal side of the angle π . From the ﬁgure below, it is clear that the point P (x, y) we seek can be obtained by reﬂecting that point about the y-axis. Hence, cos 5π 6 6 , when plotted in standard position, is 2 and sin 5π = x, y) θ = 5π = 5π is called the reference angle for the angle 5π In the above scenario, the angle π 6 . In general, for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x-axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (cos(θ), sin(θ)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle 722 Foundations of Trigonometry Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. In light of Theorem 10.2, it pays to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) cos(θ) sin(θ Example 10.2.3. Find the cosine and sine of the following angles. 1. θ = 225◦ 2. θ = 11π 6 3. θ = − 5π 4 4. θ = 7π 3 Solution. 1. We begin by plotting θ = 225◦ in standard position and ﬁnd its terminal side overshoots the negative x-axis to land in Quadrant III. Hence, we obtain θ’s reference angle α by subtracting: α = θ − 180◦ = 225◦ − 180◦ = 45◦. Since θ is a Quadrant III angle, both cos(θ) < 0 and 10.2 The Unit Circle: Cosine and Sine sin(θ) < 0. The Reference Angle Theorem yields: cos (225◦) = − cos (45◦) = − sin (225◦) = − sin (45◦) = − √ 2 2 . 723 √ 2 2 and 2. The terminal side of θ = 11π 6 , when plotted in standard position, lies in Quadrant IV, just shy of the positive x-axis. To ﬁnd θ’s reference angle α, we subtract: α = 2π − θ = 2π − 11π 6 = π 6 . Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem gives: cos 11π 6 2 and sin 11π = − sin π 6 = cos = 225◦ 45 = 11π 6 Finding cos (225◦) and sin (225◦) Finding cos 11π 6 and sin 11π 6 3. To plot θ = − 5π 4 , we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5π 4 − π = π 4 radians with respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos − 5π 4 = − cos π 4 = sin . Since the angle θ = 7π one full revolution followed by an additional α = 7π coterminal, cos 7π 3 2 and sin 7π = cos π 3 = 1 = sin π 3 3 3 , we ﬁnd the terminal side of θ by rotating 3 − 2π = π 3 radians. Since θ and α are √ = 3 2 . = − 2 and sin − 5π 3 measures more than 2π = 6π 4 y 1 π 4 1 x θ = − 5π 4 y 1 θ = 7π 3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian measure, the reference angle for a non-quadrantal angle is easy to spot. Reduced fraction multiples of π with a denominator of 6 have π 4 as their reference angle, and those with a denominator of 3 have π 3 as their reference angle.6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 722 can be used to generate the following ﬁgure, which the authors feel should be committed to memory. y (0, 1 5π 6 2π 3 3π (−1, 0) 0, 2π (1, 0) x 7π 6 11π 5π 4 4π 3 − 1 2 , − √ 3 2 7π 4 5π 3π 2 (0, −1) Important Points on the Unit Circle 6For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas discussed thus far in the section. Example 10.2.4. Suppose α is an acute angle with cos(α) = 5 13 . 1. Find sin(α) and use this to plot α in standard position. 2. Find the sine and cosine of the following angles: (a) θ = π + α (b) θ = 2π − α (c) θ = 3π − α (d) θ = π 2 + α Solution. 1. Proceeding as in Example 10.2.2, we substitute cos(α) = 5 ﬁnd sin(α) = ± 12 Hence, sin(α) = 12 x-axis to the ray which contains the point (cos(α), sin(α)) = 5 13 into cos2(α) + sin2(α) = 1 and 13 . Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. 13 . To plot α in standard position, we begin our rotation on the positive . 13 , 12 13 y 1 5 13 , 12 13 α 1 x Sketching α 2. (a) To ﬁnd the cosine and sine of θ = π + α, we ﬁrst plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians.7 We see that α is the reference angle for θ, so by 13 and sin(θ) = ± sin(α) = ± 12 The Reference Angle Theorem, cos(θ) = ± cos(α) = ± 5 13 . Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are negative, hence, cos(θ) = − 5 13 and sin(θ) = − 12 13 . 7Since π + α = α + π, θ may be plotted by reversing the order of rotations given here. You should do this. 726 Foundations of Trigonometry Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cos(θ) = 5 13 and sin(θ) = − 12 13 . y 1 y 1 θ θ 2π −α 1 x 1 x α Visualizing θ = 2π − α θ has reference angle α (c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cos(θ) = − 5 13 and sin(θ) = 12 13 . 10.2 The Unit Circle: Cosine and Sine 727 y 1 −α 3π y 1 θ α 1 x 1 x Visualizing 3π − α θ has reference angle α (d) To plot θ = π 2 + α, we ﬁrst rotate π 2 radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the ﬁgure below are equal. Hence, x = cos(θ) = − 12 13 . Similarly, we ﬁnd y = sin(θ) = 5 13 . 13 , 12 13 Q (x, y) α α 1 x 1 x Visualizing θ = π 2 + α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justiﬁed later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian measure will be the only appropriate angle measure so it is worth the time to become “ﬂuent in radians” now. 1. If cos(θ) = 1 Unit Circle at x = 1 2 , then the terminal side of θ, when plotted in standard position, intersects the 2 . This means θ is a Quadrant I or IV angle with reference angle One solution in Quadrant I is θ = π coterminal with π IV case, we ﬁnd the solution to cos(θ) = 1 θ = 5π 3 , we ﬁnd θ = π 3 + 2πk for integers k. 3 , and since all other Quadrant I solutions must be 3 + 2πk for integers k.9 Proceeding similarly for the Quadrant 3 , so our answer in this Quadrant is 2 here is 5π 2. If sin(θ) = − 1 2 , then when θ is plotted in standard position, its terminal side intersects the Unit Circle at y = − 1 2 . From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle π 6 . 8We will study trigonometric equations more formally in Section 10.7. Enjoy these relatively straightforward exercises while they last! 9Recall in Section 10.1, two angles in radian measure are coterminal if and only if they diﬀer by an integer multiple of 2π. Hence to describe all angles coterminal with a given angle, we add 2πk for integers k = 0, ±1, ±2, . . . . 10.2 The Unit Circle: Cosine and Sine 729 In Quadrant III, one solution is 7π multiples of 2π: θ = 7π are of the form θ = 11π 6 , so we capture all Quadrant III solutions by adding integer so all the solutions here 6 + 2πk. In Quadrant IV, one solution is 11π 6 + 2πk for integers k. 6 3. The angles with cos(θ) = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the y-axis While, technically speaking, π answers. If we follow the procedure set
forth in the previous examples, we ﬁnd θ = π and θ = 3π θ = π 2 isn’t a reference angle we can nonetheless use it to ﬁnd our 2 + 2πk 2 + 2πk for integers, k. While this solution is correct, it can be shortened to 2 + πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Example 10.2.5 is that, in general, solutions to trigonometric equations consist of inﬁnitely many answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, ‘When in doubt, write it out!’ This is especially important when checking answers to the exercises. For example, another Quadrant IV solution to 2 is θ = − π sin(θ) = − 1 6 . Hence, the family of Quadrant IV answers to number 2 above could just have easily been written θ = − π 6 + 2πk for integers k. While on the surface, this family may look 730 Foundations of Trigonometry diﬀerent than the stated solution of θ = 11π they represent the same list of angles. 6 + 2πk for integers k, we leave it to the reader to show 10.2.1 Beyond the Unit Circle We began the section with a quest to describe the position of a particle experiencing circular motion. In deﬁning the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x2 + y2 = r2, and let P (x, y) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,10 thus it follows that x 1 = r, so x = rx and, similarly, we ﬁnd y = ry. Since, by deﬁnition, x = cos(θ) and y = sin(θ), we get the coordinates of Q to be x = r cos(θ) and y = r sin(θ). By reﬂecting these points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. x = r y r 1 Q (x, y) P (x, y) θ y 1 1 x r P (x, y) Q(x, y) = (r cos(θ), r sin(θ)) θ O A(x, 0) B(x, 0) x Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of x2 + y2, we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. the circle is r = These results are summarized in the following theorem. Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle x2 + y2 = r2 then x = r cos(θ) and y = r sin(θ). Moreover, cos(θ) = x r = x x2 + y2 and sin(θ) = y r = y x2 + y2 10Do you remember why? 10.2 The Unit Circle: Cosine and Sine 731 Note that in the case of the Unit Circle we have r = our deﬁnitions of cos(θ) and sin(θ). x2 + y2 = 1, so Theorem 10.3 reduces to Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628◦ north latitude to be 2960 miles. Justify this approximation if the radius of the Earth at the Equator is approximately 3960 miles. Solution. 1. Using Theorem 10.3 with x = 4 and y = −2, we ﬁnd r = √ √ 5 5 5 . 5 and sin(θ) = y that cos(θ) = x r = −4)2 + (−2)2 = √ 20 = 2 √ 5 so 2. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 3960 miles. Viewing the Equator as the x-axis, the value we seek is the x-coordinate of the point Q(x, y) indicated in the ﬁgure below. y 4 2 −4 −2 2 4 x Q(4, −2) −2 −4 y 3960 Q (x, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we ﬁnd 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately 2960 miles. 732 Foundations of Trigonometry Theorem 10.3 gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocity ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point (r, 0) when t = 0, the angle θ is in standard position. By deﬁnition, ω = θ t which we rewrite as θ = ωt. According to Theorem 10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocity ω. If t = 0 corresponds to the point (r, 0), then the x and y coordinates of the object are functions of t and are given by x = r cos(ωt) and y = r sin(ωt). Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. y r 1 Q (x, y) = (r cos(ωt), r sin(ωt)) θ = ωt 1 x r Equations for Circular Motion Example 10.2.7. Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. Solution. From Example 10.1.5, we take r = 2960 miles and and ω = π of motion are x = r cos(ωt) = 2960 cos π measured in miles and t is measured in hours. 12 hours . Hence, the equations 12 t, where x and y are 12 t and y = r sin(ωt) = 2960 sin π In addition to circular motion, Theorem 10.3 is also the key to developing what is usually called ‘right triangle’ trigonometry.11 As we shall see in the sections to come, many applications in trigonometry involve ﬁnding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to ﬁnd the exact values of the cosine and sine of many of the angles in Example 10.2.1, so the following development shouldn’t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the 11You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. y c θ P (a, b) x c b c a θ According to the Pythagorean Theorem, a2 + b2 = c2, so that the point P (a, b) lies on a circle of radius c. Theorem 10.3 tells us that cos(θ) = a c , so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. c and sin(θ) = b Theorem 10.4. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then cos(θ) = and sin(θ) = a c b c . Example 10.2.8. Find the measure of the missing angle and the lengths of the missing sides of: 30◦ 7 Solution. The ﬁrst and easiest task is to ﬁnd the measure of the missing angle. Since the sum of angles of a triangle is 180◦, we know that the missing angle has measure 180◦ − 30◦ − 90◦ = 60◦. We now proceed to ﬁnd the lengths of the remaining two sides of the triangle. Let c denote the 7 length of the hypotenuse of the triangle. By Theorem 10.4, we have cos (30◦) = 7 cos(30◦) . 2 , we have, after the usual fraction gymnastics, c = 14 Since cos (30◦) = . At this point, we have two ways to proceed to ﬁnd the length of the side opposite the 30◦ angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14 , so we 3 c , or c = √ √ √ 3 3 3 3 734 Foundations of Trigonometry could use the Pythagorean Theorem to ﬁnd the missing side and solve (7)2 + b2 = Alternatively, we could use Theorem 10.4, namely that sin (30◦) = b b = c sin (30◦) = 14 3 √ 3 3 . The triangle with all of its data is recorded below. 2 = 7 · 1 √ 3 c . Choosing the latter, we ﬁnd 2 √ 14 3 3 for b. √ 3 c = 14 3 60◦ b = 7 √ 3 3 30◦ 7 We close this section by noting that we can easily extend the functions cosine and sine to real numbers by identifying a real number t with the angle θ = t radians. Using this identiﬁcation, we deﬁne cos(t) = cos(θ) and sin(t) = sin(θ). In practice this means expressions like cos(π) and sin(2) can be found by regarding the inputs as angles in radian measure or real numbers; the choice is the reader’s. If we trace the identiﬁcation of real numbers t with angles θ in radian measure to its roots on page 704, we can spell out this correspondence more precisely. For each real number t, we associate an oriented arc t units in length with initial point (1, 0) and endpoint P (cos(t), sin(t)). cos(t), sin(t)) θ = t 1 x In the same way we studied polynomial, rational, exponential, and logarithmic functions, we will study the trigonometric functions f (t) = cos(t) and g(t) = sin(t). The ﬁrst order of business is to ﬁnd the domains and ranges of these functions. Whether we think of identifying the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to ﬁnd coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are deﬁned for all real numbers t. In other words, the domain of f (t) = cos(t) and of g(t) = sin(t) is (−∞, ∞). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: C
osine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2 . As we have already mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sin(t) = − 1 2 in the exact same manner12 as we did in Example 10.2.5 number 2. Our solution is only cosmetically diﬀerent in that the variable used is t rather than θ: t = 7π 6 + 2πk for integers, k. We will study the cosine and sine functions in greater detail in Section 10.5. Until then, keep in mind that any properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure apply equally well if the inputs are regarded as real numbers. 6 + 2πk or t = 11π 12Well, to be pedantic, we would be technically using ‘reference numbers’ or ‘reference arcs’ instead of ‘reference angles’ – but the idea is the same. 736 Foundations of Trigonometry 10.2.2 Exercises In Exercises 1 - 20, ﬁnd the exact value of the cosine and sine of the given angle. 1. θ = 0 5. θ = 9. θ = 13. θ = 2π 3 5π 4 7π 4 17. θ = − 3π 4 2. θ = 6. θ = 10. θ = 14. θ = π 4 3π 4 4π 3 23π 6 18. θ = − π 6 3. θ = π 3 7. θ = π 11. θ = 3π 2 15. θ = − 13π 2 19. θ = 10π 3 4. θ = 8. θ = 12. θ = π 2 7π 6 5π 3 16. θ = − 43π 6 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to ﬁnd the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, what is cos(θ)? 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 with θ in Quadrant I, what is sin(θ)? with θ in Quadrant II, what is cos(θ)? 24. If cos(θ) = − 2 11 with θ in Quadrant III, what is sin(θ)? 25. If sin(θ) = − 2 3 with θ in Quadrant III, what is cos(θ)? 26. If cos(θ) = 27. If sin(θ) = 28. If cos(θ) = 28 53 √ 2 5 √ 10 10 with θ in Quadrant IV, what is sin(θ)? 5 and π 2 < θ < π, what is cos(θ)? and 2π < θ < 5π 2 3π 2 , what is sin(θ)? , what is cos(θ)? 29. If sin(θ) = −0.42 and π < θ < 30. If cos(θ) = −0.98 and π 2 < θ < π, what is sin(θ)? 10.2 The Unit Circle: Cosine and Sine 737 In Exercises 31 - 39, ﬁnd all of the angles which satisfy the given equation. 31. sin(θ) = 34. cos(θ) = 1 2 √ 2 2 32. cos(θ) = − √ 3 2 35. sin(θ) = √ 3 2 √ 3 2 33. sin(θ) = 0 36. cos(θ) = −1 39. cos(θ) = −1.001 37. sin(θ) = −1 38. cos(θ) = In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − 43. sin(t) = − 1 2 44. cos(t) = 1 2 √ 2 2 42. cos(t) = 3 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exercises 49 - 54, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 49. sin(78.95◦) 50. cos(−2.01) 51. sin(392.994) 52. cos(207◦) 53. sin (π◦) 54. cos(e) In Exercises 55 - 58, ﬁnd the measurement of the missing angle and the lengths of the missing sides. (See Example 10.2.8) 55. Find θ, b, and c. 56. Find θ, a, and c. c θ b 30◦ 1 45◦ c a θ 3 738 Foundations of Trigonometry 57. Find α, a, and b. 58. Find β, a, and c. b α 8 a 33◦ a 48◦ 6 c β In Exercises 59 - 64, assume that θ is an acute angle in a right triangle and use Theorem 10.4 to ﬁnd the requested side. 59. If θ = 12◦ and the side adjacent to θ has length 4, how long is the hypotenuse? 60. If θ = 78.123◦ and the hypotenuse has length 5280, how long is the side adjacent to θ? 61. If θ = 59◦ and the side opposite θ has length 117.42, how long is the hypotenuse? 62. If θ = 5◦ and the hypotenuse has length 10, how long is the side opposite θ? 63. If θ = 5◦ and the hypotenuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P (−7, 24) 66. Q(3, 4) 67. R(5, −9) 68. T (−2, −11) In Exercises 69 - 72, ﬁnd the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive x-axis. (See Equation 10.3 and Example 10.1.5.) 69. A point on the edge of the spinning yo-yo in Exercise 50 from Section 10.1. Recall: The diameter of the yo-yo is 2.25 inches and it spins at 4500 revolutions per minute. 70. The yo-yo in exercise 52 from Section 10.1. Recall: The radius of the circle is 28 inches and it completes one revolution in 3 seconds. 71. A point on the edge of the hard drive in Exercise 53 from Section 10.1. Recall: The diameter of the hard disk is 2.5 inches and it spins at 7200 revolutions per minute. 10.2 The Unit Circle: Cosine and Sine 739 72. A passenger on the Big Wheel in Exercise 55 from Section 10.1. Recall: The diameter is 128 feet and completes 2 revolutions in 2 minutes, 7 seconds. 73. Consider the numbers: 0, 1, 2, 3, 4. Take the square root of each of these numbers, then divide each by 2. The resulting numbers should look hauntingly familiar. (See the values in the table on 722.) 74. Let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sin(α) = cos(β) and sin(β) = cos(α). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 75. In the scenario of Equation 10.3, we assumed that at t = 0, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t0 > 0 is the ﬁrst time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t − t0)) and y = r sin(ω(t − t0)). 740 Foundations of Trigonometry 10.2.3 Answers 1. cos(0) = 1, sin(0) = 0 1 2 = 3. cos 5. cos π 3 2π 3 , sin π 3 = = − 1 2 , sin 2π 3 √ 3 2 √ 3 2 = 7. cos(π) = −1, sin(π) = 0 9. cos 11. cos 13. cos 5π 4 3π 2 7π 4 √ 2 2 = − , sin 5π 4 = − √ 2 2 = 0, sin 3π 2 = −1 √ 2 2 = , sin 7π 4 = − √ 2 2 2. cos 4. cos 6. cos 8. cos 10. cos 12. cos π 4 π 2 3π 4 7π 6 4π 3 5π 3 √ 2 2 = , sin π 4 = √ 2 2 = 0, sin , sin , sin 3π 4 7π , sin 4π 3 = − √ 3 2 = 1 2 , sin 5π 3 = − √ 3 2 14. cos 23π 6 √ 3 2 , sin 23π 6 = − 1 2 = 15. cos − 13π 2 = 0, sin − 13π 2 = −1 16. cos − 43π 6 √ 3 2 , sin − 43π 6 = 1 2 = − 17. cos 19. cos − 3π 4 10π 3 = − √ 2 2 , sin − 3π 4 √ 2 2 = − 18. cos − π 6 = √ 3 2 , sin − , sin 10π 3 = − √ 3 2 20. cos(117π) = −1, sin(117π) = 0 21. If sin(θ) = − 7 25 22. If cos(θ) = 23. If sin(θ) = 4 9 5 13 24. If cos(θ) = − 2 11 25. If sin(θ) = − 2 3 26. If cos(θ) = 28 53 with θ in Quadrant IV, then cos(θ) = with θ in Quadrant I, then sin(θ) = with θ in Quadrant II, then cos(θ) = − 12 13 . with θ in Quadrant III, then sin(θ) = − 24 25 . . √ 65 9 √ 117 11 . √ 5 3 . with θ in Quadrant III, then cos(θ) = − with θ in Quadrant IV, then sin(θ) = − 45 53 . 10.2 The Unit Circle: Cosine and Sine 741 27. If sin(θ) = 28. If cos(θ) = √ 2 5 5 √ 10 10 and π 2 < θ < π, then cos(θ) = − and 2π < θ < , then sin(θ) = √ . 5 5 √ 3 10 . 10 √ 5π 2 3π 2 29. If sin(θ) = −0.42 and π < θ < , then cos(θ) = − 0.8236 ≈ −0.9075. 30. If cos(θ) = −0.98 and < θ < π, then sin(θ) = √ 0.0396 ≈ 0.1990. π 2 31. sin(θ) = 1 2 32. cos(θ) = − √ 3 2 when θ = π 6 + 2πk or θ = 5π 6 + 2πk for any integer k. when θ = 5π 6 + 2πk or θ = 7π 6 + 2πk for any integer k. 33. sin(θ) = 0 when θ = πk for any integer k. 34. cos(θ) = 35. sin(θ) = √ 2 2 √ 3 2 when θ = when θ = π 4 π 3 + 2πk or θ = + 2πk or θ = 7π 4 2π 3 + 2πk for any integer k. + 2πk for any integer k. 36. cos(θ) = −1 when θ = (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 38. cos(θ) = √ 3 2 when θ = 3π 2 π 6 + 2πk for any integer k. + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 + πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any integer k. 42. cos(t) = 3 never happens. 43. sin(t) = − 1 2 when t = 7π 6 + 2πk or t = 11π 6 + 2πk for any integer k. 44. cos(t) = 1 2 when t = π 3 + 2πk or t = 5π 3 + 2πk for any integer k. 45. sin(t) = −2 never happens 46. cos(t) = 1 when t = 2πk for any integer k. 742 Foundations of Trigonometry π 2 + 2πk for any integer k. 47. sin(t) = 1 when t = 48. cos(t) = − √ 2 2 when t = 3π 4 + 2πk or t = 5π 4 + 2πk for any integer k. 49. sin(78.95◦) ≈ 0.981 50. cos(−2.01) ≈ −0.425 51. sin(392.994) ≈ −0.291 52. cos(207◦) ≈ −0.891 53. sin (π◦) ≈ 0.055 54. cos(e) ≈ −0.912 55. θ = 60◦, b = √ 3 3 , c = 56. θ = 45◦, a = 3 57. α = 57◦, a = 8 cos(33◦) ≈ 6.709, b = 8 sin(33◦) ≈ 4.357 58. β = 42◦, c = 6 sin(48◦) ≈ 8.074, a = √ c2 − 62 ≈ 5.402 59. The hypotenuse has length 4 cos(12◦) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123◦) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59◦) ≈ 136.99. 62. The side opposite θ has length 10 sin(5◦) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5◦) ≈ 9.962. 64. The hypotenuse has length c = √ c2 − 3062 ≈ 398.797. 306 sin(37.5◦) 65. cos(θ) = − 7 25 , sin(θ) = 24 25 ≈ 502.660, so the side adjacent to θ has length , sin(θ) = 4 5 66. cos(θ) = 67. cos(θ) = 3 5 √ 5 , sin(θ) = − 68. cos(θ) = − , sin(θ) = − 106 106 √ 5 2 25 √ 106 9 106 √ 5 11 25 69. r = 1.125 inches, ω = 9000π radians minute , x = 1.125 cos(9000π t), y = 1.125 sin(9000π t). Here x and y are measured in inches and t is measured in minutes. 10.2 The Unit Circle: Cosine and Sine 743 70. r = 28 inches, ω = 2π 3 radians second , x = 28 cos 2π 3 t, y = 28 sin 2π 3 t. Here x and y are measured in inches and t is measured in seconds. 71. r = 1.25 inches, ω = 14400π radians minute , x = 1.25 cos(14400π t), y = 1.25 sin(14400π t). Here x and y are measured in inches and t is measured in minutes. 127 t, y = 64 sin 4π second , x = 64 cos 4π 72. r = 64 feet, ω = 4π 12
7 in feet and t is measured in seconds radians 127 t. Here x and y are measured 744 Foundations of Trigonometry 10.3 The Six Circular Functions and Fundamental Identities In section 10.2, we deﬁned cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we deﬁne below. Deﬁnition 10.2. The Circular Functions: Suppose θ is an angle plotted in standard position and P (x, y) is the point on the terminal side of θ which lies on the Unit Circle. The cosine of θ, denoted cos(θ), is deﬁned by cos(θ) = x. The sine of θ, denoted sin(θ), is deﬁned by sin(θ) = y. The secant of θ, denoted sec(θ), is deﬁned by sec(θ) = 1 x , provided x = 0. The cosecant of θ, denoted csc(θ), is deﬁned by csc(θ) = The tangent of θ, denoted tan(θ), is deﬁned by tan(θ) = 1 y y x The cotangent of θ, denoted cot(θ), is deﬁned by cot(θ) = , provided y = 0. , provided x = 0. x y , provided y = 0. While we left the history of the name ‘sine’ as an interesting research project in Section 10.2, the names ‘tangent’ and ‘secant’ can be explained using the diagram below. Consider the acute angle θ below in standard position. Let P (x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y) denote the point on the terminal side of θ which lies on the vertical line x = 1. Q(1, y) = (1, tan(θ)) P (x, y) y 1 θ O A(x, 0) B(1, 0) x 1In Theorem 10.4 we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will ﬁnd that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 y = 1 x which gives y = y The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles ∆OP A and ∆OQB. Thus y x = tan(θ), where this last equality comes from applying Deﬁnition 10.2. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OP A is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar triangles that h x = sec(θ). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts oﬀ’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. x , or h = 1 1 = 1 Of the six circular functions, only cosine and sine are deﬁned for all angles. Since cos(θ) = x and sin(θ) = y in Deﬁnition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Deﬁnition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: sec(θ) = 1 cos(θ) , provided cos(θ) = 0; if cos(θ) = 0, sec(θ) is undeﬁned. csc(θ) = 1 sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, csc(θ) is undeﬁned. tan(θ) = sin(θ) cos(θ) , provided cos(θ) = 0; if cos(θ) = 0, tan(θ) is undeﬁned. cot(θ) = cos(θ) sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, cot(θ) is undeﬁned. It is high time for an example. Example 10.3.1. Find the indicated value, if it exists. 1. sec (60◦) 2. csc 7π 4 3. cot(3) 4. tan (θ), where θ is any angle coterminal with 3π 2 . 5. cos (θ), where csc(θ) = − √ 5 and θ is a Quadrant IV angle. 6. sin (θ), where tan(θ) = 3 and π < θ < 3π 2 . 2Compare this with the deﬁnition given in Section 2.1. 746 Solution. 1. According to Theorem 10.6, sec (60◦) = Foundations of Trigonometry 1 cos(60◦) . Hence, sec (60◦) = 1 (1/2) = 2. 2. Since sin 7π 4 = − √ 2 2 , csc 7π 4 = 1 sin( 7π 4 ) = 1 √ − 2/2 = − 2√ 2 √ = − 2. 3. Since θ = 3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we ﬁnd cot(3) = cos(3) sin(3) ≈ −7.015. = 0 and sin(θ) = sin 3π 2 = −1. Attempting 4. If θ is coterminal with 3π to compute tan(θ) = sin(θ) 2 , then cos(θ) = cos 3π cos(θ) results in −1 √ 2 0 , so tan(θ) is undeﬁned. 5. We are given that csc(θ) = 1 √ 5 5 so sin(θ) = − 1√ 5 . As we saw in Section 10.2, 5 we can use the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, to ﬁnd cos(θ) by knowing sin(θ). Substituting, we get cos2(θ) + θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2 = 1, which gives cos2(θ) = 4 √ 5 5 . 5 , or cos(θ) = ± 2 √ 5 5 . Since sin(θ. If tan(θ) = 3, then sin(θ) cos(θ) = 3. Be careful - this does NOT mean we can take sin(θ) = 3 and cos(θ) = 1. Instead, from sin(θ) cos(θ) = 3 we get: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we once again employ the Pythagorean Identity, cos2(θ) + sin2(θ) = 1. Solving sin(θ) = 3 cos(θ) for cos(θ), we ﬁnd cos(θ) = 1 3 sin(θ). Substituting this into the Pythagorean Identity, we ﬁnd sin2(θ) + 1 3 sin(θ)2 10 so sin(θ) = ± 3 10 . Since π < θ < 3π 2 , θ is a Quadrant III angle. This means sin(θ) < 0, so our ﬁnal answer is sin(θ) = − 3 = 1. Solving, we get sin2(θ) = 9 10 √ √ 10 10 . While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not always convenient to do so.3 It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. 3As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we’re dealt. 10.3 The Six Circular Functions and Fundamental Identities 747 Tangent and Cotangent Values of Common Angles θ(degrees) 0◦ 30◦ 45◦ 60◦ 90◦ θ(radians) tan(θ) cot(θ undeﬁned undeﬁned √ 3 1 √ 3 3 0 Coupling Theorem 10.6 with the Reference Angle Theorem, Theorem 10.2, we get the following. Theorem 10.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More speciﬁcally, if α is the reference angle for θ, then: cos(θ) = ± cos(α), sin(θ) = ± sin(α), sec(θ) = ± sec(α), csc(θ) = ± csc(α), tan(θ) = ± tan(α) and cot(θ) = ± cot(α). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. We put Theorem 10.7 to good use in the following example. Example 10.3.2. Find all angles which satisfy the given equation. 1. sec(θ) = 2 Solution. 2. tan(θ) = √ 3 3. cot(θ) = −1. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π or θ = 5π 3 + 2πk for integers k. 2 . This is the exact 3 + 2πk √ √ = 2. From the table of common values, we see tan π 3 3 must, therefore, have a reference angle of π 3. According to Theorem 10.7, we know 3 . Our next task is the solutions to tan(θ) = to determine in which quadrants the solutions to this equation lie. Since tangent is deﬁned as the ratio y x of points (x, y) on the Unit Circle with x = 0, tangent is positive when x and y have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = π 3 + 2πk for integers k, and for Quadrant III, we get θ = 4π 3 + 2πk for integers k. While these descriptions of the solutions are correct, they can be combined into one list as θ = π 3 + πk for integers k. The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.4 4See Example 10.2.5 number 3 in Section 10.2 for another example of this kind of simpliﬁcation of the solution. 748 Foundations of Trigonometry . From the table of common values, we see that π 4 has a cotangent of 1, which means the solutions to cot(θ) = −1 have a reference angle of π 4 . To ﬁnd the quadrants in which our solutions lie, we note that cot(θ) = x y for a point (x, y) on the Unit Circle where y = 0. If cot(θ) is negative, then x and y must have diﬀerent signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 + πk for integers k We have already seen the importance of identities in trigonometry. Our next task is to use use the Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(θ) = 0, we may start with cos2(θ) + sin2(θ) = 1 and divide both sides by cos2(θ) to obtain 1 + sin2(θ) cos2(θ) . Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + tan2(θ) = sec2(θ). If sin(θ) = 0, we can divide both sides of the identity cos2(θ) + sin2(θ) = 1 by sin2(θ), apply Theorem 10.6 once again, and obtain cot2(θ) + 1 = csc2(θ). These three Pythagorean Identities are
worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos2(θ) + sin2(θ) = 1. Common Alternate Forms: 1 − sin2(θ) = cos2(θ) 1 − cos2(θ) = sin2(θ) 2. 1 + tan2(θ) = sec2(θ), provided cos(θ) = 0. Common Alternate Forms: sec2(θ) − tan2(θ) = 1 sec2(θ) − 1 = tan2(θ) 3. 1 + cot2(θ) = csc2(θ), provided sin(θ) = 0. Common Alternate Forms: csc2(θ) − cot2(θ) = 1 csc2(θ) − 1 = cot2(θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to ﬁnd the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are deﬁned. 1. 1 csc(θ) = sin(θ) 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 5. 6 sec(θ) tan(θ) = 3 1 − sin(θ) − 3 1 + sin(θ) 2. tan(θ) = sin(θ) sec(θ) 4. 6. sec(θ) 1 − tan(θ) = 1 cos(θ) − sin(θ) sin(θ) 1 − cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ) , we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove. 750 Foundations of Trigonometry 2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and ﬁnd: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ). According to Theorem 10.8, sec2(θ) − tan2(θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ) = 1. 4. While both sides of our last identity contain fractions, the left side aﬀords us more opportu- nities to use our identities.5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ) , we get: sec(θ) 1 − tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) · cos(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) (cos(θ)) = (cos(θ)) (1)(cos(θ)) − 1 sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ) , which is exactly what we had set out to show. 5. The right hand side of the equation seems to hold more promise. We get common denomina- tors and add: 3 1 − sin(θ) − 3 1 + sin(θ) = = = = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − sin(θ)) 3 + 3 sin(θ) 1 − sin2(θ) − 3 − 3 sin(θ) 1 − sin2(θ) (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2(θ) 6 sin(θ) 1 − sin2(θ) 5Or, to put to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we ﬁnd 6 sec(θ) tan(θ) = 6 . In other words, we need to get cosines in our denominator. Theorem 10.8 tells us 1 − sin2(θ) = cos2(θ) so we get: sin(θ) cos(θ) 1 cos(θ) 3 1 − sin(θ) − 3 1 + sin(θ) = = 6 sin(θ) cos2(θ) 6 sin(θ) 1 − sin2(θ) 1 cos(θ) = 6 sin(θ) cos(θ) = 6 sec(θ) tan(θ) 6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ) 1 + cos(θ) sin(θ) In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a diﬀerence of squares that can be simpliﬁed to one term using Theorem 10.8. This is exactly the √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other common Pythagorean Conjugates. 2 by 1 + √ Pythagorean Conjugates 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos2(θ) = sin2(θ) 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin2(θ) = cos2(θ) sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec2(θ) − 1 = tan2(θ) sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec2(θ)−tan2(θ) = 1 csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc2(θ) − 1 = cot2(θ) csc(θ) − cot(θ) and csc(θ) + cot(θ): (csc(θ) − cot(θ))(csc(θ) + cot(θ)) = csc2(θ) − cot2(θ) = 1 752 Foundations of Trigonometry Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proﬁcient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Try working on the more complicated side of the identity. Strategies for Verifying Identities Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. Add rational expressions with unlike denominators by obtaining common denominators. Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or diﬀerences of squares to one term. Multiply numerator and denominator by Pythagorean Conjugates in order to take advan- tage of the Pythagorean Identities in Theorem 10.8. If you ﬁnd yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can ﬁnd a way to bridge the two parts of your work. 10.3.1 Beyond the Unit Circle In Section 10.2, we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 10.3 in conjunction with Theorem 10.8, we generalize the remaining circular functions in kind. Theorem 10.9. Suppose Q(x, y) is the point on the terminal side of an angle θ (plotted in standard position) which lies on the circle of radius r, x2 + y2 = r2. Then: r x r y = = x2 + y2 x x2 + y2 y , provided x = 0. , provided y = 0. , provided x = 0. , provided y = 0. sec(θ) = csc(θ) = tan(θ) = cot(θ) = y x x y 10.3 The Six Circular Functions and Fundamental Identities 753 Example 10.3.4. 1. Suppose the terminal side of θ, when plotted in standard position, contains the point Q(3, −4). Find the values of the six circular functions of θ. 2. Suppose θ is a Quadrant IV angle with cot(θ) = −4. Find the values of the ﬁve remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = x2 + y2 = (3)2 + (−4)2 = √ cos(θ) = 3 5 , sin(θ) = − 4 5 , sec(θ) = 5 3 , csc(θ) = − 5 4 , tan(θ) = − 4 25 = 5. Theorem 10.9 tells us 3 and cot(θ) = − 3 4 . 2. In order to use Theorem 10.9, we need to ﬁnd a point Q(x, y) which lies on the terminal side y , and since θ is a −1 , we may choose6 x = 4 17. Applying Theorem 10.9 once √ √ 17 17 4 , csc(θ) = − 17 17 , sec(θ) = of θ, when θ is plotted in standard position. We have that cot(θ) = −4 = x Quadrant IV angle, we also know x > 0 and y < 0. Viewing −4 = 4 x2 + y2 = and y = −1 so that r = √ = 4 more, we ﬁnd cos(θ) = 4√ 17 and tan(θ) = − 1 4 . (4)2 + (−1)2 = = − 17 , sin(θ) = − 1√ √ √ 17 17 We may also specialize Theorem 10.9 to the case of acute angles θ which reside in a right triangle, as visualized below. b c a θ Theorem 10.10. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, then tan(θ) = sec(θ) = csc(θ) = cot(θ The following example uses Theorem 10.10 as well as the concept of an ‘angle of inclination.’ The angle of inclination (or angle of elevation) of an object refers to the angle whose initial side is some kind of base-line (say, the ground), and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematically below. 6We may choose any values x and y so long as x > 0, y < 0 and x y = −4. For example, we could choose x = 8 and y = −2. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope − 1 4 which extends from the origin into Quadrant IV. 754 Foundations of Trigonometry object θ ‘base line’ The angle of inclination from the base line to the object is θ Example 10.3.5. 1. The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower7 is 60◦. Find the height of the Clocktower to the nearest foot. 2. In order to determine the height of a California Redwood tree, two sightings from the ground, one 200 feet directly behind the other, are made. If the angles of inclination were 45◦ and 30◦, respectively, how tall is the tree to the nearest foot? Solution. 1. We can represent the problem situation using a right triangle as shown below. If we let h 30 . From this we get denote the height of the tower, then Theorem 10.10 gives tan (60◦) = h h = 30 tan (60◦) = 30 3 ≈ 51.96. Hence, the Clocktower is approximately 52 feet tall. √ h ft. 60◦ 30 ft. Finding the height of t
he Clocktower 2. Sketching the problem situation below, we ﬁnd ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the ﬁrst observation point. 7Named in honor of Raymond Q. Armington, Lakeland’s Clocktower has been a part of campus since 1972. 10.3 The Six Circular Functions and Fundamental Identities 755 h ft. 30◦ 200 ft. 45◦ x ft. Finding the height of a California Redwood Using Theorem 10.10, we get a pair of equations: tan (45◦) = h x+200 . Since tan (45◦) = 1, the ﬁrst equation gives h x = 1, or x = h. Substituting this into the second equation gives 3. The result is a linear equation for h, so we proceed to expand the right hand side and gather all the terms involving h to one side. √ 3 3 . Clearing fractions, we get 3h = (h + 200) x and tan (30◦) = h h+200 = tan (30◦) = √ h 3h = (h + 200) √ 3h = h √ 3 + 200 √ 3 = 200 3 √ 3 √ 3 3h − h √ (3 − 3)h = 200 √ 3 √ 3 √ 3 ≈ 273.20 Hence, the tree is approximately 273 feet tall. h = 200 3 − As we did in Section 10.2.1, we may consider all six circular functions as functions of real numbers. At this stage, there are three equivalent ways to deﬁne the functions sec(t), csc(t), tan(t) and cot(t) for real numbers t. First, we could go through the formality of the wrapping function on page 704 and deﬁne these functions as the appropriate ratios of x and y coordinates of points on the Unit Circle; second, we could deﬁne them by associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastly, we could simply deﬁne them using the Reciprocal and Quotient Identities as combinations of the functions f (t) = cos(t) and g(t) = sin(t). Presently, we adopt the last approach. We now set about determining the domains and ranges of the remaining four circular functions. Consider the function F (t) = sec(t) deﬁned as F (t) = sec(t) = 1 cos(t) . We know F is undeﬁned whenever cos(t) = 0. From Example 10.2.5 number 3, we know cos(t) = 0 whenever t = π 2 + πk for integers k. Hence, our domain for F (t) = sec(t), in set builder notation is {t : t = π 2 + πk, for integers k}. To get a better understanding what set of real numbers we’re dealing with, it pays to write out and graph this set. Running through a few values of k, we ﬁnd the domain to be {t : t = ± π 2 , . . .}. Graphing this set on the number line we get 2 , ± 5π 2 , ± 3π 756 Foundations of Trigonometry − 5π 2 − 3π 2 − π 2 0 π 2 3π 2 5π 2 Using interval notation to describe this set, we get . . . ∪ − 5π 2 , − 3π 2 ∪ − 3π , 3π 2 ∪ 3π 2 , 5π 2 ∪ . . . This is cumbersome, to say the least! In order to write this in a more compact way, we note that from the set-builder description of the domain, the kth point excluded from the domain, which we’ll call xk, can be found by the formula xk = π 2 +πk. (We are using sequence notation from Chapter 9.) Getting a common denominator and factoring out the π in the numerator, we get xk = (2k+1)π . The . domain consists of the intervals determined by successive points xk: (xk, xk + 1) = In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±1, ±2, . . . . The way we denote taking the union of inﬁnitely many intervals like this is to use what we call in this text extended interval notation. The domain of F (t) = sec(t) can now be written as 2 , (2k+3)π 2 (2k+1)π 2 ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 The reader should compare this notation with summation notation introduced in Section 9.2, in particular the notation used to describe geometric series in Theorem 9.2. In the same way the index k in the series ∞ k=1 ark−1 can never equal the upper limit ∞, but rather, ranges through all of the natural numbers, the index k in the union ∞ (2k + 1)π 2 , (2k + 3)π 2 k=−∞ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F (t) = sec(t), it is time to discuss the range. Once again, we appeal to the deﬁnition F (t) = sec(t) = 1 cos(t) . The range of f (t) = cos(t) is [−1, 1], and since F (t) = sec(t) is undeﬁned when cos(t) = 0, we split our discussion into two cases: when 0 < cos(t) ≤ 1 and when −1 ≤ cos(t) < 0. If 0 < cos(t) ≤ 1, then we can divide the inequality cos(t) ≤ 1 by cos(t) to obtain sec(t) = 1 cos(t) ≥ 1. Moreover, using the notation introduced in Section 4.2, we have that as cos(t) → 0+, sec(t) = 1 very small (+) ≈ very big (+). In other words, as cos(t) → 0+, sec(t) → ∞. If, on the other hand, if −1 ≤ cos(t) < 0, then dividing by cos(t) causes a reversal of the inequality so that sec(t) = 1 sec(t) ≤ −1. In this case, as cos(t) → 0−, sec(t) = 1 very small (−) ≈ very big (−), so that as cos(t) → 0−, we get sec(t) → −∞. Since cos(t) ≈ cos(t) ≈ 1 1 10.3 The Six Circular Functions and Fundamental Identities 757 f (t) = cos(t) admits all of the values in [−1, 1], the function F (t) = sec(t) admits all of the values in (−∞, −1] ∪ [1, ∞). Using set-builder notation, the range of F (t) = sec(t) can be written as {u : u ≤ −1 or u ≥ 1}, or, more succinctly,8 as {u : \|u\| ≥ 1}.9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csc(t), tan(t) and cot(t). The reader is encouraged to do so. (See the Exercises.) For now, we gather these facts into the theorem below. Theorem 10.11. Domains and Ranges of the Circular Functions • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] • The function F (t) = sec(t) = 1 cos(t) – has domain {t : t = π 2 + πk, for integers k} = – has range {u : \|u\| ≥ 1} = (−∞, −1] ∪ [1, ∞) ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 • The function G(t) = csc(t) = 1 sin(t) – has domain {t : t = πk, for integers k} = ∞ k=−∞ (kπ, (k + 1)π) – has range {u : \|u\| ≥ 1} = (−∞, −1] ∪ [1, ∞) • The function J(t) = tan(t) = sin(t) cos(t) – has domain {t : t = π 2 + πk, for integers k} = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range (−∞, ∞) • The function K(t) = cot(t) = cos(t) sin(t) – has domain {t : t = πk, for integers k} = – has range (−∞, ∞) ∞ k=−∞ (kπ, (k + 1)π) 8Using Theorem 2.4 from Section 2.4. 9Notice we have used the variable ‘u’ as the ‘dummy variable’ to describe the range elements. While there is no mathematical reason to do this (we are describing a set of real numbers, and, as such, could use t again) we choose u to help solidify the idea that these real numbers are the outputs from the inputs, which we have been calling t. 758 Foundations of Trigonometry We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 735 in Section 10.2.1 concerning solving equations applies to all six circular functions, not just f (t) = cos(t) and g(t) = sin(t). In particular, to solve the equation cot(t) = −1 for real numbers t, we can use the same thought process we used in Example 10.3.2, number 3 to solve cot(θ) = −1 for angles θ in radian measure – we just need to remember to write our answers using the variable t as opposed to θ. Next, it is critical that you know the domains and ranges of the six circular functions so that you know which equations have no solutions. For example, sec(t) = 1 2 is not in the range of secant. Finally, you will need to review the notions of reference angles and coterminal angles so that you can see why csc(t) = −42 has an inﬁnite set of solutions in Quadrant III and another inﬁnite set of solutions in Quadrant IV. 2 has no solution because 1 10.3 The Six Circular Functions and Fundamental Identities 759 10.3.2 Exercises In Exercises 1 - 20, ﬁnd the exact value or state that it is undeﬁned. 1. tan π 4 5. tan − 11π 6 9. tan (117π) 13. tan 17. tan 31π 2 2π 3 2. sec π 6 6. sec − 10. sec − 3π 2 5π 3 14. sec π 4 18. sec (−7π) 3. csc 7. csc 5π 6 − π 3 4. cot 8. cot 4π 3 13π 2 11. csc (3π) 12. cot (−5π) 15. csc − 7π 4 19. csc π 2 16. cot 20. cot 7π 6 3π 4 In Exercises 21 - 34, use the given the information to ﬁnd the exact values of the remaining circular functions of θ. 21. sin(θ) = 3 5 with θ in Quadrant II 22. tan(θ) = 12 5 with θ in Quadrant III with θ in Quadrant I 24. sec(θ) = 7 with θ in Quadrant IV 25. csc(θ) = − with θ in Quadrant III 26. cot(θ) = −23 with θ in Quadrant II 27. tan(θ) = −2 with θ in Quadrant IV. 28. sec(θ) = −4 with θ in Quadrant II. 29. cot(θ) = √ 5 with θ in Quadrant III. 30. cos(θ) = 1 3 with θ in Quadrant I. 31. cot(θ) = 2 with 0 < θ < π 2 . 33. tan(θ) = √ 10 with π < θ < 3π 2 . 32. csc(θ) = 5 with π 2 < θ < π. 34. sec(θ) = 2 √ 5 with 3π 2 < θ < 2π. In Exercises 35 - 42, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode! 35. csc(78.95◦) 36. tan(−2.01) 37. cot(392.994) 38. sec(207◦) 39. csc(5.902) 40. tan(39.672◦) 41. cot(3◦) 42. sec(0.45) 23. csc(θ) = 25 24 10 91 √ 91 760 Foundations of Trigonometry In Exercises 43 - 57, ﬁnd all of the angles which satisfy the equation. 43. tan(θ) = √ 3 44. sec(θ) = 2 45. csc(θ) = −1 46. cot(θ) = √ 3 3 47. tan(θ) = 0 48. sec(θ) = 1 49. csc(θ) = 2 50. cot(θ) = 0 51. tan(θ) = −1 52. sec(θ) = 0 53. csc(θ) = − 55. tan(θ) = − √ 3 56. csc(θ) = −2 57. cot(θ) = −1 1 2 54. sec(θ) = −1 In Exercises 58 - 65, solve the equation for t. Give exact values. 58. cot(t) = 1 59. tan(t) = 62. cot(t) = − √ 3 63. tan(t 60. sec(t) = − 64. sec(t) = √ 2 3 61. csc(t) = 0 65. csc(t) = √ 2 3 3 In Exercises 66 - 69, use Theorem 10.10 to ﬁnd the requested quantities. 66. Find θ, a, and c. 67. Find α, b, and c. 60◦ a c θ 9 b α c 12 34◦ 68. Find θ, a, and c. 69. Find β, b, and c. 47◦ c a θ 6 b β c 2.5 50◦ 10.3 The Six Circular Functions and Fundamental Identities 761 In Exercises 70 - 75, use Theorem 10.10 to answer the question. Assume that θ is an angle in a right triangle. 70. If θ = 30◦ and the side opposite θ has length 4, how long is the side adja
cent to θ? 71. If θ = 15◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2◦ and the side opposite θ has lengh 14, how long is the hypoteneuse? 74. If θ = 2.05◦ and the hypotenuse has length 3.98, how long is the side adjacent to θ? 75. If θ = 42◦ and the side adjacent to θ has length 31, how long is the side opposite θ? 76. A tree standing vertically on level ground casts a 120 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is 21.4◦. Find the height of the tree to the nearest foot. With the help of your classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ (Home of “Algebra in the Morning with Carl and Jeﬀ”) has two enormous ﬂashing red lights on it: one at the very top and one a few feet below the top. From a point 5000 feet away from the base of the tower on level ground the angle of elevation to the top light is 7.970◦ and to the second light is 7.125◦. Find the distance between the lights to the nearest foot. 78. On page 753 we deﬁned the angle of inclination (also known as the angle of elevation) and in this exercise we introduce a related angle - the angle of depression (also known as the angle of declination). The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematically below. horizontal θ observer object The angle of depression from the horizontal to the object is θ (a) Show that if the horizontal is above and parallel to level ground then the angle of depression (from observer to object) and the angle of inclination (from object to observer) will be congruent because they are alternate interior angles. 762 Foundations of Trigonometry (b) From a ﬁretower 200 feet above level ground in the Sasquatch National Forest, a ranger spots a ﬁre oﬀ in the distance. The angle of depression to the ﬁre is 2.5◦. How far away from the base of the tower is the ﬁre? (c) The ranger in part 78b sees a Sasquatch running directly from the ﬁre towards the ﬁretower. The ranger takes two sightings. At the ﬁrst sighting, the angle of depression from the tower to the Sasquatch is 6◦. The second sighting, taken just 10 seconds later, gives the the angle of depression as 6.5◦. How far did the Saquatch travel in those 10 seconds? Round your answer to the nearest foot. How fast is it running in miles per hour? Round your answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the ﬁretower from his location at the second sighting? Round your answer to the nearest minute. 79. When I stand 30 feet away from a tree at home, the angle of elevation to the top of the tree is 50◦ and the angle of depression to the base of the tree is 10◦. What is the height of the tree? Round your answer to the nearest foot. 80. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directly toward the lighthouse. The ﬁrst sighting had an angle of depression of 8.2◦ and the second sighting had an angle of depression of 25.9◦. How far had the boat traveled between the sightings? 81. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 43◦ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground? In Exercises 82 - 128, verify the identity. Assume that all quantities are deﬁned. 82. cos(θ) sec(θ) = 1 84. sin(θ) csc(θ) = 1 86. csc(θ) cos(θ) = cot(θ) 88. 90. 92. cos(θ) sin2(θ) = csc(θ) cot(θ) 1 − cos(θ) sin(θ) = csc(θ) − cot(θ) sin(θ) 1 − cos2(θ) = csc(θ) 83. tan(θ) cos(θ) = sin(θ) 85. tan(θ) cot(θ) = 1 87. 89. 91. 93. sin(θ) cos2(θ) = sec(θ) tan(θ) 1 + sin(θ) cos(θ) = sec(θ) + tan(θ) cos(θ) 1 − sin2(θ) sec(θ) 1 + tan2(θ) = sec(θ) = cos(θ) 10.3 The Six Circular Functions and Fundamental Identities 763 94. 96. csc(θ) 1 + cot2(θ) cot(θ) csc2(θ) − 1 = sin(θ) = tan(θ) 95. tan(θ) sec2(θ) − 1 = cot(θ) 97. 4 cos2(θ) + 4 sin2(θ) = 4 98. 9 − cos2(θ) − sin2(θ) = 8 99. tan3(θ) = tan(θ) sec2(θ) − tan(θ) 100. sin5(θ) = 1 − cos2(θ)2 sin(θ) 101. sec10(θ) = 1 + tan2(θ)4 sec2(θ) 102. cos2(θ) tan3(θ) = tan(θ) − sin(θ) cos(θ) 103. sec4(θ) − sec2(θ) = tan2(θ) + tan4(θ) 104. 106. cos(θ) + 1 cos(θ) − 1 1 − cot(θ) 1 + cot(θ) = = 1 + sec(θ) 1 − sec(θ) tan(θ) − 1 tan(θ) + 1 105. sin(θ) + 1 sin(θ) − 1 = 1 + csc(θ) 1 − csc(θ) 107. 1 − tan(θ) 1 + tan(θ) = cos(θ) − sin(θ) cos(θ) + sin(θ) 108. tan(θ) + cot(θ) = sec(θ) csc(θ) 109. csc(θ) − sin(θ) = cot(θ) cos(θ) 110. cos(θ) − sec(θ) = − tan(θ) sin(θ) 111. cos(θ)(tan(θ) + cot(θ)) = csc(θ) 112. sin(θ)(tan(θ) + cot(θ)) = sec(θ) 113. 1 1 − cos(θ) + 1 1 + cos(θ) = 2 csc2(θ) 114. 1 sec(θ) + 1 + 1 sec(θ) − 1 = 2 csc(θ) cot(θ) 115. 1 csc(θ) + 1 + 1 csc(θ) − 1 = 2 sec(θ) tan(θ) 116. 1 csc(θ) − cot(θ) − 1 csc(θ) + cot(θ) = 2 cot(θ) 117. cos(θ) 1 − tan(θ) + sin(θ) 1 − cot(θ) = sin(θ) + cos(θ) 118. 120. 122. 124. 126. 128. 1 sec(θ) + tan(θ) 1 csc(θ) − cot(θ) = sec(θ) − tan(θ) = csc(θ) + cot(θ) 1 1 − sin(θ) 1 1 − cos(θ) = sec2(θ) + sec(θ) tan(θ) = csc2(θ) + csc(θ) cot(θ) cos(θ) 1 + sin(θ) = 1 − sin(θ) cos(θ) 1 − sin(θ) 1 + sin(θ) = (sec(θ) − tan(θ))2 119. 121. 123. 125. 1 sec(θ) − tan(θ) 1 csc(θ) + cot(θ) = sec(θ) + tan(θ) = csc(θ) − cot(θ) 1 1 + sin(θ) 1 1 + cos(θ) = sec2(θ) − sec(θ) tan(θ) = csc2(θ) − csc(θ) cot(θ) 127. csc(θ) − cot(θ) = sin(θ) 1 + cos(θ) 764 Foundations of Trigonometry In Exercises 129 - 132, verify the identity. You may need to consult Sections 2.2 and 6.2 for a review of the properties of absolute value and logarithms before proceeding. 129. ln \| sec(θ)\| = − ln \| cos(θ)\| 130. − ln \| csc(θ)\| = ln \| sin(θ)\| 131. − ln \| sec(θ) − tan(θ)\| = ln \| sec(θ) + tan(θ)\| 132. − ln \| csc(θ) + cot(θ)\| = ln \| csc(θ) − cot(θ)\| 133. Verify the domains and ranges of the tangent, cosecant and cotangent functions as presented in Theorem 10.11. 134. As we did in Exercise 74 in Section 10.2, let α and β be the two acute angles of a right triangle. (Thus α and β are complementary angles.) Show that sec(α) = csc(β) and tan(α) = cot(β). The fact that co-functions of complementary angles are equal in this case is not an accident and a more general result will be given in Section 10.4. 135. We wish to establish the inequality cos(θ) < < 1 for 0 < θ < . Use the diagram from the beginning of the section, partially reproduced below, to answer the following. sin(θ(1, 0) x (a) Show that triangle OP B has area 1 2 sin(θ). (b) Show that the circular sector OP B with central angle θ has area (c) Show that triangle OQB has area 1 2 tan(θ). (d) Comparing areas, show that sin(θ) < θ < tan(θ) for 0 < θ < π 2 . (e) Use the inequality sin(θ) < θ to show that sin(θ) θ < 1 for 0 < θ < 1 2 θ. π 2 . (f) Use the inequality θ < tan(θ) to show that cos(θ) < with the previous part to complete the proof. sin(θ) θ for 0 < θ < π 2 . Combine this 10.3 The Six Circular Functions and Fundamental Identities 765 136. Show that cos(θ) < sin(θ) θ < 1 also holds for − π 2 < θ < 0. 137. Explain why the fact that tan(θ) = 3 = 3 solution to number 6 in Example 10.3.1.) 1 does not mean sin(θ) = 3 and cos(θ) = 1? (See the 766 Foundations of Trigonometry 10.3.3 Answers 1. tan 4. cot π 4 4π 3 = 1 = 7. csc − π 3 = − = 2 √ 3 = 10. sec 13. tan 16. cot 19. csc − 5π 3 31π 2 7π 6 π 2 = 1 2. sec 5. tan 8. cot 3 = π 6 − 11π 6 13π . csc 5π 6 = 2 6. sec − 3π 2 is undeﬁned 9. tan (117π) = 0 11. csc (3π) is undeﬁned 12. cot (−5π) is undeﬁned √ 3 3 √ 2 3 is undeﬁned 14. sec 15. csc − √ 2 = 7π 4 18. sec (−7π) = −1 = π 4 2π 3 3π 4 √ 2 √ = − 3 = −1 17. tan 20. cot 21. sin(θ) = 3 5 , cos(θ) = − 4 5 , tan(θ) = − 3 4 , csc(θ) = 5 3 , sec(θ) = − 5 4 , cot(θ) = − 4 3 22. sin(θ) = − 12 13 , cos(θ) = − 5 13 , tan(θ) = 12 5 , csc(θ) = − 13 12 , sec(θ) = − 13 5 , cot(θ) = 5 12 23. sin(θ) = 24 25 , cos(θ) = 7 25 , tan(θ) = 24 24 , sec(θ) = 25 7 , cot(θ) = 7 24 7 , csc(θ) = 25 √ 3 , cos(θ) = 1 7 , tan(θ) = −4 3, csc(θ) = − 7 √ 12 , sec(θ) = 7, cot(θ) = − 3 √ 3 12 √ 24. sin(θ) = −4 7 √ 25. sin(θ) = − √ 26. sin(θ) = 91 10 , cos(θ) = − 3 530 , cos(θ) = − 23 530 10 , tan(θ) = √ 91 3 , csc(θ) = − 10 91 √ 530 , tan(θ) = − 1 530 23 , csc(θ) = 3 , cot(θ) = 3 , sec(θ) = − 10 √ 530 23 , cot(θ) = −23 91 91 √ √ 91 √ 27. sin(θ) = − 2 √ 28. sin(θ) = √ 5 5 , cos(θ) = 4 , cos(θ) = − 1 15 29. sin(θ) = − √ 30. sin(θ) = 2 √ 31. sin(θ) = 32. sin(θ) = 1 2 √ 6 6 , cos(θ) = − 3 , cos(θ) = 1 5 , cos(θ) = 2 5 , cos(θ) = − 2 5 √ 5 5 , tan(θ) = −2, csc(θ) = − √ 4 , tan(θ) = − √ 30 6 , tan(θ) = √ 15, csc(θ) = 4 √ 5 5 , csc(θ) = − 530, sec(θ) = − √ 5, cot(θ) = − 1 2 √ 5 2 , sec(θ) = √ 15 15 , sec(θ) = −4, cot(θ) = − √ √ 30 5 , cot(θ) = 6, sec(θ) = − √ 15 15 √ 5 3 , tan(θ) = 2 √ 5 , tan(θ) = 1 5 2, csc(θ) = 3 √ 2 , csc(θ) = √ 2 4 , sec(θ) = 3, cot(θ) = √ 5 2 , cot(θ) = 2 5, sec(θ) = √ 2 4 √ 6 5 , tan(θ) = − √ 6 12 , csc(θ) = 5, sec(θ) = − 5 √ 12 , cot(θ) = −2 6 √ 6 33. sin(θ) = − 34. sin(θ) = − √ 110 11 , cos(θ) = − √ √ 95 5 10 , tan(θ) = − 10 , cos(θ) = √ 11 11 , tan(θ) = √ √ 10, csc(θ) = − √ 95 19 , sec(θ) = 2 √ 110 10 , sec(θ) = − √ 19, csc(θ) = − 2 √ 11, cot(θ) = √ 5, cot(θ) = − 19 19 √ 10 10 10.3 The Six Circular Functions and Fundamental Identities 767 35. csc(78.95◦) ≈ 1.019 37. cot(392.994) ≈ 3.292 39. csc(5.902) ≈ −2.688 41. cot(3◦) ≈ 19.081 43. tan(θ) = √ 3 when θ = π 3 36. tan(−2.01) ≈ 2.129 38. sec(207◦) ≈ −1.122 40. tan(39.672◦) ≈ 0.829 42. sec(0.45) ≈ 1.111 + πk for any integer k 44. sec(θ) = 2 when θ = π 3 + 2πk or θ = 5π 3 + 2πk for any integer k 45. csc(θ) = −1 when θ = 46. cot(θ) = √ 3 3 when θ = 3π 2 π 3 + 2πk for any integer k. + πk for any integer k 47. tan(θ) = 0 when θ = πk for any integer k 48. sec(θ) = 1 when θ = 2πk for any integer k 49. csc(θ) = 2 whe
n θ = 50. cot(θ) = 0 when θ = π 6 π 2 51. tan(θ) = −1 when θ = + 2πk or θ = 5π 6 + 2πk for any integer k. + πk for any integer k 3π 4 + πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 + πk for any integer k 56. csc(θ) = −2 when θ = 57. cot(θ) = −1 when θ = 7π 6 3π 4 + 2πk or θ = 11π 6 + 2πk for any integer k + πk for any integer k 58. cot(t) = 1 when t = π 4 + πk for any integer k 59. tan(t) = √ 3 3 when t = π 6 + πk for any integer k 768 Foundations of Trigonometry 60. sec(t) = − √ 2 3 3 when t = 5π 6 + 2πk or t = 7π 6 61. csc(t) = 0 never happens + 2πk for any integer k 62. cot(t) = − √ 3 when t = 5π 6 5π 6 + πk for any integer k + πk for any integer k when t = √ 3 3 63. tan(t) = − 64. sec(t) = 65. csc(t when t = π 6 + 2πk or t = 11π 6 + 2πk for any integer k when t = + 2πk or t = 2π 3 + 2πk for any integer k π 3 √ 66. θ = 30◦, a = 3 √ 3, c = √ 108 = 6 3 67. α = 56◦, b = 12 tan(34◦) = 8.094, c = 12 sec(34◦) = 12 cos(34◦) ≈ 14.475 68. θ = 43◦, a = 6 cot(47◦) = 6 tan(47◦) ≈ 5.595, c = 6 csc(47◦) = 6 sin(47◦) ≈ 8.204 69. β = 40◦, b = 2.5 tan(50◦) ≈ 2.979, c = 2.5 sec(50◦) = 2.5 cos(50◦) ≈ 3.889 70. The side adjacent to θ has length 4 √ 3 ≈ 6.928 71. The side opposite θ has length 10 sin(15◦) ≈ 2.588 72. The side opposite θ is 2 tan(87◦) ≈ 38.162 73. The hypoteneuse has length 14 csc(38.2◦) = 14 sin(38.2◦) ≈ 22.639 74. The side adjacent to θ has length 3.98 cos(2.05◦) ≈ 3.977 75. The side opposite θ has length 31 tan(42◦) ≈ 27.912 76. The tree is about 47 feet tall. 77. The lights are about 75 feet apart. 78. (b) The ﬁre is about 4581 feet from the base of the tower. (c) The Sasquatch ran 200 cot(6◦) − 200 cot(6.5◦) ≈ 147 feet in those 10 seconds. This translates to ≈ 10 miles per hour. At the scene of the second sighting, the Sasquatch was ≈ 1755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about 2 minutes. 10.3 The Six Circular Functions and Fundamental Identities 769 79. The tree is about 41 feet tall. 80. The boat has traveled about 244 feet. 81. The tower is about 682 feet tall. The guy wire hits the ground about 731 feet away from the base of the tower. 770 Foundations of Trigonometry 10.4 Trigonometric Identities In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our ﬁrst set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, cos(−θ) = cos(θ) sin(−θ) = − sin(θ) tan(−θ) = − tan(θ) sec(−θ) = sec(θ) csc(−θ) = − csc(θ) cot(−θ) = − cot(θ) In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suﬃces to show cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θ0 be the angle coterminal with θ with 0 ≤ θ0 < 2π. (We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ) = cos(θ0) and sin(θ) = sin(θ0). y 1 θ0 y 1 θ0 P (cos(θ0), sin(θ0)) 1 x θ Q(cos(−θ0), sin(−θ0)) 1 x −θ0 We now consider the angles −θ and −θ0. Since θ is coterminal with θ0, there is some integer k so that θ = θ0 + 2π · k. Therefore, −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By deﬁnition, the coordinates of P are (cos(θ0), sin(θ0)) and the coordinates of Q are (cos(−θ0), sin(−θ0)). Since θ0 and −θ0 sweep out congruent central sectors of the Unit Circle, it 1As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section 1.6.) 10.4 Trigonometric Identities 771 follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θ0) = cos(θ0) and sin(−θ0) = − sin(θ0). Since the cosines and sines of θ0 and −θ0 are the same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. Theorem 10.13. Sum and Diﬀerence Identities for Cosine: For all angles α and β, cos(α + β) = cos(α) cos(β) − sin(α) sin(β) cos(α − β) = cos(α) cos(β) + sin(α) sin(β) We ﬁrst prove the result for diﬀerences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α0 and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α − β is coterminal with α0 − β0. Consider the case below where α0 ≥ β0. y P (cos(α0), sin(α0)) α0 − β0 Q(cos(β0), sin(β0)) y 1 A(cos(α0 − β0), sin(α0 − β0)) α0 β0 α0 − β0 O 1 x O B(1, 0) x Since the angles P OQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B.2 The distance formula, Equation 1.1, yields (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 Squaring both sides, we expand the left hand side of this equation as (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = cos2(α0) − 2 cos(α0) cos(β0) + cos2(β0) + sin2(α0) − 2 sin(α0) sin(β0) + sin2(β0) = cos2(α0) + sin2(α0) + cos2(β0) + sin2(β0) −2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) 2In the picture we’ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α0 − β0 could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about those three cases. 772 Foundations of Trigonometry From the Pythagorean Identities, cos2(α0) + sin2(α0) = 1 and cos2(β0) + sin2(β0) = 1, so (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) Turning our attention to the right hand side of our equation, we ﬁnd (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = cos2(α0 − β0) − 2 cos(α0 − β0) + 1 + sin2(α0 − β0) = 1 + cos2(α0 − β0) + sin2(α0 − β0) − 2 cos(α0 − β0) Once again, we simplify cos2(α0 − β0) + sin2(α0 − β0) = 1, so that (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = 2 − 2 cos(α0 − β0) Putting it all together, we get 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) = 2 − 2 cos(α0 − β0), which simpliﬁes to: cos(α0 − β0) = cos(α0) cos(β0) + sin(α0) sin(β0). Since α and α0, β and β0 and α − β and α0 − β0 are all coterminal pairs of angles, we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). For the case where α0 ≤ β0, we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0) = cos(β0) cos(α0) + sin(β0) sin(α0). Applying the Even Identity of cosine, we get cos(β0 − α0) = cos(−(α0 − β0)) = cos(α0 − β0), and we get the identity in this case, too. To get the sum identity for cosine, we use the diﬀerence formula along with the Even/Odd Identities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β) We put these newfound identities to good use in the following example. Example 10.4.1. 1. Find the exact value of cos (15◦). 2. Verify the identity: cos π 2 − θ = sin(θ). Solution. 1. In order to use Theorem 10.13 to ﬁnd cos (15◦), we need to write 15◦ as a sum or diﬀerence of angles whose cosines and sines we know. One way to do so is to write 15◦ = 45◦ − 30◦. cos (15◦) = cos (45◦ − 30◦) = cos (45◦) cos (30◦) + sin (45◦) sin (30◦) √ √ √ 10.4 Trigonometric Identities 773 2. In a straightforward application of Theorem 10.13, we ﬁnd cos − θ π 2 = cos π 2 π 2 = (0) (cos(θ)) + (1) (sin(θ)) cos (θ) + sin sin (θ) = sin(θ) The identity veriﬁed in Example 10.4.1, namely, cos π 2 − θ = sin(θ), is the ﬁrst of the celebrated ‘cofunction’ identities. These identities were ﬁrst hinted at in Exercise 74 in Section 10.2. From sin(θ) = cos π 2 − θ, we get: π 2 which says, in words, that the ‘co’sine of an angle is the sine of its ‘co’mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises. = cos(θ), π 2 π 2 = cos − θ − θ sin − Theorem 10.14. Cofunction Identities: For all applicable angles θ, cos sin − θ − θ π 2 π 2 = sin(θ) = cos(θ) sec csc − θ − θ π 2 π 2 = csc(θ) = sec(θ) tan cot − θ − θ π 2 π 2 = cot(θ) = tan(θ) With the Cofunction Identities in place, we are now in the position to derive the sum and diﬀerence formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the diﬀerence formula for cosine sin(α + β) = cos = cos − (α + β = cos π 2 = sin(α) cos(β) + cos(α) sin(β) cos(β) + sin − α − α sin(β) We can derive the diﬀerence formula for sine by rewriting sin(α − β) as sin(α + (−β)) and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Theore
m 10.15. Sum and Diﬀerence Identities for Sine: For all angles α and β, sin(α + β) = sin(α) cos(β) + cos(α) sin(β) sin(α − β) = sin(α) cos(β) − cos(α) sin(β) 774 Example 10.4.2. 1. Find the exact value of sin 19π 12 Foundations of Trigonometry 2. If α is a Quadrant II angle with sin(α) = 5 13 , and β is a Quadrant III angle with tan(β) = 2, ﬁnd sin(α − β). 3. Derive a formula for tan(α + β) in terms of tan(α) and tan(β). Solution. 1. As in Example 10.4.1, we need to write the angle 19π 12 as a sum or diﬀerence of common angles. The denominator of 12 suggests a combination of angles with denominators 3 and 4. One such combination is 19π 4 . Applying Theorem 10.15, we get 12 = 4π 3 + π 19π 12 sin = sin 4π 3 4π 3 √ 3 2 π 4 + cos √ π 4 2 2 √ 2 6 − 4 = sin − √ − = = π 4 + cos + − 1 2 4π 3 √ 2 2 sin 13 2 = 1, or cos(α) = ± 12 2. In order to ﬁnd sin(α − β) using Theorem 10.15, we need to ﬁnd cos(α) and both cos(β) and sin(β). To ﬁnd cos(α), we use the Pythagorean Identity cos2(α) + sin2(α) = 1. Since 13 , we have cos2(α) + 5 sin(α) = 5 13 . Since α is a Quadrant II angle, cos(α) = − 12 13 . We now set about ﬁnding cos(β) and sin(β). We have several ways to proceed, but the Pythagorean Identity 1 + tan2(β) = sec2(β) is a quick way to get sec(β), and hence, √ cos(β). With tan(β) = 2, we get 1 + 22 = sec2(β) so that sec(β) = ± 5. Since β is a √ 5 = − Quadrant III angle, we choose sec(β) = − 5 . We now need to determine sin(β). We could use The Pythagorean Identity cos2(β) + sin2(β) = 1, but we opt instead to use a quotient identity. From tan(β) = sin(β) cos(β) , we have sin(β) = tan(β) cos(β) √ 5 5 . We now have all the pieces needed to ﬁnd sin(α − β): so we get sin(β) = (2) 5 so cos(β) = 1 sec(β sin(α − β) = sin(α) cos(β) − cos(α) sin(β) √ 5 5 − − 12 13 √ 2 5 5 − = 5 13 29 − 5 √ = − 65 10.4 Trigonometric Identities 775 3. We can start expanding tan(α + β) using a quotient identity and our sum formulas 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) Since tan(α) = sin(α) cos(α) and tan(β) = sin(β) denominator by cos(α) cos(β) we will have what we want cos(β) , it looks as though if we divide both numerator and tan(α + β) = sin(α) cos(β) + cos(α) sin(β) cos(α) cos(β) − sin(α) sin(β) · 1 cos(α) cos(β) 1 cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) sin(α) cos(β) cos(α) cos(β) cos(α) cos(β) cos(α) cos(β) + − + − cos(α) sin(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) tan(α) + tan(β) 1 − tan(α) tan(β) = = = Naturally, this formula is limited to those cases where all of the tangents are deﬁned. The formula developed in Exercise 10.4.2 for tan(α +β) can be used to ﬁnd a formula for tan(α −β) by rewriting the diﬀerence as a sum, tan(α+(−β)), and the reader is encouraged to ﬁll in the details. Below we summarize all of the sum and diﬀerence formulas for cosine, sine and tangent. Theorem 10.16. Sum and Diﬀerence Identities: For all applicable angles α and β, cos(α ± β) = cos(α) cos(β) ∓ sin(α) sin(β) sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β) tan(α ± β) = tan(α) ± tan(β) 1 ∓ tan(α) tan(β) In the statement of Theorem 10.16, we have combined the cases for the sum ‘+’ and diﬀerence ‘−’ of angles into one formula. The convention here is that if you want the formula for the sum ‘+’ of 776 Foundations of Trigonometry two angles, you use the top sign in the formula; for the diﬀerence, ‘−’, use the bottom sign. For example, tan(α − β) = tan(α) − tan(β) 1 + tan(α) tan(β) If we specialize the sum formulas in Theorem 10.16 to the case when α = β, we obtain the following ‘Double Angle’ Identities. Theorem 10.17. Double Angle Identities: For all applicable angles θ, cos(2θ) =    cos2(θ) − sin2(θ) 2 cos2(θ) − 1 1 − 2 sin2(θ) sin(2θ) = 2 sin(θ) cos(θ) tan(2θ) = 2 tan(θ) 1 − tan2(θ) The three diﬀerent forms for cos(2θ) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity cos2(θ) + sin2(θ) = 1 and we leave the details to the reader. It is interesting to note that to determine the value of cos(2θ), only one piece of information is required: either cos(θ) or sin(θ). To determine sin(2θ), however, it appears that we must know both sin(θ) and cos(θ). In the next example, we show how we can ﬁnd sin(2θ) knowing just one piece of information, namely tan(θ). Example 10.4.3. 1. Suppose P (−3, 4) lies on the terminal side of θ when θ is plotted in standard position. Find cos(2θ) and sin(2θ) and determine the quadrant in which the terminal side of the angle 2θ lies when it is plotted in standard position. 2. If sin(θ) = x for − π 2 ≤ θ ≤ π 2 , ﬁnd an expression for sin(2θ) in terms of x. 3. Verify the identity: sin(2θ) = 2 tan(θ) 1 + tan2(θ) . 4. Express cos(3θ) as a polynomial in terms of cos(θ). Solution. 1. Using Theorem 10.3 from Section 10.2 with x = −3 and y = 4, we ﬁnd r = x2 + y2 = 5. Hence, cos(θ) = − 3 5 . Applying Theorem 10.17, we get cos(2θ) = cos2(θ) − sin2(θ) = − 3 25 , and sin(2θ) = 2 sin(θ) cos(θ) = 2 4 2 − 4 25 . Since both 5 5 cosine and sine of 2θ are negative, the terminal side of 2θ, when plotted in standard position, lies in Quadrant III. 5 and sin(θ) = 4 = − 7 = − 24 − 3 5 2 5 10.4 Trigonometric Identities 777 2. If your ﬁrst reaction to ‘sin(θ) = x’ is ‘No it’s not, cos(θ) = x!’ then you have indeed learned something, and we take comfort in that. However, context is everything. Here, ‘x’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, x represents the quantity sin(θ), and what we wish to know is how to express sin(2θ) in terms of x. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to ﬁnish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2(θ) + sin2(θ) = 1, 2 ≤ θ ≤ π to get cos2(θ) + x2 = 1, or cos(θ) = ± 2 , cos(θ) ≥ 0, and thus √ 1 − x2. cos(θ) = 1 − x2. Our ﬁnal answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2. Since − π √ √ 3. We start with the right hand side of the identity and note that 1 + tan2(θ) = sec2(θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan2(θ) = 2 tan(θ) sec2(θ) = = 2 sin(θ) cos(θ) 2 sin(θ) cos(θ) 1 cos2(θ) = 2 sin(θ) cos(θ) cos2(θ) cos(θ) cos(θ) = 2 sin(θ) cos(θ) = sin(2θ) 4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos2(θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to ﬁnd such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos2(θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) = 2 cos2(θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) Finally, we exchange sin2(θ) for 1 − cos2(θ) courtesy of the Pythagorean Identity, and get cos(3θ) = 2 cos3(θ) − cos(θ) − 2 sin2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 1 − cos2(θ) cos(θ) = 2 cos3(θ) − cos(θ) − 2 cos(θ) + 2 cos3(θ) = 4 cos3(θ) − 3 cos(θ) and we are done. 778 Foundations of Trigonometry In the last problem in Example 10.4.3, we saw how we could rewrite cos(3θ) as sums of powers of cos(θ). In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity cos(2θ) = 2 cos2(θ) − 1 for cos2(θ) and the identity cos(2θ) = 1 − 2 sin2(θ) for sin2(θ) results in the aptly-named ‘Power Reduction’ formulas below. Theorem 10.18. Power Reduction Formulas: For all angles θ, cos2(θ) = sin2(θ) = 1 + cos(2θ) 2 1 − cos(2θ) 2 Example 10.4.4. Rewrite sin2(θ) cos2(θ) as a sum and diﬀerence of cosines to the ﬁrst power. Solution. We begin with a straightforward application of Theorem 10.18 1 − cos(2θ) 2 1 − cos2(2θ) 1 + cos(2θ) 2 sin2(θ) cos2(θ cos2(2θ) Next, we apply the power reduction formula to cos2(2θ) to ﬁnish the reduction sin2(θ) cos2(θ cos2(2θ) 1 + cos(2(2θ)) 2 − 1 8 cos(4θ) cos(4θ) Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to cos2 θ 2 cos2 θ 2 We can obtain a formula for cos θ by extracting square roots. In a similar fashion, we may obtain 2 a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below. 1 + cos(θ) 2 1 + cos 2 θ 2 2 = = . 10.4 Trigonometric Identities 779 Theorem 10.19. Half Angle Formulas: For all applicable angles θ, cos sin θ 2 θ 2 = ± 1 + cos(θ) 2 = ± 1 − cos(θ) 2 tan = ± θ 2 1 − cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ 2 lies. Example 10.4.5. 1. Use a half angle formula to ﬁnd the exact value of cos (15◦). 2. Suppose −π ≤ θ ≤ 0 with cos(θ) = − 3 5 . Find sin θ 2 . 3. Use the identity given in number 3 of Example 10.4.3 to derive the identity tan θ 2 = sin(θ) 1 + cos(θ) Solution. 1. To use the half angle formula, we note that 15◦ = 30◦ 2 and since 15◦ is a Quadrant I angle, its cosine is positive. Thus we have cos (15◦) = + 1 + cos (30◦) = = Back in Example 10.4.1, we found cos (15◦) by using the diﬀerence formula for cosine. In that case, we determined cos (15◦) = . The reader is encouraged to prove that these two expressions are equal. 6+ 4 √ √ 2 2. If −π ≤ θ ≤ 0, then − π < 0. Theorem 10.19 gives 2 sin 2 ≤ 0, which means sin θ 2 ≤ θ θ 1 − cos (θ 10 √ 2 5 5 780 Foundations of Trigonometry 3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number 3 of Example 10.4.3 and manipulate i
t into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan2(θ) . If we are to use this to derive an identity for tan θ , it seems reasonable to proceed by replacing each occurrence of θ with θ 2 2 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 1 + tan2 θ 2 sin 2 θ 2 sin(θ) = = We now have the sin(θ) we need, but we somehow need to get a factor of 1 + cos(θ) involved. . We continue to manipulate our To get cosines involved, recall that 1 + tan2 θ 2 given identity by converting secants to cosines and using a power reduction formula = sec2 θ 2 sin(θ) = 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 sec2 θ 2 sin(θ) = 2 tan θ 2 sin(θ) = sin(θ) = 2 tan θ 2 sin(θ) = tan θ 2 θ sin(θ) 1 + cos(θ) 2 = tan cos2 θ 2 1 + cos 2 θ 2 2 (1 + cos(θ)) Our next batch of identities, the Product to Sum Formulas,3 are easily veriﬁed by expanding each of the right hand sides in accordance with Theorem 10.16 and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference. Theorem 10.20. Product to Sum Formulas: For all angles α and β, cos(α) cos(β) = 1 2 [cos(α − β) + cos(α + β)] sin(α) sin(β) = 1 2 [cos(α − β) − cos(α + β)] sin(α) cos(β) = 1 2 [sin(α − β) + sin(α + β)] 3These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows. 10.4 Trigonometric Identities 781 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 10.7. These are easily veriﬁed using the Product to Sum Formulas, and as such, their proofs are left as exercises. Theorem 10.21. Sum to Product Formulas: For all angles α and β, cos(α) + cos(β) = 2 cos cos(α) − cos(β) = −2 sin sin(α) ± sin(β) = 2 sin Example 10.4.6 cos sin cos 1. Write cos(2θ) cos(6θ) as a sum. 2. Write sin(θ) − sin(3θ) as a product. Solution. 1. Identifying α = 2θ and β = 6θ, we ﬁnd cos(2θ) cos(6θ) = 1 = 1 = 1 2 [cos(2θ − 6θ) + cos(2θ + 6θ)] 2 cos(−4θ) + 1 2 cos(4θ) + 1 2 cos(8θ) 2 cos(8θ), where the last equality is courtesy of the even identity for cosine, cos(−4θ) = cos(4θ). 2. Identifying α = θ and β = 3θ yields sin(θ) − sin(3θ) = 2 sin θ − 3θ 2 cos θ + 3θ 2 = 2 sin (−θ) cos (2θ) = −2 sin (θ) cos (2θ) , where the last equality is courtesy of the odd identity for sine, sin(−θ) = − sin(θ). The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises 38 - 43 in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs. 782 Foundations of Trigonometry 10.4.1 Exercises In Exercises 1 - 6, use the Even / Odd Identities to verify the identity. Assume all quantities are deﬁned. 1. sin(3π − 2θ) = − sin(2θ − 3π) 2. cos − π 4 − 5t = cos 5t + π 4 3. tan(−t2 + 1) = − tan(t2 − 1) 4. csc(−θ − 5) = − csc(θ + 5) 5. sec(−6t) = sec(6t) 6. cot(9 − 7θ) = − cot(7θ − 9) In Exercises 7 - 21, use the Sum and Diﬀerence Identities to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 7. cos(75◦) 10. csc(195◦) 8. sec(165◦) 11. cot(255◦) 13. cos 16. cos 19. cot 13π 12 7π 12 11π 12 14. sin 17. tan 20. csc 11π 12 17π 12 5π 12 9. sin(105◦) 12. tan(375◦) 15. tan 13π 12 18. sin π 12 21. sec − π 12 22. If α is a Quadrant IV angle with cos(α) = √ 5 5 , and sin(β) = √ 10 10 , where π 2 < β < π, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 23. If csc(α) = 3, where 0 < α < π 2 , and β is a Quadrant II angle with tan(β) = −7, ﬁnd (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 24. If sin(α) = 3 5 , where 0 < α < π 2 , and cos(β) = 12 13 where 3π 2 < β < 2π, ﬁnd (a) sin(α + β) (b) cos(α − β) (c) tan(α − β) 10.4 Trigonometric Identities 783 25. If sec(α) = − 5 3 , where π 2 < α < π, and tan(β) = 24 7 , where π < β < 3π 2 , ﬁnd (a) csc(α − β) (b) sec(α + β) (c) cot(α + β) In Exercises 26 - 38, verify the identity. 26. cos(θ − π) = − cos(θ) 27. sin(π − θ) = sin(θ) 28. tan θ + π 2 = − cot(θ) 29. sin(α + β) + sin(α − β) = 2 sin(α) cos(β) 30. sin(α + β) − sin(α − β) = 2 cos(α) sin(β) 31. cos(α + β) + cos(α − β) = 2 cos(α) cos(β) 32. cos(α + β) − cos(α − β) = −2 sin(α) sin(β) 33. sin(α + β) sin(α − β) = 1 + cot(α) tan(β) 1 − cot(α) tan(β) 34. 36. 37. cos(α + β) cos(α − β) = 1 − tan(α) tan(β) 1 + tan(α) tan(β) 35. tan(α + β) tan(α − β) = sin(α) cos(α) + sin(β) cos(β) sin(α) cos(α) − sin(β) cos(β) sin(t + h) − sin(t) h = cos(t) sin(h) h + sin(t) cos(h) − 1 h cos(t + h) − cos(t) h = cos(t) cos(h) − 1 h − sin(t) sin(h) h 38. tan(t + h) − tan(t) h = tan(h) h sec2(t) 1 − tan(t) tan(h) In Exercises 39 - 48, use the Half Angle Formulas to ﬁnd the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. 39. cos(75◦) (compare with Exercise 7) 40. sin(105◦) (compare with Exercise 9) 41. cos(67.5◦) 43. tan(112.5◦) 45. sin 47. sin π 12 5π 8 42. sin(157.5◦) 44. cos (compare with Exercise 16) 7π 12 π 8 7π 8 (compare with Exercise 18) 46. cos 48. tan 784 Foundations of Trigonometry In Exercises 49 - 58, use the given information about θ to ﬁnd the exact values of sin(2θ) θ 2 sin cos(2θ) θ 2 cos tan(2θ) θ 2 tan 49. sin(θ) = − 7 25 where 3π 2 < θ < 2π 50. cos(θ) = 28 53 where 0 < θ < π 2 51. tan(θ) = 53. cos(θ) = 55. cos(θ) = 57. sec(θ) = 12 5 3 5 12 13 √ where π < θ < 3π 2 where 0 < θ < π 2 52. csc(θ) = 4 where π 2 < θ < π 54. sin(θ) = − 4 5 where π < θ < 3π 2 where 5 where 3π 2 3π 2 < θ < 2π 56. sin(θ) = 5 13 where < θ < 2π 58. tan(θ) = −2 where In Exercises 59 - 73, verify the identity. Assume all quantities are deﬁned. 59. (cos(θ) + sin(θ))2 = 1 + sin(2θ) 60. (cos(θ) − sin(θ))2 = 1 − sin(2θ) 61. tan(2θ) = 1 1 − tan(θ) − 1 1 + tan(θ) 62. csc(2θ) = cot(θ) + tan(θ) 2 63. 8 sin4(θ) = cos(4θ) − 4 cos(2θ) + 3 64. 8 cos4(θ) = cos(4θ) + 4 cos(2θ) + 3 65. sin(3θ) = 3 sin(θ) − 4 sin3(θ) 66. sin(4θ) = 4 sin(θ) cos3(θ) − 4 sin3(θ) cos(θ) 67. 32 sin2(θ) cos4(θ) = 2 + cos(2θ) − 2 cos(4θ) − cos(6θ) 68. 32 sin4(θ) cos2(θ) = 2 − cos(2θ) − 2 cos(4θ) + cos(6θ) 69. cos(4θ) = 8 cos4(θ) − 8 cos2(θ) + 1 70. cos(8θ) = 128 cos8(θ) − 256 cos6(θ) + 160 cos4(θ) − 32 cos2(θ) + 1 (HINT: Use the result to 69.) 71. sec(2θ) = cos(θ) cos(θ) + sin(θ) + sin(θ) cos(θ) − sin(θ) 72. 73. 1 cos(θ) − sin(θ) 1 cos(θ) − sin(θ) + − 1 cos(θ) + sin(θ) 1 cos(θ) + sin(θ) = = 2 cos(θ) cos(2θ) 2 sin(θ) cos(2θ) 10.4 Trigonometric Identities 785 In Exercises 74 - 79, write the given product as a sum. You may need to use an Even/Odd Identity. 74. cos(3θ) cos(5θ) 75. sin(2θ) sin(7θ) 76. sin(9θ) cos(θ) 77. cos(2θ) cos(6θ) 78. sin(3θ) sin(2θ) 79. cos(θ) sin(3θ) In Exercises 80 - 85, write the given sum as a product. You may need to use an Even/Odd or Cofunction Identity. 80. cos(3θ) + cos(5θ) 81. sin(2θ) − sin(7θ) 82. cos(5θ) − cos(6θ) 83. sin(9θ) − sin(−θ) 84. sin(θ) + cos(θ) 85. cos(θ) − sin(θ) 86. Suppose θ is a Quadrant I angle with sin(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 − x2 (b) sin(2θ) = 2x √ 1 − x2 (c) cos(2θ) = 1 − 2x2 87. Discuss with your classmates how each of the formulas, if any, in Exercise 86 change if we change assume θ is a Quadrant II, III, or IV angle. 88. Suppose θ is a Quadrant I angle with tan(θ) = x. Verify the following formulas (a) cos(θ) = √ 1 x2 + 1 (c) sin(2θ) = 2x x2 + 1 (b) sin(θ) = √ x x2 + 1 (d) cos(2θ) = 1 − x2 x2 + 1 89. Discuss with your classmates how each of the formulas, if any, in Exercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. for − for − 90. If sin(θ) = 91. If tan(θ) = 92. If sec(θ) = x 2 x 7 x 4 , ﬁnd an expression for cos(2θ) in terms of x. , ﬁnd an expression for sin(2θ) in terms of x. for 0 < θ < π 2 , ﬁnd an expression for ln \| sec(θ) + tan(θ)\| in terms of x. 93. Show that cos2(θ) − sin2(θ) = 2 cos2(θ) − 1 = 1 − 2 sin2(θ) for all θ. 94. Let θ be a Quadrant III angle with cos(θ) = − . Show that this is not enough information to 1 5 by ﬁrst assuming 3π < θ < 7π 2 and then assuming π < θ < 3π 2 determine the sign of sin θ 2 and computing sin in both cases. θ 2 786 Foundations of Trigonometry 95. Without using your calculator, show that 96. In part 4 of Example 10.4.3, we wrote cos(3θ) as a polynomial in terms of cos(θ). In Exercise 69, we had you verify an identity which expresses cos(4θ) as a polynomial in terms of cos(θ). Can you ﬁnd a polynomial in terms of cos(θ) for cos(5θ)? cos(6θ)? Can you ﬁnd a pattern so that cos(nθ) could be written as a polynomial in cosine for any natural number n? 97. In Exercise 65, we has you verify an identity which expresses sin(3θ) as a polynomial in terms of sin(θ). Can you do the same for sin(5θ)? What about for sin(4θ)? If not, what goes wrong? 98. Verify the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verify the Cofunction Identities for tangent, secant, cosecant and cotangent. 100. Verify the Diﬀerence Identities for sine and tangent. 101. Verify the Product to Sum Identities. 102. Verify the Sum to Product Identities. 10.4 Trigonometric Identities 787 = 2 − √ 3 12. tan(375◦) = 10.4.2 Answers 7. cos(75◦) = √ 2 √ 6 − 4 9. sin(105◦) = √ 2 √ 6 + 4 11. cot(255◦) = √ √ 13. cos 15. tan 17. tan 19. cot 21. sec 13π 12 13π 12 17π 12 11π 12 − π 12 = 2 + √ 3 = −(2 + √ 3) √ √ 2 6 − = 18. sin 20. csc π 12 5π 12 8. sec(165◦) = − √ 4 2 + √ 6 √ √ 6 2 − = 10. csc(195◦) = √ √ √ = −( √ 6 + √ 14. sin 11π 12 = 16. cos √ 7π 12 = √ − = 22. (a) cos(α + β) = − √ 2 10 (b) sin(α + β) = √ 2 7 10 √ 2 2 (c) tan(α + β) = −7 (d) cos(α − β) = − (e) sin(α − β) = √ 2 2 23. (a) cos(α + β) = − √ 4 + 7 30 2 (c) tan(α + β) = (e) sin(α − β28 + 4 + 7 28 + 30 (f) tan(α − β) = −1 63 − 100 √ 2 41 = (b)
sin(α + β) = (d) cos(α − β) = (f) tan(α − β) = √ 2 √ 28 − 30 2 −4 + 7 30 √ √ 28 + 4 − 7 2 2 = − 24. (a) sin(α + β) = 16 65 (b) cos(α − β) = 33 65 (c) tan(α − β) = 63 + 100 √ 2 41 56 33 788 Foundations of Trigonometry 25. (a) csc(α − β) = − 5 4 (b) sec(α + β) = 125 117 (c) cot(α + β) = 117 44 39. cos(75◦) = 41. cos(67.5◦) = √ 40. sin(105◦) = 42. sin(157.5◦) = √ 1 − √ 2 44. cos 7π 12 = − √ 3 2 − 2 46. cos 48. tan 2 + 2 = − π 8 = 7π 43. tan(112.5◦) = − 45. sin 47. sin π 12 5π 49. sin(2θ) = − √ sin θ 2 = 50. sin(2θ) = sin θ 2 = 51. sin(2θ) = sin θ 2 = 52. sin(2θ) = − 336 625 2 10 2520 2809 √ 5 106 106 120 169 √ 3 13 13 √ 15 8 cos(2θ) = cos θ 2 = − cos(2θ) = − cos θ 2 = 9 527 625 √ 2 7 10 1241 2809 √ 106 106 119 169 √ 2 13 13 cos(2θ) = − cos θ 2 = − cos(2θ) = 7 8 sin θ 2 = √ 15 8 + 2 4 cos θ 2 = √ 15 8 − 2 4 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 9 tan(2θ) = − tan θ 2 = − 336 527 1 7 2520 1241 120 119 3 2 √ 15 7 tan(2θ) = − tan θ 2 tan √ 15 √ √ 15 15 53. sin(2θ) = sin θ 2 = 24 25 √ 5 5 cos(2θ) = − cos θ 2 = 2 7 25 √ 5 5 24 7 tan(2θ) = − tan θ 2 = 1 2 10.4 Trigonometric Identities 789 54. sin(2θ) = sin θ 2 = 24 25 √ 2 5 sin(2θ) = − √ sin θ 2 = sin(2θ) = − sin θ 2 = 5 5 120 169 26 26 120 169 √ 26 26 4 5 55. 56. 57. cos(2θ) = − cos θ 2 = − cos(2θ) = cos θ 2 = − 26 26 cos(2θ) = cos θ 2 = 7 25 √ 5 5 119 169 5 √ 119 169 √ 26 26 3 5 tan(2θ) = − 24 7 = −2 tan θ 2 120 119 1 5 120 119 tan(2θ) = − tan θ 2 = − tan(2θ) = − tan θ 2 = 5 tan(2θ) = 4 3 tan θ 2 = − sin(2θ) = − cos(2θ) = − sin θ 2 = 50 − 10 10 √ 5 cos θ 2 = − √ 5 50 + 10 10 58. sin(2θ) = − 4 5 cos(2θ) = − 3 5 sin θ 2 = √ 5 50 + 10 10 cos θ 2 = 50 − 10 10 √ tan θ 2 = tan(2θ) = 5 tan θ 2 = tan 10 10 74. 77. cos(2θ) + cos(8θ) 2 cos(4θ) + cos(8θ) 2 80. 2 cos(4θ) cos(θ) 83. 2 cos(4θ) sin(5θ) 90. 1 − x2 2 75. 78. cos(5θ) − cos(9θ) 2 cos(θ) − cos(5θ) 2 9 2 sin θ 81. −2 cos θ 5 2 76. 79. sin(8θ) + sin(10θ) 2 sin(2θ) + sin(4θ) 2 11 2 sin θ 1 2 θ 82. 2 sin √ θ − 2 cos 84. π 4 91. 14x x2 + 49 √ 85. − θ − 2 sin π 4 √ 92. ln \|x + x2 + 16\| − ln(4) 790 Foundations of Trigonometry 10.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular (trigonometric) functions as functions of real numbers and pick up where we left oﬀ in Sections 10.2.1 and 10.3.1. As usual, we begin our study with the functions f (t) = cos(t) and g(t) = sin(t). 10.5.1 Graphs of the Cosine and Sine Functions From Theorem 10.5 in Section 10.2.1, we know that the domain of f (t) = cos(t) and of g(t) = sin(t) is all real numbers, (−∞, ∞), and the range of both functions is [−1, 1]. The Even / Odd Identities in Theorem 10.12 tell us cos(−t) = cos(t) for all real numbers t and sin(−t) = − sin(t) for all real numbers t. This means f (t) = cos(t) is an even function, while g(t) = sin(t) is an odd function.1 Another important property of these functions is that for coterminal angles α and β, cos(α) = cos(β) and sin(α) = sin(β). Said diﬀerently, cos(t+2πk) = cos(t) and sin(t+2πk) = sin(t) for all real numbers t and any integer k. This last property is given a special name. Deﬁnition 10.3. Periodic Functions: A function f is said to be periodic if there is a real number c so that f (t + c) = f (t) for all real numbers t in the domain of f . The smallest positive number p for which f (t + p) = f (t) for all real numbers t in the domain of f , if it exists, is called the period of f . We have already seen a family of periodic functions in Section 2.1: the constant functions. However, despite being periodic a constant function has no period. (We’ll leave that odd gem as an exercise for you.) Returning to the circular functions, we see that by Deﬁnition 10.3, f (t) = cos(t) is periodic, since cos(t + 2πk) = cos(t) for any integer k. To determine the period of f , we need to ﬁnd the smallest real number p so that f (t + p) = f (t) for all real numbers t or, said diﬀerently, the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. We know that cos(t + 2π) = cos(t) for all real numbers t but the question remains if any smaller real number will do the trick. Suppose p > 0 and cos(t + p) = cos(t) for all real numbers t. Then, in particular, cos(0 + p) = cos(0) so that cos(p) = 1. From this we know p is a multiple of 2π and, since the smallest positive multiple of 2π is 2π itself, we have the result. Similarly, we can show g(t) = sin(t) is also periodic with 2π as its period.2 Having period 2π essentially means that we can completely understand everything about the functions f (t) = cos(t) and g(t) = sin(t) by studying one interval of length 2π, say [0, 2π].3 One last property of the functions f (t) = cos(t) and g(t) = sin(t) is worth pointing out: both of these functions are continuous and smooth. Recall from Section 3.1 that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, 1See section 1.6 for a review of these concepts. 2Alternatively, we can use the Cofunction Identities in Theorem 10.14 to show that g(t) = sin(t) is periodic with period 2π since g(t) = sin(t) = cos . 3Technically, we should study the interval [0, 2π),4since whatever happens at t = 2π is the same as what happens at t = 0. As we will see shortly, t = 2π gives us an extra ‘check’ when we go to graph these functions. 4In some advanced texts, the interval of choice is [−π, π). 10.5 Graphs of the Trigonometric Functions 791 corners or cusps. As we shall see, the graphs of both f (t) = cos(t) and g(t) = sin(t) meander nicely and don’t cause any trouble. We summarize these facts in the following theorem. Theorem 10.22. Properties of the Cosine and Sine Functions • The function f (x) = cos(x) • The function g(x) = sin(x) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] – is continuous and smooth – is continuous and smooth – is even – has period 2π – is odd – has period 2π In the chart above, we followed the convention established in Section 1.6 and used x as the independent variable and y as the dependent variable.5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph y = cos(x), we make a table as we did in Section 1.6 using some of the ‘common values’ of x in the interval [0, 2π]. This generates a portion of the cosine graph, which we call the ‘fundamental cycle’ of y = cos(x). x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cos(xx, cos(x)) (0, 1 √ 3π 4 , − 2 2 (π, −1 3π √ 7π 4 , (2π, 1) 2 2 5π y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = cos(x). A few things about the graph above are worth mentioning. First, this graph represents only part of the graph of y = cos(x). To get the entire graph, we imagine ‘copying and pasting’ this graph end to end inﬁnitely in both directions (left and right) on the x-axis. Secondly, the vertical scale here has been greatly exaggerated for clarity and aesthetics. Below is an accurate-to-scale graph of y = cos(x) showing several cycles with the ‘fundamental cycle’ plotted thicker than the others. The 5The use of x and y in this context is not to be confused with the x- and y-coordinates of points on the Unit Circle which deﬁne cosine and sine. Using the term ‘trigonometric function’ as opposed to ‘circular function’ can help with that, but one could then ask, “Hey, where’s the triangle?” 792 Foundations of Trigonometry graph of y = cos(x) is usually described as ‘wavelike’ – indeed, many of the applications involving the cosine and sine functions feature modeling wavelike phenomena. y x An accurately scaled graph of y = cos(x). We can plot the fundamental cycle of the graph of y = sin(x) similarly, with similar results. x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(xx, sin(x)) (0, 0 √ 3π 4 , 2 2 0 √ 2 2 −1 √ 2 2 − − 5π 7π (π, 01 3π √ 4 , − 2 2 y 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x 0 (2π, 0) The ‘fundamental cycle’ of y = sin(x). As with the graph of y = cos(x), we provide an accurately scaled graph of y = sin(x) below with the fundamental cycle highlighted. y x An accurately scaled graph of y = sin(x). It is no accident that the graphs of y = cos(x) and y = sin(x) are so similar. Using a cofunction identity along with the even property of cosine, we have sin(x) = cos π 2 − x = cos − x − = cos x − π 2 π 2 Recalling Section 1.7, we see from this formula that the graph of y = sin(x) is the result of shifting the graph of y = cos(x) to the right π Now that we know the basic shapes of the graphs of y = cos(x) and y = sin(x), we can use Theorem 1.7 in Section 1.7 to graph more complicated curves. To do so, we need to keep track of 2 units. A visual inspection conﬁrms this. 10.5 Graphs of the Trigonometric Functions 793 the movement of some key points on the original graphs. We choose to track the values x = 0, π 2 , π, 3π 2 and 2π. These ‘quarter marks’ correspond to quadrantal angles, and as such, mark the location of the zeros and the local extrema of these functions over exactly one period. Before we begin our next example, we need to review the concept of the ‘argument’ of a function as ﬁrst introduced in Section 1.4. For the function f (x) = 1 − 5 cos(2x − π), the argument of f is x. We shall have occasion, however, to refer to the argument of the cosine, which in this case is 2x − π. Loosely stated, the argument of a trigonometric function is the expression ‘inside’ the function. Example 10.5.1. Graph one cycle of the following functions. State the period of each. 1. f (x) = 3 cos πx−π 2 + 1 Solution. 1. We set the argument of the cosine, πx−π 2 2. g(x) = 1 2 sin(π − 2x) + 3 2 , equal to each of the values: 0, π 2 , π, 3π 2 , 2π and solve for x. We summarize the results below. πx−π a πx−π πx− = 3π 4 2 2 = 2π 5 0 π 2 π 3π 2 2π πx−π πx−π πx−π 2 Next, we substitute each of these x values into f (x) = 3 cos πx−π corresponding y-values and connect the dots in a pleasing wavelike fashion. 2 + 1 to determine the x 1 2 3 4 5 f (x) (x, f (x)) 4 1 (1, 4) (2, 1) −2 (3, −2) 1 4 (4, 1)
(5, 4) y 4 3 2 1 −1 −2 1 2 3 4 5 x One cycle is graphed on [1, 5] so the period is the length of that interval which is 4. One cycle of y = f (x). 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π − 2x = a x π π − 2x = 0 0 2 π π π − 2x = π 2 4 2 π − 2x = π π 0 3π π − 2x = 3π 2 − π 2 4 2π π − 2x = 2π − π 2 We now ﬁnd the corresponding y-values on the graph by substituting each of these x-values into g(x) = 1 2 . Once again, we connect the dots in a wavelike fashion. 2 sin(π − 2x(xx, g(x)) π 2 , 3 π 4 , 2 0 One cycle of y = g(x). One cycle was graphed on the interval − π 2 , π 2 so the period is π 2 − − π 2 = π. The functions in Example 10.5.1 are examples of sinusoids. Roughly speaking, a sinusoid is the result of taking the basic graph of f (x) = cos(x) or g(x) = sin(x) and performing any of the transformations6 mentioned in Section 1.7. Sinusoids can be characterized by four properties: period, amplitude, phase shift and vertical shift. We have already discussed period, that is, how long it takes for the sinusoid to complete one cycle. The standard period of both f (x) = cos(x) and g(x) = sin(x) is 2π, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how ‘tall’ the wave is, as indicated in the ﬁgure below. The amplitude of the standard cosine and sine functions is 1, but vertical scalings can alter this. 6We have already seen how the Even/Odd and Cofunction Identities can be used to rewrite g(x) = sin(x) as a transformed version of f (x) = cos(x), so of course, the reverse is true: f (x) = cos(x) can be written as a transformed version of g(x) = sin(x). The authors have seen some instances where sinusoids are always converted to cosine functions while in other disciplines, the sinusoids are always written in terms of sine functions. We will discuss the applications of sinusoids in greater detail in Chapter 11. Until then, we will keep our options open. 10.5 Graphs of the Trigonometric Functions 795 amplitude baseline period = sin(x). As the reader can verify, a phase shift of π The phase shift of the sinusoid is the horizontal shift experienced by the fundamental cycle. We have seen that a phase (horizontal) shift of π 2 to the right takes f (x) = cos(x) to g(x) = sin(x) since cos x − π 2 to the left takes g(x) = sin(x) to 2 f (x) = cos(x). The vertical shift of a sinusoid is exactly the same as the vertical shifts in Section 1.7. In most contexts, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem 1.7 in Section 1.7, shows how to ﬁnd these four fundamental quantities from the formula of the given sinusoid. Theorem 10.23. For ω > 0, the functions C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B have period 2π ω have amplitude \|A\| have phase shift − φ ω have vertical shift B We note that in some scientiﬁc and engineering circles, the quantity φ mentioned in Theorem 10.23 is called the phase of the sinusoid. Since our interest in this book is primarily with graphing sinusoids, we focus our attention on the horizontal shift − φ The proof of Theorem 10.23 is a direct application of Theorem 1.7 in Section 1.7 and is left to the reader. The parameter ω, which is stipulated to be positive, is called the (angular) frequency of the sinusoid and is the number of cycles the sinusoid completes over a 2π interval. We can always ensure ω > 0 using the Even/Odd Identities.7 We now test out Theorem 10.23 using the functions f and g featured in Example 10.5.1. First, we write f (x) in the form prescribed in Theorem 10.23, ω induced by φ. f (x) = 3 cos πx − π 2 + 1 = 3 cos π 2 x + − π 2 + 1, 7Try using the formulas in Theorem 10.23 applied to C(x) = cos(−x + π) to see why we need ω > 0. 796 Foundations of Trigonometry 2 , φ = − π π/2 = 4, the amplitude is \|A\| = \|3\| = 3, the phase shift is − φ so that A = 3, ω = π 2 and B = 1. According to Theorem 10.23, the period of f is ω = 2π 2π π/2 = 1 (indicating a shift to the right 1 unit) and the vertical shift is B = 1 (indicating a shift up 1 unit.) All of these match with our graph of y = f (x). Moreover, if we start with the basic shape of the cosine graph, shift it 1 unit to the right, 1 unit up, stretch the amplitude to 3 and shrink the period to 4, we will have reconstructed one period of the graph of y = f (x). In other words, instead of tracking the ﬁve ‘quarter marks’ through the transformations to plot y = f (x), we can use ﬁve other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function g in Example 10.5.1, we ﬁrst need to use the odd property of the sine function to write it in the form required by Theorem 10.23 ω = − −π/2 g(x) = 1 2 sin(π − 2x) + 3 2 = 1 2 sin(−(2x − π)) + 3 2 = − 1 2 sin(2x − π) + 3 2 = − 1 2 sin(2x + (−π)) + 3 2 = 1 2 , the phase shift is − −π 2 . The period is then 2π Instead of the graph starting at x = π 2 , ω = 2, φ = −π and B = 3 2 = π We ﬁnd A = − 1 2 = π, the amplitude is − 1 2 (indicating a shift right π 2 units) and the vertical shift is up 2 3 2 . Note that, in this case, all of the data match our graph of y = g(x) with the exception of the phase shift. 2 , it ends there. Remember, however, that the graph presented in Example 10.5.1 is only one portion of the graph of y = g(x). Indeed, another complete cycle begins at x = π 2 , and this is the cycle Theorem 10.23 is detecting. The reason for the discrepancy is that, in order to apply Theorem 10.23, we had to rewrite the formula for g(x) using the odd property of the sine function. Note that whether we graph y = g(x) using the ‘quarter marks’ approach or using the Theorem 10.23, we get one complete cycle of the graph, which means we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f (x). −1, 5 2 y 3 2 1 51 1 2 3 4 5 x −1 −2 2, − 3 2 One cycle of y = f (x). 1. Find a cosine function whose graph matches the graph of y = f (x). 10.5 Graphs of the Trigonometric Functions 797 2. Find a sine function whose graph matches the graph of y = f (x). Solution. ω , so that ω = π 1. We ﬁt the data to a function of the form C(x) = A cos(ωx + φ) + B. Since one cycle is graphed over the interval [−1, 5], its period is 5 − (−1) = 6. According to Theorem 10.23, 3 . Next, we see that the phase shift is −1, so we have − φ 6 = 2π ω = −1, or . As a result the amplitude A = 1 2 (4) = 2. Finally, to determine the vertical shift, we 2 average the endpoints of the range to ﬁnd B = 1 2 . Our ﬁnal answer is 2 + 1 C(x) = 2 cos π 2 . 3 . To ﬁnd the amplitude, note that the range of the sinusoid is − 3 2 + − 3 5 2 (1. Most of the work to ﬁt the data to a function of the form S(x) = A sin(ωx + φ) + B is done. The period, amplitude and vertical shift are the same as before with ω = π 3 , A = 2 and B = 1 2 . The trickier part is ﬁnding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the ﬁgure below so that we can identify a cycle beginning at 7 = − 7π 2 , we get − φ 6 . + 1 Hence, our answer is S(x) = 2 sin π 2 . . Taking the phase shift to be 7 3 x − 7π 2 , or , 5 2 − 10 x 7 2 , 1 2 13 2 , 1 2 19 2 , 5 2 −1 −2 8, − 3 2 Extending the graph of y = f (x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. taking For example, when ﬁtting a sine function to the data, we could have chosen to start at 1 + 1 6 for an answer of S(x) = −2 sin π A = −2. In this case, the phase shift is 1 2 . Alternatively, we could have extended the graph of y = f (x) to the left and considered a sine , and so on. Each of these formulas determine the same sinusoid curve function starting at − 5 2 and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identity for cosine, we can expand the formula to yield 2 , 1 3 x − π 2 so (x) = A cos(ωx + φ) + B = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B. 798 Foundations of Trigonometry Similarly, using the sum identity for sine, we get S(x) = A sin(ωx + φ) + B = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B. Making these observations allows us to recognize (and graph) functions as sinusoids which, at ﬁrst glance, don’t appear to ﬁt the forms of either C(x) or S(x). Example 10.5.3. Consider the function f (x) = cos(2x) − √ 3 sin(2x). Find a formula for f (x): 1. in the form C(x) = A cos(ωx + φ) + B for ω > 0 2. in the form S(x) = A sin(ωx + φ) + B for ω > 0 Check your answers analytically using identities and graphically using a calculator. Solution. 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up 3 sin(2x) with the expanded form of corresponding coeﬃcients. Equating f (x) = cos(2x) − C(x) = A cos(ωx + φ) + B, we get √ √ cos(2x) − 3 sin(2x) = A cos(ωx) cos(φ) − A sin(ωx) sin(φ) + B It should be clear that we can take ω = 2 and B = 0 to get √ cos(2x) − 3 sin(2x) = A cos(2x) cos(φ) − A sin(2x) sin(φ) To determine A and φ, a bit more work is involved. We get started by equating the coeﬃcients of the trigonometric functions on either side of the equation. On the left hand side, the coeﬃcient of cos(2x) is 1, while on the right hand side, it is A cos(φ). Since this equation is to hold for all real numbers, we must have8 that A cos(φ) = 1. Similarly, we ﬁnd by equating the coeﬃcients of sin(2x) that A sin(φ) = 3. What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the Pythagorean Identity. We know cos2(φ) + sin2(φ) = 1, so multiplying this by A2 gives A2 cos2(φ)+A2 sin2(φ) = A2. Since A cos(φ) = 1 and A sin(φ) = 3)2 = 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or, after some √ 3 rearrangement, cos(φ) = 1 2 . One such angle φ which satisﬁes t
his criteria is . We can easily φ = π check our answer using the sum formula for cosine 3 . Hence, one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π 3, we get A2 = 12+( √ 2 and sin(φ) = √ √ √ 3 f (x) = 2 cos 2x + π 3 = 2 cos(2x) cos π 3 cos(2x) 1 − sin(2x) 2 √ 3 sin(2x) = 2 = cos(2x) − − sin(2x) sin π 3 √ 3 2 8This should remind you of equation coeﬃcients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 799 2. Proceeding as before, we equate f (x) = cos(2x) − S(x) = A sin(ωx + φ) + B to get √ 3 sin(2x) with the expanded form of √ cos(2x) − 3 sin(2x) = A sin(ωx) cos(φ) + A cos(ωx) sin(φ) + B Once again, we may take ω = 2 and B = 0 so that √ cos(2x) − 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) We equate9 the coeﬃcients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3. Using A2 cos2(φ) + A2 sin2(φ) = A2 as before, we get A = ±2, and again we choose A = 2. √ 3 This means 2 sin(φ) = 1, or sin(φ) = 1 2 . . One such angle which meets these criteria is φ = 5π 6 Checking our work analytically, we have 3, which means cos(φ) = − 6 . Hence, we have f (x) = 2 sin 2x + 5π 2 , and 2 cos(φ) = − √ √ f (x) = 2 sin 2x + 5π 6 + cos(2x) sin 5π = 2 sin(2x) cos 5π 6 6 √ + cos(2x) 1 3 2 2 − = 2 sin(2x) = cos(2x) − √ 3 sin(2x) Graphing the three formulas for f (x) result in the identical curve, verifying our analytic work. √ It is important to note that in order for the technique presented in Example 10.5.3 to ﬁt a function into one of the forms in Theorem 10.23, the arguments of the cosine and sine function much match. 3 sin(3x) is not.10 It That is, while f (x) = cos(2x) − is also worth mentioning that, had we chosen A = −2 instead of A = 2 as we worked through Example 10.5.3, our ﬁnal answers would have looked diﬀerent. The reader is encouraged to rework Example 10.5.3 using A = −2 to see what these diﬀerences are, and then for a challenging exercise, use identities to show that the formulas are all equivalent. The general equations to ﬁt a function of the form f (x) = a cos(ωx) + b sin(ωx) + B into one of the forms in Theorem 10.23 are explored in Exercise 35. 3 sin(2x) is a sinusoid, g(x) = cos(2x) − √ 9Be careful here! 10This graph does, however, exhibit sinusoid-like characteristics! Check it out! 800 Foundations of Trigonometry 10.5.2 Graphs of the Secant and Cosecant Functions We now turn our attention to graphing y = sec(x). Since sec(x) = 1 cos(x) , we can use our table of values for the graph of y = cos(x) and take reciprocals. We know from Section 10.3.1 that the domain of F (x) = sec(x) excludes all odd multiples of π 2 , and sure enough, we run into trouble at 2 and x = 3π x = π 2 since cos(x) = 0 at these values. Using the notation introduced in Section 4.2, −, cos(x) → 0+, so sec(x) → ∞. (See Section 10.3.1 for a more detailed we have that as x → π 2 analysis.) Similarly, we ﬁnd that as x → π , sec(x) → −∞; and as 2 x → 3π , sec(x) → ∞. This means we have a pair of vertical asymptotes to the graph of y = sec(x), 2 x = π 2 . Since cos(x) is periodic with period 2π, it follows that sec(x) is also.11 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’12 +, sec(x) → −∞; as x → 3π 2 2 and x = 3π − + y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π √ 2 − − √ √ sec(x) 1 cos(x) 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned 2 2 1 −1 √ √ − − 2 1 √ (x, sec(x)) (0, 1) √ 2 π 4 , 2 3π 2 5π √ 2 4 , − (π, −1) √ 2 4 , − √ 2 7π 4 , (2π, 1) 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The ‘fundamental cycle’ of y = sec(x). y x The graph of y = sec(x). 11Provided sec(α) and sec(β) are deﬁned, sec(α) = sec(β) if and only if cos(α) = cos(β). Hence, sec(x) inherits its period from cos(x). 12In Section 10.3.1, we argued the range of F (x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-values. Here, we encounter issues at x = 0, x = π and x = 2π. Proceeding with the usual analysis, we graph the fundamental cycle of y = csc(x) below along with the dotted graph of y = sin(x) for reference. Since y = sin(x) and y = cos(x) are merely phase shifts of each other, so too are y = csc(x) and y = sec(x). y x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π sin(x) √ 1 2 √ √ csc(x) 0 undeﬁned 2 2 1 √ 2 2 0 undeﬁned 2 2 −1 √ 2 2 0 undeﬁned −x, csc(x)) √ 3π 4 , 2 5π 2 7π √ 2 4 , − 2 , −1 3π √ 2 4 , − 3 2 1 −1 −2 −3 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x Once again, our domain and range work in Section 10.3.1 is veriﬁed geometrically in the graph of y = G(x) = csc(x). The ‘fundamental cycle’ of y = csc(x). y x The graph of y = csc(x). Note that, on the intervals between the vertical asymptotes, both F (x) = sec(x) and G(x) = csc(x) are continuous and smooth. In other words, they are continuous and smooth on their domains.13 The following theorem summarizes the properties of the secant and cosecant functions. Note that 13Just like the rational functions in Chapter 4 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everywhere. 802 Foundations of Trigonometry all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively. Theorem 10.24. Properties of the Secant and Cosecant Functions The function F (x) = sec(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is even – has period 2π The function G(x) = csc(x) ∞ k=−∞ (kπ, (k + 1)π) – has domain {x : x = πk, k is an integer} = – has range {y : \|y\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – is continuous and smooth on its domain – is odd – has period 2π In the next example, we discuss graphing more general secant and cosecant curves. Example 10.5.4. Graph one cycle of the following functions. State the period of each. 1. f (x) = 1 − 2 sec(2x) Solution. 2. g(x) = csc(π − πx) − 5 3 1. To graph y = 1 − 2 sec(2x), we follow the same procedure as in Example 10.5.1. First, we set 2 and 2π and solve for x. the argument of secant, 2x, equal to the ‘quarter marks’ 0, π 2 , π, 3π a 2x = a 0 2x = 0 π 2x = π 2 2 π 2x = π 3π 2x = 3π 2 2 2π 2x = 2π x 0 π 4 π 2 3π 4 π 10.5 Graphs of the Trigonometric Functions 803 Next, we substitute these x values into f (x). If f (x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in the picture below. Since one cycle is graphed over the interval [0, π], the period is π − 0 = π. y 3 2 1 −1 π 4 π 2 3π 4 π x x 0 π 4 π 2 3π 4 π f (x) −1 (x, f (x)) (0, −1) undeﬁned 3 π 2 , 3 undeﬁned −1 (π, −1) 2. Proceeding as before, we set the argument of cosecant in g(x) = csc(π−πx)−5 3 equal to the quarter marks and solve for x. One cycle of y = 1 − 2 sec(2x). a π − πx = a x π − πx = 0 1 0 π − πx = π 1 π 2 2 2 π − πx = π 0 π 2 − 1 π − πx = 3π 3π 2 2 2π π − πx = 2π −1 Substituting these x-values into g(x), we generate the graph below and ﬁnd the period to be 1 − (−1) = 2. The associated sine curve, y = sin(π−πx)−5 , is dotted in as a reference1 g(x) undeﬁned − 4 3 undeﬁned (x, g(x)) 1 2 , − 4 3 −2 − 1 2 , −2 undeﬁned y −1 − 1 2 1 1 2 x −1 −2 One cycle of y = csc(π−πx)−5 3 . 804 Foundations of Trigonometry Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem 10.23. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out explicitly. Finally, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 10.5.3 Graphs of the Tangent and Cotangent Functions 2 and x = 3π Finally, we turn our attention to the graphs of the tangent and cotangent functions. When constructing a table of values for the tangent function, we see that J(x) = tan(x) is undeﬁned at −, sin(x) → 1− x = π 2 , in accordance with our ﬁndings in Section 10.3.1. As x → π and cos(x) → 0+, so that tan(x) = sin(x) 2 . Using a + similar analysis, we get that as x → π , 2 tan(x) → −∞. Plotting this information and performing the usual ‘copy and paste’ produces: cos(x) → ∞ producing a vertical asymptote at x = π +, tan(x) → −∞; as x → 3π 2 , tan(x) → ∞; and as x → 3π 3π 4 π 5π 4 3π 2 7π 4 2π tan(x) 0 1 (x, tan(x)) (0, 0) 4 , 1 π undeﬁned −1 0 1 undeﬁned −1 0 4 , −1 3π (π, 0) 4 , 1 5π 4 , −1 7π (2π, 0) 1 −1 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = tan(x) over [0, 2π]. y x The graph of y = tan(x). 10.5 Graphs of the Trigonometric Functions 805 From the graph, it appears as if the tangent function is periodic with period π. To prove that this is the case, we appeal to the sum formula for tangents. We have: tan(x + π) = tan(x) + tan(π) 1 − tan(x) tan(π) = tan(x) + 0 1 − (tan(x))(0) = tan(x), which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 + p) = tan(0) = 0, which means p is a multiple of π. The smallest positive multiple of π is π itself, so we have established the result. We take as our fundamental cycle for y = tan(x) the interval − π 2 , − π 2 . From the graph, we see conﬁrmation of our domain and range work in Section 10.3.1. , and use as our ‘quarter marks’ x = − π 4 and π 4 , 0, π 2 , π 2 It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). Plotting cot(x) over the interval [0, 2π] results in the graph below
. y 1 −1 x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cot(x) undeﬁned (x, cot(x)) 1 0 −1 undeﬁned 1 0 −1 undeﬁned 1 3π 4 , 1 5π 3π 2 , 0 4 , −1 7π π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader to prove this.14 We take as one fundamental cycle the interval (0, π) with quarter marks: x = 0, π 4 , π 4 and π. A more complete graph of y = cot(x) is below, along with the fundamental cycle highlighted as usual. Once again, we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section 10.3.1. 2 , 3π 14Certainly, mimicking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities. 806 Foundations of Trigonometry y x The graph of y = cot(x). The properties of the tangent and cotangent functions are summarized below. As with Theorem 10.24, each of the results below can be traced back to properties of the cosine and sine functions and the deﬁnition of the tangent and cotangent functions as quotients thereof. Theorem 10.25. Properties of the Tangent and Cotangent Functions The function J(x) = tan(x) – has domain x : x = π 2 + πk, k is an integer = ∞ k=−∞ (2k + 1)π 2 , (2k + 3)π 2 – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π The function K(x) = cot(x) – has domain {x : x = πk, k is an integer} = – has range (−∞, ∞) – is continuous and smooth on its domain – is odd – has period π ∞ k=−∞ (kπ, (k + 1)π) 10.5 Graphs of the Trigonometric Functions 807 Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f (x) = 1 − tan x 2 . 2. g(x) = 2 cot π 2 x + π + 1. Solution. 1. We proceed as we have in all of the previous graphing examples by setting the argument of 4 , 0, π 2 , equal to each of the ‘quarter marks’ − π , namely x 2 , − π 4 tangent in f (x) = 1 − tan x 2 and π 2 , and solving for x Substituting these x-values into f (x), we ﬁnd points on the graph and the vertical asymptotes. x −x) undeﬁned (x, f (x)) 2 1 0 − π 2 , 2 (0, 1) 2 , 0 π undeﬁned y 2 1 −1 −2 −π − π 2 π 2 π x We see that the period is π − (−π) = 2π. One cycle of y = 1 − tan x 2 . 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π To graph g(x) = 2 cot π and solving for x. 2 x + π + 1, we begin by setting π 4 and π. 2 x + π equal to each quarter mark 4 , π 2 , 3π 808 Foundations of Trigonometry a 0 π 4 π 2 3π 1 4 − 1 2 x + π = 3π We now use these x-values to generate our graph. x −2 − 3 2 −1 − 1 2 0 g(x) undeﬁned (x, g(x)) 3 1 2 , 3 − 3 (−1, 1) 2 , −1 −1 − 1 undeﬁned y 3 2 1 −2 −1 x −1 One cycle of y = 2 cot π 2 x + π + 1. We ﬁnd the period to be 0 − (−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however. 10.5 Graphs of the Trigonometric Functions 809 10.5.4 Exercises In Exercises 1 - 12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function. 1. y = 3 sin(x) 2. y = sin(3x) 3. y = −2 cos(x) 4. y = cos x − 7. y = − 1 3 cos π 2 1 2 x + π 3 5. y = − sin x + π 3 6. y = sin(2x − π) 8. y = cos(3x − 2π) + 4 9. y = sin −x − π 4 − 2 10. y = 2 3 cos π 2 − 4x + 1 11. y = − cos 2x + 3 2 π 3 − 1 2 12. y = 4 sin(−2πx + π) In Exercises 13 - 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) + 4 21. y = csc −x − tan(−2x − π) + 1 1 3 sec 22. y = cot x + π 6 23. y = −11 cot x 1 5 24. y = cot 2x + 3π 2 + 1 1 3 In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0 and 0 ≤ φ < 2π. 25. f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) 27. f (x) = − sin(x) + cos(x) − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) 31. f (x) = − 1 2 cos(5x) − √ 3 2 sin(5x) 30. f (x) = 3 2 cos(2x) − √ 3 3 2 sin(2x) + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 810 Foundations of Trigonometry 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 35. In Exercises 25 - 34, you should have noticed a relationship between the phases φ for the S(x) and C(x). Show that if f (x) = A sin(ωx + α) + B, then f (x) = A cos(ωx + β) + B where β = α − . π 2 36. Let φ be an angle measured in radians and let P (a, b) be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in Theorem 10.15 to show that f (x) = a sin(ωx) + b cos(ωx) + B (with ω > 0) can be rewritten as f (x) = a2 + b2 sin(ωx + φ) + B. √ 37. With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation. (We will revisit this in Section 10.7.) In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator. 38. sin2(x) + cos2(x) = 1 39. sec2(x) − tan2(x) = 1 40. cos(x) = sin 41. tan(x + π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tan x 2 = − x π 2 sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates. 44. f (x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period? 45. f (x) = sin(x) x . What appears to be the horizontal asymptote of the graph? 46. f (x) = x sin(x). Graph y = ±x on the same set of axes and describe the behavior of f . 47. f (x) = sin 1 x . What’s happening as x → 0? 48. f (x) = x − tan(x). Graph y = x on the same set of axes and describe the behavior of f . 49. f (x) = e−0.1x (cos(2x) + sin(2x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 50. f (x) = e−0.1x (cos(2x) + 2 sin(x)). Graph y = ±e−0.1x on the same set of axes and describe the behavior of f . 51. Show that a constant function f is periodic by showing that f (x + 117) = f (x) for all real numbers x. Then show that f has no period by showing that you cannot ﬁnd a smallest number p such that f (x + p) = f (x) for all real numbers x. Said another way, show that f (x + p) = f (x) for all real numbers x for ALL values of p > 0, so no smallest value exists to satisfy the deﬁnition of ‘period’. 10.5 Graphs of the Trigonometric Functions 811 10.5.5 Answers 1. y = 3 sin(x) Period: 2π Amplitude: 3 Phase Shift: 0 Vertical Shift: 0 2. y = sin(3x) 2π 3 Period: Amplitude: 1 Phase Shift: 0 Vertical Shift: 0 3. y = −2 cos(x) Period: 2π Amplitude: 2 Phase Shift: 0 Vertical Shift: 0 4. y = cos x − π 2 Period: 2π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 y 3 −3 y 1 −1 y 2 −2 y 1 −1 π 2 π 3π 2 x 2π π 6 π 3 π 2 2π 3 x π 2 π 3π 2 2π x π 2 π 3π 2 2π 5π 2 x 812 5. y = − sin x + Period: 2π Amplitude: 1 π Phase Shift: − 3 Vertical Shift: 0 π 3 Foundations of Trigonometry y 1 − π 3 π 6 2π 3 7π 6 5π 3 x 6. y = sin(2x − π) Period: π Amplitude: 1 Phase Shift: Vertical Shift: 0 π 2 7. y = − 1 3 cos 1 2 x + π 3 Period: 4π Amplitude: 1 3 2π Phase Shift: − 3 Vertical Shift: 0 8. y = cos(3x − 2π) + 4 2π 3 Period: Amplitude: 1 2π Phase Shift: 3 Vertical Shift: 4 −1 y 1 −1 y 1 3 π 2 3π 4 π 5π 4 3π 2 x − 2π 3 π 3 4π 3 7π 3 x 10π 3 − 1 3 y 5 4 3 2π 3 5π 6 π 7π 6 4π 3 x 10.5 Graphs of the Trigonometric Functions 813 9. y = sin −x − π 4 − 2 Period: 2π Amplitude: 1 Phase Shift: − π 4 π x + 4 Vertical Shift: −2 y = − sin (You need to use − 2 to ﬁnd this.)15 − 4x + 1 10. y = 2 3 Period: π 2 cos π 2 Amplitude: 2 3 π 8 4x − cos y = Vertical Shift: 1 Phase Shift: 2 3 (You need to use π 2 + 1 to ﬁnd this.)16 − 9π 4 − 7π 4 − 5π 4 − 3π 4 y π 4 3π 4 5π 4 x 7π 4 − π 4 −1 −2 − 3π 8 π 2 5π 8 x − 3π 8 − π 4 − π 8 11. y = − cos 2x + 3 2 π 3 − 1 2 y 1 Period: π Amplitude: 3 2 Phase Shift: − π 6 Vertical Shift: − 1 2 12. y = 4 sin(−2πx + π) Period: 1 Amplitude: 4 1 2 (You need to use Phase Shift: y = −4 sin(2πx − π) to ﬁnd this.)17 Vertical Shift: 0 − π 6 − 1 2 π 12 π 3 7π 12 x 5π 6 −4 15Two cycles of the graph are shown to illustrate the discrepancy discussed on page 796. 16Again, we graph two cycles to illustrate the discrepancy discussed on page 796. 17This will be the last time we graph two cycles to illustrate the discrepancy discussed on page 796. 814 Foundations of Trigonometry 13. y = tan x − Period: π π 3 14. y = 2 tan x − 3 1 4 Period: 4π y 1 − π 6 −1 π 12 π 3 7π 12 5π 6 x −2π −π y −1 −3 −5 π 2π x 15. y = tan(−2x − π) + 1 1 3 is equivalent to 1 3 tan(2x + π) + 1 y = − via the Even / Odd identity for tangent. Period − 3π 8 − π 4 x − 3π 4 − 5π 8 10.5 Graphs of the Trigonometric Functions 815 16. y = sec x − π 2 Start with y = cos x − π 2 Period: 2π 17. y = − csc x + π 3 Start with y = − sin Period: 2π x + π 3 y 1 −1 y 1 π 2 π 3π 2 2π x 5π 2 − π 3 −1 π 6 2π 3 7π 6 x 5π 3 18. y = − 1 3 sec 1 2 x + Start with y = − Period: 4π 1 3 π 3 cos 1 2 x + π 3 y 1 3 − 2π 3 − 1 3 π 3 4π 3 7π 3 x 10π 3 816 Foundations of Trigonometry 19. y = csc(2x − π) Start with y = sin(2x − π) Period: π 20. y = sec(3x − 2π) + 4 Start with y = cos(3x − 2π) + 4 Period: 2π 3 21. y = csc −x − π 4 Start with y = sin Period: 2π − 2 −x − π 4 − 2 y 1 −1 −2 −3 π 2 3π 4 π 5π 4 x 3π 2 2π 3 5π 6 π 7π 6 4π 3 x π 4 3π 4 5π 4 x 7π 4 10.5 Graphs of the Trigonometric Functions 817 22. y = cot x + Period: π π 6 23. y = −11 cot Period: 5π 1 5 x y 1 − π 6 π 12 −1 π 3 7π 12 x 5π 6 y 11 −11 5π 4 5π 2 15π 4 5π x 24. y = 1 3 Period: cot π 2 2x + 3π − 3π 8 − π 4 x − 3π 4 − 5π 8 818 Foundations of Trigonometry 25.
f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 − 2 = √ x + 2 cos π 4 − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin 2x + 4π 3 = cos 2x + 5π 6 29. f (x) = 2 √ 3 cos(x) − 2 sin(x) = 4 sin x + 30. f (x) = 3 2 cos(2x) − 31. f (x) = − 1 2 cos(5x) − 3 2 √ 3 3 2 √ 2π 3 = 4 cos x + π 6 5π 6 + 6 = 3 cos = cos 5x + 2π 3 sin(2x) + 6 = 3 sin 2x + sin(5x) = sin 5x + 7π 6 2x + π 3 + 6 32. f (x) = −6 √ 3 cos(3x) − 6 sin(3x) − 3 = 12 sin 3x + 4π 3 − 3 = 12 cos 3x + 5π 6 − 3 33. f (x) = √ 5 2 2 sin(x) − √ 5 2 2 cos(x) = 5 sin x + 34. f (x) = 3 sin x 6 √ − 3 3 cos x 6 = 6 sin x 6 + 7π 4 5π 3 = 5 cos x + 5π 4 = 6 cos x 6 + 7π 6 10.6 The Inverse Trigonometric Functions 819 10.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with ﬁnding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in Section 5.2 to obtain a one-to-one function. We ﬁrst consider f (x) = cos(x). Choosing the interval [0, π] allows us to keep the range as [−1, 1] as well as the properties of being smooth and continuous. y x Restricting the domain of f (x) = cos(x) to [0, π]. 1 Recall from Section 5.2 that the inverse of a function f is typically denoted f −1. For this reason, some textbooks use the notation f −1(x) = cos−1(x) for the inverse of f (x) = cos(x). The obvious pitfall here is our convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is far too easy to confuse cos−1(x) with cos(x) = sec(x) so we will not use this notation in our text.1 Instead, we use the notation f −1(x) = arccos(x), read ‘arc-cosine of x’. To understand the ‘arc’ in ‘arccosine’, recall that an inverse function, by deﬁnition, reverses the process of the original function. The function f (t) = cos(t) takes a real number input t, associates it with the angle θ = t radians, and returns the value cos(θ). Digging deeper,2 we have that cos(θ) = cos(t) is the x-coordinate of the terminal point on the Unit Circle of an oriented arc of length \|t\| whose initial point is (1, 0). Hence, we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as x-coordinates on the Unit Circle. The function f −1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of f (x) = cos(x) and f −1(x) = arccos(x), where we obtain the latter from the former by reﬂecting it across the line y = x, in accordance with Theorem 5.3. y 1 −x) = cos(x), 0 ≤ x ≤ π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arccos(x). 1But be aware that many books do! As always, be sure to check the context! 2See page 704 if you need a review of how we associate real numbers with angles in radian measure. 820 Foundations of Trigonometry We restrict g(x) = sin(x) in a similar manner, although the interval of choice is − π 2 , π 2 y . x Restricting the domain of f (x) = sin(x) to − π 2 , π 2 . It should be no surprise that we call g−1(x) = arcsin(x), which is read ‘arc-sine of x’. y 1 − π 2 x π 2 −1 g(x) = sin(x), − 1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arcsin(x). We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 10.26. Properties of the Arccosine and Arcsine Functions Properties of F (x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range – arcsin(x) = t if and only if − π – sin(arcsin(x)) = x provided − – arcsin(sin(x)) = x provided − π – additionally, arcsine is odd 2 2 and sin(t) = x 10.6 The Inverse Trigonometric Functions 821 Everything in Theorem 10.26 is a direct consequence of the facts that f (x) = cos(x) for 0 ≤ x ≤ π and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for − π 2 and G(x) = arcsin(x). It’s about time for an example. 2 ≤ x ≤ π Example 10.6.1. 1. Find the exact values of the following. (a) arccos 1 2 (c) arccos − √ 2 2 (e) arccos cos π 6 (g) cos arccos − 3 5 (b) arcsin √ 2 2 (d) arcsin − 1 2 (f) arccos cos 11π 6 (h) sin arccos − 3 5 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan (arccos (x)) (b) cos (2 arcsin(x)) Solution. 1. (a) To ﬁnd arccos 1 2 , we need to ﬁnd the real number t (or, equivalently, an angle measuring 3 meets these 2 . We know t = π 2 and π 2 with sin(t) = √ 2 2 . The (b) The value of arcsin t radians) which lies between 0 and π with cos(t) = 1 = π criteria, so arccos 1 3 . 2 √ 2 is a real number t between − π 2 √ 2 number we seek is t = π 4 . Hence, arcsin 2 = π 4 . √ (c) The number t = arccos √ − 2 2 lies in the interval [0, π] with cos(t) = − √ 2 2 . Our answer 2 2 − is arccos (d) To ﬁnd arcsin − 1 2 answer is t = − π 6 so that arcsin − 1 = 3π 4 . , we seek the number t in the interval − , we could simply invoke Theorem 10.26 to get arccos cos π 6 . However, in order to make sure we understand why this is the case, we choose to work the example through using the deﬁnition of arccosine. Working from the inside out, arccos cos π is the real number t with 0 ≤ t ≤ π 6 with sin(t) = − 1 (e) Since . The 2 , π 2 6 2 = arccos √ 3 2 . We ﬁnd t = π . Now, arccos 6 , so that arccos cos π 6 = π 6 . and cos(t) = 822 Foundations of Trigonometry (f) Since 11π 6 does not fall between 0 and π, Theorem 10.26 does not apply. We are forced to . From work through from the inside out starting with arccos cos 11π 6 = arccos 6 . Hence, arccos cos 11π is to use Theorem 10.26 directly. Since − the previous problem, we know arccos (g) One way to simplify cos arccos − 3 5 between −1 and 1, we have that cos arccos − 3 5 before, to really understand why this cancellation occurs, we let t = arccos − 3 5 by deﬁnition, cos(t) = − 3 in (nearly) the same amount of time. 5 is 5 and we are done. However, as . Then, 5 , and we are ﬁnished 5 . Hence, cos arccos − 3 = cos(t) = − 3 = − 3 5 2. (h) As in the previous example, we let t = arccos − 3 5 5 for some t where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to ﬁnd = sin(t). Using the Pythagorean Identity cos2(t) + sin2(t) = 1, we get sin arccos − 3 5 − 3 5 . Since t corresponds to a Quadrants II angle, we 5 choose sin(t) = 4 + sin2(t) = 1 or sin(t) = ± 4 5 . Hence, sin arccos − 3 so that cos(ta) We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos(x), so our goal is to ﬁnd a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 < t ≤ π so that, geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach3 to ﬁnding tan(t) is to use the quotient identity tan(t) = sin(t) cos(t) . Substituting cos(t) = x into the Pythagorean Identity cos2(t) + sin2(t) = 1 gives x2 + sin2(t) = 1, from which we 1 − x2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0, get sin(t) = ± so we choose sin(t) = 2 from consideration. Hence, either 0 ≤ t < π 1 − x2. Thus, 2 or π √ √ tan(t) = sin(t) cos(t) = √ 1 − x2 x To determine the values of x for which this equivalence is valid, we consider our substitution t = arccos(x). Since the domain of arccos(x) is [−1, 1], we know we must restrict −1 ≤ x ≤ 1. Additionally, since we had to discard t = π 2 , we need to discard x = cos π 2 = 0. Hence, tan (arccos (x)) = is valid for x in [−1, 0) ∪ (0, 1]. 1−x2 x √ 2 , π (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms interval − π of x. Since cos(2t) is deﬁned everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos2(t) − sin2(t), 2 cos2(t) − 1 and 1 − 2 sin2(t). Since we know x = sin(t), it is easiest to use the last form: 2 cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2(t) = 1 − 2x2 3Alternatively, we could use the identity: 1 + tan2(t) = sec2(t). Since x = cos(t), sec(t) = 1 cos(t) = 1 x . The reader is invited to work through this approach to see what, if any, diﬃculties arise. 10.6 The Inverse Trigonometric Functions 823 To ﬁnd the restrictions on x, we once again appeal to our substitution t = arcsin(x). Since arcsin(x) is deﬁned only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1−2x2 is valid only on [−1, 1]. A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that arccos cos 11π 6 as opposed to 11π 6 . This is the exact same phenomenon discussed in Section 6 (−2)2 = 2 as opposed to −2. Additionally, even though the expression we 5.2 when we saw arrived at in part 2b above, namely 1 − 2x2, is deﬁned for all real numbers, the equivalence cos (2 arcsin(x)) = 1 − 2x2 is valid for only −1 ≤ x ≤ 1. This is akin to the fact that while the x)2 = x is valid only for x ≥ 0. For expression x is deﬁned for all real numbers, the equivalence ( this reason, it pays to be careful when we determine the intervals where such equivalences are valid. = π √ The ne
xt pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2 , π 2 vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the horizontal 2 and y = π asymptotes y = − π 2 of the graph of f −1(x) = arctan(x). y 1 −x) = tan(x), − π 2 < x < π 2 . reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates y π 2 π 4 −1(x) = arctan(x). Next, we restrict g(x) = cot(x) to its fundamental cycle on (0, π) to obtain g−1(x) = arccot(x). Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the horizontal asymptotes y = 0 and y = π of the graph of g−1(x) = arccot(x). We show these graphs on the next page and list some of the basic properties of the arctangent and arccotangent functions. 824 Foundations of Trigonometry y 1 −1 π 4 π 2 3π 4 π x y π 3π 4 π 2 π 4 g(x) = cot(x), 0 < x < π. reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates g−1(x) = arccot(x). −1 1 x Theorem 10.27. Properties of the Arctangent and Arccotangent Functions Properties of F (x) = arctan(x) – Domain: (−∞, ∞) – Range: − π 2 , π 2 – as x → −∞, arctan(x) → − π 2 – arctan(x) = t if and only if − π for x > 0 – arctan(x) = arccot 1 x – tan (arctan(x)) = x for all real numbers x 2 < x < π – arctan(tan(x)) = x provided − π – additionally, arctangent is odd 2 +; as x → ∞, arctan(x and tan(t) = x − Properties of G(x) = arccot(x) – Domain: (−∞, ∞) – Range: (0, π) – as x → −∞, arccot(x) → π−; as x → ∞, arccot(x) → 0+ – arccot(x) = t if and only if 0 < t < π and cot(t) = x – arccot(x) = arctan 1 x – cot (arccot(x)) = x for all real numbers x for x > 0 – arccot(cot(x)) = x provided 0 < x < π 10.6 The Inverse Trigonometric Functions 825 Example 10.6.2. 1. Find the exact values of the following. √ (a) arctan( 3) (c) cot(arccot(−5)) √ 3) (b) arccot(− (d) sin arctan − 3 4 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(2 arctan(x)) (b) cos(arccot(2x)) Solution. 1. 2. (a) We know arctan( 3 , so arctan( t = π 3) is the real number t between − π 3) = π 3 . √ √ √ (b) The real number t = arccot(− 3) = 5π 6 . arccot(− √ 2 and π 2 with tan(t) = √ 3. We ﬁnd 3) lies in the interval (0, π) with cot(t) = − 3. We get √ (c) We can apply Theorem 10.27 directly and obtain cot(arccot(−5)) = −5. However, working it through provides us with yet another opportunity to understand why this is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5. (a) If we let t = arctan(x), then − π (d) We start simplifying sin arctan − 3 4 . Then tan(t) = − 3 by letting t = arctan − , we choose csc(t) = − 5 2 3 , so sin(t) = − 3 3 . Substituting, we get 1 + − 4 2 . Since tan(t) < 0, we know, in fact, − π 4 for some − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot2(t) = csc2(t), since this relates the reciprocals of tan(t) and sin(t) and is valid for all t under consideration.4 From tan(t) = − 3 4 , we get cot(t) = − 4 3 . Since = − 3 − π 5 . = csc2(t) so that csc(t) = ± 5 5 . Hence, sin arctan − 3 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note that tan(2t) is undeﬁned when 2t = π 2 + πk for integers k. Dividing both sides of this equation by 2 tells us we need to exclude values of t where t = π 4 + π 2 k, where k is are t = ± π an integer. The only members of this family which lie in − π 2 , π 4 , which 2 . Returning ∪ − π means the values of t under consideration are − to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan2(t) , is valid for all the values of t under consideration, hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do you expect from this book? 826 Foundations of Trigonometry To ﬁnd where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the values of t we discarded earlier, t = ± π 4 . Since x = tan(t), this means we exclude x = tan ± π 1−x2 holds for all x in 4 (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). = ±1. Hence, the equivalence tan(2 arctan(x)) = 2x (b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always deﬁned, there are no additional restrictions on t, so we can begin using identities to relate cot(t) to cos(t). The identity cot(t) = cos(t) sin(t) is valid for t in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1 + cot2(t) = csc2(t) holds for all t in (0, π) and relates cot(t) and csc(t) = 1 sin(t) . Substituting cot(t) = 2x, we get 1 + (2x)2 = csc2(t), or csc(t) = ± 4x2 + 1. Since t is . Hence, between 0 and π, csc(t) > 0, so csc(t) = 4x2 + 1 which gives sin(t) = 1√ √ √ 4x2+1 cos(arccot(2x)) = cos(t) = cot(t) sin(t) = √ 2x 4x2 + 1 Since arccot(2x) is deﬁned for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = 2x√ for all real numbers x. 4x2+1 The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were ﬁrst discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted. y y x x The graph of y = sec(x). The graph of y = csc(x). It is clear from the graph of secant that we cannot ﬁnd one single continuous piece of its graph which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to deﬁne the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simpliﬁes the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so. We present both points of view. 10.6 The Inverse Trigonometric Functions 827 10.6.1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For f (x) = sec(x), we restrict the domain to 0x) = sec(x) on 0, π 2 ∪ π 2 , π reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to − π 2 , 0 ∪ 0, π 2 . y 1 −1 − π 2 x π 2 g(x) = csc(x) on − π 2 , 0 ∪ 0, π 2 y π 2 −1 1 x reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates − π 2 g−1(x) = arccsc(x) Note that for both arcsecant and arccosecant, the domain is (−∞, −1] ∪ [1, ∞). Taking a page from Section 2.2, we can rewrite this as {x : \|x\| ≥ 1}. This is often done in Calculus textbooks, so we include it here for completeness. Using these deﬁnitions, we get the following properties of the arcsecant and arccosecant functions. 828 Foundations of Trigonometry Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 , π 2 – as x → −∞, arcsec(x) → π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π provided \|x\| ≥ 1 2 or π 2 or π 2 < x ≤ π +; as x → ∞, arcsec(x and sec(t) = x Properties of G(x) = arccsc(x) 2 2 , 0 ∪ 0, π – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: − π – as x → −∞, arccsc(x) → 0−; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if − π – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided − π – additionally, arccosecant is odd 2 ≤ x < 0 or or 0 < t ≤ π provided \|x\| ≥ 1 2 2 and csc(t) = x a. . . assuming the “Trigonometry Friendly” ranges are used. Example 10.6.3. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 10.6 The Inverse Trigonometric Functions 829 Solution. 1. 2. (a) Using Theorem 10.28, we have arcsec(2) = arccos 1 2 (b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin − 1 2 (c) Since 5π 4 doesn’t fall between 0 and π 2 and π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which yields arcsec sec 5π 4 = arcsec(− = − π 6 . 2) = arccos = π 3 . 2 or π = 3π 4 . 2 2 √ − √ (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3 and, since this is negative, we have that t lies in the interval − π 2 , 0. We are after cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot2(t) = csc2(t). Substituting, we have 1 + cot2(t) = (−3)2, or cot(t) = ± 2 ≤ t < 0, cot(t) < 0, so we get cot (arccsc (−3)) = −2 2. Since − π 8 = ±2 √ √ √ 2. ∪ π (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π 2 , π, and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restric
tions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, √ and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2 , π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 tan(t) ≤ 0. As a result, we get a piecewise deﬁned function for tan(t) tan(t) = √ √ − x2 − 1, x2 − 1, if 0 ≤ t < π 2 if π 2 < t ≤ π Now we need to determine what these conditions on t mean for x. Since x = sec(t), when , x ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞). 2 , x ≥ 1, and when π tan(arcsec(x)) = √ √ − x2 − 1, if x ≥ 1 x2 − 1, if x ≤ −1 2 , 0∪0, π (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in , and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). − π Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| Since t belongs to − π . (The absolute values here are necessary, since x could be negative.) To ﬁnd the values for , we know cos(t) ≥ 0, so we choose cos(t) = 2 , 0 ∪ 0, π 16−x2 4\|x\| 2 √ 830 Foundations of Trigonometry which this equivalence is valid, we look back at our original substution, t = arccsc(4x). Since the domain of arccsc(x) requires its argument x to satisfy \|x\| ≥ 1, the domain of arccsc(4x) requires \|4x\| ≥ 1. Using Theorem 2.4, we rewrite this inequality and solve to get x ≤ − 1 4 . Since we had no additional restrictions on t, the equivalence ∪ 1 cos(arccsc(4x)) = holds for all x in −∞, − 1 4 4 or x ≥ 1 16x2−1 4\|x\| 4 , ∞. √ Inverses of Secant and Cosecant: Calculus Friendly Approach 10.6.2 In this subsection, we restrict f (x) = sec(x) to 0, π 2 y 1 −1 π 2 π 3π 2 x ∪ π, 3π 2 y 3π 2 π π 2 f (x) = sec(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x f −1(x) = arcsec(x) and we restrict g(x) = csc(x) to 0, π 2 ∪ π, 3π 2 . y 1 −1 π 2 π 3π 2 x y 3π 2 π π 2 g(x) = csc(x) on 0, π 2 ∪ π, 3π 2 reﬂect across y = x −−−−−−−−−−−−→ switch x and y coordinates −1 1 x g−1(x) = arccsc(x) Using these deﬁnitions, we get the following result. 10.6 The Inverse Trigonometric Functions 831 Theorem 10.29. Properties of the Arcsecant and Arccosecant Functionsa Properties of F (x) = arcsec(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arcsec(x) → 3π 2 – arcsec(x) = t if and only if 0 ≤ t < π – arcsec(x) = arccos 1 for x ≥ 1 onlyb x – sec (arcsec(x)) = x provided \|x\| ≥ 1 – arcsec(sec(x)) = x provided 0 ≤ x < π − 2 or π ≤ x < 3π 2 ; as x → ∞, arcsec(x) → π 2 2 or π ≤ t < 3π − 2 and sec(t) = x Properties of G(x) = arccsc(x) ∪ π, 3π 2 – Domain: {x : \|x\| ≥ 1} = (−∞, −1] ∪ [1, ∞) – Range: 0, π 2 – as x → −∞, arccsc(x) → π+; as x → ∞, arccsc(x) → 0+ – arccsc(x) = t if and only if 0 < t ≤ π for x ≥ 1 onlyc – arccsc(x) = arcsin 1 x – csc (arccsc(x)) = x provided \|x\| ≥ 1 – arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < t ≤ 3π 2 or π < x ≤ 3π 2 2 and csc(t) = x a. . . assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28. Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare and contrast the solution to each. Example 10.6.4. 1. Find the exact values of the following. (a) arcsec(2) (b) arccsc(−2) (c) arcsec sec 5π 4 (d) cot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the deﬁnition. and satisﬁes csc(t) = −2. The t The real number t = arccsc(−2) lies in 0, π 2 we’re after is t = 7π ∪ π, 3π 2 6 , so arccsc(−2) = 7π 6 . = π 3 . lies between π and 3π (c) Since 5π 4 arcsec sec 5π 4 tion as we have done in the previous examples to see how it goes. 2 , we may apply Theorem 10.29 directly to simplify 4 . We encourage the reader to work this through using the deﬁni- = 5π (d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t). . Using the identity 2. Since √ 2. We know csc(t) = −3, and since this is negative, t lies in π, 3π 2 1 + cot2(t) = csc2(t), we ﬁnd 1 + cot2(t) = (−3)2 so that cot(t) = ± t is in the interval π, 3π 2 , we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2 8 = ±2 √ √ ∪ π, 3π 2 (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0, π , and we seek a formula for tan(t). Since tan(t) is deﬁned for all t values 2 under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. Since t lies in 0, π , tan(t) ≥ 0, so we choose tan(t) = x2 − 1. Since we found √ 2 x2 − 1 holds for all x no additional restrictions on t, the equivalence tan(arcsec(x)) = in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞). ∪ π, 3π 2 √ √ ∪ π, 3π 2 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in , and we now set about ﬁnding an expression for cos(arccsc(4x)) = cos(t). 0, π 2 Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = 1 4x , so to ﬁnd cos(t), we can make use if the identity cos2(t) + sin2(t) = 1. Substituting sin(t) = 1 = 1. Solving, we get 2 cos(t) = ± 4x gives cos2(t) + 1 √ 4x 16x2 − 1 16x2 = ± 16x2 − 1 4\|x\| , then cos(t) ≥ 0, and we choose cos(t) = in which case cos(t) ≤ 0, so, we choose cos(t) = − If t lies in 0, π 2 to π, 3π 2 (momentarily) piecewise deﬁned function for cos(t) √ 16x2−1 4\|x\| √ . Otherwise, t belongs 16x2−1 4\|x\| This leads us to a cos(t) = √ √ −    16x2 − 1 4\|x\| 16x2 − 1 4\|x\| , , if 0 ≤ t ≤ π 2 if π < t ≤ 3π 2 10.6 The Inverse Trigonometric Functions 833 We now see what these restrictions mean in terms of x. Since 4x = csc(t), we get that for 0 ≤ t ≤ π 2 , 4x ≥ 1, or x ≥ 1 4 . In this case, we can simplify \|x\| = x so √ √ cos(t) = 16x2 − 1 4\|x\| = 16x2 − 1 4x Similarly, for π < t ≤ 3π get 2 , we get 4x ≤ −1, or x ≤ − 1 4 . In this case, \|x\| = −x, so we also cos(t) = − √ 16x2 − 1 4\|x\| = − √ 16x2 − 1 4(−x) = √ 16x2 − 1 4x 16x2−1 Hence, in all cases, cos(arccsc(4x)) = 4x the domain of t = arccsc(4x), namely −∞, − 1 4 , and this equivalence is valid for all x in ∪ 1 4 , ∞ √ 10.6.3 Calculators and the Inverse Circular Functions. In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin−1, cos−1 and tan−1, respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates. Example 10.6.5. 1. Use a calculator to approximate the following values to four decimal places. (a) arccot(2) (b) arcsec(5) (c) arccot(−2) (d) arccsc − 3 2 2. Find the domain and range of the following functions. Check your answers using a calculator. (a) f (x) = π 2 − arccos x 5 (b) f (x) = 3 arctan (4x). (c) f (x) = arccot x 2 + π Solution. 1. (a) Since 2 > 0, we can use the property listed in Theorem 10.27 to rewrite arccot(2) as arccot(2) = arctan 1 2 . In ‘radian’ mode, we ﬁnd arccot(2) = arctan 1 2 ≈ 0.4636. (b) Since 5 ≥ 1, we can use the property from either Theorem 10.28 or Theorem 10.29 to write arcsec(5) = arccos 1 5 ≈ 1.3694. 834 Foundations of Trigonometry (c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us ﬁnd arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach. Consider α, the reference angle for θ, as pictured below. By deﬁnition, α is an acute angle so 0 < α < π 2 , and the Reference Angle Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get α = arccot(2) = arctan 1 ≈ 2.6779 radians, 2 we get arccot(−2) ≈ 2.6779. radians. Since θ = π − α = π − arctan 1 2 y 1 θ = arccot(−2) radians α 1 x . By deﬁnition, the real number Another way to attack the problem is to use arctan − 1 2 t = arctan − 1 satisﬁes tan(t) = − 1 2 . Since tan(t) < 0, we know 2 more speciﬁcally that − π 2 < t < 0, so t corresponds to an angle β in Quadrant IV. To ﬁnd the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians and note that it is a Quadrant II angle with tan(θ) = − 1 2 . This means it is exactly π units away from β, and we get θ = π + β = π + arctan − 1 ≈ 2.6779 radians. Hence, 2 as before, arccot(−2) ≈ 2.6779. 2 < t < π 2 with − π 10.6 The Inverse Trigonometric Functions 835 y 1 θ = arccot(−2) radians , π (d) If the range of arccosecant is taken to be − π = arcsin − 2 3 ∪ π, 3π 2 , we can use Theorem 10.28 to ≈ −0.7297. If, on the other hand, the range of arccosecant get arccsc − 3 2 is taken to be 0, π , then we proceed as in the previo
us problem by letting 2 . Then t is a real number with csc(t) = − 3 t = arccsc − 3 2 . Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2 , so t corresponds to a Quadrant III angle, θ. As above, we let α be 2 , which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is now positive, we may use Theorem 10.29 to get α = arccsc 3 ≈ 3.8713 = arcsin 2 2 3 ≈ 3.8713. radians, arccsc − 3 2 radians. Since θ = π + α = π + arcsin 2 3 2 and csc(α) = 3 y 1 θ = arccsc − 3 2 radians α 1 x 836 2. Foundations of Trigonometry 5 by setting the argument of the arccosine, in this case x (a) Since the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can ﬁnd the domain of 2 − arccos x f (x) = π 5 , between −1 and 1. Solving −1 ≤ x 5 ≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determine the range of f , we take a cue from Section 1.7. Three ‘key’ points on the graph of and (1, 0) . Following the procedure outlined in F (x) = arccos(x) are (−1, π), 0, π 2 Theorem 1.7, we track these points to −5, − π . Plotting these values 2 tells us that the range5 of f is − π , (0, 0) and 5, π 2 . Our graph conﬁrms our results. 2 , π 2 (b) To ﬁnd the domain and range of f (x) = 3 arctan (4x), we note that since the domain of F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain of f (x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since 4x is deﬁned for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f , we can, once again, appeal to Theorem 1.7. Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2 and y = π 2 , we ﬁnd that the graph of y = f (x) = 3 arctan (4x) diﬀers from the graph of y = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretch by a factor of 3. It is the latter which aﬀects the range, producing a range of − 3π . We conﬁrm our ﬁndings on the calculator below. 2 , 3π 2 y = f (x) = − arccos y = f (x) = 3 arctan (4x) π 2 x 5 (c) To ﬁnd the domain of g(x) = arccot x + π, we proceed as above. Since the domain of 2 G(x) = arccot(x) is (−∞, ∞), and x 2 is deﬁned for all x, we get that the domain of g is (−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like that of F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0 and y = π. Following Theorem 1.7, we graph y = g(x) = arccot x + π starting with 2 y = G(x) = arccot(x) and ﬁrst performing a horizontal expansion by a factor of 2 and following that with a vertical shift upwards by π. This latter transformation is the one which aﬀects the range, making it now (π, 2π). To check this graphically, we encounter a bit of a problem, since on many calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number 1c, we attempt to rewrite +π in terms of the arctangent function. Using Theorem 10.27, we have g(x) = arccot x 2 when x = arctan 2 that arccot x 2 > 0, or, in this case, when x > 0. Hence, for x > 0, x 2 + π. When x we have g(x) = arctan 2 2 < 0, we can use the same argument in number x . 1c that gave us arccot(−2) = π + arctan − 1 2 to give us arccot x 2 = π + arctan 2 x 5It also conﬁrms our domain! 10.6 The Inverse Trigonometric Functions 837 Hence, for x < 0, g(x) = π + arctan 2 + 2π. What about x = 0? We x know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent will produce this result.6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise deﬁned function: + π = arctan 2 x g(x) = arccot x 2 + π =    We show the input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 + π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following example. Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we ﬁnd that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we 6Without Calculus, of course . . . 7The authors would like to thank Dan Stitz for this problem and associated graphics. 838 Foundations of Trigonometry ﬁnd the angle of inclination, labeled θ below, satisﬁes tan(θ) = 6 we can use the arctangent function and we ﬁnd θ = arctan 1 2 12 = 1 radians ≈ 26.56◦. 2 . Since θ is an acute angle, 6 feet θ 12 feet 10.6.4 Solving Equations Using the Inverse Trigonometric Functions. In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2 for angles θ and tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on the other hand, we had been asked to ﬁnd all angles with sin(θ) = 1 3 or solve tan(t) = −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation x2 = 4 is a lot like sin(θ) = 1 2 in that it has friendly, ‘common value’ answers x = ±2. The equation x2 = 7, on the other hand, is a lot like sin(θ) = 1 3 . We know8 there are answers, but we can’t express them using ‘friendly’ numbers.9 To solve x2 = 7, we make use of the square root function and write x = ± 7. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as x = ± 7. In the same way, we will use the arcsine function to solve sin(θ) = 1 3 , as seen in the following example. √ √ Example 10.6.7. Solve the following equations. 1. Find all angles θ for which sin(θ) = 1 3 . 2. Find all real numbers t for which tan(t) = −2 3. Solve sec(x) = − 5 3 for x. Solution. 1. If sin(θ) = 1 3 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 1 3 . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8How do we know this again? 9This is all, of course, a matter of opinion. For the record, the authors ﬁnd ± √ 7 just as ‘nice’ as ±2. 10.6 The Inverse Trigonometric Functions 839 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II = arcsin 1 3 radians α 1 x Since 1 3 isn’t the sine of any of the ‘common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number t = arcsin 1 is deﬁned so it satisﬁes 3 radians. Since the solutions in Quadrant I 0 < t < π + 2πk are all coterminal with α, we get part of our solution to be θ = α + 2πk = arcsin 1 3 for integers k. Turning our attention to Quadrant II, we get one solution to be π − α. Hence, the Quadrant II solutions are θ = π − α + 2πk = π − arcsin 1 3 3 . Hence, α = arcsin 1 + 2πk, for integers k. 2 with sin(t) = 1 3 2. We may visualize the solutions to tan(t) = −2 as angles θ with tan(θ) = −2. Since tangent is negative only in Quadrants II and IV, we focus our eﬀorts there = arctan(−2) radians π β Since −2 isn’t the tangent of any of the ‘common angles’, we need to use the arctangent function to express our answers. The real number t = arctan(−2) satisﬁes tan(t) = −2 and − π 2 < t < 0. If we let β = arctan(−2) radians, we see that all of the Quadrant IV solutions 840 Foundations of Trigonometry to tan(θ) = −2 are coterminal with β. Moreover, the solutions from Quadrant II diﬀer by exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) = −2 are of the form θ = β + πk = arctan(−2) + πk for some integer k. Switching back to the variable t, we record our ﬁnal answer to tan(t) = −2 as t = arctan(−2) + πk for integers k. 3. The last equation we are asked to solve, sec(x) = − 5 3 , poses two immediate problems. First, we are not told whether or not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universally accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem cos(x) = − 3 5 . Adopting an angle approach, we consider the equation cos(θ) = − 3 5 and note that our solutions lie in Quadrants II and III. Since − 3 5 isn’t the cosine of any of the ‘common angles’, we’ll need to express our solutions in terms of the arccosine function. The real number 5 . If we let β = arccos − 3 t = arccos − 3 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use −β = − arccos − 3 . Since all angle solutions are coterminal with β 5 + 2πk or 5 to be θ = β + 2πk = arccos − 3 or −β, we get our solutions to cos(θ) = − 3 + 2πk for integers k. Switching back to the variable x, we θ = −β + 2πk = − arccos − 3 5 + 2πk record our ﬁnal answer to sec(x) = − 5 for integers k. + 2πk or x = − arccos − 3 5 2 < t < π with cos(t) = − 3 3 as x = arccos − 3 is deﬁned so that π 5 5 5 y 1 y 1 β = arccos − 3 5 radians β = arccos − 3 5 radians 1 x 1 −β = − arccos − 3 5 x radians The reader is encouraged to check the answers found in Example 10.6.7 - both analytically and with the calculator
(see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice . . . 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises In Exercises 1 - 40, ﬁnd the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsin 1 2 3. arcsin − √ 2 2 7. arcsin √ 2 2 4. arcsin − 1 2 8. arcsin √ 3 2 9. arcsin (1) 10. arccos (−1) 11. arccos − √ 3 2 12. arccos − √ 2 2 13. arccos − 1 2 17. arccos √ 3 2 21. arctan − √ 3 3 14. arccos (0) 15. arccos 1 2 16. arccos √ 2 2 18. arccos (1) 19. arctan − √ 3 20. arctan (−1) 22. arctan (0) 23. arctan 25. arctan √ 3 26. arccot − √ 3 29. arccot (0) 30. arccot √ 3 3 33. arcsec (2) 34. arccsc (2) 37. arcsec √ 2 3 3 38. arccsc √ 2 3 3 √ 3 3 24. arctan (1) 27. arccot (−1) 28. arccot − √ 3 3 31. arccot (1) 32. arccot √ 3 35. arcsec √ 2 36. arccsc √ 2 39. arcsec (1) 40. arccsc (1) In Exercises 41 - 48, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 ∪ π, 3π 2 and that the range of 41. arcsec (−2) 45. arccsc (−2) 42. arcsec − √ 46. arccsc − √ 2 2 43. arcsec − 47. arccsc − √ 2 3 3 √ 2 3 3 44. arcsec (−1) 48. arccsc (−1) 842 Foundations of Trigonometry In Exercises 49 - 56, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2 , 0 ∪ 0, π 2 ∪ π 2 , π and that the range of 49. arcsec (−2) 53. arccsc (−2) 50. arcsec − √ 54. arccsc − √ 2 2 51. arcsec − 55. arccsc − √ 2 3 3 √ 2 3 3 52. arcsec (−1) 56. arccsc (−1) In Exercises 57 - 86, ﬁnd the exact value or state that it is undeﬁned. 57. sin arcsin 1 2 58. sin arcsin − √ 2 2 60. sin (arcsin (−0.42)) 61. sin arcsin 5 4 63. cos arccos − 1 2 64. cos arccos 5 13 66. cos (arccos (π)) 67. tan (arctan (−1)) 59. sin arcsin 3 5 √ 62. cos arccos 2 2 65. cos (arccos (−0.998)) 68. tan arctan √ 3 70. tan (arctan (0.965)) 71. tan (arctan (3π)) 69. tan arctan 5 12 72. cot (arccot (1)) 73. cot arccot − √ 3 75. cot (arccot (−0.001)) 76. cot arccot 78. sec (arcsec (−1)) 79. sec arcsec 17π 4 1 2 82. csc arccsc √ 2 81. sec (arcsec (117π)) 84. csc arccsc √ 2 2 74. cot arccot − 7 24 77. sec (arcsec (2)) 80. sec (arcsec (0.75)) 83. csc arccsc − √ 2 3 3 85. csc (arccsc (1.0001)) 86. csc arccsc π 4 In Exercises 87 - 106, ﬁnd the exact value or state that it is undeﬁned. 87. arcsin sin π 6 88. arcsin − sin π 3 89. arcsin sin 3π 4 10.6 The Inverse Trigonometric Functions 843 90. arcsin sin 93. arccos 96. arccos cos cos 11π 6 2π 3 5π 4 91. arcsin sin 94. arccos 97. arctan 4π 3 3π 2 cos 92. arccos 95. arccos cos π 4 cos − π 6 tan tan π 3 π 2 98. arctan tan 101. arctan tan − π 4 2π 3 99. arctan (tan (π)) 100. arctan 102. arccot 105. arccot cot cot π 3 π 2 103. arccot cot 106. arccot cot − π 4 2π 3 104. arccot (cot (π)) In Exercises 107 - 118, assume that the range of arcsecant is 0, π 2 arccosecant is 0, π 2 when ﬁnding the exact value. ∪ π, 3π 2 117. arcsec sec 118. arccsc csc In Exercises 119 - 130, assume that the range of arcsecant is 0, π 2 arccosecant is − π when ﬁnding the exact value. 2 , 0 ∪ 0, π 2 sec π 4 107. arcsec 110. arcsec sec 113. arccsc csc 116. arccsc csc − π 2 5π 4 11π 6 sec π 4 119. arcsec 122. arcsec sec 125. arccsc csc 128. arccsc csc − π 2 5π 4 11π 6 108. arcsec sec 111. arcsec sec 114. arccsc csc 120. arcsec sec 123. arcsec sec 126. arccsc csc 4π 3 5π 3 2π 3 11π 12 4π 3 5π 3 2π 3 11π 12 ∪ π, 3π 2 and that the range of 109. arcsec sec 5π 6 csc π 6 112. arccsc 115. arccsc csc − π 2 9π 8 ∪ π 2 , π and that the range of 121. arcsec sec 5π 6 csc π 6 124. arccsc 127. arccsc csc − π 2 9π 8 129. arcsec sec 130. arccsc csc 844 Foundations of Trigonometry In Exercises 131 - 154, ﬁnd the exact value or state that it is undeﬁned. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin (arctan (−2)) 136. cos arcsin − 5 13 142. tan arcsec 145. cot arccos 5 3 √ 3 2 √ 3 2 146. cot arccsc √ 5 147. cot (arctan (0.25)) 148. sec arccos 149. sec arcsin − 12 13 150. sec (arctan (10)) 151. sec arccot − √ 10 10 152. csc (arccot (9)) 153. csc arcsin 3 5 154. csc arctan − 2 3 In Exercises 155 - 164, ﬁnd the exact value or state that it is undeﬁned. 155. sin arcsin 5 13 + π 4 156. cos (arcsec(3) + arctan(2)) 157. tan arctan(3) + arccos − 3 5 158. sin 2 arcsin − 4 5 159. sin 2arccsc 13 5 161. cos 2 arcsin 3 5 163. cos 2arccot − √ 5 160. sin (2 arctan (2)) 162. cos 2arcsec 25 7 164. sin arctan(2) 2 10.6 The Inverse Trigonometric Functions 845 In Exercises 165 - 184, rewrite the quantity as algebraic expressions of x and state the domain on which the equivalence is valid. 165. sin (arccos (x)) 166. cos (arctan (x)) 167. tan (arcsin (x)) 168. sec (arctan (x)) 169. csc (arccos (x)) 170. sin (2 arctan (x)) 171. sin (2 arccos (x)) 172. cos (2 arctan (x)) 173. sin(arccos(2x)) 174. sin arccos x 5 175. cos arcsin x 2 176. cos (arctan (3x)) 177. sin(2 arcsin(7x)) 178. sin 2 arcsin √ x 3 3 179. cos(2 arcsin(4x)) 180. sec(arctan(2x)) tan(arctan(2x)) 181. sin (arcsin(x) + arccos(x)) 182. cos (arcsin(x) + arctan(x)) 183. tan (2 arcsin(x)) 184. sin arctan(x) 1 2 185. If sin(θ) = 186. If tan(θ) = 187. If sec(θ) = x 2 x 7 x 4 for − π 2 for − π 2 < θ < π 2 , ﬁnd an expression for θ + sin(2θ) in terms of x. < θ < π 2 , ﬁnd an expression for 1 2 θ − 1 2 sin(2θ) in terms of x. for 0 < θ < π 2 , ﬁnd an expression for 4 tan(θ) − 4θ in terms of x. In Exercises 188 - 207, solve the equation using the techniques discussed in Example 10.6.7 then approximate the solutions which lie in the interval [0, 2π) to four decimal places. 188. sin(x) = 7 11 189. cos(x) = − 2 9 190. sin(x) = −0.569 191. cos(x) = 0.117 192. sin(x) = 0.008 193. cos(x) = 194. tan(x) = 117 195. cot(x) = −12 196. sec(x) = 197. csc(x) = − 200. cos(x) = − 90 17 7 16 198. tan(x) = − √ 10 199. sin(x) = 201. tan(x) = 0.03 202. sin(x) = 0.3502 359 360 3 2 3 8 846 Foundations of Trigonometry 203. sin(x) = −0.721 204. cos(x) = 0.9824 205. cos(x) = −0.5637 206. cot(x) = 1 117 207. tan(x) = −0.6109 In Exercises 208 - 210, ﬁnd the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 208. 3, 4 and 5 209. 5, 12 and 13 210. 336, 527 and 625 211. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 212. At Cliﬀs of Insanity Point, The Great Sasquatch Canyon is 7117 feet deep. From that point, a ﬁre is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the ﬁre? Express your answer using degree measure rounded to one decimal place. 213. Shelving is being built at the Utility Muﬃn Research Library which is to be 14 inches deep. An 18-inch rod will be attached to the wall and the underside of the shelf at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 214. A parasailor is being pulled by a boat on Lake Ippizuti. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 215. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the riﬂe with a tranquilizer dart is 300 feet, ﬁnd the smallest value of θ for which the corresponding point on the ground is in range of the riﬂe. Round your answer to the nearest hundreth of a degree. In Exercises 216 - 221, rewrite the given function as a sinusoid of the form S(x) = A sin(ωx + φ) using Exercises 35 and 36 in Section 10.5 for reference. Approximate the value of φ (which is in radians, of course) to four decimal places. 216. f (x) = 5 sin(3x) + 12 cos(3x) 217. f (x) = 3 cos(2x) + 4 sin(2x) 218. f (x) = cos(x) − 3 sin(x) 219. f (x) = 7 sin(10x) − 24 cos(10x) 10.6 The Inverse Trigonometric Functions 847 220. f (x) = − cos(x) − 2 √ 2 sin(x) 221. f (x) = 2 sin(x) − cos(x) In Exercises 222 - 233, ﬁnd the domain of the given function. Write your answers in interval notation. 222. f (x) = arcsin(5x) 223. f (x) = arccos 3x − 1 2 224. f (x) = arcsin 2x2 225. f (x) = arccos 1 x2 − 4 226. f (x) = arctan(4x) 227. f (x) = arccot 2x x2 − 9 228. f (x) = arctan(ln(2x − 1)) 229. f (x) = arccot( √ 2x − 1) 230. f (x) = arcsec(12x) 231. f (x) = arccsc(x + 5) 232. f (x) = arcsec x3 8 233. f (x) = arccsc e2x 234. Show that arcsec(x) = arccos of f (x) = arcsec(x). 1 x for \|x\| ≥ 1 as long as we use 0, ∪ π 2 , π π 2 as the range 235. Show that arccsc(x) = arcsin of f (x) = arccsc(x). 1 x for \|x\| ≥ 1 as long as we use − π 2 , 0 0, ∪ π 2 as the range 236. Show that arcsin(x) + arccos(x) = π 2 for −1 ≤ x ≤ 1. 237. Discuss with your classmates why arcsin 1 2 = 30◦. 238. Use the following picture and the series of exercises on the next page to show that arctan(1) + arctan(2) + arctan(3) = π y D(2, 3) A(0, 1) α β γ B(1, 0) x C(2, 0) O(0, 0) 848 Foundations of Trigonometry (a) Clearly AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the x-axis. Use the distance formula to show that BAD is also a right triangle (with ∠BAD being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem. (b) Use AOB to show that α = arctan(1)
(c) Use BAD to show that β = arctan(2) (d) Use BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α + β + γ = π. Thus arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3. arcsin − √ 2 2 = − π 4 4. arcsin 7. arcsin − . arcsin (0) = 0 8. arcsin √ 3 2 10. arccos (−1) = π 11. arccos − √ 3 2 = π 3 = 5π 6 13. arccos 16. arccos = 2π 14. arccos (0) = 17. arccos √ 3 2 π 2 = π 6 19. arctan − √ 3 = − π 3 20. arctan (−1) = − π 4 22. arctan (0) = 0 23. arctan √ 3 3 = π 6 25. arctan √ 3 = √ π 3 28. arccot − = 3 3 26. arccot − √ 3 = 5π 6 2π 3 29. arccot (0) = π 2 31. arccot (1) = 34. arccsc (2) = 37. arcsec √ 2 3 3 π 4 π 6 32. arccot √ 35. arcsec √ 38. arccsc . arcsin 1 2 = π 6 9. arcsin (1) = π 2 12. arccos − √ 2 2 = 3π 4 15. arccos 1 2 = π 3 18. arccos (1) = 0 21. arctan − √ 3 3 = − π 6 24. arctan (1) = π 4 27. arccot (−1) = 3π 4 30. arccot √ 3 3 = π 3 33. arcsec (2) = 36. arccsc √ π 3 2 = π 4 39. arcsec (1) = 0 40. arccsc (1) = π 2 41. arcsec (−2) = 4π 3 42. arcsec − √ 2 = 5π 4 850 Foundations of Trigonometry 44. arcsec (−1) = π 45. arccsc (−2) = 47. arccsc − √ 3 2 3 = 4π 3 48. arccsc (−1) = 7π 6 3π 2 50. arcsec − 2 = √ 51. arcsec 2 − √ 3 3 = 5π 6 54. arccsc − √ 2 = − π 4 3π 4 π 6 π 2 52. arcsec (−1) = π 53. arccsc (−2) = − 55. arccsc − √ 3 2 = − π 3 56. arccsc (−1) = − 43. arcsec − √ 3 2 3 = 7π 6 46. arccsc − √ 2 = 5π 4 49. arcsec (−2) = 2π 3 57. sin arcsin 59. sin arcsin 61. sin arcsin 58. sin arcsin − √ 2 2 √ 2 2 = − 60. sin (arcsin (−0.42)) = −0.42 √ 2 2 √ 2 2 = is undeﬁned. 62. cos arccos 63. cos arccos − 1 2 = − 1 2 65. cos (arccos (−0.998)) = −0.998 67. tan (arctan (−1)) = −1 69. tan arctan 5 12 = 5 12 64. cos arccos 5 13 = 5 13 66. cos (arccos (π)) is undeﬁned. 68. tan arctan √ 3 = √ 3 70. tan (arctan (0.965)) = 0.965 71. tan (arctan (3π)) = 3π 72. cot (arccot (1)) = 1 73. cot arccot − √ √ 3 = − 3 75. cot (arccot (−0.001)) = −0.001 74. cot arccot 76. cot arccot − 7 24 17π 4 = − 7 24 = 17π 4 77. sec (arcsec (2)) = 2 78. sec (arcsec (−1)) = −1 10.6 The Inverse Trigonometric Functions 851 79. sec arcsec 1 2 is undeﬁned. 81. sec (arcsec (117π)) = 117π 83. csc arccsc 2 − √ 3 3 √ 3 2 3 = − 85. csc (arccsc (1.0001)) = 1.0001 87. arcsin sin 89. arcsin sin 91. arcsin sin 93. arccos cos π 6 3π 4 4π 3 2π 3 95. arccos − cos π 6 = π 4 = − π 3 = 2π 3 = π 6 = π 6 97. arctan tan π 3 = π 3 99. arctan (tan (π)) = 0 101. arctan tan 2π 3 = − π 3 80. sec (arcsec (0.75)) is undeﬁned. 82. csc arccsc √ √ 2 = √ 2 84. csc arccsc is undeﬁned. 2 2 86. csc arccsc π 4 is undeﬁned. 88. arcsin sin 90. arcsin sin − π 3 11π 6 = − π 3 = − π 6 92. arccos cos 94. arccos cos 96. arccos cos π 4 3π 2 5π 4 = − tan 98. arctan 100. arctan tan π 4 π 4 = = π 2 3π 4 = − π 4 102. arccot cot π 2 π 3 is undeﬁned = π 3 103. arccot cot 105. arccot cot − π 4 3π 2 = 107. arcsec sec 109. arcsec sec 111. arcsec sec π 4 5π 6 5π 3 = 3π 4 π 2 7π 6 π 3 104. arccot (cot (π)) is undeﬁned 106. arccot cot 108. arcsec sec 2π 3 4π 3 = = 2π 3 4π 3 110. arcsec 112. arccsc sec − π 2 is undeﬁned. csc π 6 = π 6 = π 4 = = 852 Foundations of Trigonometry 113. arccsc csc 5π 4 = 5π 4 115. arccsc csc 117. arcsec sec 119. arcsec sec 121. arcsec sec 123. arcsec sec 125. arccsc csc 127. arccsc csc 129. arcsec sec − π 2 11π 12 = 3π 2 = 13π 12 π 4 = = 5π 5π 6 5π 3 5π 4 − π 2 11π 12 = 11π 12 131. sin arccos − = 1 2 133. sin (arctan (−2)) = − 135. sin (arccsc (−3)) = − 137. cos arctan √ 7 = 139. cos (arcsec (5)) = 1 5 114. arccsc csc 116. arccsc csc 118. arccsc csc 120. arcsec sec 2π 3 11π 6 9π 8 4π 3 = π 3 = 7π 6 = = 9π 8 2π 3 122. arcsec sec − π 2 is undeﬁned csc csc csc 126. arccsc 128. arccsc 124. arccsc π 6 2π 3 11π 6 9π 8 3 5 134. sin arccot √ 130. arccsc 132. sin arccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 12 13 = 5 12 10.6 The Inverse Trigonometric Functions 853 145. cot arccos √ 3 2 √ 3 = 147. cot (arctan (0.25)) = 4 = 13 5 149. sec arcsin − 12 13 √ 151. sec arccot − 10 10 153. csc arcsin 155. sin arcsin 3 5 = 5 3 5 13 + π 4 = 157. tan arctan(3) + arccos − 146. cot arccsc √ 5 = 2 148. sec arccos √ 3 2 √ 3 2 3 = 150. sec (arctan (10)) = √ 101 √ = − 11 152. csc (arccot (9)) = √ 82 154. csc arctan − 2 3 = − √ 13 2 156. cos (arcsec(3) + arctan(2)) = √ 10 √ 5 − 4 15 √ 17 2 26 3 5 = 1 3 158. sin 2 arcsin − 4 5 = − 24 25 159. sin 2arccsc 161. cos 2 arcsin 13 5 3 5 = 120 169 = 7 25 160. sin (2 arctan (2)) = 4 5 162. cos 2arcsec 25 7 = − 163. cos 2arccot − √ 5 = 2 3 165. sin (arccos (x)) = √ 1 − x2 for −1 ≤ x ≤ 1 164. sin arctan(2) 2 = 5 − 10 527 625 √ 5 166. cos (arctan (x)) = √ 167. tan (arcsin (x)) = √ 1 1 + x2 x 1 − x2 for all x for −1 < x < 1 168. sec (arctan (x)) = √ 1 + x2 for all x 169. csc (arccos (x)) = √ 1 1 − x2 for −1 < x < 1 170. sin (2 arctan (x)) = for all x 2x x2 + 1 √ 171. sin (2 arccos (x)) = 2x 1 − x2 for −1 ≤ x ≤ 1 854 Foundations of Trigonometry 172. cos (2 arctan (x)) = 173. sin(arccos(2x)) = 174. sin arccos 175. cos arcsin x 5 x 2 = = √ 1 − x2 1 + x2 for all x √ 1 − 4x2 for − 1 √ 25 − x2 5 2 ≤ x ≤ 1 2 for −5 ≤ x ≤ 5 4 − x2 2 for −2 ≤ x ≤ 2 176. cos (arctan (3x)) = √ for all x 1 1 + 9x2 √ 177. sin(2 arcsin(7x)) = 14x 1 − 49x2 for − 1 7 ≤ x ≤ 1 7 178. sin 2 arcsin √ x 3 3 √ 2x = 3 − x2 3 √ for − 3 ≤ x ≤ √ 3 179. cos(2 arcsin(4x)) = 1 − 32x2 for − 1 4 1 4 ≤ x ≤ √ 180. sec(arctan(2x)) tan(arctan(2x)) = 2x 1 + 4x2 for all x 181. sin (arcsin(x) + arccos(x)) = 1 for −1 ≤ x ≤ 1 182. cos (arcsin(x) + arctan(x)) = √ 1 − x2 − x2 √ 1 + x2 for −1 ≤ x ≤ 1 183. 10 tan (2 arcsin(x)) = √ 2x 1 − x2 1 − 2x2 for x in −1 184. sin arctan(x) = 1 2    √ x2 + 1 − 1 √ x2 + 1 2 √ x2 + 1 − 1 √ x2 + 1 2 − for x ≥ 0 for x < 0 185. If sin(θ) = 186. If tan(θ) = x 2 x 7 for − π 2 for − π 2 < θ < π 2 , then θ + sin(2θ) = arcsin x 2 x + √ 4 − x2 2 < θ < π 2 , then 1 2 θ − 1 2 sin(2θ) = 1 2 arctan x 7 − 7x x2 + 49 10The equivalence for x = ±1 can be veriﬁed independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those x values. You’ll see what we mean when you work through the details of the identity for tan(2t). For now, we exclude x = ±1 from our answer. 10.6 The Inverse Trigonometric Functions 855 187. If sec(θ) = 188. x = arcsin x 4 7 11 for 0 < θ < π 2 , then 4 tan(θ) − 4θ = √ x2 − 16 − 4arcsec x 4 + 2πk or x = π − arcsin 7 11 + 2πk, in [0, 2π), x ≈ 0.6898, 2.4518 189. x = arccos − 2 9 + 2πk or x = − arccos − 2 9 + 2πk, in [0, 2π), x ≈ 1.7949, 4.4883 190. x = π + arcsin(0.569) + 2πk or x = 2π − arcsin(0.569) + 2πk, in [0, 2π), x ≈ 3.7469, 5.6779 191. x = arccos(0.117) + 2πk or x = 2π − arccos(0.117) + 2πk, in [0, 2π), x ≈ 1.4535, 4.8297 192. x = arcsin(0.008) + 2πk or x = π − arcsin(0.008) + 2πk, in [0, 2π), x ≈ 0.0080, 3.1336 193. x = arccos 359 360 + 2πk or x = 2π − arccos 359 360 + 2πk, in [0, 2π), x ≈ 0.0746, 6.2086 194. x = arctan(117) + πk, in [0, 2π), x ≈ 1.56225, 4.70384 195. x = arctan − 1 12 + πk, in [0, 2π), x ≈ 3.0585, 6.2000 196. x = arccos 2 3 + 2πk or x = 2π − arccos 2 3 + 2πk, in [0, 2π), x ≈ 0.8411, 5.4422 197. x = π + arcsin 198. x = arctan − 3 8 199. x = arcsin 200. x = arccos − 7 16 + 2πk or x = 2π − arcsin 17 90 10 + πk, in [0, 2π), x ≈ 1.8771, 5.0187 17 90 √ + 2πk or x = π − arcsin 3 8 + 2πk, in [0, 2π), x ≈ 0.3844, 2.7572 7 16 + 2πk, in [0, 2π), x ≈ 2.0236, 4.2596 + 2πk or x = − arccos − + 2πk, in [0, 2π), x ≈ 3.3316, 6.0932 201. x = arctan(0.03) + πk, in [0, 2π), x ≈ 0.0300, 3.1716 202. x = arcsin(0.3502) + 2πk or x = π − arcsin(0.3502) + 2πk, in [0, 2π), x ≈ 0.3578, 2.784 203. x = π + arcsin(0.721) + 2πk or x = 2π − arcsin(0.721) + 2πk, in [0, 2π), x ≈ 3.9468, 5.4780 204. x = arccos(0.9824) + 2πk or x = 2π − arccos(0.9824) + 2πk, in [0, 2π), x ≈ 0.1879, 6.0953 205. x = arccos(−0.5637) + 2πk or x = − arccos(−0.5637) + 2πk, in [0, 2π), x ≈ 2.1697, 4.1135 206. x = arctan(117) + πk, in [0, 2π), x ≈ 1.5622, 4.7038 207. x = arctan(−0.6109) + πk, in [0, 2π), x ≈ 2.5932, 5.7348 856 Foundations of Trigonometry 208. 36.87◦ and 53.13◦ 209. 22.62◦ and 67.38◦ 210. 32.52◦ and 57.48◦ 211. 68.9◦ 212. 7.7◦ 213. 51◦ 216. f (x) = 5 sin(3x) + 12 cos(3x) = 13 sin 3x + arcsin 214. 19.5◦ 12 13 ≈ 13 sin(3x + 1.1760) 215. 41.81◦ 217. f (x) = 3 cos(2x) + 4 sin(2x) = 5 sin 2x + arcsin 3 5 ≈ 5 sin(2x + 0.6435) 218. f (x) = cos(x) − 3 sin(x) = √ 10 sin x + arccos − 10 219. f (x) = 7 sin(10x) − 24 cos(10x) = 25 sin 10x + arcsin − √ 3 10 √ ≈ 10 sin(x + 2.8198) 24 25 ≈ 25 sin(10x − 1.2870) 220. f (x) = − cos(x) − 2 √ 2 sin(x) = 3 sin x + π + arcsin 221. f (x) = 2 sin(x) − cos(x) = √ 5 sin x + arcsin − 1 3 ≈ 3 sin(x + 3.4814) √ ≈ 5 sin(x − 0.4636) √ 5 5 223. − , 1 1 3 1 5 222. − 224. − , 1 5 √ √ 2 2 , 225. (−∞, − √ √ 5] ∪ [− √ 3, √ 3] ∪ [ 5, ∞) 2 2 226. (−∞, ∞) 227. (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 228. , ∞ 1 2 230. −∞, − 1 12 ∪ 1 12 , ∞ 229. , ∞ 1 2 231. (−∞, −6] ∪ [−4, ∞) 232. (−∞, −2] ∪ [2, ∞) 233. [0, ∞) 10.7 Trigonometric Equations and Inequalities 857 10.7 Trigonometric Equations and Inequalities In Sections 10.2, 10.3 and most recently 10.6, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we’ve employed thus far. Note that we use the neutral letter ‘u’ as the argument1 of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions To solve cos(u) = c or sin(u) = c for −1 ≤ c ≤ 1, ﬁrst solve for u in the interval [0, 2π) and add integer multiples of the period 2π. If c < −1 or of c > 1, there are no real solutions. To solve sec(u) = c or csc(u) = c for c ≤ −1 or c ≥ 1, convert to cosine or sine, respectively, and solve as above. If −1 < c < 1, there are no real solutions. To solve tan(u) = c for any real number c, ﬁrst solve for u in the interval − π 2 , π 2 and add integer multiples of the period π. To solv
e cot(u) = c for c = 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 + πk for integers k. 2 and ﬁnd the solution x = π 2 , we know the solutions take the form u = π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 Using the above guidelines, we can comfortably solve sin(x) = 1 or x = 5π has the form sin(u) = 1 integers k. Since the argument of sine here is 3x, we have 3x = π integers k. To solve for x, we divide both sides2 of these equations by 3, and obtain x = π or x = 5π 6 + 2πk 2 ? Since this equation 6 + 2πk for 6 + 2πk for 18 + 2π 3 k 3 k for integers k. This is the technique employed in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them using a graphing utility. 2. csc 1 6 + 2πk or u = 5π 6 + 2πk or 3x = 5π 3. cot (3x) = 0 18 + 2π 1. cos(2x3 5. tan x 2 6. sin(2x) = 0.87 4. sec2(x) = 4 Solution. 1. The solutions to cos(u) = − √ the argument of cosine here is 2x, this means 2x = 5π k. Solving for x gives x = 5π analytically, we substitute them into the original equation. For any integer k we have 6 + 2πk for integers k. Since 6 + 2πk for integers 12 + πk for integers k. To check these answers 6 + 2πk or 2x = 7π 12 + πk or x = 7π 6 + 2πk or u = 7π 2 are u = 5π 3 cos 2 5π 12 + πk = cos 5π = cos 5π 6 √ = − 3 2 6 + 2πk (the period of cosine is 2π) 1See the comments at the beginning of Section 10.5 for a review of this concept. 2Don’t forget to divide the 2πk by 3 as well! 858 Foundations of Trigonometry 12 + πk = cos 7π √ Similarly, we ﬁnd cos 2 7π 3 2 . To determine which of our solutions lie in [0, 2π), we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = 1 and are x = 5π 12 , 17π 12 . Using a √ 3 calculator, we graph y = cos(2x) and y = − 2 over [0, 2π) and examine where these two graphs intersect. We see that the x-coordinates of the intersection points correspond to the decimal representations of our exact answers. 6 + 2πk = cos 7π 12 and 19π = − 12 , 7π 6 2. Since this equation has the form csc(u) = 2, we rewrite this as sin(u) = u = π 4 + 2πk or u = 3π 4 + 2πk for integers k. Since the argument of cosecant here is 1 √ 2 2 and ﬁnd 3 x − π + 2πk or 1 3 x − π = 3π 4 + 2πk To solve 1 3 x − π = π 4 + 2πk, we ﬁrst add π to both sides 1 3 x = π 4 + 2πk + π A common error is to treat the ‘2πk’ and ‘π’ terms as ‘like’ terms and try to combine them when they are not.3 We can, however, combine the ‘π’ and ‘ π 4 ’ terms to get We now ﬁnish by multiplying both sides by 3 to get 1 3 x = 5π 4 + 2πk x = 3 5π 4 + 2πk = 15π 4 + 6πk Solving the other equation, 1 check the ﬁrst family of answers, we substitute, combine line terms, and simplify. 4 + 2πk produces x = 21π 3 x − π = 3π 4 + 6πk for integers k. To csc 1 3 15π 4 + 6πk − π = csc 5π = csc π = csc π √ 4 2 = 4 + 2πk − π 4 + 2πk (the period of cosecant is 2π) The family x = 21π 4 + 6πk checks similarly. Despite having inﬁnitely many solutions, we ﬁnd that none of them lie in [0, 2π). To verify this graphically, we use a reciprocal identity to 2 do not intersect at rewrite the cosecant as a sine and we ﬁnd that y = and y = √ 1 sin( 1 3 x−π) all over the interval [0, 2π). 3Do you see why? 10.7 Trigonometric Equations and Inequalities 859 y = cos(2x) and y = − √ 3 2 y = 1 sin( 1 3 x−π) and y = √ 2 3. Since cot(3x) = 0 has the form cot(u) = 0, we know u = π 2 + πk, so, in this case, 3x = π 2 + πk for integers k. Solving for x yields x = π 3 k. Checking our answers, we get 6 + π 2 + πk cot 3 π 6 + π 3 k = cot π = cot π 2 = 0 (the period of cotangent is π) As k runs through the integers, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 2 and 11π 6 . To conﬁrm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose4 to use sin(3x) . Graphing y = cos(3x) the quotient identity cot(3x) = cos(3x) sin(3x) and y = 0 (the x-axis), we see that the x-coordinates of the intersection points approximately match our solutions. 6 , 3π 6 , 7π 2 , 5π 6 , π 4. The complication in solving an equation like sec2(x) = 4 comes not from the argument of secant, which is just x, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we extract square roots to get sec(x) = ±2. Converting to cosines, we have cos(x) = ± 1 3 + 2πk or x = 5π 3 + 2πk for integers k. For cos(x) = − 1 3 + 2πk for integers k. If we take a step back and think of these families of solutions geometrically, we see we are ﬁnding the measures of all angles with a reference angle of π 3 . As a result, these solutions can be combined and we may write our solutions as x = π 3 + πk for integers k. To check the ﬁrst family of solutions, we note that, depending on the integer . However, it is true that for all integers k, k, sec π sec π 3 = ±2. (Can you show this?) As a result, 3 + πk doesn’t always equal sec π 2 . For cos(x) = 1 2 , we get x = 2π 2 , we get x = π 3 + 2πk or x = 4π 3 + πk and x = 2π 3 + πk = ± sec π 3 sec2 π 3 + πk = ± sec π 3 2 = (±2)2 = 4 The same holds for the family x = 2π the values k = 0 and k = 1, namely x = π 3 + πk. The solutions which lie in [0, 2π) come from 3 . To conﬁrm graphically, we use 3 and 5π 3 , 4π 3 , 2π 4The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = 1 tan(3x) instead. The graph on the calculator appears identical, but what happens when you try to ﬁnd the intersection points? 860 Foundations of Trigonometry a reciprocal identity to rewrite the secant as cosine. The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. 1 y = cos(3x) sin(3x) and y = 0 y = 1 cos2(x) and y = 4 5. The equation tan x 2 Hence, x = −3 has the form tan(u) = −3, whose solution is u = arctan(−3)+πk. 2 = arctan(−3) + πk, so x = 2 arctan(−3) + 2πk for integers k. To check, we note tan 2 arctan(−3)+2πk 2 = tan (arctan(−3) + πk) = tan (arctan(−3)) = −3 (the period of tangent is π) (See Theorem 10.27) To determine which of our answers lie in the interval [0, 2π), we ﬁrst need to get an idea of the value of 2 arctan(−3). While we could easily ﬁnd an approximation using a calculator,5 we proceed analytically. Since −3 < 0, it follows that − π 2 < arctan(−3) < 0. Multiplying through by 2 gives −π < 2 arctan(−3) < 0. We are now in a position to argue which of the solutions x = 2 arctan(−3) + 2πk lie in [0, 2π). For k = 0, we get x = 2 arctan(−3) < 0, so we discard this answer and all answers x = 2 arctan(−3) + 2πk where k < 0. Next, we turn our attention to k = 1 and get x = 2 arctan(−3) + 2π. Starting with the inequality −π < 2 arctan(−3) < 0, we add 2π and get π < 2 arctan(−3) + 2π < 2π. This means x = 2 arctan(−3) + 2π lies in [0, 2π). Advancing k to 2 produces x = 2 arctan(−3) + 4π. Once again, we get from −π < 2 arctan(−3) < 0 that 3π < 2 arctan(−3) + 4π < 4π. Since this is outside the interval [0, 2π), we discard x = 2 arctan(−3) + 4π and all solutions of the form x = 2 arctan(−3) + 2πk for k > 2. Graphically, we see y = tan x and y = −3 intersect only 2 once on [0, 2π) at x = 2 arctan(−3) + 2π ≈ 3.7851. 6. To solve sin(2x) = 0.87, we ﬁrst note that it has the form sin(u) = 0.87, which has the family of solutions u = arcsin(0.87) + 2πk or u = π − arcsin(0.87) + 2πk for integers k. Since the argument of sine here is 2x, we get 2x = arcsin(0.87) + 2πk or 2x = π − arcsin(0.87) + 2πk 2 − 1 which gives x = 1 2 arcsin(0.87) + πk for integers k. To check, 2 arcsin(0.87) + πk or x = π 5Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 10.7 Trigonometric Equations and Inequalities 861 sin 2 1 2 arcsin(0.87) + πk = sin (arcsin(0.87) + 2πk) = sin (arcsin(0.87)) (the period of sine is 2π) = 0.87 (See Theorem 10.26) For the family x = π 2 − 1 2 arcsin(0.87) + πk , we get sin 2 π 2 − 1 2 arcsin(0.87) + πk = sin (π − arcsin(0.87) + 2πk) = sin (π − arcsin(0.87)) = sin (arcsin(0.87)) = 0.87 (the period of sine is 2π) (sin(π − t) = sin(t)) (See Theorem 10.26) 2 arcsin(0.87) < π 2 so that multiplying through by 1 4 . Starting with the family of solutions x = 1 To determine which of these solutions lie in [0, 2π), we ﬁrst need to get an idea of the value of x = 1 2 arcsin(0.87). Once again, we could use the calculator, but we adopt an analytic route here. By deﬁnition, 0 < arcsin(0.87) < π 2 gives us 0 < 1 2 arcsin(0.87) + πk, we use the same kind of arguments as in our solution to number 5 above and ﬁnd only the solutions corresponding to k = 0 and k = 1 lie in [0, 2π): x = 1 2 arcsin(0.87) + π. Next, we move to the family x = π 2 arcsin(0.87) + πk for integers k. Here, we need to get a better estimate of π 2 arcsin(0.87) < π 4 , we ﬁrst multiply through by −1 and then add π 4 , or 4 < π π 2 . Proceeding with the usual arguments, we ﬁnd the only solutions which lie in [0, 2π) correspond to k = 0 and k = 1, namely x = π 2 arcsin(0.87) and x = 3π 2 arcsin(0.87). All told, we have found four solutions to sin(2x) = 0.87 in [0, 2π): x = 1 2 arcsin(0.87). By graphing y = sin(2x) and y = 0.87, we conﬁrm our results. 2 arcsin(0.87). From the inequality 0 < 1 2 − 1 2 arcsin(0.87), x = 1 2 arcsin(0.87) and x = 3π 2 arcsin(0.87) + π, x = π 2 arcsin(0.87) and x = 1 2 arcsin(0.87) > π 2 arcsin(0.87) < π 2 to get = tan x 2 and y = −3 y = sin(2x) and y = 0.87 862 Foundations of Trigonometry Each of the problems in Example 10.7.1 featured one trigonometric function. If an equation involves two diﬀerent trigonometric functions or if the equation contains the same trigonometric function but with diﬀerent arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Example 10.7.2. Solve the following equations and list the solutions which lie
in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin3(x) = sin2(x) 3. cos(2x) = 3 cos(x) − 2 5. cos(3x) = cos(5x) 7. sin(x) cos x 2 + cos(x) sin x 2 = 1 Solution. 2. sec2(x) = tan(x) + 3 4. cos(3x) = 2 − cos(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to divide both sides of 3 sin3(x) = sin2(x) by sin2(x) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. 3 sin3(x) = sin2(x) 3 sin3(x) − sin2(x) = 0 sin2(x)(3 sin(x) − 1) = 0 Factor out sin2(x) from both terms. We get sin2(x) = 0 or 3 sin(x) − 1 = 0. Solving for sin(x), we ﬁnd sin(x) = 0 or sin(x) = 1 3 . The solution to the ﬁrst equation is x = πk, with x = 0 and x = π being the two solutions +2πk which lie in [0, 2π). To solve sin(x) = 1 + 2πk for integers k. We ﬁnd the two solutions here which lie in [0, 2π) or x = π − arcsin 1 3 . To check graphically, we plot y = 3(sin(x))3 and and x = π − arcsin 1 to be x = arcsin 1 3 3 y = (sin(x))2 and ﬁnd the x-coordinates of the intersection points of these two curves. Some extra zooming is required near x = 0 and x = π to verify that these two curves do in fact intersect four times.6 3 , we use the arcsine function to get x = arcsin 1 3 2. Analysis of sec2(x) = tan(x) + 3 reveals two diﬀerent trigonometric functions, so an identity is in order. Since sec2(x) = 1 + tan2(x), we get sec2(x) = tan(x) + 3 1 + tan2(x) = tan(x) + 3 (Since sec2(x) = 1 + tan2(x).) tan2(x) − tan(x) − 2 = 0 u2 − u − 2 = 0 (u + 1)(u − 2) = 0 Let u = tan(x). 6Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π). In the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among the solutions in [0, 2π). 10.7 Trigonometric Equations and Inequalities 863 This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From tan(x) = −1, we get x = − π 4 + πk for integers k. To solve tan(x) = 2, we employ the arctangent function and get x = arctan(2) + πk for integers k. From the ﬁrst set of solutions, we get x = 3π 4 as our answers which lie in [0, 2π). Using the same sort of argument we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our second set of solutions which lie in [0, 2π). Using a reciprocal identity, we rewrite the secant as a cosine and graph y = (cos(x))2 and y = tan(x) + 3 to ﬁnd the x-values of the points where they intersect. 4 and x = 7π 1 y = 3(sin(x))3 and y = (sin(x))2 y = 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments diﬀer. Using the identity cos(2x) = 2 cos2(x) − 1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) = 3 cos(x) − 2 2 cos2(x) − 1 = 3 cos(x) − 2 (Since cos(2x) = 2 cos2(x) − 1.) 2 cos2(x) − 3 cos(x) + 1 = 0 2u2 − 3u + 1 = 0 (2u − 1)(u − 1) = 0 Let u = cos(x). 2 , we get x = π 2 or u = 1. Since u = cos(x), we get cos(x) = 1 This gives u = 1 2 or cos(x) = 1. Solving cos(x) = 1 3 + 2πk or x = 5π 3 + 2πk for integers k. From cos(x) = 1, we get x = 2πk for integers k. The answers which lie in [0, 2π) are x = 0, π 3 . Graphing y = cos(2x) and y = 3 cos(x) − 2, we ﬁnd, after a little extra eﬀort, that the curves intersect in three places on [0, 2π), and the x-coordinates of these points conﬁrm our results. 3 , and 5π 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos3(x) − 3 cos(x). This transforms the equation into a polynomial in terms of cos(x). cos(3x) = 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k. The only solution which lies in [0, 2π) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the same set of axes over [0, 2π) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x) − 2 y = cos(3x) and y = 2 − cos(x) 5. While we could approach cos(3x) = cos(5x) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. From cos(3x) = cos(5x), we get cos(5x) − cos(3x) = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move.7 Using Theorem 10.21, we have that cos(5x) − cos(3x) = −2 sin 5x+3x = −2 sin(4x) sin(x). Hence, 2 the equation cos(5x) = cos(3x) is equivalent to −2 sin(4x) sin(x) = 0. From this, we get sin(4x) = 0 or sin(x) = 0. Solving sin(4x) = 0 gives x = π 4 k for integers k, and the solution to sin(x) = 0 is x = πk for integers k. The second set of solutions is contained in the ﬁrst set of solutions,8 so our ﬁnal solution to cos(5x) = cos(3x) is x = π 4 k for integers k. There are eight of these answers which lie in [0, 2π): x = 0, π 4 , 3π 4 , π, 5π 4 . Our plot of the graphs of y = cos(3x) and y = cos(5x) below (after some careful zooming) bears this out. sin 5x−3x 2 and 7π 2 , 3π 4 , π 2 √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have diﬀerent circular functions involved, namely sine and cosine, we also have diﬀerent arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) = 2 sin(x) cos(x) = √ 2 sin(x) cos(x) − cos(x)(2 sin(x) − 3 cos(x) = 0 3) = 0 √ √ √ 3 cos(x) 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) from which we get cos(x) = 0 or sin(x) = 2 , we get x = π integers k. From sin(x) = √ 3 3 2 . From cos(x) = 0, we obtain x = π 2 + πk for 3 +2πk for integers k. The answers 3 +2πk or x = 2π √ 7As always, experience is the greatest teacher here! 8As always, when in doubt, write it out! 10.7 Trigonometric Equations and Inequalities which lie in [0, 2π) are x = π after some careful zooming, verify our answers. 3 and 2π 2 , 3π 2 , π 3 . We graph y = sin(2x) and y = 865 √ 3 cos(x) and, y = cos(3x) and y = cos(5x) y = sin(2x) and y = √ 3 cos(x) 7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation sin(x) cos x = 1. If we stare at 2 it long enough, however, we realize that the left hand side is the expanded form of the sum formula for sin x + x 2 x = 1. Solving, 2 we ﬁnd x = π 3 and x = 5π 3 . Graphing y = sin(x) cos x 2 . Hence, our original equation is equivalent to sin 3 3 k for integers k. Two of these solutions lie in [0, 2π): x = π and y = 1 validates our solutions. + cos(x) sin x 2 + cos(x) sin x 2 3 + 4π 8. With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid.9 To ﬁt f (x) = cos(x) − 3 sin(x) to the form A sin(ωt + φ) + B, we use what we learned in Example 10.5.3 and ﬁnd A = 2, B = 0, ω = 1 = 2, 3 sin(x) = 2 as 2 sin x + 5π and φ = 5π 6 or sin x + 5π 3 + 2πk for integers k. Only one of 6 these solutions, x = 5π 3 , which corresponds to k = 1, lies in [0, 2π). Geometrically, we see that y = cos(x) − 6 . Hence, we can rewrite the equation cos(x) − = 1. Solving the latter, we get x = − π 3 sin(x) and y = 2 intersect just once, supporting our answer. √ √ √ y = sin(x) cos x 2 + cos(x) sin x 2 and y = 1 y = cos(x) − √ 3 sin(x) and y = 2 We repeat here the advice given when solving systems of nonlinear equations in section 8.7 – when it comes to solving equations involving the trigonometric functions, it helps to just try something. 9We are essentially ‘undoing’ the sum / diﬀerence formula for cosine or sine, depending on which form we use, so this problem is actually closely related to the previous one! 866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.10 Example 10.7.3. Solve the following inequalities on [0, 2π). Express your answers using interval notation and verify your answers graphically. 1. 2 sin(x) ≤ 1 2. sin(2x) > cos(x) 3. tan(x) ≥ 3 Solution. 1. We begin solving 2 sin(x) ≤ 1 by collecting all of the terms on one side of the equation and zero on the other to get 2 sin(x) − 1 ≤ 0. Next, we let f (x) = 2 sin(x) − 1 and note that our original inequality is equivalent to solving f (x) ≤ 0. We now look to see where, if ever, f is undeﬁned and where f (x) = 0. Since the domain of f is all real numbers, we can immediately set about ﬁnding the zeros of f . Solving f (x) = 0, we have 2 sin(x) − 1 = 0 or sin(x) = 1 2 . The solutions here are x = π 6 + 2πk for integers k. Since we are restricting our attention to [0, 2π), only x = π 6 are of concern to us. Next, we choose test values in [0, 2π) other than the zeros and determine if f is positive or negative there. For = 1 and for x = π we get f (π) = −1. 2 we get f π x = 0 we have f (0) = −1, for x = π Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function is negative (−) or 0, and we get the intervals 0, π 6 , 2π. We can conﬁrm our answer 6 graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. 6 + 2πk and x = 5π 6 and x = 5π ∪ 5π 2 (−) 0 (+) π 6 0 (−) 5π 6 2π 0 y = 2 sin(x) and y = 1 2. We ﬁrst rewrite sin(2x) > co
s(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to ﬁnding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or 2 +πk for integers k of which only x = π cos(x)(2 sin(x)−1) = 0. From cos(x) = 0, we get x = π 2 and x = 3π 2 which gives x = π 6 + 2πk or x = 5π 6 lie in [0, 2π). Next, we choose 2 lie in [0, 2π). For 2 sin(x) − 1 = 0, we get sin(x) = 1 6 + 2πk for integers k. Of those, only x = π 6 and x = 5π 10See page 214, Example 3.1.5, page 321, page 399, Example 6.3.2 and Example 6.4.2 for discussion of this technique. 10.7 Trigonometric Equations and Inequalities 867 4 we get f 3π our test values. For x = 0 we ﬁnd f (0) = −1; when x = π = −1 + for x = 3π ; when x = π we have f (π) = 1, and lastly, for √ , so this is 2 x = 7π 6 , π our answer. We can use the calculator to check that the graph of y = sin(2x) is indeed above the graph of y = cos(x) on those intervals. 2−2 2 . We see f (x) > 0 on π 2 √ 2 2 = 2 = −2− 4 we get f 7π = −1 − 2 = 2− ∪ 5π 6 , 3π 4 we get ; √ √ (−) 0 (+) π 6 0 (−) π 2 0 (+) 5π 6 0 (−) 3π 2 2π 0 y = sin(2x) and y = cos(x) 2 and 3π 3. Proceeding as in the last two problems, we rewrite tan(x) ≥ 3 as tan(x) − 3 ≥ 0 and let f (x) = tan(x) − 3. We note that on [0, 2π), f is undeﬁned at x = π 2 , so those values will need the usual disclaimer on the sign diagram.11 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We ﬁnd x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π). Since 3 > 0, we know 0 < arctan(3) < π 2 which allows us to position these zeros correctly on the sign diagram. To choose test values, we begin with x = 0 and ﬁnd f (0) = −3. Finding a is a bit more challenging. Keep in mind convenient test value in the interval arctan(3), π 2 that the arctangent function is increasing and is bounded above by π 2 . This means that the number x = arctan(117) is guaranteed12 to lie between arctan(3) and π 2 . We see that f (arctan(117)) = tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and ﬁnd f (π) = −3. To ﬁnd our next test value, we note that since arctan(3) < arctan(117) < π 2 , it follows13 that arctan(3) + π < arctan(117) + π < 3π 2 . Evaluating f at x = arctan(117) + π yields f (arctan(117) + π) = tan(arctan(117) + π) − 3 = tan(arctan(117)) − 3 = 114. We = −4. Since we want f (x) ≥ 0, we 4 and ﬁnd f 7π choose our last test value to be x = 7π . Using the graphs of y = tan(x) ∪ arctan(3) + π, 3π see that our answer is arctan(3), π 2 2 and y = 3, we see when the graph of the former is above (or meets) the graph of the latter. 4 11See page 321 for a discussion of the non-standard character known as the interrobang. 12We could have chosen any value arctan(t) where t > 3. 13. . . by adding π through the inequality . . . 868 Foundations of Trigonometry (−) 0 (+) 0 arctan(3) (−) π 2 (arctan(3) + π) 0 (+) (−) 3π 2 2π y = tan(x) and y = 3 Our next example puts solving equations and inequalities to good use – ﬁnding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.14 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution. 1. To ﬁnd the domain of f (x) = csc 2x + π 3 , we rewrite f in terms of sine as f (x) = 1 sin(2x+ π 3 ) Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π 2 k for integers k. In set-builder notation, 3 our domain is x : x = − π 2 k for integers k. To help visualize the domain, we follow the 6 , . . ., old mantra ‘When in doubt, write it out!’ We get x : x = − π where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6 , − 7π 6 , − 4π 6 + π 6 , 8π 6 , 5π 6 , 2π . − 7π 6 − 4π 6 − π 6 2π 6 5π 6 8π 6 Proceeding as we did on page 756 in Section 10.3.1, we let xk denote the kth number excluded from the domain and we have xk = − π for integers k. The intervals which comprise the domain are of the form (xk, xk + 1) = integers. Using extended interval notation, we have that the domain is as k runs through the 2 k = (3k−1)π 6 (3k−1)π 6 , (3k+2)π 6 6 + π ∞ k=−∞ (3k − 1)π 6 , (3k + 2)π 6 We can check our answer by substituting in values of k to see that it matches our diagram. 14See page 756 for details about this notation. 10.7 Trigonometric Equations and Inequalities 869 2. Since the domains of sin(x) and cos(x) are all real numbers, the only concern when ﬁnding 2 cos(x)−1 is division by zero so we set the denominator equal to zero and 3 +2πk for integers 3 + 2πk for integers k, the domain of f (x) = sin(x) solve. From 2 cos(x)−1 = 0 we get cos(x) = 1 k. Using set-builder notation, the domain is x : x = π or x : x = ± π 3 , ± 11π 3 , . . ., so we have 3 + 2πk and x = 5π 3 +2πk or x = 5π 2 so that x = π 3 , ± 7π 3 , ± 5π − 11π 3 − 7π 3 − 5π 3 − π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two diﬀerent families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals. To that end, we let ak = π for integers k. The goal now is to write the domain in terms of the a’s an b’s. We ﬁnd a0 = π 3 , a1 = 7π 3 and b−2 = − 7π 3 . Hence, in terms of the a’s and b’s, our domain is 3 + 2πk = (6k+5)π 3 + 2πk = (6k+1)π 3 , a−2 = − 11π 3 , a−1 = − 5π 3 , b−1 = − π 3 , a2 = 13π 3 , b1 = 11π 3 , b0 = 5π 3 , b2 = 17π and bk = 5π 3 3 . . . (a−2, b−2) ∪ (b−2, a−1) ∪ (a−1, b−1) ∪ (b−1, a0) ∪ (a0, b0) ∪ (b0, a1) ∪ (a1, b1) ∪ . . . If we group these intervals in pairs, (a−2, b−2)∪(b−2, a−1), (a−1, b−1)∪(b−1, a0), (a0, b0)∪(b0, a1) and so forth, we see a pattern emerge of the form (ak, bk) ∪ (bk, ak + 1) for integers k so that our domain can be written as ∞ k=−∞ (ak, bk) ∪ (bk, ak + 1) = ∞ k=−∞ (6k + 1)π 3 , (6k + 5)π 3 ∪ (6k + 5)π 3 , (6k + 7)π 3 A second approach to the problem exploits the periodic nature of f . Since cos(x) and sin(x) have period 2π, it’s not too diﬃcult to show the function f repeats itself every 2π units.15 This means if we can ﬁnd a formula for the domain on an interval of length 2π, we can express the entire domain by translating our answer left and right on the x-axis by adding integer multiples of 2π. One such interval that arises from our domain work is π . The portion . Adding integer multiples of 2π, we get the family of of the domain here is π 3 , 7π 3 + 2πk for integers k. We leave it to the reader intervals π 3 + 2πk, 7π to show that getting common denominators leads to our previous answer. ∪ 5π 3 + 2πk ∪ 5π 3 + 2πk, 5π 3 , 5π 3 , 7π 3 3 3 15This doesn’t necessarily mean the period of f is 2π. The tangent function is comprised of cos(x) and sin(x), but its period is half theirs. The reader is invited to investigate the period of f . 870 Foundations of Trigonometry 3. To ﬁnd the domain of f (x) = 1 − cot(x), we ﬁrst note that, due to the presence of the cot(x) term, x = πk for integers k. Next, we recall that for the square root to be deﬁned, we need 1−cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted here to a given interval. Our strategy is to solve this inequality over (0, π) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, π. We let g(x) = 1 − cot(x) and set about making a sign diagram for g over the interval (0, π) to ﬁnd where g(x) ≥ 0. We note that g is undeﬁned for x = πk for integers k, in particular, at the endpoints of our interval x = 0 and x = π. Next, we look for the zeros of g. Solving g(x) = 0, we get cot(x) = 1 or x = π 4 + πk for integers k and 6 and x = π only one of these, x = π 2 , we get √ 3, and g π g π 2 6 4 , lies in (0, π). Choosing the test values x = π = 1. = 1 − (−) 0 0 (+) π 4 π We ﬁnd g(x) ≥ 0 on π the intervals π express our ﬁnal answer as 4 + πk, π + πk = 4 , π. Adding multiples of the period we get our solution to consist of . Using extended interval notation, we , (k + 1)π (4k+1)π 4 ∞ k=−∞ (4k + 1)π 4 , (k + 1)π We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions. Example 10.7.5. Solve the following equations and inequalities analytically. Check your answers using a graphing utility. 1. arcsin(2x) = π 3 2. 4 arccos(x) − 3π = 0 3. 3 arcsec(2x − 1) + π = 2π 4. 4 arctan2(x) − 3π arctan(x) − π2 = 0 5. π2 − 4 arccos2(x) < 0 6. 4 arccot(3x) > π Solution. 1. To solve arcsin(2x) = π 3 is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26 3 , we ﬁrst note that π 10.7 Trigonometric Equations and Inequalities 871 arcsin(2x) = π 3 sin (arcsin(2x)) = sin π 3 2x = x = √ 3 2 √ 3 4 Since sin(arcsin(u)) = u Graphing y = arcsin(2x) and the horizontal line y = π √ 3 4 ≈ 0.4430. 2. Our ﬁrst step in solving 4 arccos(x) − 3π = 0 is to isolate the arccosine. Doing so, we get 3 , we see they intersect at arccos(x) = 3π 4 . Since 3π 4 is in the range of arccosine, we may apply Theorem 10.26 arccos(x) = 3π 4 cos (arccos(x)) = cos 3π 4 x = − √ 2 2 Since cos(arccos(u)) = u The calculator conﬁrms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − √ 2 2 ≈ −0.7071. y = arcsin(2x) and y = π 3 y = 4 arccos(x) − 3π 3. From 3 arcsec(2x − 1) + π = 2π, we get arcsec(2x − 1) = π 3 . As we saw in Section 10.6, there are two possible ranges for the arcsecant function. Fortunately, both ranges contain π 3 . Applying Theorem 10.28 / 10.29, we get arcsec(2x − 1) = π 3 sec(arcsec(2x − 1)) = sec π 3 2x − 1 = 2 x = 3 2 Since sec(arcsec(u))
= u To check using our calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcsecant is positive, we know u = 2x − 1 ≥ 1, and so the identity applies in both cases. 872 Foundations of Trigonometry 4. With the presence of both arctan2(x) ( = (arctan(x))2) and arctan(x), we substitute u = arctan(x). The equation 4 arctan2(x) − 3π arctan(x) − π2 = 0 becomes 4u2 − 3πu − π2 = 0. Factoring,17 we get (4u + π)(u − π) = 0, so u = arctan(x) = − π 4 or u = arctan(x) = π. Since − π 4 is in the range of arctangent, but π is not, we only get solutions from the ﬁrst equation. Using Theorem 10.27, we get arctan(x) = − π 4 tan(arctan(x)) = tan − π 4 x = −1 Since tan(arctan(u)) = u. The calculator veriﬁes our result. y = 3 arcsec(2x − 1) + π and y = 2π y = 4 arctan2(x) − 3π arctan(x) − π2 5. Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of π2 − 4 arccos2(x) < 0 are on one side of the inequality, we let f (x) = π2 − 4 arccos2(x) and note the domain of f is limited by the arccos(x) to [−1, 1]. Next, we ﬁnd the zeros of f by setting f (x) = π2 − 4 arccos2(x) = 0. We get arccos(x) = ± π 2 , and since the range of arccosine is = 0 [0, π], we focus our attention on arccos(x) = π as our only zero. Hence, we have two test intervals, [−1, 0) and (0, 1]. Choosing test values x = ±1, we get f (−1) = −3π2 < 0 and f (1) = π2 > 0. Since we are looking for where f (x) = π2 − 4 arccos2(x) < 0, our answer is [−1, 0). The calculator conﬁrms that for these values of x, the graph of y = π2 − 4 arccos2(x) is below y = 0 (the x-axis.) 2 . Using Theorem 10.26, we get x = cos π 2 17It’s not as bad as it looks... don’t let the π throw you! 10.7 Trigonometric Equations and Inequalities 873 (−) (+) 0 0 1 −1 y = π2 − 4 arccos2(x) 6. To begin, we rewrite 4 arccot(3x) > π as 4 arccot(3x) − π > 0. We let f (x) = 4 arccot(3x) − π, and note the domain of f is all real numbers, (−∞, ∞). To ﬁnd the zeros of f , we set f (x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 is in the range of arccotangent, we may apply Theorem 10.27 and solve 4 , and since π arccot(3x) = π 4 cot(arccot(3x)) = cot π 4 3x = 1 x = 1 3 Since cot(arccot(u)) = u. 3 , we have two test intervals, −∞, 1 Next, we make a sign diagram for f . Since the domain of f is all real numbers, and there is 3 , ∞. Ideally, we wish only one zero of f , x = 1 to ﬁnd test values x in these intervals so that arccot(4x) corresponds to one of our oft-used ‘common’ angles. After a bit of computation,18 we choose x = 0 for x < 1 3 , we √ 3 = − π choose x = 3 < 0. Since we are looking for where 3 . We ﬁnd f (0) = π > 0 and f . To check graphically, we use the f (x) = 4 arccot(3x) − π > 0, we get our answer −∞, 1 3 technique in number 2c of Example 10.6.5 in Section 10.6 to graph y = 4 arccot(3x) and we see it is above the horizontal line y = π on −∞, 1 3 3 and for x > 1 = −∞, 0.3. and 1 √ 3 3 3 (−) (+) 0 1 3 y = 4 arccot(3x) and y = π 18Set 3x equal to the cotangents of the ‘common angles’ and choose accordingly. 874 Foundations of Trigonometry 10.7.1 Exercises In Exercises 1 - 18, ﬁnd all of the exact solutions of the equation and then list those solutions which are in the interval [0, 2π). 1. sin (5x) = 0 2. cos (3x) = 1 2 4. tan (6x) = 1 5. csc (4x) = −1 √ 3 3 = 0 7. cot (2x) = − 10. cos x + 5π 6 13. csc(x) = 0 16. sec2 (x) = 4 3 8. cos (9x) = 9 11. sin 2x − π 3 = − 1 2 14. tan (2x − π) = 1 17. cos2 (x. sin (−2x) = 6. sec (3x) = 9. sin = x 3 12. 2 cos x + √ 3 = 7π 4 15. tan2 (x) = 3 18. sin2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0, 2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) 22. cos (2x) = sin (x) 24. cos(2x) = 2 − 5 cos(x) 25. 3 cos(2x) + cos(x) + 2 = 0 26. cos(2x) = 5 sin(x) − 2 27. 3 cos(2x) = sin(x) + 2 29. tan2(x) = 1 − sec(x) 31. sec(x) = 2 csc(x) 33. sin(2x) = tan(x) 35. cos(2x) + csc2(x) = 0 37. tan2 (x) = 3 2 sec (x) 39. tan(2x) − 2 cos(x) = 0 28. 2 sec2(x) = 3 − tan(x) 30. cot2(x) = 3 csc(x) − 3 32. cos(x) csc(x) cot(x) = 6 − cot2(x) 34. cot4(x) = 4 csc2(x) − 7 36. tan3 (x) = 3 tan (x) 38. cos3 (x) = − cos (x) 40. csc3(x) + csc2(x) = 4 csc(x) + 4 41. 2 tan(x) = 1 − tan2(x) 42. tan (x) = sec (x) 10.7 Trigonometric Equations and Inequalities 875 In Exercises 43 - 58, solve the equation, giving the exact solutions which lie in [0, 2π) 43. sin(6x) cos(x) = − cos(6x) sin(x) 44. sin(3x) cos(x) = cos(3x) sin(x) 45. cos(2x) cos(x) + sin(2x) sin(x) = 1 46. cos(5x) cos(3x) − sin(5x) sin(3x) = √ 3 2 47. sin(x) + cos(x) = 1 √ 49. 2 cos(x) − √ 2 sin(x) = 1 51. cos(2x) − √ 3 sin(2x) = √ 2 48. sin(x) + √ 3 cos(x) = 1 √ 50. 3 sin(2x) + cos(2x) = 1 √ 52. 3 3 sin(3x) − 3 cos(3x) = 3 √ 3 53. cos(3x) = cos(5x) 54. cos(4x) = cos(2x) 55. sin(5x) = sin(3x) 56. cos(5x) = − cos(2x) 57. sin(6x) + sin(x) = 0 58. tan(x) = cos(x) In Exercises 59 - 68, solve the equation. 59. arccos(2x) = π 60. π − 2 arcsin(x) = 2π 61. 4 arctan(3x − 1) − π = 0 62. 6 arccot(2x) − 5π = 0 63. 4 arcsec x 2 = π 64. 12 arccsc x 3 = 2π 65. 9 arcsin2(x) − π2 = 0 66. 9 arccos2(x) − π2 = 0 67. 8 arccot2(x) + 3π2 = 10π arccot(x) 68. 6 arctan(x)2 = π arctan(x) + π2 In Exercises 69 - 80, solve the inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2π. 69. sin (x) ≤ 0 72. cos2 (x) > 75. cot2 (x) ≥ 1 2 1 3 78. cos(3x) ≤ 1 70. tan (x) ≥ √ 3 73. cos (2x) ≤ 0 76. 2 cos(x) ≥ 1 79. sec(x) ≤ √ 2 71. sec2 (x) ≤ 4 74. sin x + π 3 > 1 2 77. sin(5x) ≥ 5 80. cot(x) ≤ 4 876 Foundations of Trigonometry In Exercises 81 - 86, solve the inequality. Express the exact answer in interval notation, restricting your attention to −π ≤ x ≤ π. 81. cos (x) > 84. sin2 (x) < √ 3 2 3 4 82. sin(x) > 1 3 83. sec (x) ≤ 2 85. cot (x) ≥ −1 86. cos(x) ≥ sin(x) In Exercises 87 - 92, solve the inequality. Express the exact answer in interval notation, restricting your attention to −2π ≤ x ≤ 2π. 87. csc (x) > 1 90. tan2 (x) ≥ 1 88. cos(x) ≤ 5 3 89. cot(x) ≥ 5 91. sin(2x) ≥ sin(x) 92. cos(2x) ≤ sin(x) In Exercises 93 - 98, solve the given inequality. 93. arcsin(2x) > 0 94. 3 arccos(x) ≤ π 95. 6 arccot(7x) ≥ π 96. π > 2 arctan(x) 97. 2 arcsin(x)2 > π arcsin(x) 98. 12 arccos(x)2 + 2π2 > 11π arccos(x) In Exercises 99 - 107, express the domain of the function using the extended interval notation. (See page 756 in Section 10.3.1 for details.) 99. f (x) = 1 cos(x) − 1 100. f (x) = cos(x) sin(x) + 1 101. f (x) = tan2(x) − 1 102. f (x) = 2 − sec(x) 103. f (x) = csc(2x) 104. f (x) = sin(x) 2 + cos(x) 105. f (x) = 3 csc(x) + 4 sec(x) 106. f (x) = ln (\| cos(x)\|) 107. f (x) = arcsin(tan(x)) 108. With the help of your classmates, determine the number of solutions to sin(x) = 1 2 and sin(4x) = 1 2 in [0, 2π). Then ﬁnd the number of solutions to sin(2x) = 1 2 in [0, 2π). A pattern should emerge. Explain how this pattern would help you solve equations like sin(11x) = 1 2 . What do you ﬁnd? 2 , sin(3x) = 1 2 and sin 5x = 1 = 1 2 2 2 = 1 2 . Now consider sin x with −1 and repeat the whole exploration. 2 , sin 3x Replace 1 2 10.7 Trigonometric Equations and Inequalities 877 10.7.2 Answers 1. x = πk 5 ; x = 0, π 5 , 2π 5 , 3π 5 , 4π 5 , π, 6π 5 , 7π 5 , 8π 5 , 9π 5 2. x = π 9 + 2πk 3 or x = + πk or x = 5π 9 5π 6 + 2πk 3 ; x = π 9 , 5π 9 , 7π 9 , 11π 9 , 13π 9 , 17π 9 + πk; x = 2π 3 , 5π 6 , 5π 3 , 11π 6 + + πk 6 πk 2 ; x = π 24 , 5π 24 , 3π 8 , 13π 24 , 17π 24 , 7π 8 , 25π 24 , 29π 24 , 11π 8 , 37π 24 , 41π 24 , 15π 8 ; x = 3π 8 , 7π 8 , 11π 8 , 15π 8 + 2πk 3 or x = 7π 12 + 2πk 3 ; x = π 12 , 7π 12 , 3π 4 , 5π 4 , 17π 12 , 23π 12 3. x = 4. x = 5. x = 6. x = 2π 3 π 24 3π 8 π 12 7. x = π 3 + πk 2 ; x = π 3 , 5π 6 , 4π 3 , 11π 6 8. No solution 9. x = 3π 4 10. x = − π 3 11. x = 3π 4 + 6πk or x = 9π 4 + 6πk; x = 3π 4 + πk; x = 2π 3 , 5π 3 + πk or x = 13π 12 + πk; x = π 12 , 3π 4 , 13π 12 , 7π 4 12. x = − 19π 12 + 2πk or x = π 12 + 2πk; x = π 12 , 5π 12 13. No solution 14. x = 5π 8 + πk 2 ; x = π 8 , 5π 8 , 9π 8 , 13π 8 15. x = 16. x = 17. x = 18 + πk or x = + πk or x = 2π 3 5π 6 + πk; x = + πk; x = π 3 π 6 , , 2π 3 5π 6 , , 4π 3 7π 6 , , 5π 3 11π 6 + πk 2 ; x = π 4 , 3π 4 , 5π 4 , 7π 4 + πk or x = 2π 3 + πk; x = π 3 , 2π 3 , 4π 3 , 5π 3 878 19. x = 21. x = π 4 π 6 23. x = 0, , , 5π 4 π 2 2π 3 , , 3π 2 , 5π 6 4π 3 25. x = 27. x = 2π 3 7π 6 , , 4π 3 , arccos 11π 6 , arcsin 1 3 1 3 , 2π − arccos , π − arcsin 1 3 1 3 29. x = 0, 2π 3 , 4π 3 31. x = arctan(2), π + arctan(2) Foundations of Trigonometry 20. x = 0, 22. x = 24. x = 26, 5π 3 3π 2 , π 3 5π 6 5π 3 5π 6 , , , 28. x = 3π 4 , 7π 4 , arctan 1 2 , π + arctan 1 2 30. x = 32. x = 34. x = π 6 π 6 π 6 36. x = 0, 38. x = 40. x = π 2 π 6 , , , , , , 5π 6 7π 6 π 4 π 3 3π 2 5π 6 , , , π 2 5π 6 3π 4 2π 3 , , 11π 6 5π 6 4π 3 , , 7π 6 5π 3 , , π, , 7π 6 , 3π 2 , 11π 6 5π 4 , 7π 4 , 11π 6 33. x = 0, π, , 3π 4 , 5π 4 , 7π 4 π 4 3π 2 5π 3 π 2 5π 8 , 35. x = 37. x = 39. x = 41 , , , , , 5π 6 9π 8 , 3π 2 13π 8 4π 7 , , , 3π 7 3π 2 13π 48 , 43. x = 0, 44. x = 0, π 7 π 2 , 2π 7 , π, 46. x = π 48 , 11π 48 47. x = 0, π 2 49. x = 51. x = , π 12 17π 24 17π 12 41π 24 , , 23π 24 , 47π 24 42. No solution , 5π 7 , 6π 7 , π, 8π 7 , 9π 7 , 10π 7 , 11π 7 , 12π 7 , 13π 7 45. x = 0 , 23π 48 , 25π 48 , 35π 48 , 37π 48 , 47π 48 , 49π 48 , 59π 48 , 48. x = , 71π 48 , 73π 48 , 83π 48 , 85π 48 , 95π 48 61π 48 π 2 , 11π 6 π 3 5π 18 , , 50. x = 0, π, 52. x = π 6 , 4π 3 5π 6 , 17π 18 , 3π 2 , 29π 18 10.7 Trigonometric Equations and Inequalities 879 53. x = 0, 55. x = 0 3π 8 3π 4 5π 8 , , π, , 5π 4 7π 8 , 3π 2 9π 8 7π 4 11π 8 , , 13π 8 , 15π 8 54. x = 0, π 3 , 2π 3 , π, 4π 3 , 5π 3 , , π, 56. x = π 7 , π 3 , 3π 7 , 5π 7 , π, 9π 7 , 11π 7 , 5π 3 , 13π 7 57. x = 0, 2π 7 , 58. x = arcsin 5
9. x = − 1 2 61. x = 2 3 63. x = 2 √ 2 65. x = ± √ 3 2 67. x = −1, 0 , 6π 7 , 4π 7 −1 + 2 8π 7 √ 5 , 10π 7 , 12π 7 , π 5 , 3π 5 , π, , 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4 , 3π 4 , 2π 3 5π 4 ∪ , 7π 4 4π 3 ∪ 5π 3 , 2π 0, 75. π 3 ∪ 2π 3 , π ∪ π, 4π 3 ∪ 5π 3 , 2π 70. 72. 74. 76. π 3 , π 2 ∪ 0, 0, 0 , , 4π 3 3π 4 11π 6 5π 3 , 2π 3π 2 5π 4 ∪ 7π 4 , 2π , 2π 77. No solution 78. [0, 2π] 0, 79. π 4 ∪ π 2 , 3π 2 ∪ 7π 4 , 2π 81. − π 6 , π 6 −π, − 83 80. [arccot(4), π) ∪ [π + arccot(4), 2π) 82. arcsin 1 3 , π − arcsin 1 3 84. − 2π 3 , − π 3 ∪ π 3 , 2π 3 880 Foundations of Trigonometry −π, − 85. π 4 ∪ 0, 3π 4 −2π, − 87. 3π 2 ∪ − 3π 2 , −π ∪ 0, π 2 ∪ π 2 , π 86. − 3π 4 , π 4 88. [−2π, 2π] 89. (−2π, arccot(5) − 2π] ∪ (−π, arccot(5) − π] ∪ (0, arccot(5)] ∪ (π, π + arccot(5)] − 3π 2 , − 5π 4 − ∪ 3π 4 , − −π, − 0 , 3π 4 ∪ 5π 4 , 3π 2 ∪ 3π 2 , 7π 4 ∪ π, 5π 3 90. − 7π 4 , − 3π 2 −2π, − 91. 5π 3 ∪ 92. − 93. 0, 1 2 , − 7π 6 11π 6 ∪ ∪ π 6 , 5π 6 ∪, , − π 3π 2 2 2 , 1 94. 1 96. (−∞, ∞) 97. [−1, 0) 99. ∞ k=−∞ (2kπ, (2k + 2)π) 100. ∞ k=−∞ (4k − 1)π 2 , (4k + 3)π 2 −∞, √ 3 7 95. 98. −14k + 1)π k=−∞ 4 ∞ (6k − 1)π 3 (2k + 1)π 2 ∪ (2k + 1)π 2 , (4k + 3)π 4 (6k + 1)π 3 ∪ (4k + 1)π 2 , (4k + 3)π 2 , , kπ 2 kπ 2 , , (k + 1)π 2 (k + 1)π 2 (4k − 1)π 4 , (4k + 1)π 4 104. (−∞, ∞) 106. ∞ k=−∞ (2k − 1)π 2 , (2k + 1)π 2 101. 102. 103. 105. 107. k=−∞ ∞ k=−∞ ∞ k=−∞ ∞ k=−∞ Chapter 11 Applications of Trigonometry 11.1 Applications of Sinusoids In the same way exponential functions can be used to model a wide variety of phenomena in nature,1 the cosine and sine functions can be used to model their fair share of natural behaviors. In section 10.5, we introduced the concept of a sinusoid as a function which can be written either in the form C(x) = A cos(ωx+φ)+B for ω > 0 or equivalently, in the form S(x) = A sin(ωx+φ)+B for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarity of exposition we focus on the sine function2 in this section and switch to the independent variable t, since the applications in this section are time-dependent. We reintroduce and summarize all of the important facts and deﬁnitions about this form of the sinusoid below. Properties of the Sinusoid S(t) = A sin(ωt + φ) + B The amplitude is \|A\| The angular frequency is ω and the ordinary frequency is f = ω 2π The period is T = 1 f = 2π ω The phase is φ and the phase shift is − φ ω The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in context. The amplitude measures the maximum displacement of the sine wave from its baseline (determined by the vertical shift), the period is the length of time it takes to complete one cycle of the sinusoid, the angular frequency tells how many cycles are completed over an interval of length 2π, and the ordinary frequency measures how many cycles occur per unit of time. The phase indicates what 1See Section 6.5. 2Sine haters can use the co-function identity cos π 2 − θ = sin(θ) to turn all of the sines into cosines. 882 Applications of Trigonometry angle φ corresponds to t = 0, and the phase shift represents how much of a ‘head start’ the sinusoid has over the un-shifted sine function. The ﬁgure below is repeated from Section 10.5. amplitude baseline period In Section 10.1.1, we introduced the concept of circular motion and in Section 10.2.1, we developed formulas for circular motion. Our ﬁrst foray into sinusoidal motion puts these notions to good use. Example 11.1.1. Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, ﬁnd a sinusoid which describes the height of the passengers above the ground t seconds after they pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation.3 θ Q h P O 3Otherwise, we could just observe the motion of the wheel from the other side. 11.1 Applications of Sinusoids 883 We know from the equations given on page 732 in Section 10.2.1 that the y-coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocity (frequency) ω is given by y = r sin(ωt). Here, t = 0 corresponds to the point (r, 0) so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is r = 64 feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period T = 1 seconds. Hence, the angular frequency is ω = 2π T = 4π 127 radians per second. Putting these two pieces of information together, we have that y = 64 sin 4π 127 t describes the y-coordinate on the Giant Wheel after t seconds, assuming it is centered at (0, 0) with t = 0 corresponding to the point Q. In order to ﬁnd an expression for h, we take the point O in the ﬁgure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical shift upward,4 we add 72 to our formula for y to obtain the new formula h = y + 72 = 64 sin 4π 127 t + 72. Next, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into play. Geometrically, we need to shift the angle θ in the ﬁgure back π 2 radians. From Section 10.2.1, we know θ = ωt = 4π 127 t, so we (temporarily) write the height in terms of θ as h = 64 sin (θ) + 72. + 72. We 127 t − π Subtracting π 2 . can check the reasonableness of our answer by graphing y = h(t) over the interval 0, 127 2 2 from θ gives the ﬁnal answer h(t) = 64 sin θ − π + 72 = 64 sin 4π 2 (127) = 127 2 2 y 136 72 8 t 127 2 A few remarks about Example 11.1.1 are in order. First, note that the amplitude of 64 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stray more than 64 feet vertically from the center of the Wheel, which makes sense. 8 = 15.875. This represents the Second, the phase shift of our answer works out to be ‘time delay’ (in seconds) we introduce by starting the motion at the point P as opposed to the point Q. Said diﬀerently, passengers which ‘start’ at P take 15.875 seconds to ‘catch up’ to the point Q. 4π/127 = 127 π/2 Our next example revisits the daylight data ﬁrst introduced in Section 2.5, Exercise 6b. 4We are readjusting our ‘baseline’ from y = 0 to y = 72. 884 Applications of Trigonometry Example 11.1.2. According to the U.S. Naval Observatory website, the number of hours H of daylight that Fairbanks, Alaska received on the 21st day of the nth month of 2009 is given below. Here t = 1 represents January 21, 2009, t = 2 represents February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. 2. Compare your answer to part 1 to one obtained using the regression feature of a calculator. Solution. 1. To get a feel for the data, we plot it below. H 22 20 18 16 14 12 10 10 11 12 6 The data certainly appear sinusoidal,5 but when it comes down to it, ﬁtting a sinusoid to data manually is not an exact science. We do our best to ﬁnd the constants A, ω, φ and B so that the function H(t) = A sin(ωt + φ) + B closely matches the data. We ﬁrst go after the vertical shift B whose value determines the baseline. In a typical sinusoid, the value of B is the average of the maximum and minimum values. So here we take B = 3.3+21.8 = 12.55. Next is the amplitude A which is the displacement from the baseline to the maximum (and minimum) values. We ﬁnd A = 21.8 − 12.55 = 12.55 − 3.3 = 9.25. At this point, we have H(t) = 9.25 sin(ωt + φ) + 12.55. Next, we go after the angular frequency ω. Since the data collected is over the span of a year (12 months), we take the period T = 12 months.6 This 2 5Okay, it appears to be the ‘∧’ shape we saw in some of the graphs in Section 2.2. Just humor us. 6Even though the data collected lies in the interval [1, 12], which has a length of 11, we need to think of the data point at t = 1 as a representative sample of the amount of daylight for every day in January. That is, it represents H(t) over the interval [0, 1]. Similarly, t = 2 is a sample of H(t) over [1, 2], and so forth. 11.1 Applications of Sinusoids 885 12 = π T = 2π means ω = 2π 6 . The last quantity to ﬁnd is the phase φ. Unlike the previous example, it is easier in this case to ﬁnd the phase shift − φ ω . Since we picked A > 0, the phase shift corresponds to the ﬁrst value of t with H(t) = 12.55 (the baseline value).7 Here, we choose t = 3, since its corresponding H value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to t = 4. Hence, − φ 2 . We have H(t) = 9.25 sin π + 12.55. Below is a graph of our data with the curve y = H(t). ω = 3, so φ = −3ω = −. Using the ‘SinReg’ command, we graph the calculator’s regression below. While both models seem to be reasonable ﬁts to the data, the calculator model is possibly the better ﬁt. The calculator does not give us an r2 value like it did for linear regressions in Section 2.5, nor does it give us an R2 value like it did for quadratic, cubic and quartic regressions as in Section 3.1. The reason for this, much like the reason for the absence of R2 for the logistic model in Section 6.5, is beyond the scope of this course. We’ll just have to use our own good judgment when choosing the best sinusoid model. 11.1.1 Harmonic Motion One of the major
applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring system. Before we jump into the Mathematics, there are some Physics terms and concepts we need to discuss. In Physics, ‘mass’ is deﬁned as a measure of an object’s resistance to straight-line motion whereas ‘weight’ is the amount of force (pull) gravity exerts on an object. An object’s mass cannot change,8 while its weight could change. 7See the ﬁgure on page 882. 8Well, assuming the object isn’t subjected to relativistic speeds . . . 886 Applications of Trigonometry An object which weighs 6 pounds on the surface of the Earth would weigh 1 pound on the surface of the Moon, but its mass is the same in both places. In the English system of units, ‘pounds’ (lbs.) is a measure of force (weight), and the corresponding unit of mass is the ‘slug’. In the SI system, the unit of force is ‘Newtons’ (N) and the associated unit of mass is the ‘kilogram’ (kg). We convert between mass and weight using the formula9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravity. In the English system, g = 32 feet second2 , and in the SI system, g = 9.8 meters second2 . Hence, on Earth a mass of 1 slug weighs 32 lbs. and a mass of 1 kg weighs 9.8 N.10 Suppose we attach an object with mass m to a spring as depicted below. The weight of the object will stretch the spring. The system is said to be in ‘equilibrium’ when the weight of the object is perfectly balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring’s ‘spring constant’. Usually denoted by the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke’s Law11 F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocity, the object will bounce up and down on the end of the spring until some external force stops it. If we let x(t) denote the object’s displacement from the equilibrium position at time t, then x(t) = 0 means the object is at the equilibrium position, x(t) < 0 means the object is above the equilibrium position, and x(t) > 0 means the object is below the equilibrium position. The function x(t) is called the ‘equation of motion’ of the object.12 x(t) = 0 at the equilibrium position x(t) < 0 above the equilibrium position x(t) > 0 below the equilibrium position If we ignore all other inﬂuences on the system except gravity and the spring force, then Physics tells us that gravity and the spring force will battle each other forever and the object will oscillate indeﬁnitely. In this case, we describe the motion as ‘free’ (meaning there is no external force causing the motion) and ‘undamped’ (meaning we ignore friction caused by surrounding medium, which in our case is air). The following theorem, which comes from Diﬀerential Equations, gives x(t) as a function of the mass m of the object, the spring constant k, the initial displacement x0 of the 9This is a consequence of Newton’s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused by the acceleration due to gravity. 10Note that 1 pound = 1 slug foot 11Look familiar? We saw Hooke’s Law in Section 4.3.1. 12To keep units compatible, if we are using the English system, we use feet (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2 . are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is released from the equilibrium position, x0 < 0 means the object is released above the equilibrium position and x0 > 0 means the object is released below the equilibrium position. As far as the initial velocity v0 is concerned, v0 = 0 means the object is released ‘from rest,’ v0 < 0 means the object is heading upwards and v0 > 0 means the object is heading downwards.13 Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is x0 and the initial velocity of the object is v0, then the displacement x from the equilibrium position at time t is given by x(t) = A sin(ωt + φ) where ω = k m and A = x2 0 + 2 v0 ω A sin(φ) = x0 and Aω cos(φ) = v0. It is a great exercise in ‘dimensional analysis’ to verify that the formulas given in Theorem 11.1 work out so that ω has units 1 s and A has units ft. or m, depending on which system we choose. Example 11.1.3. Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, ﬁnd the equation of motion of the object, x(t). When does the object ﬁrst pass through the equilibrium position? Is the object heading upwards or downwards at this instant? 2. If the object is attached to the spring and released 3 feet below the equilibrium position with an upward velocity of 8 feet per second, ﬁnd the equation of motion of the object, x(t). What is the longest distance the object travels above the equilibrium position? When does this ﬁrst happen? Conﬁrm your result using a graphing utility. Solution. In order to use the formulas in Theorem 11.1, we ﬁrst need to determine the spring constant k and the mass of the object m. To ﬁnd k, we use Hooke’s Law F = kd. We know the object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or k = 8 lbs. s2 . We get m = 2 slugs. We can now proceed to apply Theorem 11.1. ft. . To ﬁnd m, we use w = mg with w = 64 lbs. and g = 32 ft. k 8 1. With k = 8 and m = 2, we get ω = 2 = 2. We are told that the object is released 3 feet below the equilibrium position ‘from rest.’ This means x0 = 3 and v0 = 0. Therefore, 32 + 02 = 3. To determine the phase φ, we have A sin(φ) = x0, 2 and angles coterminal to it A = which in this case gives 3 sin(φ) = 3 so sin(φ) = 1. Only φ = π 0 + v0 x2 m = 2 √ = ω 13The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the ‘natural’ or ‘positive’ direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered ‘negative’. 888 Applications of Trigonometry = 0. Going through the usual analysis we ﬁnd t = − π satisfy this condition, so we pick14 the phase to be φ = π 2 . Hence, the equation of motion is x(t) = 3 sin 2t + π . To ﬁnd when the object passes through the equilibrium position we 2 solve x(t) = 3 sin 2t + π 2 k for 2 integers k. Since we are interested in the ﬁrst time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = π 4 ≈ 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it ﬁrst passes through it. To check this answer, we graph one cycle of x(t). Since our applied domain in this situation is t ≥ 0, and the period of x(t) is T = 2π 2 = π, we graph x(t) over the interval [0, π]. Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-axis from positive x to negative x at t = π ω = 2π 4 + π 4 conﬁrms our answer. ω = 2 0 + v0 x2 5 . From Aω cos(φ) = v0, we get 10 cos(φ) = −8, or cos(φ) = − 4 2. The only diﬀerence between this problem and the previous problem is that we now release the object with an upward velocity of 8 ft s . We still have ω = 2 and x0 = 3, but now we have v0 = −8, the negative indicating the velocity is directed upwards. Here, we get 32 + (−4)2 = 5. From A sin(φ) = x0, we get 5 sin(φ) = 3 which gives A = sin(φ) = 3 5 . This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3 . Hence, 5 . Since the amplitude of x(t) is 5, the object will travel x(t) = 5 sin 2t + π − arcsin 3 5 at most 5 feet above the equilibrium position. To ﬁnd when this happens, we solve the equation x(t) = 5 sin 2t + π − arcsin 3 = −5, the negative once again signifying that 5 the object is above the equilibrium position. Going through the usual machinations, we get 2 arcsin 3 t = 1 4 + πk for integers k. The smallest of these values occurs when k = 0, 5 that is, t = 1 4 ≈ 1.107 seconds after the start of the motion. To check our answer using the calculator, we graph y = 5 sin 2x + π − arcsin 3 on a graphing utility 5 and conﬁrm the coordinates of the ﬁrst relative minimum to be approximately (1.107, −5). + π 2 arcsin 3 + π 5 x π 4 π 2 3π 4 π t 3 2 1 −1 −2 −3 x(t) = 3 sin 2t + π 2 y = 5 sin 2x + π − arcsin 3 5 It is possible, though beyond the scope of this course, to model the eﬀects of friction and other external forces acting on the system.15 While we may not have the Physics and Calculus background 14For conﬁrmation, we note that Aω cos(φ) = v0, which in this case reduces to 6 cos(φ) = 0. 15Take a good Diﬀerential Equations class to see this! 11.1 Applications of Sinusoids 889 to derive equations of motion for these scenarios, we can certainly analyze them. We examine three cases in the following example. Example 11.1.4. 1. Write x(t) = 5e−t/5 cos(t) + 5e−t/5 using a graphing utility. √ 3 sin(t) in the form x(t) = A(t) sin(ωt + φ). Graph x(t) 2. Write x(t) = (t
+ 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both terms to get x(t) = 5e−t/5 cos(t) + 3 sin(t). We convert what’s left in parentheses to the required form using the formulas introduced in Exercise 36 from Section 10.5. We ﬁnd so that x(t) = 10e−t/5 sin t + π cos(t) + . Graphing this on the 3 sin(t) = 2 sin t + π 3 3 calculator as y = 10e−x/5 sin x + π reveals some interesting behavior. The sinusoidal nature 3 continues indeﬁnitely, but it is being attenuated. In the sinusoid A sin(ωx + φ), the coeﬃcient , we can think A of the sine function is the amplitude. In the case of y = 10e−x/5 sin x + π 3 of the function A(x) = 10e−x/5 as the amplitude. As x → ∞, 10e−x/5 → 0 which means the amplitude continues to shrink towards zero. Indeed, if we graph y = ±10e−x/5 along with y = 10e−x/5 sin x + π , we see this attenuation taking place. This equation corresponds to 3 the motion of an object on a spring where there is a slight force which acts to ‘damp’, or slow the motion. An example of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and smaller amplitude. y = 10e−x/5 sin x + π 3 y = 10e−x/5 sin x + π 3 , y = ±10e−x/5 √ 2. Proceeding as in the ﬁrst example, we factor out (t + 3) 2 from each term in the function 2(cos(2t) + sin(2t)). We ﬁnd x(t) = (t + 3) . Graphing this on the (cos(2t) + sin(2t)) = calculator as y = 2(x + 3) sin 2x + π , we ﬁnd the sinusoid’s amplitude growing. Since our 4 amplitude function here is A(x) = 2(x + 3) = 2x + 6, which continues to grow without bound , so x(t) = 2(t + 3) sin 2t + π 4 √ 2 sin 2t + π 4 2 sin(2t) to get x(t) = (t + 3) 2 cos(2t) + (t + 3) √ √ √ 890 Applications of Trigonometry as x → ∞, this is hardly surprising. The phenomenon illustrated here is ‘forced’ motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a ‘resonance’ eﬀect – the frequency of the external oscillation matches the frequency of the motion of the object on the spring.16 y = 2(x + 3) sin 2x + π 4 y = 2(x + 3) sin 2x + π 4 y = ±2(x + 3) 8 = 3 3. Last, but not least, we come to x(t) = 5 sin(6t) − 5 sin(8t). To ﬁnd the period of this function, we need to determine the length of the smallest interval on which both f (t) = 5 sin(6t) and g(t) = 5 sin(8t) complete a whole number of cycles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 4 . This tells us that for every 3 cycles f makes, In other words, the period of x(t) is three times the period of f (t) (which is g makes 4. four times the period of g(t)), or π. We graph y = 5 sin(6x) − 5 sin(8x) over [0, π] on the calculator to check this. This equation of motion also results from ‘forced’ motion, but here the frequency of the external oscillation is diﬀerent than that of the object on the spring. Since the sinusoids here have diﬀerent frequencies, they are ‘out of sync’ and do not amplify each other as in the previous example. Taking things a step further, we can use a sum to product identity to rewrite x(t) = 5 sin(6t) − 5 sin(8t) as x(t) = −10 sin(t) cos(7t). The lower frequency factor in this expression, −10 sin(t), plays an interesting role in the graph of x(t). Below we graph y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π]. This is an example of the ‘beat’ phenomena, and the curious reader is invited to explore this concept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resonance. 17A good place to start is this article on beats. 11.1 Applications of Sinusoids 891 11.1.2 Exercises 1. The sounds we hear are made up of mechanical waves. The note ‘A’ above the note ‘middle second . Find a sinusoid which C’ is a sound wave with ordinary frequency f = 440 Hertz = 440 cycles models this note, assuming that the amplitude is 1 and the phase shift is 0. 2. The voltage V in an alternating current source has amplitude 220 √ 2 and ordinary frequency f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0. 3. The London Eye is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counterclockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and oﬀ of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level. 4. On page 732 in Section 10.2.1, we found the x-coordinate of counter-clockwise motion on a circle of radius r with angular frequency ω to be x = r cos(ωt), where t = 0 corresponds to the point (r, 0). Suppose we are in the situation of Exercise 3 above. Find a sinsusoid which models the horizontal displacement x of the passenger from the center of the Eye in meters t minutes after they board the Eye. Here we take x(t) > 0 to mean the passenger is to the right of the center, while x(t) < 0 means the passenger is to the left of the center. 5. In Exercise 52 in Section 10.1, we introduced the yo-yo trick ‘Around the World’ in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, ﬁnd a sinsuoid which models the height h of the yo-yo above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant k in lbs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the ﬁrst time the object passes through the equilibrium position? In which direction is it heading? ft. and the mass of the object in slugs. (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time. 892 Applications of Trigonometry 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pen- dulum from the vertical position, θ, can be modeled as a sinusoid.18 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ0, of the pendulum and the period of the motion is given by T = 2π l g where l is the length of the pendulum and g is the acceleration due to gravity. (a) Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ(0) = θ0. (b) In Exercise 40 section 5.3, you found the length of the pendulum needed in Jeﬀ’s antique Seth-Thomas clock to ensure the period of the pendulum is 1 2 of a second. Assuming the initial displacement of the pendulum is 15◦, ﬁnd a sinusoid which models the displacement of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the ﬁrst of the month for each month during the years 1971 – 2000.19 For example, t = 3 represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000. Month Number, t Temperature (◦ F), 10 11 12 36 33 34 38 47 57 67 74 73 67 56 46 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data. (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. 18Provided θ is kept ‘small.’ Carl remembers the ‘Rule of Thumb’ as being 20◦ or less. Check with your friendly neighborhood physicist to make sure. 19See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html. 11.1 Applications of Sinusoids 893 (c) Use the model you found in part 8a to predict the average temperature recorded for Lake Erie on April 15th and September 15th during the years 1971–2000.20 (d) Compare your results to those obtained using a graphing utility. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the tth day of June, 2009 is given in the table below.21 Day of June, t Fraction Illuminated, F 3 6 9 12 15 18 21 24 27 30 0.81 0.98 0.98 0.83 0.57 0.27 0.04 0.03 0.26 0.58 (a) Using the techniques discussed in Example 11.1.2, ﬁt a sinusoid to these data.22 (b) Using a graphing utility, graph your model along with the data set to judge the reason- ableness of the ﬁt. (c) Use the model you found in part 9a to predict the fraction of the moon illuminated on June 1, 2009. 23 (d) Compare your results to those obtained using a graphing utility. 10. With the help of your classmates, research the phenomena mentioned in Example 11.1.4, namely resonance and beats. 11. With the help of your classmates, research Amplitude Modulation and Frequency Modulation. 12. What other things in the world might be roughly sinusoidal? Look to see what models you can ﬁnd for them and share your results with your class. 20The computed average is 41◦F for April 15th and 71◦F for September 15th. 21See this website: http://www.usno.navy.mil/USNO/astronomical-applications/data-services/frac-moon-ill. 22You may want to plot the data before you ﬁnd the phase shift. 23The listed fraction is 0.62. 894 Applications of Trigonometry 11.1.3 Answers 1. S(t) = sin (880πt) 2. V (t) = 220 √ 2 sin (120πt) 3. h(t) = 67.5 sin π 15 t − π 2 + 67.5
4. x(t) = 67.5 cos π 15 t − π 2 = 67.5 sin π 15 t 5. h(t) = 28 sin 2π + 30 2 3 t − π ft. and m = 5 16 slugs 6. (a) k = 5 lbs. (b) x(t) = sin 4t + π 2 . The object ﬁrst passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object is heading upwards. (c) x(t) = √ 2 2 sin 4t + 7π 4 . The object passes through the equilibrium point heading down- wards for the third time when t = 17π 16 ≈ 3.34 seconds. 7. (a) θ(t) = θ0 sin g l t + π 2 (b) θ(t) = π 12 sin 4πt + π 2 8. (a) T (t) = 20.5 sin π (b) Our function and the data set are graphed below. The sinusoid seems to be shifted to 6 t − π + 53.5 the right of our data. (c) The average temperature on April 15th is approximately T (4.5) ≈ 39.00◦F and the average temperature on September 15th is approximately T (9.5) ≈ 73.38◦F. (d) Using a graphing calculator, we get the following This model predicts the average temperature for April 15th to be approximately 42.43◦F and the average temperature on September 15th to be approximately 70.05◦F. This model appears to be more accurate. 11.1 Applications of Sinusoids 895 9. (a) Based on the shape of the data, we either choose A < 0 or we ﬁnd the second value of t which closely approximates the ‘baseline’ value, F = 0.505. We choose the latter to obtain F (t) = 0.475 sin π 15 t − 2π + 0.505 = 0.475 sin π 15 t + 0.505 (b) Our function and the data set are graphed below. It’s a pretty good ﬁt. (c) The fraction of the moon illuminated on June 1st, 2009 is approximately F (1) ≈ 0.60 (d) Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately 0.59. This appears to be a better ﬁt to the data than our ﬁrst model. 896 Applications of Trigonometry 11.2 The Law of Sines Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, ﬁnd the length of each side of a triangle and the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, we’ve had some experience solving right triangles. The following example reviews what we know. Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, ﬁnd the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For deﬁnitiveness, we label the triangle below √ To ﬁnd the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72 which then yields a = 33 units. Now that all three sides of the triangle are known, there are several ways we can ﬁnd α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to α. According to Theorem 10.4, cos(α) = 4 7 . Since α is an acute angle, α = arccos 4 radians. Converting to degrees, we ﬁnd α ≈ 55.15◦. Now 7 that we have the measure of angle α, we could ﬁnd the measure of angle β using the fact that α and β are complements so α + β = 90◦. Once again, we opt to use the data given to us in the radians and we problem. According to Theorem 10.4, we have that sin(β) = 4 have β ≈ 34.85◦. 7 so β = arcsin 4 7 A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1as well as the measure of said angle 2as well as the length of said side 11.2 The Law of Sines 897 minimizes the chances of propagated error.3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following ratios hold or, equivalently, sin(α) a = sin(β) b = sin(γ) c a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our ﬁrst case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles: ABQ and BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4 that sin(α) = h a so that h = c sin(α) = a sin(γ). After some rearrangement of the last equation, we get sin(α) . If we drop an altitude from vertex A, we can proceed as above using the triangles ABQ and ACQ to get sin(β) B , completing the proof for this case. c and sin(γ) = h a = sin(γ) b = sin(γ For our next case consider the triangle ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sin(γ) and h = c sin(β) so that sin(β) . b = sin(γ 3Your Science teachers should thank us for this. 4Don’t worry! Radians will be back before you know it! 898 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles, ABQ and BCQ. We know that sin(α) = h c so that h = c sin(α). Since α = 180◦ − α, sin(α) = sin(α), so in fact, we have h = c sin(α). Proceeding to BCQ, we get sin(γ) = h a so h = a sin(γ). Putting this together with the previous equation, we get sin(γ) , and we are ﬁnished with this case. c = sin(α The remaining case is when ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120◦, a = 7 units, β = 45◦ 2. α = 85◦, β = 30◦, c = 5.25 units 3. α = 30◦, a = 1 units, c = 4 units 4. α = 30◦, a = 2 units, c = 4 units 5. α = 30◦, a = 3 units, c = 4 units 6. α = 30◦, a = 4 units, c = 4 units Solution. b sin(45◦) = 1. Knowing an angle-side opposite pair, namely α and a, we may proceed in using the Law of √ 6 Sines. Since β = 45◦, we use 3 ≈ 5.72 units. Now that we have two angle-side pairs, it is time to ﬁnd the third. To ﬁnd γ, we use the fact that the sum of the measures of the angles in a triangle is 180◦. Hence, γ = 180◦ − 120◦ − 45◦ = 15◦. To ﬁnd c, we have no choice but to used the derived value γ = 15◦, yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines gives us sin(120◦) so b = 7 sin(45◦) sin(120◦) so that c = 7 sin(15◦) sin(120◦) ≈ 2.09 units.5 sin(120◦) = 7 c sin(15◦) = 7 7 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦. As in the previous example, we are forced to use a derived value in our computations since the only 5The exact value of sin(15◦) could be found using the diﬀerence identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15◦) sin(120◦) . 11.2 The Law of Sines 899 angle-side pair available is (γ, c). The Law of Sines gives rearrangement, we get a = 5.25 sin(85◦) which yields sin(65◦) . After the usual sin(65◦) ≈ 5.77 units. To ﬁnd b we use the angle-side pair (γ, c) sin(65◦) hence b = 5.25 sin(30◦) sin(65◦) ≈ 2.90 units. sin(30◦) = 5.25 b a sin(85◦) = 5.25 β = 45◦ a = 7 c ≈ 2.09 α = 120◦ γ = 15◦ b ≈ 5.72 Triangle for number 1 β = 30◦ c = 5.25 a ≈ 5.77 α = 85◦ γ = 65◦ b ≈ 2.90 Triangle for number 2 1 4 = sin(30◦) 3. Since we are given (α, a) and c, we use the Law of Sines to ﬁnd the measure of γ. We start with sin(γ) and get sin(γ) = 4 sin (30◦) = 2. Since the range of the sine function is [−1, 1], there is no real number with sin(γ) = 2. Geometrically, we see that side a is just too short to make a triangle. The next three examples keep the same values for the measure of α and the length of c while varying the length of a. We will discuss this case in more detail after we see what happens in those examples. 2 4 = sin(30◦) 4. In this case, we have the measure of α = 30◦, a = 2 and c = 4. Using the Law of Sines, we get sin(γ) so sin(γ) = 2 sin (30◦) = 1. Now γ is an angle in a triangle which also contains α = 30◦. This means that γ must measure between 0◦ and 150◦ in order to ﬁt inside the triangle with α. The only angle that satisﬁes this requirement and has sin(γ) = 1 is γ = 90◦. In other words, we have a right triangle. We ﬁnd the measure of β to be β = 180◦ − 30◦ − 90◦ = 60◦ and then determine b using the Law of Sines. We ﬁnd b = 2 sin(60◦) 3 ≈ 3.46 units. In this case, the side a is precisely long enough to form a sin(30◦) = 2 unique right triangle. √ c = 4 a = 1 α = 30◦ c = 4 β = 60◦ a = 2 α = 30◦ b ≈ 3.46 Diagram for number 3 Triangle for number 4 5. Proceeding as we have in the previous two examples, we use the Law of Sines to ﬁnd γ. In this 3 . Since γ lies in a triangle
with α = 30◦, or sin(γ) = 4 sin(30◦) case, we have sin(γ) 4 = sin(30◦) = 2 3 3 900 Applications of Trigonometry 3 3 : γ = arcsin 2 we must have that 0◦ < γ < 150◦. There are two angles γ that fall in this range and have radians ≈ 138.19◦. At radians ≈ 41.81◦ and γ = π − arcsin 2 sin(γ) = 2 3 this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair (α, a) = (30◦, 3) in that both choices for γ can ﬁt in a triangle with α and both have a sine of 2 3 . The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as radians ≈ 41.81◦, we well. Thus have two triangles on our hands. In the case γ = arcsin 2 3 ﬁnd6 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦. Using the Law of Sines with the angle-side opposite radians pair (α, a) and β, we ﬁnd b ≈ 3 sin(108.19◦) ≈ 138.19◦, we repeat the exact same steps and ﬁnd β ≈ 11.81◦ and b ≈ 1.23 units.7 Both triangles are drawn below. sin(30◦) ≈ 5.70 units. In the case γ = π − arcsin 2 3 β ≈ 11.81◦ c = 4 β ≈ 108.19◦ a = 3 α = 30◦ γ ≈ 41.81◦ α = 30◦ c = 4 a = 3 γ ≈ 138.19◦ b ≈ 5.70 b ≈ 1.23 6. For this last problem, we repeat the usual Law of Sines routine to ﬁnd that sin(γ) so that sin(γ) = 1 2 . Since γ must inhabit a triangle with α = 30◦, we must have 0◦ < γ < 150◦. Since the measure of γ must be strictly less than 150◦, there is just one angle which satisﬁes both required conditions, namely γ = 30◦. So β = 180◦ − 30◦ − 30◦ = 120◦ and, using the √ Law of Sines one last time, b = 4 sin(120◦) 3 ≈ 6.93 units. 4 = sin(30◦) 4 sin(30◦) = 4 c = 4 β = 120◦ a = 4 α = 30◦ γ = 30◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We ﬁrst note that if we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely 6 radians, γ = arcsin 2 6To ﬁnd an exact expression for β, we convert everything back to radians: α = 30◦ = π 3 radians and 180◦ = π radians. Hence, β = π − π 7An exact answer for β in this case is β = arcsin 2 6 − arcsin 2 − π 3 6 − arcsin 2 = 5π 6 radians ≈ 11.81◦. 3 3 radians ≈ 108.19◦. 11.2 The Law of Sines 901 determined using the equation γ = 180◦ − α − β. Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to ﬁnd the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case.8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case.9 In number 3, the length of the one given side a was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem 11.3. Suppose (α, a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) If a < h, then no triangle exists which satisﬁes the given criteria. If a = h, then γ = 90◦ so exactly one (right) triangle exists which satisﬁes the criteria. If h < a < c, then two distinct triangles exist which satisfy the given criteria. If a ≥ c, then γ is acute and exactly one triangle exists which satisﬁes the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were c = sin(α) to exist, the Law of Sines would have sin(γ) a = 1, which is impossible. In the ﬁgure below, we see geometrically why this is the case. so that sin(γ) = c sin(α) > a a a a h = c sin(α sin(α) a < h, no triangle a = h, γ = 90◦ Simply put, if a < h the side a is too short to connect to form a triangle. This means if a ≥ h, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem 8If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisﬁes the given criteria. 9In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 902 Applications of Trigonometry c a a = a a = sin(γ) < 1 which means there are two solutions to sin(γ) = c sin(α) tell us what kind and how many triangles to expect in each case. If a = h, then a = c sin(α) and so that sin(γ) = c sin(α) the Law of Sines gives sin(α) a = 1. Here, γ = 90◦ as required. Moving along, now suppose h < a < c. As before, the Law of Sines10 gives sin(γ) = c sin(α) . Since h < a, c sin(α) < a or c sin(α) : an acute angle which we’ll call γ0, and its supplement, 180◦ − γ0. We need to argue that each of these angles ‘ﬁt’ into a triangle with α. Since (α, a) and (γ0, c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0, giving us one of the triangles promised in the theorem. If we manipulate the inequality γ0 > α a bit, we have 180◦ −γ0 < 180◦ −α which gives (180◦ − γ0) + α < 180◦. This proves a triangle can contain both of the angles α and (180◦ − γ0), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a ≥ c. Then α ≥ γ, which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof. a a c a a h α γ0 γ0 h < a < c, two triangles c α h a γ a ≥ c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the AngleSide-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Example 11.2.3. Sasquatch Island lies oﬀ the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the ﬁrst observation point is 30◦ and at the second point the angle is 45◦. Assuming a straight coastline, ﬁnd the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the ﬁrst observation point labeled as P and the second as Q. In order to use the Law of Sines to ﬁnd the distance d from Q to the island, we ﬁrst need to ﬁnd the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦. We can now 5 ﬁnd β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦. By the Law of Sines, we have sin(15◦) which gives d = 5 sin(30◦) sin(15◦) ≈ 9.66 miles. Next, to ﬁnd the point on the coast closest to the island, which we’ve labeled as C, we need to ﬁnd the perpendicular distance from the island to the coast.11 d sin(30◦) = 10Remember, we have already argued that a triangle exists in this case! 11Do you see why C must lie to the right of Q? 11.2 The Law of Sines 903 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using Theorem 10.4, we get sin (45◦) = y d . After some rearranging, we ﬁnd y = d sin (45◦) ≈ 9.66 ≈ 6.83 miles. Hence, the island is approximately 6.83 miles from the coast. To ﬁnd the distance from Q to C, we note that β = 180◦ − 90◦ − 45◦ = 45◦ so by symmetry,12 we get x = y ≈ 6.83 miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. √ 2 2 Sasquatch Island Sasquatch Island β β d ≈ 9.66 miles d ≈ 9.66 miles y miles γ Q 30◦ 5 miles P 45◦ Shoreline 45◦ Q C x miles We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 11.4. Suppose (α, a), (β, b) and (γ, c) are the angle-side opposite pairs of a triangle. Then the area A enclosed by the triangle is given by A = 1 2 bc sin(α) = 1 2 ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the most pieces of given information. We are given a = 7 and β = 45◦, and we calculated c = 7 sin(15◦) sin (45◦) =≈ 5.18 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4. sin(120◦) . Using these values, we ﬁnd A = 1 7 sin(15◦) sin(120◦) 2 (7) 12Or by Theorem 10.4 again . . . 904 Applications of Trigonometry 11.2.1 Exercises In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, (α, a), (β, b) and (γ, c) are angle-side opposite pairs. 1. α = 13◦, β = 17◦, a = 5 2. α = 73.2◦, β = 54.1◦, a = 117 3. α = 95◦, β = 85◦
, a = 33.33 4. α = 95◦, β = 62◦, a = 33.33 5. α = 117◦, a = 35, b = 42 6. α = 117◦, a = 45, b = 42 7. α = 68.7◦, a = 88, b = 92 8. α = 42◦, a = 17, b = 23.5 9. α = 68.7◦, a = 70, b = 90 10. α = 30◦, a = 7, b = 14 11. α = 42◦, a = 39, b = 23.5 12. γ = 53◦, α = 53◦, c = 28.01 13. α = 6◦, a = 57, b = 100 14. γ = 74.6◦, c = 3, a = 3.05 15. β = 102◦, b = 16.75, c = 13 16. β = 102◦, b = 16.75, c = 18 17. β = 102◦, γ = 35◦, b = 16.75 18. β = 29.13◦, γ = 83.95◦, b = 314.15 19. γ = 120◦, β = 61◦, c = 4 20. α = 50◦, a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we ﬁrst have you change road grades into angles and then use the Law of Sines in an application. 22. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4◦ angle with the horizontal. (It will not be exactly 4◦, but it’s pretty close.) 23. What grade is given by a 9.65◦ angle made by the road and the horizontal?13 13I have friends who live in Paciﬁca, CA and their road is actually this steep. It’s not a nice road to drive. 11.2 The Law of Sines 905 24. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6◦. Use the Law of Sines to ﬁnd the height of the tree. (Hint: First show that the tree makes a 94◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must ﬁrst understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40◦E (read “40◦ east of north”) is a bearing which is rotated clockwise 40◦ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50◦. Similarly, S50◦W would point into Quadrant III along the terminal side of θ = 220◦ because we started out pointing due south (along θ = 270◦) and rotated clockwise 50◦ back to 220◦. Counter-clockwise rotations would be found in the bearings N60◦W (which is on the terminal side of θ = 150◦) and S27◦E (which lies along the terminal side of θ = 297◦). These four bearings are drawn in the plane below. N N40◦E N60◦W 60◦ 40◦ W E 50◦ 27◦ S50◦W S27◦E S The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 25. Find the angle θ in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. (a) due west (b) S83◦E (c) N5.5◦E (d) due south 14The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel spots a campﬁre at a of bearing N42◦E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the ﬁre to be N20◦W from his current position. Determine the distance from the campﬁre to each man, rounded to the nearest foot. 27. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53◦W which brings her to the Muﬃn Ridge Observatory. From there, she knows a bearing of S65◦E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muﬃn Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. The captain of the SS Bigfoot sees a signal ﬂare at a bearing of N15◦E from her current location. From his position, the captain of the HMS Sasquatch ﬁnds the signal ﬂare to be at a bearing of N75◦W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50◦E, ﬁnd the distances from the ﬂare to each vessel, rounded to the nearest tenth of a mile. 29. Carl spies a potential Sasquatch nest at a bearing of N10◦E and radios Jeﬀ, who is at a bearing of N50◦E from Carl’s position. From Jeﬀ’s position, the nest is at a bearing of S70◦W. If Jeﬀ and Carl are 500 feet apart, how far is Jeﬀ from the Sasquatch nest? Round your answer to the nearest foot. 30. A hiker determines the bearing to a lodge from her current position is S40◦W. She proceeds to hike 2 miles at a bearing of S20◦E at which point she determines the bearing to the lodge is S75◦W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ship oﬀ shore at a bearing of N70◦E. A second tower, which is 50 miles from the ﬁrst at a bearing of S80◦E from the ﬁrst tower, determines the bearing to the ship to be N25◦W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 32. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75◦ and radios Sally immediately to ﬁnd the angle of inclination from her position to the craft is 50◦. How high oﬀ the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 15See Example 10.1.1 in Section 10.1 for a review of the DMS system. 11.2 The Law of Sines 907 33. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55◦. From a point ﬁve stories below the original observer, the angle of inclination to the gargoyle is 20◦. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 34. Prove that the Law of Sines holds when ABC is a right triangle. 35. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 36. Discuss with your classmates why the Law of Sines cannot be used to ﬁnd the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30◦ and b = 10, choose four diﬀerent values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30◦, b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles. 38. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2) to prove the area formulas given in Theorem 11.4. Why do those formulas yield square units when four quantities are being multiplied together? 908 Applications of Trigonometry 11.2.2 Answers 1. 3. 5. 7. α = 13◦ β = 17◦ a = 5 γ = 150◦ b ≈ 6.50 c ≈ 11.11 Information does not produce a triangle Information does not produce a triangle α = 68.7◦ β ≈ 76.9◦ γ ≈ 34.4◦ c ≈ 53.36 a = 88 α = 68.7◦ β ≈ 103.1◦ γ ≈ 8.2◦ c ≈ 13.47 a = 88 b = 92 b = 92 9. Information does not produce a triangle 11. 13. 15. 17. 19. α = 42◦ β ≈ 23.78◦ γ ≈ 114.22◦ a = 39 c ≈ 53.15 b = 23.5 α = 6◦ β ≈ 169.43◦ γ ≈ 4.57◦ c ≈ 43.45 a = 57 b = 100 α = 6◦ β ≈ 10.57◦ γ ≈ 163.43◦ c ≈ 155.51 a = 57 b = 100 α ≈ 28.61◦ β = 102◦ a ≈ 8.20 b = 16.75 c = 13 γ ≈ 49.39◦ α = 43◦ γ = 35◦ β = 102◦ a ≈ 11.68 b = 16.75 c ≈ 9.82 Information does not produce a triangle 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. α = 73.2◦ β = 54.1◦ γ = 52.7◦ c ≈ 97.22 b ≈ 99.00 a = 117 β = 62◦ α = 95◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 γ = 23◦ α = 117◦ β ≈ 56.3◦ γ ≈ 6.7◦ c ≈ 5.89 b = 42 a = 45 b = 23.5 α = 42◦ β ≈ 67.66◦ γ ≈ 70.34◦ c ≈ 23.93 a = 17 α = 42◦ β ≈ 112.34◦ γ ≈ 25.66◦ c ≈ 11.00 a = 17 b = 23.5 α = 30◦ β = 90◦ γ = 60◦ √ 3 b = 14 a = 7 c = 7 α = 53◦ β = 74◦ a = 28.01 b ≈ 33.71 c = 28.01 γ = 53◦ α ≈ 78.59◦ β ≈ 26.81◦ γ = 74.6◦ b ≈ 1.40 a = 3.05 α ≈ 101.41◦ β ≈ 3.99◦ γ = 74.6◦ b ≈ 0.217 a = 3.05 c = 3 c = 3 Information does not produce a triangle α = 66.92◦ β = 29.13◦ γ = 83.95◦ c ≈ 641.75 a ≈ 593.69 b = 314.15 α = 50◦ β ≈ 22.52◦ γ ≈ 107.48◦ a = 25 c ≈ 31.13 b = 12.5 21. The area of the triangle from Exercise 1 is about 8.1 square units. The area of the triangle from Exercise 12 is about 377.1 square units. The area of the triangle from Exercise 20 is about 149 square units. ≈ 0.699 radians, which is equiva
lent to 4.004◦ 22. arctan 7 100 23. About 17% 24. About 53 feet 11.2 The Law of Sines 909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campﬁre. Sarge is about 2525 feet to the campﬁre. 27. The distance from the Muﬃn Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles. 28. The SS Bigfoot is about 4.1 miles from the ﬂare. The HMS Sasquatch is about 2.9 miles from the ﬂare. 29. Jeﬀ is about 371 feet from the nest. 30. She is about 3.02 miles from the lodge 31. The boat is about 25.1 miles from the second tower. 32. The UFO is hovering about 9539 feet above the ground. 33. The gargoyle is about 44 feet from the observer on the upper ﬂoor. The gargoyle is about 27 feet from the observer on the lower ﬂoor. The gargoyle is about 25 feet from the other building. 910 Applications of Trigonometry 11.3 The Law of Cosines In Section 11.2, we developed the Law of Sines (Theorem 11.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS), the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 We state and prove the theorem below. Theorem 11.5. Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following equations hold a2 = b2 + c2 − 2bc cos(α) b2 = a2 + c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. B = (c cos(α), c sin(α)) c α a A = (0, 0) b C = (b, 0) From this set-up, we immediately ﬁnd that the coordinates of A and C are A(0, 0) and C(b, 0). From Theorem 10.3, we know that since the point B(x, y) lies on a circle of radius c, the coordinates 1Here, ‘Side-Angle-Side’ means that we are given two sides and the ‘included’ angle - that is, the given angle is adjacent to both of the given sides. 11.3 The Law of Cosines 911 of B are B(x, y) = B(c cos(α), c sin(α)). (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for any angle α drawn in standard position where 0 < α < 180◦.) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation 1.1, we get a = (c cos(α) − b)2 + (c sin(α) − 0)2 2 (c cos(α) − b)2 + c2 sin2(α) a2 = a2 = (c cos(α) − b)2 + c2 sin2(α) a2 = c2 cos2(α) − 2bc cos(α) + b2 + c2 sin2(α) a2 = c2 cos2(α) + sin2(α) + b2 − 2bc cos(α) a2 = c2(1) + b2 − 2bc cos(α) a2 = c2 + b2 − 2bc cos(α) Since cos2(α) + sin2(α) = 1 The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a diﬀerent vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which γ = 90◦, then cos(γ) = cos (90◦) = 0 so we get the familiar relationship c2 = a2 + b2. What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50◦, a = 7 units, c = 2 units 2. a = 4 units, b = 7 units, c = 5 units Solution. 1. We are given the lengths of two sides, a = 7 and c = 2, and the measure of the included angle, β = 50◦. With no angle-side opposite pair to use, we apply the Law of Cosines. We get b2 = 72 + 22 − 2(7)(2) cos (50◦) which yields b = In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair (β, b) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 53 − 28 cos (50◦) ≈ 5.92 units. 2This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the deﬁnition of the circular functions along with the distance formula and hence, the Pythagorean Theorem. 912 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method ﬁrst. When using the Law of Cosines, it’s always best to ﬁnd the measure of the largest unknown angle ﬁrst, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to ﬁnd α ﬁrst. To that end, we use the formula cos(α) = b2+c2−a2 53 − 28 cos (50◦) and c = 2. We get3 and substitute a = 7, b = 2bc cos(α) = 2 − 7 cos (50◦) 53 − 28 cos (50◦) Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have α = arccos 2 − 7 cos (50◦) 53 − 28 cos (50◦) radians ≈ 114.99◦ At this point, we could ﬁnd γ using γ = 180◦ − α − β ≈ 180◦ − 114.99◦ − 50◦ = 15.01◦, that is if we trust our approximation for α. To minimize propagation of error, however, we could use the Law of Cosines again,4 in this case using cos(γ) = a2+b2−c2 . Plugging in a = 7, b = 53 − 28 cos (50◦) and c = 2, we get γ = arccos sketch the triangle below. √ 7−2 cos(50◦) 53−28 cos(50◦) 2ab radians ≈ 15.01◦. We β = 50◦ a = 7 c = 2 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to ﬁnd the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to ﬁrst ﬁnd the smallest of the unknown angles, since we are guaranteed it will be acute.5 In this case, we would ﬁnd γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair (β, b), we get sin(γ) . The usual calculations produces γ ≈ 15.01◦ and α = 180◦ − β − γ ≈ 180◦ − 50◦ − 15.01◦ = 114.99◦. sin(50◦) 53−28 cos(50◦) 2 = √ 2. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we ﬁnd β ﬁrst, since it is opposite the longest side, radians ≈ 101.54◦. As in b. We get cos(β) = a2+c2−b2 5 , so we get β = arccos − 1 = − 1 2ac 5 3after simplifying . . . 4Your instructor will let you know which procedure to use. It all boils down to how much you trust your calculator. 5There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 11.3 The Law of Cosines 913 the previous problem, now that we have obtained an angle-side opposite pair (β, b), we could proceed using the Law of Sines. The Law of Cosines, however, oﬀers us a rare opportunity to ﬁnd the remaining angles using only the data given to us in the statement of the problem. Using this, we get γ = arccos 5 7 radians ≈ 44.42◦ and α = arccos 29 35 radians ≈ 34.05◦. β ≈ 101.54◦ c = 5 a = 4 α ≈ 34.05◦ γ ≈ 44.42◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may diﬀer slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number 2 in Example 11.3.1, where the approximate values we record for the measures of the angles sum to 180.01◦, which is geometrically impossible. Next, we have an application of the Law of Cosines. Example 11.3.2. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P , he ﬁnds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60◦, ﬁnd the width of the pond. 1000 feet 950 feet 60◦ P Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to ﬁnd the length of the missing side opposite the given angle. Calling this length w (for width), we get w2 = 9502 + 10002 − 2(950)(1000) cos (60◦) = 952500 from which we get w = 952500 ≈ 976 feet. √ 914 Applications of Trigonometry In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron’s Formula. Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2 (a + b + c). Then the area A enclosed by the triangle is given by A = s(s − a)(s − b)(s − c) We prove Theorem 11.6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to
simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines tells us cos(γ) = a2+b2−c2 2ab , so substituting this into our equation for A2 gives A2 = = = = = = = = 1 − 1 − 2 1 − cos2(γ) a2 + b2 − c2 2ab a2 + b2 − c22 4a2b2 4a2b2 − a2 + b2 − c22 4a2b2 a2b2 4 a2b2 4 a2b2 4 a2b2 4 4a2b2 − a2 + b2 − c22 16 (2ab)2 − a2 + b2 − c22 16 2ab − a2 + b2 − c2 2ab + a2 + b2 − c2 16 c2 − a2 + 2ab − b2 a2 + 2ab + b2 − c2 16 diﬀerence of squares. 11.3 The Law of Cosines 915 A2 = = = = = c2 − a2 − 2ab + b2 a2 + 2ab + b2 − c2 16 c2 − (a − b)2 (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 perfect square trinomials. diﬀerence of squares. (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (a + c − b) 2 (b + c − a) 2 (a + b − c) 2 · · · (a + b + c) 2 At this stage, we recognize the last factor as the semiperimeter, s = 1 complete the proof, we note that 2 (a + b + c) = a+b+c 2 . To (s − a − 2a 2 = b + c − a 2 Similarly, we ﬁnd (s − b) = a+c−b 2 and (s − c) = a+b−c 2 . Hence, we get A2 = (b + c − a) 2 · (a + c − b) 2 · (a + b − c) 2 · (a + b + c) 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2. Solution. We are given a = 4, b = 7 and c = 5. Using these values, we ﬁnd s = 1 2 (4 + 7 + 5) = 8, (s − a) = 8 − 4 = 4, (s − b) = 8 − 7 = 1 and (s − c) = 8 − 5 = 3. Using Heron’s Formula, we get √ A = s(s − a)(s − b)(s − c) = 6 ≈ 9.80 square units. (8)(4)(1)(3) = 96 = 4 √ 916 Applications of Trigonometry 11.3.1 Exercises In Exercises 1 - 10, use the Law of Cosines to ﬁnd the remaining side(s) and angle(s) if possible. 1. a = 7, b = 12, γ = 59.3◦ 2. α = 104◦, b = 25, c = 37 3. a = 153, β = 8.2◦, c = 153 4. a = 3, b = 4, γ = 90◦ 5. α = 120◦, b = 3, c = 4 6. a = 7, b = 10, c = 13 7. a = 1, b = 2, c = 5 8. a = 300, b = 302, c = 48 9. a = 5, b = 5, c = 5 10. a = 5, b = 12, ; c = 13 In Exercises 11 - 16, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63◦, b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63◦, b = 20 14. a = 22, α = 63◦, b = 20 15. α = 42◦, b = 117, c = 88 16. β = 7◦, γ = 170◦, c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 19. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117◦, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 20. From the Pedimaxus International Airport a tour helicopter can ﬂy to Cliﬀs of Insanity Point by following a bearing of N8.2◦E for 192 miles and it can ﬂy to Bigfoot Falls by following a bearing of S68.5◦E for 207 miles.6 Find the distance between Cliﬀs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 21. Cliﬀs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliﬀs of Insanity Point? Round your angle to the nearest tenth of a degree. 6Please refer to Page 905 in Section 11.2 for an introduction to bearings. 11.3 The Law of Cosines 917 22. A naturalist sets oﬀ on a hike from a lodge on a bearing of S80◦W. After 1.5 miles, she changes her bearing to S17◦W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23◦E and travels for 5 miles. It then changes course and follows a heading of S41◦E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 24. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32◦E. A storm moves in and after 100 miles, the captain of the Bigfoot ﬁnds he has drifted oﬀ course. If his bearing to the harbor is now S70◦W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 25. From a point 300 feet above level ground in a ﬁretower, a ranger spots two ﬁres in the Yeti National Forest. The angle of depression7 made by the line of sight from the ranger to the ﬁrst ﬁre is 2.5◦ and the angle of depression made by line of sight from the ranger to the second ﬁre is 1.3◦. The angle formed by the two lines of sight is 117◦. Find the distance between the two ﬁres. Round your answer to the nearest foot. (Hint: In order to use the 117◦ angle between the lines of sight, you will ﬁrst need to use right angle Trigonometry to ﬁnd the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.) ﬁre 117◦ ﬁre ﬁretower 26. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 7See Exercise 78 in Section 10.3 for the deﬁnition of this angle. 918 Applications of Trigonometry 11.3.2 Answers 1. 3. 5. 7. 9. 11. 13. 15. α ≈ 35.54◦ β ≈ 85.16◦ γ = 59.3◦ c ≈ 10.36 a = 7 b = 12 α ≈ 85.90◦ β = 8.2◦ a = 153 b ≈ 21.88 c = 153 γ ≈ 85.90◦ α = 120◦ β ≈ 25.28◦ γ ≈ 34.72◦ a = c = 4 b = 3 37 √ Information does not produce a triangle α = 60◦ β = 60◦ γ = 60 = 20 α = 63◦ β ≈ 98.11◦ γ ≈ 18.89◦ a = 18 α = 63◦ β ≈ 81.89◦ γ ≈ 35.11◦ c ≈ 11.62 a = 18 c ≈ 6.54 b = 20 Information does not produce a triangle α = 42◦ a ≈ 78.30 b = 117 β ≈ 89.23◦ γ ≈ 48.77◦ c = 88 α = 104◦ a ≈ 49.41 b = 25 β ≈ 29.40◦ γ ≈ 46.60◦ c = 37 α ≈ 36.87◦ β ≈ 53.13◦ γ = 90 ≈ 32.31◦ β ≈ 49.58◦ γ ≈ 98.21◦ a = 7 c = 13 b = 10 α ≈ 83.05◦ β ≈ 87.81◦ γ ≈ 9.14◦ b = 302 a = 300 c = 48 α ≈ 22.62◦ β ≈ 67.38◦ γ = 90◦ c = 13 b = 12 a = 5 α ≈ 55.30◦ β ≈ 89.40◦ γ ≈ 35.30◦ a = 37 c = 26 b = 45 α = 63◦ β ≈ 54.1◦ γ ≈ 62.9◦ c ≈ 21.98 a = 22 b = 20 α ≈ 3◦ γ = 170◦ β = 7◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 √ 2. 4. 6. 8. 10. 12. 14. 16. √ √ 17. The area of the triangle given in Exercise 6 is The area of the triangle given in Exercise 8 is The area of the triangle given in Exercise 10 is exactly 30 square units. 1200 = 20 51764375 ≈ 7194.75 square units. 3 ≈ 34.64 square units. 18. The distance between the ends of the hands at four o’clock is about 8.26 inches. 19. The diameter of the crater is about 5.22 miles. 20. About 313 miles 21. N31.8◦W 22. She is about 3.92 miles from the lodge and her bearing to the lodge is N37◦E. 23. It is about 4.50 miles from port and its heading to port is S47◦W. 24. It is about 229.61 miles from the island and the captain should set a course of N16.4◦E to reach the island. 25. The ﬁres are about 17456 feet apart. (Try to avoid rounding errors.) 11.4 Polar Coordinates 919 11.4 Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We deﬁned the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P (−3, 4), we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, (r, θ), where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis. Roughly speaking, the polar coordinates (r, θ) of a point measure ‘how far out’ the point is from the pole (that’s r), and ‘how far to rotate’ from the polar axis, (that’s θ). y P (−3, 4) P (r, θ) 3 2 1 r θ −4 −3 −2 −1 −1 1 2 3 4 x Pole r Polar Axis −2 −3 −4 For example, if we wished to plot the point P with polar coordinates 4, 5π 6 move out along the polar axis 4 units, then rotate 5π , we’d start at the pole, 6 radians counter-clockwise. P 4, 5π 6 r = 4 Pole θ = 5π 6 Pole Pole We may also visualize this process by thinking of the rotation ﬁrst.3 To plot P 4, 5π 6 we rotate 5π this way, 6 counter-clockwise from the polar axis, then move outwards from the pole 4 units. 1Excluding, of course, the points in which one or both coordinates are 0. 2We will explain more about this momentarily. 3As
with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 920 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 units away from the pole. θ = 5π 6 Pole θ = 5π 6 Pole P 4, 5π 6 Pole If r < 0, we begin by moving in the opposite direction on the polar axis from the pole. For example, to plot Q −3.5, π 4 we have r = −3.5 Pole θ = π 4 Pole Pole Q −3.5, π 4 If we interpret the angle ﬁrst, we rotate π Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 radians, then move back through the pole 3.5 units. 4 , not π 4 . θ = π 4 Pole θ = π 4 Pole Pole Q −3.5, π 4 As you may have guessed, θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. Hence, to plot R 3.5, − 3π 4 we have the following. r = 3.5 Pole Pole θ = − 3π 4 Pole R 3.5, − 3π 4 From an ‘angles ﬁrst’ approach, we rotate − 3π R is the point on the terminal side of θ = − 3π 4 then move out 3.5 units from the pole. We see that 4 which is 3.5 units from the pole. Pole θ = − 3π 4 Pole θ = − 3π 4 Pole R 3.5, − 3π 4 11.4 Polar Coordinates 921 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are diﬀerent. Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by inﬁnitely many polar coordinate pairs. We explore this notion more in the following example. Example 11.4.1. For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r > 0 and the other with r < 0. 4. P −3, − π 4 3. P 117, − 5π 2 2. P −4, 7π 6 1. P (2, 240◦) Solution. 1. Whether we move 2 units along the polar axis and then rotate 240◦ or rotate 240◦ then move out 2 units from the pole, we plot P (2, 240◦) below. θ = 240◦ Pole Pole P (2, 240◦) We now set about ﬁnding alternate descriptions (r, θ) for the point P . Since P is 2 units from the pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240◦) so the angle θ we choose for the r = 2 case must be coterminal with 240◦. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2, −120◦). For the case r = −2, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate θ = 60◦ to arrive at location coterminal with 240◦. Hence, our answer here is (−2, 60◦). We check our answers by plotting them. Pole θ = −120◦ θ = 60◦ Pole P (2, −120◦) P (−2, 60◦) 2. We plot −4, 7π 6 6 radians. Since r = −4 < 0, we ﬁnd our point lies 4 units from the pole on the terminal side of π 6 . by ﬁrst moving 4 units to the left of the pole and then rotating 7π P −4, 7π 6 Pole Pole θ = 7π 6 922 Applications of Trigonometry To ﬁnd alternate descriptions for P , we note that the distance from P to the pole is 4 units, so any representation (r, θ) for P must have r = ±4. As we noted above, P lies on the terminal as one of our answers. To ﬁnd a diﬀerent side of π representation for P with r = −4, we may choose any angle coterminal with the angle in the as our second answer. original representation of P −4, 7π 6 6 , so this, coupled with r = 4, gives us 4, π . We pick − 5π 6 and get −4, − 5π 6 6 P 4, π 6 θ = π 6 Pole θ = − 5π 6 Pole P −4, − 5π 6 , we move along the polar axis 117 units from the pole and rotate 3. To plot P 117, − 5π 2 clockwise 5π 2 radians as illustrated below. Pole θ = − 5π 2 Pole P 117, − 5π 2 Since P is 117 units from the pole, any representation (r, θ) for P satisﬁes r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2 . In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an angle θ to reach P . We ﬁnd that θ = π 2 satisﬁes this requirement, so our second answer is −117, π 2 2 , and get 117, 3π . 2 Pole θ = 3π 2 Pole θ = π 2 P 117, 3π 2 P −117, π 2 11.4 Polar Coordinates 923 4. We move three units to the left of the pole and follow up with a clockwise rotation of π 4 radians to plot P −3, − π 4 . We see that P lies on the terminal side of 3π 4 . P −3, − π 4 θ = − π 4 Pole Pole . To Since P lies on the terminal side of 3π ﬁnd a diﬀerent representation for P with r = −3, we may choose any angle coterminal with − π 4 , one alternative representation for P is 3, 3π 4 for our ﬁnal answer −3, 7π 4 . We choose θ = 7π . 4 4 P 3, 3π 4 P −3, 7π 4 θ = 3π 4 Pole θ = 7π 4 Pole Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has inﬁnitely many other representations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a deﬁnition or a theorem, depending on your point of view. We state it as a property of the polar coordinate system. Equivalent Representations of Points in Polar Coordinates Suppose (r, θ) and (r, θ) are polar coordinates where r = 0, r = 0 and the angles are measured in radians. Then (r, θ) and (r, θ) determine the same point P if and only if one of the following is true: r = r and θ = θ + 2πk for some integer k r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that (r, θ) means (directed distance from pole, angle of rotation). If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus (0, θ) is the pole for all 924 Applications of Trigonometry values of θ. Now let’s assume that neither r nor r is zero. If (r, θ) and (r, θ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the ﬁrst coordinate, we have that either r = r or r = −r. If r = r, then when plotting (r, θ) and (r, θ), the angles θ and θ have the same initial side. Hence, if (r, θ) and (r, θ) determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + 2πk for some integer k, as required. If, on the other hand, r = −r, then when plotting (r, θ) and (r, θ), the initial side of θ is rotated π radians away from the initial side of θ. In this case, θ must be coterminal with π + θ. Hence, θ = π + θ + 2πk which we rewrite as θ = θ + (2k + 1)π for some integer k. Conversely, if r = r and θ = θ + 2πk for some integer k, then the points P (r, θ) and P (r, θ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = −r and θ = θ + (2k + 1)π for some integer k. To plot P , we ﬁrst move a directed distance r from the pole; to plot P , our ﬁrst step is to move the same distance from the pole as P , but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ + (2k + 1)π = (θ + π) + 2πk for some integer k, we see that θ is coterminal to (θ + π) and it is this extra π radians of rotation which aligns the points P and P . Next, we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. To do so, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. We get the following result. Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as (x, y) and in polar coordinates as (r, θ). Then x = r cos(θ) and y = r sin(θ) x2 + y2 = r2 and tan(θ) = y x (provided x = 0) In the case r > 0, Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity tan(θ) = sin(θ) If r < 0, then we know an alternate representation for (r, θ) cos(θ) . is (−r, θ + π). Since cos(θ + π) = − cos(θ) and sin(θ + π) = − sin(θ), applying the theorem to (−r, θ + π) gives x = (−r) cos(θ + π) = (−r)(− cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = (−r)(− sin(θ)) = r sin(θ). Moreover, x2 + y2 = (−r)2 = r2, and y x = tan(θ + π) = tan(θ), so the theorem is true in this case, too. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identiﬁed with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. 1. P 2, −2 √ 3 2. Q(−3, −3) 3. R(0, −3) 4. S(−3, 4) 11.4 Polar Coordinates 925 Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking 3 shows that the time to plot the points before we do any calculations. Plotting P 2, −2 √ it lies in Quadrant IV. With x = 2 and y = −2 = 4 + 12 = 16 so r = ±4. Since we are asked for r ≥ 0, we choose r = 4. To ﬁnd θ, we have that tan(θ) = y 3 , and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π, . To check, we convert (r, θ) = 4, 5π 3 . Hence, our answer is 4, 5π so we choose θ = 5π 3 back to rectangular coordinates and we ﬁnd x = r cos(θ) = 4 cos 5π = 2 and 3 y = r sin(θ) = 4 sin 5π 3 3. This tells us θ has a reference angle of π 3, we get r2 = x2 + y2 = (2)2 + −, as required. x = −2 = 4 = −2 32 √ √ 18 = ±3 2. The point Q(−3, −3) lies in Quadrant III. Using x = y = −3, we get
r2 = (−3)2 + (−3)2 = 18 2. We ﬁnd 4 . Since Q lies in Quadrant III, 4 , which satisﬁes the requirement that 0 ≤ θ < 2π. Our ﬁnal answer is . To check, we ﬁnd x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choose r = 3 = (3 2) √ √ √ √ − √ 2) cos 5π 4 2, 5π 4 2 2 and y = r sin(θ) = (3 √ 2) sin 5π 4 √ = (3 − √ 2 2 2) = −3, so we are done. y y θ = 5π 3 x θ = 5π 4 x P Q P has rectangular coordinates (2, −2 P has polar coordinates 4, 5π 3 √ 3) Q has rectangular coordinates (−3, −3) Q has polar coordinates 3 √ 2, 5π 4 3. The point R(0, −3) lies along the negative y-axis. While we could go through the usual computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 3π 2 satisﬁes 0 ≤ θ < 2π 4Since x = 0, we would have to determine θ geometrically. 926 Applications of Trigonometry with its terminal side along the negative y-axis, so our answer is 3, 3π 2 x = r cos(θ) = 3 cos 3π 2 = (3)(0) = 0 and y = r sin(θ) = 3 sin 3π 2 = 3(−1) = −3. . To check, we note −3 = − 4 4. The point S(−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 4 3 , and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π, we choose θ = π − arctan 4 ≈ (5, 2.21). To radians. Hence, our answer is (r, θ) = 5, π − arctan 4 3 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4 . These are good review exercises and are hence 3 left to the reader. We ﬁnd cos π − arctan 4 5 , so that 3 x = r cos(θ) = (5) − 3 5 = − 3 = −3 and y = r sin(θ) = (5) 4 5 = 4 which conﬁrms our answer. and sin π − arctan 4 3 5 and sin π − arctan 4 = 4 3 y y S θ = 3π 2 x R θ = π − arctan 4 3 x R has rectangular coordinates (0, −3) R has polar coordinates 3, 3π 2 S has rectangular coordinates (−3, 4) S has polar coordinates 5, π − arctan 4 3 Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we now set about converting equations from one system to another. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two systems using Theorem 11.7 as the next example illustrates. Example 11.4.3. 1. Convert each equation in rectangular coordinates into an equation in polar coordinates. (a) (x − 3)2 + y2 = 9 (b) y = −x (c) y = x2 2. Convert each equation in polar coordinates into an equation in rectangular coordinates. (a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) 11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2+y2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.5 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2(θ) − 6r cos(θ) + 9 + r2 sin2(θ) = 9 r2 cos2(θ) + sin2(θ) − 6r cos(θ) = 0 Subtract 9 from both sides. Since cos2(θ) + sin2(θ) = 1 Factor. r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our ﬁnal answer.6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r sin(θ) = −r cos(θ). Rearranging, we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4 . In this equation, the variable r is free,7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our ﬁnal answer θ = − π 4 here.8 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x2 and get r sin(θ) = (r cos(θ))2, or r2 cos2(θ) − r sin(θ) = 0. Factoring, we get r(r cos2(θ) − sin(θ)) = 0 so that either r = 0 or r cos2(θ) = sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos2(θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos2(θ) = 0, then cos(θ) = 0, and for the equation r cos2(θ) = sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeﬀ’s. 2 into r = 6 cos(θ), we recover the point r = 0, so we aren’t losing anything 6Note that when we substitute θ = π by disregarding r = 0. 7See Section 8.1. 8We could take it to be any of θ = − π 4 + πk for integers k. 928 Applications of Trigonometry information by dividing both sides of r cos2(θ) = sin(θ) by cos2(θ). Doing so, we get r = sin(θ) cos2(θ) , or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our ﬁnal answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We could solve r2 = x2 + y2 for r to get r = ± x2 + y2 + πk for and solving tan(θ) = y integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y2, r cos(θ) = x, r sin(θ) = y and/or tan(θ) = y x requires the arctangent function to get θ = arctan y x present themselves. x (a) Starting with r = −3, we can square both sides to get r2 = (−3)2 or r2 = 9. We may now substitute r2 = x2 + y2 to get the equation x2 + y2 = 9. As we have seen,9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r2 = 9 might be satisﬁed by more points than r = −3. On the surface, this appears to be the case since r2 = 9 is equivalent to r = ±3, not just r = −3. However, any point with polar coordinates (3, θ) can be represented as (−3, θ + π), which means any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent10 representation which satisﬁes r = −3. (b) We take the tangent of both sides the equation θ = 4π Since tan(θ) = y if, geometrically, the equations θ = 4π 3 and y = x The same argument presented in number 1b applies equally well here so we are done. 3 to get tan(θ) = tan 4π 3. 3. Of course, we pause a moment to wonder 3 generate the same set of points.11 x , we get y 3 or c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r2 = r − r cos(θ). We now have an r2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r2 + r cos(θ) and squaring both sides yields r2 = r2 + r cos(θ)2 . Substituting r2 = x2 + y2 and r cos(θ) = x gives x2 + y2 = x2 + y2 + x2 . Once again, we have performed some 9Exercise 5.3.1 in Section 5.3, for instance . . . 10Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3, π) are diﬀerent, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = −3 represent diﬀerent relations since they correspond to diﬀerent sets of ordered pairs. Since polar coordinates were deﬁned geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we ﬁrst deﬁned relations as sets of points in the plane in Section 1.2. Back then, a point in the plane was identiﬁed with a unique ordered pair given by its Cartesian coordinates. 11In addition to taking the tangent of both sides of an equation (There are inﬁnitely many solutions to tan(θ) = √ √ 3 is only one of them!), we also went from y x = 3, in which x cannot be 0, to y = x √ 3, 3 in which we assume and θ = 4π x can be 0. 11.4 Polar Coordinates 929 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1 − cos(θ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a
reason to pause. Are there points with coordinates (r, θ) which satisfy r2 = r2 + r cos(θ)2 but do not satisfy r = r2 + r cos(θ)? Suppose (r, θ) satisﬁes r2 = r2 + r cos(θ)2 . Then r = ± (r)2 + r cos(θ). If we have that r = (r)2+r cos(θ), we are done. What if r = − (r)2 + r cos(θ) = −(r)2−r cos(θ)? We claim that the coordinates (−r, θ + π), which determine the same point as (r, θ), satisfy r = r2 + r cos(θ). We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to see if we get a true statement. −r − −(r)2 − r cos(θ) (r)2 + r cos(θ) (r)2 + r cos(θ) ? = (−r)2 + (−r cos(θ + π)) ? = (r)2 − r cos(θ + π) ? = (r)2 − r(− cos(θ)) = (r)2 + r cos(θ) Since r = −(r)2 − r cos(θ) Since cos(θ + π) = − cos(θ) Since both sides worked out to be equal, (−r, θ + π) satisﬁes r = r2 + r cos(θ) which means that any point (r, θ) which satisﬁes r2 = r2 + r cos(θ)2 has a representation which satisﬁes r = r2 + r cos(θ), and we are done. In practice, much of the pedantic veriﬁcation of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r2 = 9 happens without a second thought. Your instructor will ultimately decide how much, if any, justiﬁcation is warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things in rectangular coordinates, such as y = x2, can turn ugly in polar coordinates, and vice-versa. In the next section, we devote our attention to graphing equations like the ones given in Example 11.4.3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates. If nothing else, number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. 930 Applications of Trigonometry 11.4.1 Exercises In Exercises 1 - 16, plot the point given in polar coordinates and then give three diﬀerent expressions (c) r > 0 and θ ≥ 2π for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π, (b) r > 0 and θ ≤ 0 2, 1. π 3 5. 12, − 7π 6 9. (−20, 3π) 2. 5, 7π 4 6. 3, − 10. −4, 5π 4 5π 4 3. 1 3 , 3π 2 √ 7. 2 2, −π 11. −1, 2π 3 13. −3, − 11π 6 14. −2.5, − π 4 15. √ − 5, − 4π 3 4. 8. 5 2 7 2 , 5π 6 , − 13π 6 −3, 12. π 2 16. (−π, −π) In Exercises 17 - 36, convert the point from polar coordinates into rectangular coordinates. 18. 2, π 3 22. −4, 5π 6 7π 6 19. 11, − 23. 9, 7π 2 20. (−20, 3π) 24. −5, − 9π 4 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 17. 5, 7π 4 21. 3 5 , π 2 25. 42, 13π 6 29. −3, arctan 31. 33. 2, π − arctan −1, π + arctan 4 3 1 2 3 4 34. 2 3 35. (π, arctan(π)) 36. 13, arctan 30. 5, arctan − 4 3 32. − 1 2 , π − arctan (5) , π + arctan 2 √ 2 12 5 In Exercises 37 - 56, convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 37. (0, 5) 41. (−3, 0) 38. (3, √ 3) 39. (7, −7) √ 42. − √ 2 2, 43. −4, −4 √ 3 √ 3) 40. (−3, − √ 44. 3 4 , − 1 4 11.4 Polar Coordinates 931 45. − √ 3 3 10 3 10 , − √ 46. − √ 5, − 5 47. (6, 8) √ √ 5) 5, 2 48. ( 49. (−8, 1) √ 50. (−2 √ 10, 6 10) 53. (24, −7) 54. (12, −9) 51. (−5, −12) √ 2 4 √ 6 4 , 55. 52. − √ 5 15 √ 5 2 15 , − 56. − √ 65 5 2 , √ 65 5 In Exercises 57 - 76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #63. In Exercises 60 - 63, you need to solve for θ 57. x = 6 61. y = −x 58. x = −3 √ 3 62. y = x 59. y = 7 60. y = 0 63. y = 2x 64. x2 + y2 = 25 65. x2 + y2 = 117 66. y = 4x − 19 67. x = 3y + 1 68. y = −3x2 69. 4x = y2 70. x2 + y2 − 2y = 0 71. x2 − 4x + y2 = 0 72. x2 + y2 = x 73. y2 = 7y − x2 75. x2 + (y − 3)2 = 9 74. (x + 2)2 + y2 = 4 76. 4x2 + 4 y − 2 1 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78. r = −3 82. θ = π 79. r = 83. θ = √ 2 3π 2 80. θ = π 4 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88. r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) √ 91. r = − 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = − csc(θ) cot(θ) 94. r2 = sin(2θ) 95. r = 1 − 2 cos(θ) 96. r = 1 + sin(θ) 97. Convert the origin (0, 0) into polar coordinates in four diﬀerent ways. 98. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 932 Applications of Trigonometry 11.4.2 Answers 1. 2, , π 3 2, − 5π 3 −2, , 4π 3 7π 3 2, 2. 5, 7π 4 π 4 , , −5, 5, 3π 4 15π 4 5, − 3. 1 3 1 3 4. 5 2 5 2 , , , 3π , , 5π 6 7π 7π 2 , , 11π 6 17π 6 y y 2 1 1 −1 1 2 x −1 1 2 3 x −1 −2 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 933 5. 12, − 12, − , −12, , 12, 7π 6 19π 6 11π 6 17π 6 6. 3, − 3, − , −3, , 3, 5π 4 13π 4 7π 4 11π 4 √ √ 7. 2 2 √ 2, −π , −2 √ 2, −3π , 2 2, 0 2, 3π 8. 7 2 7 2 , − , − 13π 23π 6 5π 6 y 6 3 −12 −9 −6 −3 x y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 3 2 1 y −3 −2 −1 1 2 3 x −1 −2 −3 4 3 2 1 y −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 934 Applications of Trigonometry 9. (−20, 3π), (−20, π) (20, −2π), (20, 4π) 10. −4, 4, − 5π 4 7π 4 11. −1, 1, − 2π 3 π 3 13π 4 , , −4, 4, 9π 4 8π 3 , , 1, −1, 11π 3 12. −3, 3, − , , π 2 π 2 5π 2 −3, 3, 7π 2 10 x 20 −20 −10 y y 1 −1 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 2 1 y −2 −1 1 2 x −1 −2 y 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 935 13. −3, − 11π 6 , 3, − 5π 6 π 6 −3, 19π 6 , 3, , , 14. −2.5, − 2.5, − π 4 5π 4 −2.5, 2.5, 7π 4 11π 4 √ 15. − 5, − √ 5, − π 3 4π 3 , , √ √ − 2π 3 5, 5, 11π 3 16. (−π, −π) , (−π, π) (π, −2π) , (π, 2π3 −2 −1 −1 −2 −3 2 1 −2 −1 −1 −2 2 1 −2 −1 −1 −2 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 936 Applications of Trigonometry √ 5 2 2 , − √ 2 5 2 17. 18. 1, √ 3 19. − √ 11 2 3 , 11 2 22. 2 √ 3, −2 23. (0, −9) 20. (20, 0) 24 28. √ 10, 3 √ 10 √ 26 52 √ 5 26 52 , − 32. 36. (5, 12) √ 2 3, 40. 44. 1 2 , 11π 6 7π 6 4 3 48. (5, arctan (2)) 26. (117, 0) 30. (3, −4 34. √ 2 3, 38. 42. 2, √ 46. 3π 4 10, π 6 5π 4 √ 5 6 5 5 √ 12 5 , 27. 31 35. π√ 1+π2 , √ π2 1+π2 7π 4 √ 7 2, 39. 43. 8, 4π 3 47. 10, arctan 21. 0, 3 5 √ 25. 21 3, 21 29. − 9 5 , − 12 5 33. 4 5 , 3 5 37. 5, π 2 41. (3, π) 4π 3 , 3 5 √ 51. 13, π + arctan 25, 2π − arctan 1 8 12 5 7 24 45. 49. 53. 55. 65, π − arctan 50. (20, π − arctan(3)) , π + arctan (2) 52. 1 3 54. 15, 2π − arctan 3 4 √ 2 2 , π 3 56. √ 13, π − arctan(2) 57. r = 6 sec(θ) 58. r = −3 sec(θ) 59. r = 7 csc(θ) 60. θ = 0 61. θ = 3π 4 √ 65. r = 117 62. θ = π 3 63. θ = arctan(2) 64. r = 5 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = − sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 937 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = sin(θ) 77. x2 + y2 = 49 78. x2 + y2 = 9 79. x2 + y2 = 2 80. y = x √ 81. y = − 3x 83. x = 0 82. y = 0 84. x2 + y2 = 4x or (x − 2)2 + y2 = 4 85. 5x2 + 5y2 = x or x − 2 1 10 + y2 = 1 100 86. x2 + y2 = 3y or x2 + y − 2 3 2 = 9 4 87. x2 + y2 = −2y or x2 + (y + 1)2 = 1 88. x = 7 89. y = 1 12 √ 91. y = − 5 93. y2 = −x 90. x = −2 92. x2 = 2y 94. x2 + y22 = 2xy 95. x2 + 2x + y22 = x2 + y2 96. x2 + y2 + y2 97. Any point of the form (0, θ) will work, e.g. (0, π), (0, −117), 0, = x2 + y2 23π 4 and (0, 0). 938 Applications of Trigonometry 11.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has inﬁnitely many diﬀerent representations in polar coordinates, our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. We state it below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a point P (r, θ) is on the graph of an equation if and only if there is a representation of P , say (r, θ), such that r and θ satisfy the equation. Our ﬁrst example focuses on some of the more structurally simple polar equations. Example 11.5.1. Graph the following polar equations. 1. r = 4 2. r = −3 √ 2 3. θ = 5π 4 4. θ = − 3π 2 Solution. In each of these equations, only one of the variables r and θ is present making the other variable free.1 This makes these graphs easier to visualize than others. 1. In the equation r = 4, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation (4, θ), for any choice of θ. Graphically this translates into tracing out all of the points 4 units away from the origin. This is exactly the deﬁnition of circle, centered at the origin, with a radius of 44 x 4 In r = 4, θ is free The graph of r = 4 −4 2. Once again we have θ being free in the equation r = −3 2, θ) gives us a circle of radius 3 form (−3 √ √ 2 centered at the origin. 2. Plotting all of the points of the √ 1See the discussion in Example 11.4.3 number 2a. 11.5 Graphs of Polar Equations 939 4 x 4 In r = −3 √ 2, θ is free −4 The graph of r = −3 √ 2 3. In the equation θ = 5π . 4 , r is free, so we plot all of the points with polar representation r, 5π What we ﬁnd is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. 4 y r < 0 θ = 5π 4 y 4 r = 0 x −4 x 4 r > 0 In θ = 5π 4 , r is free The graph of θ = 5π 4 −4 4. As in the previous example, the variable r is free in the equation θ = − 3π 2 . Plotting r, − 3π 2 for various values of r shows us that we are tracing out the y-axis. 940 Applications of Trigonometry y r > 0 r = 0 θ = − 3π 2 y 4 x −4 x 4 r < 0 −4 In θ = − 3π 2 , r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following result clear. Theorem 11.8. Graphs of Constant r and θ: Suppose a and α are constants, a = 0. The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius \|a\|. The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cos(θ). A reasonable way to start is to treat θ as the i
ndependent variable, r as the dependent variable, evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.2 We generate the table below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π r = 6 cos(θ) 6 √ 3 2 3 √ √ (r, θ) (6, 0) 2, π 4 0, π 2 2, 3π 4 (−6, π) √ 2, 5π 4 0, 3π 2 √ 2, 7π 4 (6, 2π) 0 2 −3 √ −3 −6 √ −3 √ 3 2 6 2 −3 0 3 y 3 −3 3 x 6 2For a review of these concepts and this process, see Sections 1.4 and 1.6. 11.5 Graphs of Polar Equations 941 Despite having nine ordered pairs, we get only four distinct points on the graph. For this reason, we employ a slightly diﬀerent strategy. We graph one cycle of r = 6 cos(θ) on the θr-plane3 and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π 2 , r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the ﬁgure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. r 6 3 −3 −6 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x Next, we repeat the process as θ ranges from π 2 to π. Here, the r values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. r 6 3 −3 −6 y θ runs from π 2 to π π 2 π 3π 2 2π θ x r < 0 so we plot here As θ ranges from π to 3π 2 , the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the \|r\| for these values of θ match the r values for θ in 3The graph looks exactly like y = 6 cos(x) in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r, θ) on the xy-plane. 942 Applications of Trigonometry , we have that the curve begins to retrace itself at this point. Proceeding further, we ﬁnd 2 ≤ θ ≤ 2π, we retrace the portion of the curve in Quadrant IV that we ﬁrst traced 2 ≤ θ ≤ π. The reader is invited to verify that plotting any range of θ outside the interval 0, π 2 that when 3π out as π [0, π] results in retracting some portion of the curve.4 We present the ﬁnal graph below. r 6 3 −3 −6 π 2 π θ y 3 −3 3 x 6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r2 = 16 cos(2θ) Solution. 1. We ﬁrst plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. To help us visualize what is going on graphically, we divide up [0, 2π] into the usual four subintervals 0, π 2 , π, 2 π, 3π 2 , r decreases from 2 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. 2 , 2π, and proceed as we did above. As θ ranges from 0 to π and 3π , π r 6 4 2 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x 4The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of Polar Equations 943 Next, as θ runs from π oﬀ, we gradually pull the graph away from the origin until we reach the negative x-axis. 2 to π, we see that r increases from 2 to 4. Picking up where we left θ runs from π 2 to 3π 2 2π θ , we see that r increases from 4 to 6. On the xy-plane, the curve Over the interval π, 3π 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis 3π 2 2π θ θ runs from π to 3π 2 Finally, as θ takes on values from 3π xy-plane pulls in from the negative y-axis to ﬁnish where we started. 2 to 2π, r decreases from 6 back to 4. The graph on the r 6 4 2 y x π 2 π 3π 2 2π θ θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions of the curve, so we are ﬁnished. 944 Applications of Trigonometry sin(θ) in the θr-plane 3π 2 2π y 2 −4 x 4 −6 r = 4 − 2 sin(θ) in the xy-plane. 2. The ﬁrst thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0, 2π] is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our ﬁrst task is to determine when this happens. Setting r = 0 we get 2 + 4 cos(θ) = 0, or cos(θ) = − 1 2 . Solving for θ in [0, 2π] gives θ = 2π 3 . Since these values of θ are important geometrically, we break the interval [0, 2π] into six subintervals: 0, π 2 , 2π. As 3 , π, π, 4π 2 θ ranges from 0 to π 2 , r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. 3 and θ = 4π and 3π , 2π , 4π 3 , 3π , π 2 , 2π 3 3 2 r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x On the interval π will eventually cross through) the origin. Not only do we reach the origin when θ = 2π theorem from Calculus5 states that the curve hugs the line θ = 2π , r decreases from 2 to 0, which means the graph is heading into (and 3 , a 3 as it approaches the origin. 2 , 2π 3 5The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 945 r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x 3 , π, r ranges from 0 to −2. Since r ≤ 0, the curve passes through the On the interval 2π origin in the xy-plane, following the line θ = 2π 3 and continues upwards through Quadrant IV towards the positive x-axis.6 Since \|r\| is increasing from 0 to 2, the curve pulls away from the origin to ﬁnish at a point on the positive x-axis. r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x Next, as θ progresses from π to 4π graph in the ﬁrst quadrant, heading into the origin along the line θ = 4π 3 . 3 , r ranges from −2 to 0. Since r ≤ 0, we continue our 6Recall that one way to visualize plotting polar coordinates (r, θ) with r < 0 is to start the rotation from the left 3 and π radians from the negative x-axis in side of the pole - in this case, the negative x-axis. Rotating between 2π this case determines the region between the line θ = 2π 3 and the x-axis in Quadrant IV. 946 Applications of Trigonometry r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x On the interval 4π line θ = 4π 3 , 3π 2 , r returns to positive values and increases from 0 to 2. We hug the 3 as we move through the origin and head towards the negative y-axis. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x As we round out the interval, we ﬁnd that as θ runs through 3π to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. 2 to 2π, r increases from 2 out y r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ x θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 947 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our ﬁnal graph is below. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 2 θ = 4π 3 −2 2 6 x r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As usual, we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane, which in this case, occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely 0, π , π 4 , r 4 increases from 0 to 5. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4 . 4 , π. As θ ranges from 0 to π and 3π , π 2 , 3π 3π 4 π θ x r 5 −5 Next, we see that r decreases from 5 to 0 as θ runs through π heading negative as θ crosses π as the curve heads to the origin. 2 . Hence, we draw the curve hugging the line θ = π , and furthermore, r is 2 (the y-axis) 4 , π 2 948 Applications of Trigonometry y π 4 π 2 3π 4 π θ x r 5 −5 As θ runs from π pulls away from the negative y-axis into Quadrant IV. 2 to 3π 4 , r becomes negative and ranges from 0 to −5. Since r ≤ 0, the curve y π 4 π 2 3π 4 π θ x r 5 −5 For 3π 4 ≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. y π 4 π 2 3π 4 π θ x r 5 −5 11.5 Graphs of Polar Equations 949 Even though we have ﬁnished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we ﬁnd that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the ﬁrst. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. r 5 π 1 5π 4 2 3π 2 3 7π 4 4 2π θ 4 1 −5 We have the ﬁnal graph below 3π 4 π 5π 4 3π 2 7π 4 2π θ −5 x 5 −5 −5 r = 5 sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. Graphing r2 = 16 cos(2θ) is complicated by the r2, so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). How do we sketch such a curve? First oﬀ, we sketch a fundamental period cos(2θ) is of r = cos(2θ) which we have dotted in the ﬁgure below. When cos(2θ) < 0, . On the intervals which remain, undeﬁned, so we don’t have any values on the interval π cos(2θ) ranges from 0 to 1 as well.7 From cos(2θ) ranges from 0 to 1, inclusive. Hence, cos(2θ) ranges continuously from 0 to ±4, respectively. Below we this, we know r = ±4 cos(2θ) on the θr plane and use them to sketch the graph both r = 4 corresponding pieces of the curve r2 = 16 cos(2θ) in the xy-plane. As we have seen in earlier cos(2θ) and r = −4 4 , 3π 4 7Owing to the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is de
ﬁned. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4 , which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ), θ = 3π 4 π 4 π 2 3π 4 4 3 π θ r = 4 cos(2θ) and r = −4cos(2θ) As we plot points corresponding to values of θ outside of the interval [0, π], we ﬁnd ourselves retracing parts of the curve,8 so our ﬁnal answer is below. r 4 θ = 3π 3π 4 π θ −4 x 4 −4 r = ±4 cos(2θ) in the θr-plane −4 r2 = 16 cos(2θ) in the xy-plane A few remarks are in order. First, there is no relation, in general, between the period of the function f (θ) and the length of the interval required to sketch the complete graph of r = f (θ) in the xyplane. As we saw on page 941, despite the fact that the period of f (θ) = 6 cos(θ) is 2π, we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 11.5.2, number 3, the period of f (θ) = 5 sin(2θ) is π, but in order to obtain the complete graph of r = 5 sin(2θ), we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8In this case, we could have generated the entire graph by using just the plot r = 4cos(2θ), but graphed over the interval [0, 2π] in the θr-plane. We leave the details to the reader. 9Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 951 coordinate system, but also prepare you for what is needed in Calculus. Second, the symmetry seen in the examples is also a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. In our next example, we are given the task of ﬁnding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 938, in order for a point P to be on the graph of a polar equation, it must have a representation P (r, θ) which satisﬁes the equation. What complicates matters in polar coordinates is that any given point has inﬁnitely many representations. As a result, if a point P is on the graph of two diﬀerent polar equations, it is entirely possible that the representation P (r, θ) which satisﬁes one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to ﬁnd our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 − 2 sin(θ) 2. r = 2 and r = 3 cos(θ) 3. r = 3 and r = 6 cos(2θ) Solution. 4. r = 3 sin θ 2 and r = 3 cos θ 2 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and ﬁnd it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 − 2 sin(θ) is a special kind of lima¸con called a ‘cardioid.’10 y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the ﬁrst quadrant, one in the second quadrant, and the origin. Our next task is to ﬁnd polar representations of these points. In 10Presumably, the name is derived from its resemblance to a stylized human heart. 952 Applications of Trigonometry 2 . From this, we get θ = π 6 into r = 2 sin(θ), we get r = 2 sin π order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P (r, θ) which satisﬁes r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly diﬀerent) representation P (r, θ) which satisﬁes r = 2 sin(θ). We ﬁrst try to see if we can ﬁnd any points which have a single representation P (r, θ) that satisﬁes both r = 2 sin(θ) and r = 2 − 2 sin(θ). Assuming such a pair (r, θ) exists, then equating11 the expressions for r gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 1 6 + 2πk = 1, which for integers k. Plugging θ = π is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). Hence, 1, π is one 6 representation for the point of intersection in the ﬁrst quadrant. For the point of intersection , so this is in the second quadrant, we try θ = 5π our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π, and as the reader can verify, we are at the origin exactly when θ = πk for integers k. On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determined the three points of intersection to be 1, π 6 6 . Both equations give us the point 1, 5π 2 , and more generally, when θ = π 6 + 2πk or θ = 5π = 2 1 2 and the origin. , 1, 5π 6 6 6 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a circle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ) is also a circle - but this one is centered at the point with rectangular coordinates 3 2 , 0 and has a radius of 3 2 . y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ) We have two intersection points to ﬁnd, one in Quadrant I and one in Quadrant IV. Proceeding as above, we ﬁrst determine if any of the intersection points P have a representation (r, θ) which satisﬁes both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 3 . To solve this equation, we need the arccosine function. We get 11We are really using the technique of substitution to solve the system of equations r = 2 sin(θ) r = 2 − 2 sin(θ) 11.5 Graphs of Polar Equations 953 θ = arccos 2 + 2πk or θ = 2π − arccos 2 3 3 2, arccos 2 as one representation for our answer in Quadrant I, and 2, 2π − arccos 2 3 3 as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraically and geometrically. + 2πk for integers k. From these solutions, we get 3. Proceeding as above, we ﬁrst graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12 y 6 3 −6 −3 3 x 6 −3 −6 r = 3 and r = 6 cos(2θ) , 3, 5π 6 , 3, 7π 6 and 3, 11π 6 6 + πk or θ = 5π 2 . Solving, we get θ = π It appears as if there are eight points of intersection - two in each quadrant. We ﬁrst look to see if there any points P (r, θ) with a representation that satisﬁes both r = 3 and r = 6 cos(2θ). For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 6 + πk for integers k. Out of all of these solutions, we obtain just four distinct points represented by 3, π . To determine the coordinates of the remaining four 6 points, we have to consider how the representations of the points of intersection can diﬀer. We know from Section 11.4 that if (r, θ) and (r, θ) represent the same point and r = 0, then either r = r or r = −r. If r = r, then θ = θ+2πk, so one possibility is that an intersection point P has a representation (r, θ) which satisﬁes r = 3 and another representation (r, θ+2πk) for some integer, k which satisﬁes r = 6 cos(2θ). At this point,13 if we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if, by equating the resulting expressions for r, we get any more solutions for θ. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k, however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r, we have that θ = θ + (2k + 1)π for integers k. We look to see if we can ﬁnd points P which have a representation (r, θ) that satisﬁes r = 3 and another, 12See Example 11.5.2 number 3. 13The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of eﬃciency. 954 Applications of Trigonometry (−r, θ + (2k + 1)π), that satisﬁes r = 6 cos(2θ). To do this, we substitute14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). Since cos(2(θ + (2k + 1)π)) = cos(2θ + (2k + 1)(2π)) = cos(2θ) for all integers k, the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 1 3 + πk. From these solutions, we obtain15 the remaining four intersection points with representations −3, π 3 , which we can readily check graphically. 3 + πk or θ = 2π 2 . We get θ = π and −3, 5π 3 , −3, 2π 3 , −3, 4π 3 . Using the techniques 4. As usual, we begin by graphing r = 3 sin θ 2 presented in Example 11.5.2, we ﬁnd that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! and r = 3 cos θ 2 y 3 −3 3 x −3 r = 3 sin θ 2 and r = 3 cos θ 2 appear to determine the same curve in the xy-plane and r = 3 cos θ 2 To verify this incredible claim,16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (
r, θ) which satisﬁes both r = 3 sin θ . Equating these two expressions for r gives 2 . While normally we discourage dividing by a variable = 3 cos θ the equation 3 sin θ 2 2 expression (in case it could be 0), we note here that if 3 cos θ = 0, then for our equation 2 = 0 as well. Since no angles have both cosine and sine equal to zero, to hold, 3 sin θ 2 to get we are safe to divide both sides of the equation 3 sin θ 2 tan θ 2 + 2πk for integers k. From these solutions, however, we 2 = 1 which gives θ = π by 3 cos θ 2 = 3 cos θ 2 14Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. 15We obtain these representations by substituting the values for θ into r = 6 cos(2θ), once again, for illustration purposes. Again, in the interests of eﬃciency, we could ‘plug’ these values for θ into r = 3 (where there is no θ) and represents the same point get the list of points: 3, π as −3, π , we still get the same set of solutions. . While it is not true that 3, π and 3, 5π , 3, 4π , 3, 2π 3 3 3 3 3 16A quick sketch of r = 3 sin θ and r = 3 cos θ in the θr-plane will convince you that, viewed as functions of r, 3 2 2 these are two diﬀerent animals. 11.5 Graphs of Polar Equations 955 √ 3 2 2 , π 2 2 + πk = 3 cos θ 2 [θ + 2πk] = 3 cos θ get only one intersection point which can be represented by . We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation (r, θ) which satisﬁes r = 3 sin θ and the same point P has a diﬀerent 2 . Substituting representation (r, θ + 2πk) for some integer k which satisﬁes r = 3 cos θ 2 into the latter, we get r = 3 cos 1 2 + πk. Using the sum formula for , since sin (πk) = ±3 cos θ cosine, we expand 3 cos θ 2 sin(πk) = 0 for all integers k, and cos (πk) = ±1 for all integers k. If k is an even integer, we get the same equation r = 3 cos θ as before. If k is odd, we get r = −3 cos θ . This 2 2 = −1. Solving, latter expression for r leads to the equation 3 sin θ 2 √ we get θ = − π 2 , − π . Next, we assume P has a representation (r, θ) which satisﬁes r = 3 sin θ and a representation 2 for some integer k. Substituting (−r) for (−r, θ + (2k + 1)π) which satisﬁes r = 3 cos θ 2 r and (θ + (2k + 1)π) in for θ into r = 3 cos θ 2 [θ + (2k + 1)π]. Once gives −r = 3 cos 1 2 again, we use the sum formula for cosine to get 2 + 2πk for integers k, which gives the intersection point cos(πk) − 3 sin θ 2 , or tan θ 2 3 = −3 cos θ 2 2 2 2 cos 1 2 [θ + (2k + 1)π] = cos 2 θ 2 + (2k+1)π cos (2k+1)π 2 = cos θ 2 = ± sin θ 2 − sin θ 2 sin (2k+1)π 2 = 0 and sin (2k+1)π (2k+1)π where the last equality is true since cos 2 2 2 [θ + (2k + 1)π] can be rewritten as r = ±3 sin θ Hence, −r = 3 cos 1 (2k+1)π = 1, and the equation −r = 3 cos 1 = sin π then sin 2 2 reduces to −r = −3 sin θ , or r = 3 sin θ 2 2 What this means is that if a polar representation (r, θ) for the point P satisﬁes r = 3 sin( θ then the representation (−r, θ + π) for P automatically satisﬁes r = 3 cos θ 2 equations r = 3 sin( θ = ±1 for integers k. . If we choose k = 0, 2 [θ + (2k + 1)π] in this case which is the other equation under consideration! 2 ), . Hence the 2 ) determine the same set of points in the plane. 2 ) and r = 3 cos( θ 2 Our work in Example 11.5.3 justiﬁes the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To ﬁnd the points of intersection of the graphs of two polar equations E1 and E2: Sketch the graphs of E1 and E2. Check to see if the curves intersect at the origin (pole). Solve for pairs (r, θ) which satisfy both E1 and E2. Substitute (θ + 2πk) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 956 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. Example 11.5.4. Sketch the region in the xy-plane described by the following sets. 1. (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 3. (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 4. (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 3 ≤ θ ≤ 4π 3 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π 6 ≤ θ ≤ π 2 Solution. Our ﬁrst step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π 2 , we are tracing out the ‘leaf’ of the rose which lies in the ﬁrst quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0, π 2 . Hence, the region we seek is the leaf itself. r 5 −5 π 4 π 2 3π 4 π θ y x (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π 6 , so the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π 6 . In other words, we are looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below and r = 6 cos(2θ) (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 11.5 Graphs of Polar Equations 957 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given values 2π 3 . Since the values r takes on in this interval are non-positive, the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense, and we are looking for all of the points between the pole r = 0 and the lima¸con as θ ranges over the interval 2π r . In other words, we shade in the inner loop of the lima¸con. 3 to 4π 3 , 4π 3 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 θ = 4π 3 x (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 4. We have two regions described here connected with the union symbol ‘∪.’ We shade each in turn and ﬁnd our ﬁnal answer by combining the two. In Example 11.5.3, number 1, we found that the curves r = 2 sin(θ) and r = 2 − 2 sin(θ) intersect when θ = π 6 . Hence, for the , we are shading the region between the origin ﬁrst region, (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π 6 , which is the angle of intersection of the two curves. For the second region, (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π , θ picks up 6 ≤ θ ≤ π where it left oﬀ at π 2 . In this case, however, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2 . Putting these two regions together gives us our ﬁnal answer. 6 and continues to sin(θ) and r = 2 sin(θ) (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π ∪ 6 ≤ θ ≤ π 2 958 Applications of Trigonometry 11.5.1 Exercises In Exercises 1 - 20, plot the graph of the polar equation by hand. Carefully label your graphs. 1. Circle: r = 6 sin(θ) 2. Circle: r = 2 cos(θ) 3. Rose: r = 2 sin(2θ) 4. Rose: r = 4 cos(2θ) 5. Rose: r = 5 sin(3θ) 6. Rose: r = cos(5θ) 7. Rose: r = sin(4θ) 8. Rose: r = 3 cos(4θ) 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ) 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) 16. Lima¸con: r = 3 − 5 cos(θ) 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) In Exercises 21 - 30, ﬁnd the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 21. r = 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 − cos(θ) 23. r = 1 − 2 sin(θ) and r = 2 24. r = 1 − 2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 27. r2 = 4 cos(2θ) and r = 2 √ 3 sin(θ) √ 26. r = 3 cos(θ) and r = sin(θ) 28. r2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In Exercises 31 - 40, sketch the region in the xy-plane described by the given set. 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π} 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 11.5 Graphs of Polar Equations 959 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 4 3 ≤ θ ≤ π 3 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 38. (r, θ) \| 1 ≤ r ≤ 2 sin(2θ), 13π √ 12 ≤ θ ≤ 17π 12 39. (r, θ) \| 0 ≤ r ≤ 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 3 sin(θ), 0 ≤ θ ≤ π 6 ∪ (r, θ) \| 0 ≤ r ≤ 1, π 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the left half of the circle r = 6 sin(θ). 44. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. 45. The region inside the top half of the cardioid r = 3 − 3 cos(θ) 46. The region inside the cardioid r = 2 − 2 sin(θ) which lies in Quadrants I and IV. 47. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. The region inside the circle r = 5 but outside the circle r = 3. 49. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to you
r understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difﬁcult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The ﬁrst thing you must do is switch the MODE of your calculator to POL, which stands for “polar”. 960 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu, we ﬁnd three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more about the θ-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [θmin, θmax] using the θstep. For most graphs, a θstep of 0.1 is ﬁne. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this. You will need to experiment with the settings in order to get a nice graph. Exercises 51 - 60 give you some curves to graph using your calculator. Notice that some of them have explicit bounds on θ and others do not. 11.5 Graphs of Polar Equations 961 51. r = θ, 0 ≤ θ ≤ 12π 52. r = ln(θ), 1 ≤ θ ≤ 12π 53. r = e.1θ, 0 ≤ θ ≤ 12π 54. r = θ3 − θ, −1.2 ≤ θ ≤ 1.2 55. r = sin(5θ) − 3 cos(θ) 57. r = arctan(θ), −π ≤ θ ≤ π 59. r = 1 2 − cos(θ) 56. r = sin3 θ 2 + cos2 θ 3 58. r = 1 1 − cos(θ) 60. r = 1 2 − 3 cos(θ) 61. How many petals does the polar rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section, it’s clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26. In Exercises 62 - 64, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves. 62. Show that if f is even17 then the graph of r = f (θ) is symmetric about the x-axis. (a) Show that f (θ) = 2 + 4 cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis. (See Example 11.5.2 number 2.) (b) Show that f (θ) = 3 sin θ 2 is not even, yet the graph of r = 3 sin θ 2 is symmetric about the x-axis. (See Example 11.5.3 number 4.) 63. Show that if f is odd18 then the graph of r = f (θ) is symmetric about the origin. (a) Show that f (θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 sin(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f (θ) = 3 cos θ 2 is not odd, yet the graph of r = 3 cos θ 2 is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric about the y-axis. (a) For f (θ) = 4 − 2 sin(θ), show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17Recall that this means f (−θ) = f (θ) for θ in the domain of f . 18Recall that this means f (−θ) = −f (θ) for θ in the domain of f . 962 Applications of Trigonometry (b) For f (θ) = 5 sin(2θ), show that f π − π 4 = f π 4 , yet the graph of r = 5 sin(2θ) is symmetric about the y-axis. (See Example 11.5.2 number 3.) In Section 1.7, we discussed transformations of graphs. classmates explore transformations of polar graphs. In Exercise 65 we have you and your 65. For Exercises 65a and 65b below, let f (θ) = cos(θ) and g(θ) = 2 − sin(θ). (a) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of . Repeat this process for g(θ). In general, how do you think the graph of r = f (θ + α) compares with the graph of r = f (θ)? and r = f θ − 3π 4 , r = f θ + 3π 4 (b) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of r = 2f (θ), r = 1 2 f (θ), r = −f (θ) and r = −3f (θ). Repeat this process for g(θ). In general, how do you think the graph of r = k · f (θ) compares with the graph of r = f (θ)? (Does it matter if k > 0 or k < 0?) 66. In light of Exercises 62 - 64, how would the graph of r = f (−θ) compare with the graph of r = f (θ) for a generic function f ? What about the graphs of r = −f (θ) and r = f (θ)? What about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research. 11.5 Graphs of Polar Equations 963 11.5.2 Answers 1. Circle: r = 6 sin(θ) y −6 6 −6 3. Rose: r = 2 sin(2θ) y −2 2 −2 2. Circle: r = 2 cos(θ) y 2 6 x −2 2 x −2 4. Rose: r = 4 cos(2θ) y 4 θ = 3π 4 θ = π 4 2 x −4 4 x 5. Rose: r = 5 sin(3θ) y 5 θ = 2π 3 θ = π 3 −5 5 x −4 6. Rose: r = cos(5θ) y 1 θ = 7π 10 θ = 3π 10 θ = 9π 10 −1 θ = π 10 1 x −5 −1 964 Applications of Trigonometry 7. Rose: r = sin(4θ) y 1 θ = 3π 4 θ = π 4 −1 1 x 8. Rose: r = 3 cos(4θ) y θ = 5π 8 3 θ = 3π 8 θ = 7π 8 −3 θ = π 8 3 x −1 −3 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) y 6 3 y 10 5 −6 −3 3 6 x −10 −5 5 10 x −3 −6 −5 −10 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) y 4 2 y 2 1 −4 −2 2 4 x −2 −1 1 2 x −2 −4 −1 −2 11.5 Graphs of Polar Equations 965 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ = 5π 6 −3 −1 1 3 x −3 −1 1 3 x −1 −3 θ = 5π 3 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) y √ 2 3 + 4 θ = 5π 6 √ 2 3 √ −2 3 − 4 θ = 7π 6 √ −2 3 √ −1 −3 16. Lima¸con: r = 3 − 5 cos(θ) y 8 3 θ = arccos 3 5 −8 −2 8 x −3 −8 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) y 8 θ = π − arcsin 3 5 θ = arcsin 3 5 y 9 5 −8 −3 3 −2 −8 8 x −9 θ = π + arcsin −2 2 7 2 9 θ = 2π − arcsin 2 7 x −9 966 Applications of Trigonometry 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) y 1 y 2 θ = 3π 4 θ = π 4 −1 1 x −2 2 x −1 −2 3 2 , π 3 , 3 2 , 5π 3 , pole √ 2 + 2 , √ 2 − 2 2 , 7π 4 2 , 3π 4 , pole 21. r = 3 cos(θ) and r = 1 + cos(θ) y 3 2 1 −3 −2 −1 1 2 3 −1 −2 −3 x 22. r = 1 + sin(θ) and r = 1 − cos(θ) y 2 1 −2 −1 1 2 x −1 −2 11.5 Graphs of Polar Equations 967 2, 7π 6 , 2, 11π 6 1, , π 2 1, 3π 2 , (−1, 0) √ 3, π 6 , pole 23. r = 1 − 2 sin(θ) and r = 2 y 3 1 −3 −1 1 3 x −1 −3 24. r = 1 − 2 cos(θ) and r = 1 y 3 1 −3 −1 1 3 x −1 −3 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) y 4 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 968 Applications of Trigonometry √ 3 10 10 , arctan(3) , pole √ 2, , π 6 √ , 5π 6 2, √ , 7π 6 2, √ 2, 11π 6 1, , π 12 1, , 5π 12 1, 13π 12 , 1, 17π 12 26. r = 3 cos(θ) and r = sin(θ) y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 27. r2 = 4 cos(2θ) and r = √ 2 y 2 −2 2 x −2 28. r2 = 2 sin(2θ) and r = 1 y √ 2 1 −1 11.5 Graphs of Polar Equations 969 29. r = 4 cos(2θ) and r = 2 y 4 2, 2, , π 6 11π 6 , 7π 6 2, 2, 5π 6 , −2, , π 3 , −2, , 2π 3 −2, 4π 3 , −2, 5π 3 −4 4 x −4 30. r = 2 sin(2θ) and r = 1 y 2 −2 2 x −2 , π 12 , 17π 12 1, 1, 1, −1, , 19π 12 , 5π 12 1, , 7π 12 −1, −1, 23π 12 , 13π 12 −1, , 11π 12 970 Applications of Trigonometry 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π3 −2 −1 1 2 3 −1 −2 −3 x x −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 3 −2 −1 1 2 3 −1 −2 −3 x −2 2 x −2 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 y 4 y 3 1 −4 4 x −3 −1 1 3 x −4 −1 −3 11.5 Graphs of Polar Equations 971 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − 3 −2 −1 1 2 3 −1 −2 −3 x 38. (r, θ) \| 1 ≤ r ≤ y 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 √ − 2 −1 √ 2 1 −r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 39. (r, θ) \| 0 ≤ r ≤ 2 3 sin(θ), 3 −2 −1 1 2 3 x −1 −2 −3 −4 972 Applications of Trigonometry ∪ (r, θ) \| 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 y −2 2 −2 2 x 41. {(r, θ) \| 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 42. (r, θ) \| 0 ≤ r ≤ 5, π ≤ θ ≤ 3π 2 43. (r, θ) \| 0 ≤ r ≤ 6 sin(θ), π 44. (r, θ) \| 4 cos(θ) ≤ r ≤ 0 45. {(r, θ) \| 0 ≤ r ≤ 3 − 3 cos(θ), 0 ≤ θ ≤ π} 46. (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 0 ≤ θ ≤ π 2 or (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 47. (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 or (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), − ≤ 5π 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 2 ≤ θ ≤ 2π ∪ (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π 48. {(r, θ) \| 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 49. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 50. (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ θ ≤ π (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 5π 2 2 ≤ θ ≤ 0 ∪ {(r, θ) \| sin(θ) ≤ r ≤ 3 cos(θ), 0 ≤ θ ≤ arctan(3)} ∪ (r, θ) \| 0 ≤ r ≤ 3, π 12 ≤ θ ≤ 5π 12 ∪ 11.6 Hooked on Conics Again 973 11.6 Hooked on Conics Again In this sect
ion, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our ﬁrst task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More speciﬁcally, it is the hyperbola obtained by rotating the graph of x2 − y2 = 4 counter-clockwise through a 45◦ angle. Armed with polar coordinates, we can generalize the process of rotating axes as shown below. 11.6.1 Rotation of Axes Consider the x- and y-axes below along with the dashed x- and y-axes obtained by rotating the xand y-axes counter-clockwise through an angle θ and consider the point P (x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to ﬁnd rectangular coordinates based on the x- and y-axes. That is, we wish to determine P (x, y). While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive x-axis and whose terminal side contains the point P . y y P (x, y) = P (x, y) x θ φ θ x We relate P (x, y) and P (x, y) by converting them to polar coordinates. Converting P (x, y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). To convert the point P (x, y) into polar coordinates, we ﬁrst match the polar axis with the positive x-axis, choose the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). Using the sum formulas for sine and cosine, we have x = r cos(θ + φ) = r cos(θ) cos(φ) − r sin(θ) sin(φ) = (r cos(φ)) cos(θ) − (r sin(φ)) sin(θ) = x cos(θ) − y sin(θ) Sum formula for cosine Since x = r cos(φ) and y = r sin(φ) 974 Applications of Trigonometry Similarly, using the sum formula for sine we get y = x sin(θ) + y cos(θ). These equations enable us to easily convert points with xy-coordinates back into xy-coordinates. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y.1 If we want equations which enable us to convert points with xy-coordinates into xy-coordinates, we need to solve the system x cos(θ) − y sin(θ) = x x sin(θ) + y cos(θ) = y for x and y. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = cos(θ) − sin(θ) cos(θ) sin(θ Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2(θ) + sin2(θ) = 1, the determinant of A is not zero so A is invertible and X = A−1X. Using the formula given in Equation 8.2 with det(A) = 1, we ﬁnd so that A−1 = cos(θ) − sin(θ) sin(θ) cos(θ) X = A−1X x y x y = = cos(θ) − sin(θ) sin(θ) cos(θ) x y x cos(θ) + y sin(θ) −x sin(θ) + y cos(θ) From which we get x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ). To summarize, Theorem 11.9. Rotation of Axes: Suppose the positive x and y axes are rotated counterclockwise through an angle θ to produce the axes x and y, respectively. Then the coordinates P (x, y) and P (x, y) are related by the following systems of equations x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) and x = x cos(θ) + y sin(θ) y = −x sin(θ) + y cos(θ) We put the formulas in Theorem 11.9 to good use in the following example. 1Sound familiar? In Section 11.4, the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates, and they make it easy to convert equations from rectangular coordinates into polar coordinates. 2We could, of course, interchange the roles of x and x, y and y and replace φ with −φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. Let P (x, y) = (2, −4) and ﬁnd P (x, y). Check your answer algebraically and graphically. 2. Convert the equation 21x2 + 10xy √ 3 + 31y2 = 144 to an equation in x and y and graph. Solution. √ 3 , Theorem 11.9 gives x = x cos(θ) + y sin(θ) = 2 cos π √ 1. If P (x, y) = (2, −4) then x = 2 and y = −4. Using these values for x and y along with which simpliﬁes 3. Similarly, y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which √ 3 3 3. To check our answer 3. Hence P (x, y to x = 1 − 2 √ gives y = − algebraically, we use the formulas in Theorem 11.9 to convert P (x, y) = 1 − 2 back into x and y coordinates. We get + (−4) sin π 3 3 − 2 = −2 − 3, −2 − 3, −2 − √ √ 3 x = x cos(θ) − y sin(θ) √ − (−2 − 3) cos ) sin π 3 = ( Similarly, using y = x sin(θ) + y cos(θ), we obtain y = −4 as required. To check our answer graphically, we sketch in the x-axis and y-axis to see if the new coordinates P (x, y) = 3 ≈ (−2.46, −3.73) seem reasonable. Our graph is below. 1 − 2 3, −x, y) = (2, −4) P (x, y) ≈ (−2.46, −3.73) 2. To convert the equation 21x2 +10xy we substitute x = x cos π 3 − y sin π 3 √ 3+31y2 = 144 to an equation in the variables x and y, = x 2 + y and y = x sin π 3 + y cos 976 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start by verifying that x2 = (x)2 4 − √ xy 2 3 + 3(y)2 4 , √ (x)2 4 3 − xy 2 − xy = √ 3 , y2 = 3(x)2 4 + √ xy 2 3 + (y)2 4 To our surprise and delight, the equation 21x2 + 10xy 3 + 31y2 = 144 in xy-coordinates 4 + (y)2 reduces to 36(x)2 + 16(y)2 = 144, or (x)2 9 = 1 in xy-coordinates. The latter is an ellipse centered at (0, 0) with vertices along the y-axis with (xy-coordinates) (0, ±3) and whose minor axis has endpoints with (xy-coordinates) (±2, 0). We graph it below. (y) 21x2 + 10xy √ 3 + 31y2 = 144 √ The elimination of the troublesome ‘xy’ term from the equation 21x2 + 10xy 3 + 31y2 = 144 in Example 11.6.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7. It is natural to wonder if we can always do this. That is, given an equation of the form Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, with B = 0, is there an angle θ so that if we rotate the x and yaxes counter-clockwise through that angle θ, the equation in the rotated variables x and y contains no xy term? To explore this conjecture, we make the usual substitutions x = x cos(θ) − y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 and set the coeﬃcient of the xy term equal to 0. Terms containing xy in this expression will come from the ﬁrst three terms of the equation: Ax2, Bxy and Cy2. We leave it to the reader to verify that x2 = (x)2 cos2(θ) − 2xy cos(θ) sin(θ) + (y)2 sin(θ) xy = (x)2 cos(θ) sin(θ) + xy cos2(θ) − sin2(θ) − (y)2 cos(θ) sin(θ) y2 = (x)2 sin2(θ) + 2xy cos(θ) sin(θ) + (y)2 cos2(θ) 11.6 Hooked on Conics Again 977 The contribution to the xy-term from Ax2 is −2A cos(θ) sin(θ), from Bxy it is B cos2(θ) − sin2(θ), and from Cy2 it is 2C cos(θ) sin(θ). Equating the xy-term to 0, we get −2A cos(θ) sin(θ) + B cos2(θ) − sin2(θ) + 2C cos(θ) sin(θ) = 0 −A sin(2θ) + B cos(2θ) + C sin(2θ) = 0 Double Angle Identities From this, we get B cos(2θ) = (A − C) sin(2θ), and our goal is to solve for θ in terms of the coeﬃcients A, B and C. Since we are assuming B = 0, we can divide both sides of this equation by B. To solve for θ we would like to divide both sides of the equation by sin(2θ), provided of course that we have assurances that sin(2θ) = 0. If sin(2θ) = 0, then we would have B cos(2θ) = 0, and since B = 0, this would force cos(2θ) = 0. Since no angle θ can have both sin(2θ) = 0 and cos(2θ) = 0, we can safely assume3 sin(2θ) = 0. We get cos(2θ) B . We have just proved the following theorem. B , or cot(2θ) = A−C sin(2θ) = A−C Theorem 11.10. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any xy terms by rotating the xand y- axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C B . We put Theorem 11.10 to good use in the following example. Example 11.6.2. Graph the following equations. 1. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 2. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. 1. Since the equation 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 is already given to us in the form required by Theorem 11.10, we identify A = 5, B = 26 and C = 5 so that cot(2θ) = A−C 26 = 0. This means cot(2θ) = 0 which gives θ = π 2 k for integers k. √ √ 2 + y 2 2 We choose θ = π 2 . The reader should verify that 4 so that our rotation equations are x = x 4 + π and y = x B = 5−5 2 − y 2 + 16y √ √ 2 2 2 √ √ x2 = (x)2 2 − xy + (y)2 2 , xy = (x)2 2 − (y)2 2 , y2 = (x)2 2 + xy + (y)2 2 Making the other substitutions, we get that 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 2 + 16y reduces to 18(x)2 − 8(y)2 + 32y − 104 = 0, or (x)2 9 = 1. The latter is the equation of a hyperbola centered at the xy-coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in xy-coordinates) and asymptotes y = ± 3 4 − (y−2)2 2 x + 2. We graph it below. √ √ 3The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 978 Applications of Trigonometry 24 2. From 16x2 + 24xy + 9y2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that cot(2θ) = 7 24 . Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to ﬁnd cos(θ) and sin(θ), which means we have two If we use the arccotangent function immediately, after the usual calculations we options. . To get cos(θ) and sin(θ) from this, we would need to use half angle 2 arccot 7 get θ = 1 identities. Alternatively, we can start with cot(2θ) = 7 24 , use a double angle identity, and then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ)
= 7 24 , we have tan(2θ) = 24 7 , which 7 . Using the double angle identity for tangent, we have gives 24 tan2(θ) + 14 tan(θ) − 24 = 0. Factoring, we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which gives tan(θ) = − 4 4 . While either of these values of tan(θ) satisﬁes the equation . To 4 , since this produces an acute angle,4 θ = arctan 3 cot(2θ) = 7 . ﬁnd the rotation equations, we need cos(θ) = cos arctan 3 4 Using the techniques developed in Section 10.6 we get cos(θ) = 4 5 . Our rotation 5 + 4y 5 − 3y equations are x = x cos(θ) − y sin(θ) = 4x 5 . As usual, we now substitute these quantities into 16x2 + 24xy + 9y2 + 15x − 20y = 0 and simplify. As a ﬁrst step, the reader can verify 3 or tan(θ) = 3 24 , we choose tan(θ) = 3 5 and y = x sin(θ) + y cos(θ) = 3x and sin(θ) = sin arctan 3 4 5 and sin(θ) = 3 1−tan2(θ) = 24 2 tan(θ) 4 x2 = 16(x)2 25 − 24xy 25 + 9(y)2 25 , xy = 12(x)2 25 + 7xy 25 − 12(y)2 25 , y2 = 9(x)2 25 + 24xy 25 + 16(y)2 25 Once the dust settles, we get 25(x)2 − 25y = 0, or y = (x)2, whose graph is a parabola opening along the positive y-axis with vertex (0, 0). We graph this equation below = arctan 3 4 x 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 16x2 + 24xy + 9y2 + 15x − 20y = 0 4As usual, there are inﬁnitely many solutions to tan(θ) = 3 . The 4 reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C B and what this means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 4 . We choose the acute angle θ = arctan 3 3 in our ﬁnal graph. (Hint: 3 = −1.) − 4 3 4 11.6 Hooked on Conics Again 979 We note that even though the coeﬃcients of x2 and y2 were both positive numbers in parts 1 and 2 of Example 11.6.2, the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola. Whereas in Chapter 7, we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coeﬃcients, Example 11.6.2 demonstrates that all bets are oﬀ when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coeﬃcients of the quadratic terms. We have the following theorem. Theorem 11.11. Suppose the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 describes a non-degenerate conic section.a If B2 − 4AC > 0 then the graph of the equation is a hyperbola. If B2 − 4AC = 0 then the graph of the equation is a parabola. If B2 − 4AC < 0 then the graph of the equation is an ellipse or circle. aRecall that this means its graph is either a circle, parabola, ellipse or hyperbola. See page 497. As you may expect, the quantity B2 −4AC mentioned in Theorem 11.11 is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’, we will at least work through the proof of Theorem 11.11 mechanically to show that it is true.5 First note that if the coeﬃcient B = 0 in the equation Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, Theorem 11.11 reduces to the result presented in Exercise 34 in Section 7.5, so we proceed here under the assumption that B = 0. We rotate the xy-axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C to produce an equation with no xy-term in accordance with B Theorem 11.10: A(x)2 + C(y)2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 34 in Section 7.5 once more using the product AC. Our goal is to ﬁnd the product AC in terms of the coeﬃcients A, B and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coeﬃcient A on (x)2 and the coeﬃcient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ) + C sin2(θ) C = A sin2(θ) − B cos(θ) sin(θ) + C cos2(θ) In order to make use of the condition cot(2θ) = A−C the power reduction formulas. After some regrouping, we get B , we rewrite our formulas for A and C using 2A = [(A + C) + (A − C) cos(2θ)] + B sin(2θ) 2C = [(A + C) − (A − C) cos(2θ)] − B sin(2θ) Next, we try to make sense of the product (2A)(2C) = {[(A + C) + (A − C) cos(2θ)] + B sin(2θ)} {[(A + C) − (A − C) cos(2θ)] − B sin(2θ)} 5We hope that someday you get to see why this works the way it does. 980 Applications of Trigonometry We break this product into pieces. First, we use the diﬀerence of squares to multiply the ‘ﬁrst’ quantities in each factor to get [(A + C) + (A − C) cos(2θ)] [(A + C) − (A − C) cos(2θ)] = (A + C)2 − (A − C)2 cos2(2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C) + (A − C) cos(2θ)] +B sin(2θ) [(A + C) − (A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B2 sin2(2θ). Putting all of this together yields 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) From cot(2θ) = A−C sin(2θ) = A−C twice along with the Pythagorean Identity cos2(2θ) = 1 − sin2(2θ) to get B , we get cos(2θ) B , or (A−C) sin(2θ) = B cos(2θ). We use this substitution 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 1 − sin2(2θ) − 2B cos(2θ)B cos(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + (A − C)2 sin2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [(A − C) sin(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [B cos(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + B2 cos2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) + sin2(2θ) = (A + C)2 − (A − C)2 − B2 = A2 + 2AC + C2 − A2 − 2AC + C2 − B2 = 4AC − B2 Hence, B2 − 4AC = −4AC, so the quantity B2 − 4AC has the opposite sign of AC. The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. 1. 21x2 + 10xy √ 3 + 31y2 = 144 2. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 3. 16x2 + 24xy + 9y2 + 15x − 20y = 0 Solution. This is a straightforward application of Theorem 11.11. 11.6 Hooked on Conics Again 981 1. We have A = 21, B = 10 3)2 − 4(21)(31) = −2304 < 0. Theorem 11.11 predicts the graph is an ellipse, which checks with our work from Example 11.6.1 number 2. 3 and C = 31 so B2 − 4AC = (10 √ √ 2. Here, A = 5, B = 26 and C = 5, so B2 − 4AC = 262 − 4(5)(5) = 576 > 0. Theorem 11.11 classiﬁes the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1. 3. Finally, we have A = 16, B = 24 and C = 9 which gives 242 − 4(16)(9) = 0. Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2. 11.6.2 The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more uniﬁed perspective. We have our ‘new’ deﬁnition below. Deﬁnition 11.1. Given a ﬁxed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricity of the conic section. We have seen the notions of focus and directrix before in the deﬁnition of a parabola, Deﬁnition 7.3. There, a parabola is deﬁned as the set of points equidistant from the focus and directrix, giving an eccentricity e = 1 according to Deﬁnition 11.1. We have also seen the concept of eccentricity before. It was introduced for ellipses in Deﬁnition 7.5 in Section 7.4, and later extended to hyperbolas in Exercise 31 in Section 7.5. There, e was also deﬁned as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the ‘old’ ideas of focus, directrix and eccentricity with the ‘new’ ones presented in Deﬁnition 11.1 is to derive equations for the conic sections using Deﬁnition 11.1 and compare these parameters with what we know from Chapter 7. We begin by assuming the conic section has eccentricity e, a focus F at the origin and that the directrix is the vertical line x = −d as in the ﬁgure below. y d r cos(θ) P (r, θ) r θ x = −d O = F x 982 Applications of Trigonometry Using a polar coordinate representation P (r, θ) for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = r d + r cos(θ) so that r = e(d + r cos(θ)). Solving this equation for r, yields r = ed 1 − e cos(θ) At this point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e = 1, the usual conversion process outlined in Section 11.4 gives6 1 − e22 e2d2 x − e2d 1 − e2 2 1 − e2 e2d2 + y2 = 1 1−e2 , so the major axis has length 2ed 1−e2 and the minor axis has length 2ed√ e2d 1−e2 , 0 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at with major axis along the x-axis. Using the notation from Section 7.4, we have a2 = e2d2 (1−e2)2 and b2 = e2d2 1−e2 . Moreover, we ﬁnd that one focus is (0, 0) and working through the formula given in Deﬁnition 7.5 gives the eccentricity 1−e2 , 0 to be e, as required. If e > 1, then the equation generates a hyperbola with center whose transverse axis lies along the x-axis. Since such hyperbolas have the form (x−h)2 b2 = 1, we need to take the opposite reciprocal of the coeﬃcient of y2 to ﬁnd b2. We get7 a2 = e2d2 (e2−1)2 and b2 = − e2d2 e2−1 and the conjugate axis has length 2ed√ . e2−1 Additionally, we verify that one focus is at (0, 0), and the formula given in Exercise 31 in Section ed 1−e cos(θ)
reduces to 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r = . This is a parabola with vertex r = − d 2 , the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented in Deﬁnition 11.1 correspond with the ‘old’ deﬁnitions given in Chapter 7. 1−cos(θ) which gives the rectangular equation y2 = 2d x + d 2 , 0 opening to the right. In the language of Section 7.3, 4p = 2d so p = d e2−1 , so the transverse axis has length 2ed a2 − y2 (1−e2)2 = e2d2 1−e2 = e2d2 e2d d 2 ed Before we summarize our ﬁndings, we note that in order to arrive at our general equation of a conic 1−e cos(θ) , we assumed that the directrix was the line x = −d for d > 0. We could have just as r = easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases 1+e sin(θ) , respectively. The key thing to 1−e sin(θ) and r = we obtain the forms r = remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is (0, 0) and the directrix also tells us which way the parabola opens. We have established the following theorem. 1+e cos(θ) , r = ed ed ed 6Turn r = e(d + r cos(θ)) into r = e(d + x) and square both sides to get r2 = e2(d + x)2. Replace r2 with x2 + y2, expand (d + x)2, combine like terms, complete the square on x and clean things up. 7Since e > 1 in this case, 1 − e2 < 0. Hence, we rewrite 1 − e22 = e2 − 12 to help simplify things later on. 11.6 Hooked on Conics Again 983 Theorem 11.12. Suppose e and d are positive numbers. Then the graph of r = ed 1−e cos(θ) is the graph of a conic section with directrix x = −d. the graph of r = ed 1+e cos(θ) is the graph of a conic section with directrix x = d. the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose minor axis has length 2ed√ 1−e2 If e = 1, the graph is a parabola whose focal diameter is 2d. If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1 . We test out Theorem 11.12 in the next example. Example 11.6.4. Sketch the graphs of the following equations. 1. r = 4 1 − sin(θ) Solution. 2. r = 12 3 − cos(θ) 3. r = 6 1 + 2 sin(θ) 4 1. From r = 1−sin(θ) , we ﬁrst note e = 1 which means we have a parabola on our hands. Since ed = 4, we have d = 4 and considering the form of the equation, this puts the directrix at y = −4. Since the focus is at (0, 0), we know that the vertex is located at the point (in rectangular coordinates) (0, −2) and must open upwards. With d = 4, we have a focal diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below. 4 12 2. We ﬁrst rewrite r = 4 1−(1/3) cos(θ) . 3−cos(θ) in the form found in Theorem 11.12, namely r = Since e = 1 3 satisﬁes 0 < e < 1, we know that the graph of this equation is an ellipse. Since ed = 4, we have d = 12 and, based on the form of the equation, the directrix is x = −12. This means that the ellipse has its major axis along the x-axis. We can ﬁnd the vertices of the ellipse by ﬁnding the points of the ellipse which lie on the x-axis. We ﬁnd r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3, 0) and (6, 0), so these are our 2 , 0.8 vertices. The center of the ellipse is the midpoint of the vertices, which in this case is 3 2 , 0 and this allows us to ﬁnd the We know one focus is (0, 0), which is 3 1−e2 = (2)(4) 2 from the center 3 8As a quick check, we have from Theorem 11.12 the major axis should have length 2ed 1−(1/3)2 = 9. 984 Applications of Trigonometry other focus (3, 0), even though we are not asked to do so. Finally, we know from Theorem 2ed√ 3 which means the endpoints 1−e2 = 11.12 that the length of the minor axis is of the minor axis are 3 1−(1/3)2 = 6 4√ 2. We now have everything we need to graph r = 12 √ √ 3−cos(θ) . 2 , ±3 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 −3 y = −4 r = 4 1−sin(θ) y 4 3 2 1 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 x = −12 r = 12 3−cos(θ) 3. From r = 6 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get d = 3, and from the form of the equation, we know the directrix is y = 3. This means the transverse axis of the hyperbola lies along the y-axis, so we can ﬁnd the vertices by looking where the hyperbola intersects the y-axis. We ﬁnd r π = −6. These two 2 points correspond to the rectangular points (0, 2) and (0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 3. Putting this together with the location of the vertices, we get that √ 3 the asymptotes of the hyperbola have slopes ± 2 3 . Since the center of the hyperbola √ 2 = 2 and r 3π 2 = (2)(6) √ 22−1 = ± 2ed√ = 4 e2−1 √ 3 is (0, 4), the asymptotes are y = ± √ 3 3 x + 4. We graph the hyperbola below5 −4 −3 −2 −+2 sin(θ) 11.6 Hooked on Conics Again 985 In light of Section 11.6.1, the reader may wonder what the rotated form of the conic sections would look like in polar form. We know from Exercise 65 in Section 11.5 that replacing θ with (θ − φ) in an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. For instance, to graph r = 1−sin(θ) , which we obtained in Example 11.6.4 number 1, counter-clockwise by π all we need to do is rotate the graph of r = 4 1−sin(θ− π 4 ) 4 radians, as shown below4 −3 −2 −1 −1 −2 −3 r = 4 1−sin(θ− π 4 ) Using rotations, we can greatly simplify the form of the conic sections presented in Theorem 11.12, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π 2 . Since rotations do not aﬀect lengths, all of the formulas for lengths Theorem 11.12 remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0. We conclude this section with the statement of the following theorem. Theorem 11.13. Given constants > 0, e ≥ 0 and φ, the graph of the equation r = 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0, 0). If e = 0, the graph is a circle centered at (0, 0) with radius . If e = 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e . minor axis has length 2ed√ 1−e2 – If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose – If e = 1, the graph is a parabola whose focal diameter is 2d. – If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1 . 986 Applications of Trigonometry 11.6.3 Exercises Graph the following equations. 1. x2 + 2xy + y2 − x √ √ 2 − 6 = 0 2. 7x2 − 4xy √ 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2y = 0 2 + y √ 5. 13x2 − 34xy √ 3 + 47y2 − 64 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 Graph the following equations. 9. r = 11. r = 2 1 − cos(θ) 3 2 − cos(θ) 13. r = 4 1 + 3 cos(θ) 15. r = 2 1 + sin(θ − π 3 ) √ √ 4. x2 + 2 6. x2 − 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 3xy − y2 + 8 = 0 8. 8x2 + 12xy + 17y2 − 20 = 0 10. r = 12. r = 3 2 + sin(θ) 2 1 + sin(θ) 14. r = 2 1 − 2 sin(θ) 16. r = 6 3 − cos θ + π 4 The matrix A(θ) = cos(θ) − sin(θ) cos(θ) sin(θ) is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A π 4 . 18. Discuss with your classmates how to use A(θ) to rotate points in the plane. 19. Using the even / odd identities for cosine and sine, show A(θ)−1 = A(−θ). Interpret this geometrically. 11.6 Hooked on Conics Again 987 11.6.4 Answers 1. x2 + 2xy + y2 − becomes (x)2 = −(y − 3) after rotating counter-clockwise through x2 + 2xy + y2 − x √ 2 + y √ 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2 − 6 = 0 √ 2y = 0 becomes (x)2 + (y+2)2 counter-clockwise through θ = π 4 . 4 = 1 after rotating . 7x2 − 4xy 3 + 3y2 − 2x − 2y 3 − 5 = 0 9 + (y)2 = 1 after rotating becomes (x−2)2 counter-clockwise through √ 7x2 − 4xy 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 √ 4. x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 becomes(x)2 = y + 4 after rotating counter-clockwise through 5x2 + 6xy + 5y2 − 4 √ 2x + 4 √ 2y = 0 √ x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 988 Applications of Trigonometry √ 5. 13x2 − 34xy 3 + 47y2 − 64 = 0 becomes (y)2 − (x)2 counter-clockwise through θ = π 6 . 16 = 1 after rotating √ 6. x2 − 2 3xy − y2 + 8 = 0 4 − (y)2 becomes (x)2 counter-clockwise through θ = π 3 4 = 1 after rotating 13x2 − 34xy √ 3 + 47y2 − 64 = 0 √ x2 − 2 3xy − y2 + 8 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 becomes (y)2 = x after rotating counter-clockwise through θ = arctan 1 2 y . y 8. 8x2 + 12xy + 17y2 − 20 = 0 becomes (x)2 + (y)2 4 = 1 after rotating counter-clockwise through θ = arctan(2) y x y x θ = arctan 1 2 x θ = arctan(2) x x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 8x2 + 12xy + 17y2 − 20 = 0 11.6 Hooked on Conics Again 989 2 9. r = 1−cos(θ) is a parabola directrix x = −2 , vertex (−1, 0) focus (0, 0), focal diameter 4 10. r = 3 2+sin(θ) = 3 2 1+ 1 2 sin(θ) is an ellipse directrix y = 3 , vertices (0, 1), (0, −3) center (0, −2) , foci (0, 0), (0, −2) minor axis length 4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 11. r = 3 2−cos(θ) = 3 2 1− 1 2 cos(θ) is an ellipse directrix x = −3 , vertices (−1, 0), (3, 0) center (1, 0) , foci (0, 0), (2
, 0) minor axis length 2 √ 3 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 12. r = 2 1+sin(θ) is a parabola directrix y = 2 , vertex (0, 1) focus (0, 0), focal diameter 4 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis length 2 3 , vertices (1, 0), (2, 0) √ 2 2 , 0, foci (0, 0), (3, 0) 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices 0, − 2 3 center 0, − 4 , foci (0, 0), 0, − 8 3 3 √ conjugate axis length 2 3 3 , (0, −24 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 15. r = 2 3 ) is 1+sin(θ− π 2 the parabola r = 1+sin(θ) rotated through 1 −2 −3 −4 is the ellipse 16. r = 6 3−cos(θ+ π 4 ) 6 3−cos(θ) = r = 3 cos(θ) rotated through φ = − π 4 14 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 φ = − π 4 x 11.7 Polar Form of Complex Numbers 991 11.7 Polar Form of Complex Numbers √ In this section, we return to our study of complex numbers which were ﬁrst introduced in Section 3.4. Recall that a complex number is a number of the form z = a + bi where a and b are real −1. The number a is called the real part of numbers and i is the imaginary unit deﬁned by i = z, denoted Re(z), while the real number b is called the imaginary part of z, denoted Im(z). From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Re(z) and Im(z) are well-deﬁned.1 To start oﬀ this section, we associate each complex number z = a + bi with the point (a, b) on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Imaginary Axis (3, 0) ←→ z = 3 0 1 2 3 4 Real Axis (−4, 2) ←→ z = −4 + 2i 4i 3i 2i i −4 −3 −2 −1 −i −2i −3i (0, −3) ←→ z = −3i −4i The Complex Plane Since the ordered pair (a, b) gives the rectangular coordinates associated with the complex number z = a + bi, the expression z = a + bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates (r, θ). Although it is not as straightforward as the deﬁnitions of Re(z) and Im(z), we can still give r and θ special names in relation to z. Deﬁnition 11.2. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Re(z) and b = Im(z). Let (r, θ) be a polar representation of the point with rectangular coordinates (a, b) where r ≥ 0. The modulus of z, denoted \|z\|, is deﬁned by \|z\| = r. The angle θ is an argument of z. The set of all arguments of z is denoted arg(z). If z = 0 and −π < θ ≤ π, then θ is the principal argument of z, written θ = Arg(z). 1‘Well-deﬁned’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. 992 Applications of Trigonometry Some remarks about Deﬁnition 11.2 are in order. We know from Section 11.4 that every point in the plane has inﬁnitely many polar coordinate representations (r, θ) which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-deﬁned. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, \|z\| = 0 is well-deﬁned. If z = 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r ≥ 0 in our deﬁnition so this pins down the value of \|z\| to one and only one number. Thus the modulus is well-deﬁned in this case, too.2 Even with the requirement r ≥ 0, there are inﬁnitely many angles θ which can be used in a polar representation of a point (r, θ). If z = 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval (−π, π], and this angle is what we call the principal argument of z, Arg(z). In fact, the set arg(z) of all arguments of z can be described using set-builder notation as arg(z) = {Arg(z) + 2πk \| k is an integer}. Note that since arg(z) is a set, we will write ‘θ ∈ arg(z)’ to mean ‘θ is in3 the set of arguments of z’. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as (0, θ) for any angle θ. In this case, we have arg(0) = (−∞, ∞) and since there is no one value of θ which lies (−π, π], we leave Arg(0) undeﬁned.4 It is time for an example. Example 11.7.1. For each of the following complex numbers ﬁnd Re(z), Im(z), \|z\|, arg(z) and Arg(z). Plot z in the complex plane. 1. z = √ 3 − i Solution. 2. z = −2 + 4i 3. z = 3i 4. z = −117 √ √ √ 1. For z = 3 − i = 3 + (−1)i, we have Re(z) = 3 and Im(z) = −1. To ﬁnd \|z\|, arg(z) and Arg(z), we need to ﬁnd a polar representation (r, θ) with r ≥ 0 for the point P ( 3, −1) associated with z. We know r2 = ( 3)2 + (−1)2 = 4, so r = ±2. Since we require r ≥ 0, we choose r = 2, so \|z\| = 2. Next, we ﬁnd a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tan(θ) = −1√ 6 + 2πk 3 6 + 2πk \| k is an integer. Of these values, only θ = − π for integers k. Hence, arg(z) = − π satisﬁes the requirement that −π < θ ≤ π, hence Arg(z) = − π 6 . 3 , so . The complex number z = −2 + 4i has Re(z) = −2, Im(z) = 4, and is associated with the point P (−2, 4). Our next task is to ﬁnd a polar representation (r, θ) for P where r ≥ 0. 5. To ﬁnd θ, we get Running through the usual calculations gives r = 2 tan(θ) = −2, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We ﬁnd θ = π + arctan(−2) + 2πk, or, more succinctly θ = π − arctan(2) + 2πk for integers k. Hence arg(z) = {π − arctan(2) + 2πk \| k is an integer}. Only θ = π − arctan(2) satisﬁes the requirement −π < θ ≤ π, so Arg(z) = π − arctan(2). 5, so \|z\| = 2 √ √ 2In case you’re wondering, the use of the absolute value notation \|z\| for modulus will be explained shortly. 3Recall the symbol being used here, ‘∈,’ is the mathematical symbol which denotes membership in a set. 4If we had Calculus, we would regard Arg(0) as an ‘indeterminate form.’ But we don’t, so we won’t. 11.7 Polar Form of Complex Numbers 993 3. We rewrite z = 3i as z = 0 + 3i to ﬁnd Re(z) = 0 and Im(z) = 3. The point in the plane which corresponds to z is (0, 3) and while we could go through the usual calculations to ﬁnd the required polar form of this point, we can almost ‘see’ the answer. The point (0, 3) lies 3 units away from the origin on the positive y-axis. Hence, r = \|z\| = 3 and θ = π 2 + 2πk for integers k. We get arg(z) = π 2 + 2πk \| k is an integer and Arg(z) = π 2 . 4. As in the previous problem, we write z = −117 = −117 + 0i so Re(z) = −117 and Im(z) = 0. The number z = −117 corresponds to the point (−117, 0), and this is another instance where we can determine the polar form ‘by eye’. The point (−117, 0) is 117 units away from the origin along the negative x-axis. Hence, r = \|z\| = 117 and θ = π + 2π = (2k + 1)πk for integers k. We have arg(z) = {(2k + 1)π \| k is an integer}. Only one of these values, θ = π, just barely lies in the interval (−π, π] which means and Arg(z) = π. We plot z along with the other numbers in this example below. Imaginary Axis z = 3i 4i 3i 2i i z = −2 + 4i z = −117 −117 −2 −1 − Real Axis Now that we’ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 11.14. Properties of the Modulus: Let z and w be complex numbers. \|z\| is the distance from z to 0 in the complex plane \|z\| ≥ 0 and \|z\| = 0 if and only if z = 0 \|z\| = Re(z)2 + Im(z)2 Product Rule: \|zw\| = \|z\|\|w\| Power Rule: \|zn\| = \|z\|n for all natural numbers, n Quotient Rule: z w = \|z\| \|w\| , provided w = 0 To prove the ﬁrst three properties in Theorem 11.14, suppose z = a + bi where a and b are real numbers. To determine \|z\|, we ﬁnd a polar representation (r, θ) with r ≥ 0 for the point (a, b). From √ Section 11.4, we know r2 = a2 + b2 so that r = ± a2 + b2. Since we require r ≥ 0, then it must be a2 + b2. Using the distance formula, we ﬁnd the distance a2 + b2, which means \|z\| = that r = √ √ 994 Applications of Trigonometry √ a2 + b2, establishing the ﬁrst property.5 For the second property, note from (0, 0) to (a, b) is also that since \|z\| is a distance, \|z\| ≥ 0. Furthermore, \|z\| = 0 if and only if the distance from z to 0 is 0, and the latter happens if and only if z = 0, which is what we were asked to show.6 For the third property, we note that since a = Re(z) and b = Im(z), z = To prove the product rule, suppose z = a + bi and w = c + di for real numbers a, b, c and d. Then zw = (a + bi)(c + di). After the usual arithmetic7 we get zw = (ac − bd) + (ad + bc)i. Therefore, Re(z)2 + Im(z)2. a2 + b2 = √ \|zwac − bd)2 + (ad + bc)2 a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 Expand a2c2 + a2d2 + b2c2 + b2d2 a2 (c2 + d2) + b2 (c2 + d2) (a2 + b2) (c2 + d2) a2 + b2 c2 + d2 Factor Factor √ Rearrange terms = = \|z\|\|w\| Product Rule for Radicals Deﬁnition of \|z\| and \|w\| Hence \|zw\| = \|z\|\|w\| as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction8 to prove the power rule. Let P (n) be the statement \|zn\| = \|z\|n. Then P (1) is true since z1 = \|z\|k for some k ≥ 1. Our job = \|z\|k+1. As is customary with induction proofs, is to show that P (k + 1) is true, namely we ﬁrst try to reduce the problem in such a way as to use the Induction Hypothesis. = \|z\| = \|z\|1. Next, assume P (k) is true. That is, assume zk+1 zk zk+1 Properties of Exponents \|z\| Product Rule zkz = zk = = \|z\|k\|z\| = \|z\|k+1 Properties of Exponents Induction Hypothesis Hence, P (k + 1) is true, which
means \|zn\| = \|z\|n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w = 0 (so \|w\| = 0) and we get (z) 1 w Product Rule. = = \|z\| z w 1 w 5Since the absolute value \|x\| of a real number x can be viewed as the distance from x to 0 on the number line, this ﬁrst property justiﬁes the notation \|z\| for modulus. We leave it to the reader to show that if z is real, then the deﬁnition of modulus coincides with absolute value so the notation \|z\| is unambiguous. 6This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is a2 + b2 = 0 if and only if a2 + b2 = 0, which is true if and only if a = b = 0. √ true. We know \|z\| = 0 if and only if The latter happens if and only if z = a + bi = 0. There. 7See Example 3.4.1 in Section 3.4 for a review of complex number arithmetic. 8See Section 9.3 for a review of this technique. 11.7 Polar Form of Complex Numbers 995 1 Hence, the proof really boils down to showing w Next, we characterize the argument of a complex number in terms of its real and imaginary parts. \|w\| . This is left as an exercise. = 1 Theorem 11.15. Properties of the Argument: Let z be a complex number. If Re(z) = 0 and θ ∈ arg(z), then tan(θ) = Im(z) Re(z) . 2 + 2πk \| k is an integer. 2 + 2πk \| k is an integer. If Re(z) = 0 and Im(z) > 0, then arg(z) = π If Re(z) = 0 and Im(z) < 0, then arg(z) = − π If Re(z) = Im(z) = 0, then z = 0 and arg(z) = (−∞, ∞). To prove Theorem 11.15, suppose z = a + bi for real numbers a and b. By deﬁnition, a = Re(z) and b = Im(z), so the point associated with z is (a, b) = (Re(z), Im(z)). From Section 11.4, we know that if (r, θ) is a polar representation for (Re(z), Im(z)), then tan(θ) = Im(z) Re(z) , provided Re(z) = 0. If Re(z) = 0 and Im(z) > 0, then z lies on the positive imaginary axis. Since we take r > 0, we have that θ is coterminal with π 2 , and the result follows. If Re(z) = 0 and Im(z) < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with − π 2 . The last property in the theorem was already discussed in the remarks following Deﬁnition 11.2. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the ﬁgure below. Imaginary Axis bi (a, b) ←→ z = a + bi ←→ (r, θ ∈ arg(z) 0 a Real Axis Polar coordinates, (r, θ) associated with z = a + bi with r ≥ 0. We know from Theorem 11.7 that a = r cos(θ) and b = r sin(θ). Making these substitutions for a and b gives z = a + bi = r cos(θ) + r sin(θ)i = r [cos(θ) + i sin(θ)]. The expression ‘cos(θ) + i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Since r = \|z\| and θ ∈ arg(z), we get Deﬁnition 11.3. A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). The expression: is called a polar form for z. \|z\|cis(θ) = \|z\| [cos(θ) + i sin(θ)] 996 Applications of Trigonometry Since there are inﬁnitely many choices for θ ∈ arg(z), there inﬁnitely many polar forms for z, so we used the indeﬁnite article ‘a’ in Deﬁnition 11.3. It is time for an example. Example 11.7.2. 1. Find the rectangular form of the following complex numbers. Find Re(z) and Im(z). (a) z = 4cis 2π 3 (b) z = 2cis − 3π 4 (c) z = 3cis(0) (d) z = cis π 2 2. Use the results from Example 11.7.1 to ﬁnd a polar form of the following complex numbers. (a) z = √ 3 − i Solution. (b) z = −2 + 4i (c) z = 3i (d) z = −117 1. The key to this problem is to write out cis(θ) as cos(θ) + i sin(θ). (a) By deﬁnition, z = 4cis 2π √ 3 z = −2 + 2i + i sin 2π = 4 cos 2π √ 3 3 3, so that Re(z) = −2 and Im(z) = 2 = 2 cos − 3π 4 . After some simplifying, we get 3. + i sin − 3π 4 . From this, we ﬁnd (b) Expanding, we get z = 2cis − 3π √ 4 2 = Im(z). 2, so Re(zc) We get z = 3cis(0) = 3 [cos(0) + i sin(0)] = 3. Writing 3 = 3 + 0i, we get Re(z) = 3 and (d) Lastly, we have z = cis π 2 Im(z) = 0, which makes sense seeing as 3 is a real number. = cos π 2 = i. Since i = 0 + 1i, we get Re(z) = 0 and Im(z) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. + i sin π 2 2. To write a polar form of a complex number z, we need two pieces of information: the modulus \|z\| and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 11.7.1 to ﬁnd what we need. (a) For z = √ 3 − i, \|z\| = 2 and θ = − π 6 , so z = 2cis − π 6 . We can check our answer by √ converting it back to rectangular form to see that it simpliﬁes to z = 3 − i. √ √ (b) For z = −2 + 4i, \|z\| = 2 5 and θ = π − arctan(2). Hence, z = 2 5cis(π − arctan(2)). It is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, \|z\| = 3 and θ = π 2 . In this case, z = 3cis π 2 geometrically. Head out 3 units from 0 along the positive real axis. Rotating π counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i. . This can be checked 2 radians (d) Last but not least, for z = −117, \|z\| = 117 and θ = π. We get z = 117cis(π). As with the previous problem, our answer is easily checked geometrically. 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = \|z\|cis(α) and w = \|w\|cis(β). Then Product Rule: zw = \|z\|\|w\|cis(α + β) Power Rule (DeMoivre’s Theorem) : zn = \|z\|ncis(nθ) for every natural number n Quotient Rule: z w = \|z\| \|w\| cis(α − β), provided \|w\| = 0 The proof of Theorem 11.16 requires a healthy mix of deﬁnition, arithmetic and identities. We ﬁrst start with the product rule. zw = [\|z\|cis(α)] [\|w\|cis(β)] = \|z\|\|w\| [cos(α) + i sin(α)] [cos(β) + i sin(β)] We now focus on the quantity in brackets on the right hand side of the equation. [cos(α) + i sin(α)] [cos(β) + i sin(β)] = cos(α) cos(β) + i cos(α) sin(β) + i sin(α) cos(β) + i2 sin(α) sin(β) = cos(α) cos(β) + i2 sin(α) sin(β) Rearranging terms + i sin(α) cos(β) + i cos(α) sin(β) = (cos(α) cos(β) − sin(α) sin(β)) Since i2 = −1 + i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α + β) + i sin(α + β) Sum identities = cis(α + β) Deﬁnition of ‘cis’ Putting this together with our earlier work, we get zw = \|z\|\|w\|cis(α + β), as required. Moving right along, we next take aim at the Power Rule, better known as DeMoivre’s Theorem.9 We proceed by induction on n. Let P (n) be the sentence zn = \|z\|ncis(nθ). Then P (1) is true, since z1 = z = \|z\|cis(θ) = \|z\|1cis(1 · θ). We now assume P (k) is true, that is, we assume zk = \|z\|kcis(kθ) for some k ≥ 1. Our goal is to show that P (k + 1) is true, or that zk+1 = \|z\|k+1cis((k + 1)θ). We have zk+1 = zkz = \|z\|kcis(kθ) (\|z\|cis(θ)) = \|z\|k\|z\| cis(kθ + θ) = \|z\|k+1cis((k + 1)θ) Properties of Exponents Induction Hypothesis Product Rule 9Compare this proof with the proof of the Power Rule in Theorem 11.14. 998 Applications of Trigonometry Hence, assuming P (k) is true, we have that P (k + 1) is true, so by the Principle of Mathematical Induction, zn = \|z\|ncis(nθ) for all natural numbers n. The last property in Theorem 11.16 to prove is the quotient rule. Assuming \|w\| = 0 we have z w = = \|z\|cis(α) \|w\|cis(β) \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) Next, we multiply both the numerator and denominator of the right hand side by (cos(β) − i sin(β)) which is the complex conjugate of (cos(β) + i sin(β)) to get z w = \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) · cos(β) − i sin(β) cos(β) − i sin(β) If we let the numerator be N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] and simplify we get N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] = cos(α) cos(β) − i cos(α) sin(β) + i sin(α) cos(β) − i2 sin(α) sin(β) Expand = [cos(α) cos(β) + sin(α) sin(β)] + i [sin(α) cos(β) − cos(α) sin(β)] Rearrange and Factor = cos(α − β) + i sin(α − β) = cis(α − β) Diﬀerence Identities Deﬁnition of ‘cis’ If we call the denominator D then we get D = [cos(β) + i sin(β)] [cos(β) − i sin(β)] = cos2(β) − i cos(β) sin(β) + i cos(β) sin(β) − i2 sin2(β) Expand = cos2(β) − i2 sin2(β) Simplify = cos2(β) + sin2(β) Again, i2 = −1 Pythagorean Identity = 1 Putting it all together, we get · cos(β) − i sin(β) cos(β) − i sin(β) z w = = = \|z\| \|w\| \|z\| \|w\| cos(α) + i sin(α) cos(β) + i sin(β) cis(α − β) 1 \|z\| \|w\| cis(α − β) and we are done. The next example makes good use of Theorem 11.16. 11.7 Polar Form of Complex Numbers 999 Example 11.7.3. Let z = 2 √ 3 + 2i and w = −1 + i √ 3. Use Theorem 11.16 to ﬁnd the following. 1. zw 2. w5 Write your ﬁnal answers in rectangular form. 3. z w √ √ √ 3)2 + (2)2 = Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 3+2i, √ 3 3 . Since 3, we ﬁnd \|z\| = (2 z lies in Quadrant I, we have θ = π 16 = 4. If θ ∈ arg(z), we know tan(θ) = Im(z) 6 + 2πk for integers k. Hence, z = 4cis π 3)2 = 2. For an argument θ of w, we have tan(θ) = 3 + 2πk for integers k and w = 2cis 2π = 8cis π 2cis 2π √ 3 After simplifying, we get zw = −4 3 + 4i. = 3 . For w = −1 + i √ 3 −1 = − we have \|w\| = w lies in Quadrant II, θ = 2π 1. We get zw = 4cis π 6 . We can now proceed. + i sin 5π 6 . = 8 cos 5π 6 = 8cis 5π 6 Re(z) = 2 (−1)2 + ( 6 + 2π 3. Since √ √ √ √ 6 2 3 3 2. We use DeMoivre’s Theorem which yields w5 = 2cis 2π 3 + i sin 4π 3 is coterminal with 4π Since 10π 3 = 25cis 5 · 2π 3 = −16 − 16i = 32cis 10π √ 3 5 3. . 3. Last, but not least, we have = 4 2 is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating π 2 radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that z 6 − 2π = 2cis − π 2 . Since − π 3 w = −2i. 3 , we get w5 = 32 cos 4π z 4cis( π 6 ) 2cis( 2π 3 ) w 2 cis π = 3 Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with.
Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out zw = (2 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of complex numbers. Consider how we computed w5 above and compare that to using the Binomial Theorem, Theorem 9.4, to accomplish the same feat by 3)5. Division is tricky in the best of times, and we saved ourselves a lot of expanding (−1 + i time and eﬀort using Theorem 11.16 to ﬁnd and simplify z w using their polar forms as opposed to starting with 2 , rationalizing the denominator, and so forth. 3 + 2i)(−1 + i √ √ √ √ 3+2i √ 3 −1+i There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11.16. Take the product rule, for instance. If z = \|z\|cis(α) and w = \|w\|cis(β), the formula zw = \|z\|\|w\|cis(α + β) can be viewed geometrically as a two step process. The multiplication of \|z\| by \|w\| can be interpreted as magnifying10 the distance \|z\| from z to 0, by the factor \|w\|. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise.11 Focusing on z and w from Example 10Assuming \|w\| > 1. 11Assuming β > 0. 1000 Applications of Trigonometry 11.7.3, we can arrive at the product zw by plotting z, doubling its distance from 0 (since \|w\| = 2), and rotating 2π 3 radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometrically. Imaginary Axis Imaginary Axis 6i 5i 4i 3i 2i i z\|w\| = 8cis π 6 z = 4cis π 6 zw = 8cis π 6 + 2π 3 6i 5i 4i 3i 2i i z\|w\| = 8cis Real Axis −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 Real Axis Multiplying z by \|w\| = 2. Rotating counter-clockwise by Arg(w) = 2π 3 radians. Visualizing zw for z = 4cis π 6 and w = 2cis 2π 3 . We may also visualize division similarly. Here, the formula z \|w\| cis(α − β) may be interpreted as shrinking12 the distance from 0 to z by the factor \|w\|, followed up by a clockwise 13 rotation of β radians. In the case of z and w from Example 11.7.3, we arrive at z w by ﬁrst halving the distance from 0 to z, then rotating clockwise 2π w = \|z\| 3 radians. Imaginary Axis Imaginary Axis 3i 2i i z = 4cis π 6 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis i −i −2i 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis zw = 2cis π 6 2π 3 Dividing z by \|w\| = 2. Rotating clockwise by Arg(w) = 2π 3 radians. Visualizing z w for z = 4cis π 6 and w = 2cis 2π 3 . Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. Deﬁnition 11.4. Let z and w be complex numbers. If there is a natural number n such that wn = z, then w is an nth root of z. Unlike Deﬁnition 5.4 in Section 5.3, we do not specify one particular prinicpal nth root, hence the use of the indeﬁnite article ‘an’ as in ‘an nth root of z’. Using this deﬁnition, both 4 and −4 are 12Again, assuming \|w\| > 1. 13Again, assuming β > 0. 11.7 Polar Form of Complex Numbers 1001 √ √ 16 means the principal square root of 16 as in square roots of 16, while 16 = 4. Suppose we wish to ﬁnd all complex third (cube) roots of 8. Algebraically, we are trying to solve w3 = 8. We √ know that there is only one real solution to this equation, namely w = 3 8 = 2, but if we take the time to rewrite this equation as w3 − 8 = 0 and factor, we get (w − 2) w2 + 2w + 4 = 0. The quadratic factor gives two more cube roots w = −1 ± i 3, for a total of three cube roots of 8. In accordance with Theorem 3.14, since the degree of p(w) = w3 − 8 is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we get z = 8cis(0). If we let w = \|w\|cis(α) be a polar form of w, the equation w3 = 8 becomes √ w3 = 8 (\|w\|cis(α))3 = 8cis(0) \|w\|3cis(3α) = 8cis(0) DeMoivre’s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates \|w\|3, 3α, while the complex number on the right hand side corresponds to the point with polar coordinates (8, 0). Since \|w\| ≥ 0, so is \|w\|3, which means \|w\|3, 3α and (8, 0) are two polar representations corresponding to the same complex number, both with positive r values. From Section 11.4, we know \|w\|3 = 8 and 3α = 0 + 2πk for integers k. Since \|w\| is a real number, √ we solve \|w\|3 = 8 by extracting the principal cube root to get \|w\| = 3 8 = 2. As for α, we get α = 2πk for integers k. This produces three distinct points with polar coordinates corresponding to 3 k = 0, 1 and 2: speciﬁcally (2, 0), 2, 2π . These correspond to the complex numbers 3 w0 = 2cis(0), w1 = 2cis 2π , respectively. Writing these out in rectangular form 3 yields w0 = 2, w1 = −1 + i 3. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to ﬁnd, for example, all of the ﬁfth roots of 32. (Try using Chapter 3 techniques on that!) If we start with a generic complex number in polar form z = \|z\|cis(θ) and solve wn = z in the same manner as above, we arrive at the following theorem. and w2 = 2cis 4π √ and 2, 4π 3 3 and w2 = −1 − i √ 3 Theorem 11.17. The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct nth roots, which we denote by w0, w1, . . . , wn − 1, and they are given by the formula √ wk = n rcis θ n + 2π n k The proof of Theorem 11.17 breaks into to two parts: ﬁrst, showing that each wk is an nth root, and second, showing that the set {wk \| k = 0, 1, . . . , (n − 1)} consists of n diﬀerent complex numbers. To show wk is an nth root of z, we use DeMoivre’s Theorem to show (wk)n = z. 1002 Applications of Trigonometry √ (wk)n = n √ = ( n rcis θ n + 2π r)n cis n · θ n kn n + 2π n k DeMoivre’s Theorem = rcis (θ + 2πk) Since k is a whole number, cos(θ + 2πk) = cos(θ) and sin(θ + 2πk) = sin(θ). Hence, it follows that cis(θ + 2πk) = cis(θ), so (wk)n = rcis(θ) = z, as required. To show that the formula in Theorem 11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note √ that the modulus of each of the wk is the same, namely n r. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments diﬀer by an integer multiple of 2π. Suppose k and j are whole numbers between 0 and (n − 1), inclusive, with k = j. Since k and j are diﬀerent, let’s assume for the sake of argument that k > j. Then θ . For this to be an integer multiple of 2π, (k − j) must be a multiple of n. But because of the restrictions on k and j. (Think this through.) Hence, (k − j) is a positive number less than n, so it cannot be a multiple of n. As a result, wk and wj are diﬀerent complex numbers, and we are done. By Theorem 3.14, we know there at most n distinct solutions to wn = z, and we have just found all of them. We illustrate Theorem 11.17 in the next example. n j = 2π n k − θ k−j n n + 2π n + 2π Example 11.7.4. Use Theorem 11.17 to ﬁnd the following: 1. both square roots of z = −2 + 2i √ 3 2. the four fourth roots of z = −16 3. the three cube roots of z = √ √ 2 + i 2 4. the ﬁve ﬁfth roots of z = 1. Solution. √ 1. We start by writing z = −2 + 2i 3 = 4cis 2π 3 θ = 2π in Theorem 11.17, we’ll call them w0 and w1. We get w0 = . To use Theorem 11.17, we identify r = 4, 3 and n = 2. We know that z has two square roots, and in keeping with the notation = 2cis π 3 . In rectangular form, the two square roots of 3. We can check our answers by squaring them and 2 + 2π 2 (1) 3 and w1 = −1 − i = 2cis 4π 3 √ (2π/3) √ 2 + 2π 2 (0) and w1 = (2π/3) 4cis 4cis √ √ z are w0 = 1 + i showing that we get z = −2 + 2i √ 3. 2. Proceeding as above, we get z = −16 = 16cis(π). With r = 16, θ = π and n = 4, we get the √ √ 4 (1) = 4 (0) = 2cis π 16cis π 16cis π , w1 = 4 4 + 2π 4 + 2π four fourth roots of z to be w0 = 4 √ √ 4 . and w3 = 4 , w2 = 4 16cis π 2cis 3π 4 (3) = 2cis 7π 4 (2) = 2cis 5π 16cis −i 2, w2 = − Converting these to rectangular form gives w0 = 2 √ and w3 = 4 + 2π √ 2+i 2, w1 = − 4 + 2π 2 − i 2+i √ 2. 11.7 Polar Form of Complex Numbers 1003 √ √ 3. For z = 2cis π 12 2, we have z = 2cis π √ 4 , w1 = 3 2cis 9π 12 4 and n = 3 the usual computations 2+i √ 2cis 17π yield w0 = 3 If we were 12 to convert these to rectangular form, we would need to use either the Sum and Diﬀerence Identities in Theorem 10.16 or the Half-Angle Identities in Theorem 10.19 to evaluate w0 and w2. Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. . With r = 2, θ = π √ = 3 √ and w2 = 3 2cis 3π 4 . 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 2π 4. To ﬁnd the ﬁve ﬁfth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. and . The situation here is even graver than in the previous example, since we have 5 . At this stage, we √ Since 5 w4 = cis 8π not developed any identities to help us determine the cosine or sine of 2π could approximate our answers using a calculator, and we leave this as an exercise. , w3 = cis 6π , w2 = cis 4π 5 5 5 5 Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Essentially, Theorem 11.17 says that to ﬁnd the nth roots of a complex number, we ﬁrst take the nth root of the modulus and divide the argument by n. This gives the ﬁrst root w0. Each succeessive root is found by adding 2π n to the argument, which amounts to rotating w0 by 2π n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 11.7.4 below. Imaginary Ax
is w1 2i i w0 −2 −1 0 1 2 Real Axis w2 −i −2i w3 The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications.14 For now, the following exercises will have to suﬃce. 14For more on this, see the beautifully written epilogue to Section 3.4 found on page 294. 1004 Applications of Trigonometry 11.7.1 Exercises In Exercises 1 - 20, ﬁnd a polar representation for the complex number z and then identify Re(z), Im(z), \|z\|, arg(z) and Arg(z). 2. z = 5 + 5i √ 3 3. z = 6i 4. z = −3 √ √ 2 2+3i 1. z = 9 + 9i 5. z = −6 √ 3 + 6i 9. z = −5i 13. z = 3 + 4i 6. z = −2 7. z = − √ √ 2 − 2i 2 11. z = 6 10. z = 2 √ 14. z = 2 + i 15. z = −7 + 24i 16. z = −2 + 6i √ 3 2 − 1 2 i 8. z = −3 − 3i √ 12. z = i 3 7 17. z = −12 − 5i 18. z = −5 − 2i 19. z = 4 − 2i 20. z = 1 − 3i In Exercises 21 - 40, ﬁnd the rectangular form of the given complex number. Use whatever identities are necessary to ﬁnd the exact values. 21. z = 6cis(0) 22. z = 2cis π 6 3π 4 3π 2 √ 26. z = 6cis √ 30. z = 13cis √ 23. z = 7 2cis π 4 24. z = 3cis 27. z = 9cis (π) 28. z = 3cis π 2 4π 3 31. z = 1 2 cis 32. z = 12cis − π 3 7π 4 7π 8 34. z = 2cis 10cis arctan √ 3 arctan − √ 1 3 2 5 12 4 3 √ 36. z = 37. z = 15cis (arctan (−2)) 38. z = 39. z = 50cis π − arctan 7 24 40. z = 1 2 cis π + arctan For Exercises 41 - 52, use z = − your answers in polar form using the principal argument. i and w = 3 + 2 2 − 3i √ 3 3 3 2 √ √ 2 to compute the quantity. Express 41. zw 45. w3 42. z w 46. z5w2 43. w z 47. z3w2 44. z4 48. z2 w 25. z = 4cis 2π 3 29. z = 7cis − 3π 4 33. z = 8cis π 12 35. z = 5cis arctan 11.7 Polar Form of Complex Numbers 1005 49. w z2 50. z3 w2 51. w2 z3 52. 6 w z In Exercises 53 - 64, use DeMoivre’s Theorem to ﬁnd the indicated power of the given complex number. Express your ﬁnal answers in rectangular form. 53. −2 + 2i √ 33 √ 54. (− 3 − i)3 57 + 61 58. 62. (2 + 2i)5 55. (−3 + 3i)4 59)5 63. ( √ 56. ( 3 + i)4 4 √ 3 3 − 1 3 i 60. 64. (1 − i)8 In Exercises 65 - 76, ﬁnd the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. 65. the two square roots of z = 4i 66. the two square roots of z = −25i 67. the two square roots of z = 1 + i √ 3 68. the two square roots of 5 2 − 5 √ 3 2 i 69. the three cube roots of z = 64 70. the three cube roots of z = −125 71. the three cube roots of z = i 72. the three cube roots of z = −8i 73. the four fourth roots of z = 16 74. the four fourth roots of z = −81 75. the six sixth roots of z = 64 76. the six sixth roots of z = −729 77. Use the Sum and Diﬀerence Identities in Theorem 10.16 or the Half Angle Identities in 2 in rectangular form. (See 2 + i Theorem 10.19 to express the three cube roots of z = Example 11.7.4, number 3.) √ √ 78. Use a calculator to approximate the ﬁve ﬁfth roots of 1. (See Example 11.7.4, number 4.) 79. According to Theorem 3.16 in Section 3.4, the polynomial p(x) = x4 + 4 can be factored into the product linear and irreducible quadratic factors. In Exercise 28 in Section 8.7, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))(x − (−1 + i))(x − (−1 − i)). By multiplying the ﬁrst two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that p(x) = (x2 − 2x + 2)(x2 + 2x + 2). Use the 12 complex 12th roots of 4096 to factor p(x) = x12 − 4096 into a product of linear and irreducible quadratic factors. 1006 Applications of Trigonometry 80. Complete the proof of Theorem 11.14 by showing that if w = 0 than 1 w = 1 \|w\| . 81. Recall from Section 3.4 that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a − bi. (a) Prove that \|z\| = \|z\|. √ (b) Prove that \|z\| = (c) Show that Re(z) = zz z + z 2 and Im(z) = z − z 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). Interpret this result geometrically. (e) Is it always true that Arg (z) = −Arg(z)? 82. Given any natural number n ≥ 2, the n complex nth roots of the number z = 1 are called the In the following exercises, assume that n is a ﬁxed, but arbitrary, nth Roots of Unity. natural number such that n ≥ 2. (a) Show that w = 1 is an nth root of unity. (b) Show that if both wj and wk are nth roots of unity then so is their product wjwk. (c) Show that if wj is an nth root of unity then there exists another nth root of unity wj such that wjwj = 1. Hint: If wj = cis(θ) let wj = cis(2π − θ). You’ll need to verify that wj = cis(2π − θ) is indeed an nth root of unity. 83. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula deﬁnes eit = cos(t) + i sin(t). (a) Use Theorem 11.16 to show that eixeiy = ei(x+y) for all real numbers x and y. (b) Use Theorem 11.16 to show that eixn = ei(nx) for any real number x and any natural number n. (c) Use Theorem 11.16 to show that eix eiy = ei(x−y) for all real numbers x and y. (d) If z = rcis(θ) is the polar form of z, show that z = reit where θ = t radians. (e) Show that eiπ + 1 = 0. (This famous equation relates the ﬁve most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that cos(t) = eit + e−it 2 and that sin(t) = eit − e−it 2i for all real numbers t. 11.7 Polar Form of Complex Numbers 1007 11.7.2 Answers √ 2cis π 4 , Re(z) = 9, 1. z = 9 + 9i = 9 4 + 2πk \| k is an integer and Arg(z) = π arg(z) = π 4 . √ √ , Re(z) = 5, 3 = 10cis π 3, 3 3 + 2πk \| k is an integer and Arg(z) = π 3 . arg(z) = π 2. z = 5 + 5i Im(z) = 9, Im(z) = 5 √ \|z\| = 9 2 \|z\| = 10 3. z = 6i = 6cis π 2 arg(z) = π , Re(z) = 0, \|z\| = 6 2 + 2πk \| k is an integer and Arg(z) = π 2 . Im(z) = 6, 4. z = −3 √ √ 2 + 3i 2 = 6cis 3π 4 , Re(z) = −3 √ 2, Im(z) = 3 √ 2, \|z\| = 6 arg(z) = 3π √ 5. z = −6 arg(z) = 5π 4 + 2πk \| k is an integer and Arg(z) = 3π 4 . , Re(z) = −6 6 + 2πk \| k is an integer and Arg(z) = 5π 6 . 3 + 6i = 12cis 5π 6 √ 3, Im(z) = 6, \|z\| = 12 6. z = −2 = 2cis (π), Re(z) = −2, Im(z) = 0, \|z\| = 2 arg(z) = {(2k + 1)π \| k is an integer} and Arg(z) = π. √ 3 7. z = − 2 − 1 arg(z) = 7π 8. z = −3 − 3i = 3 arg(z) = 5π √ 3 2 , 6 , Re(z) = − 2 i = cis 7π Im(z) = − 1 2 , 6 + 2πk \| k is an integer and Arg(z) = − 5π 6 . , Re(z) = −3, Im(z) = −3, 4 + 2πk \| k is an integer and Arg(z) = − 3π 4 . 2cis 5π 4 √ \|z\| = 1 √ \|z\| = 3 2 9. z = −5i = 5cis 3π 2 arg(z) = 3π √ , Re(z) = 0, Im(z) = −5, \|z\| = 5 2 + 2πk \| k is an integer and Arg(z) = − π 2 . √ , Re(z) = 2 4 + 2πk \| k is an integer and Arg(z) = − π 4 . 2 = 4cis 7π 4 √ 2, 10. z = 2 2 − 2i arg(z) = 7π √ Im(z) = −2 2, \|z\| = 4 11. z = 6 = 6cis (0), Re(z) = 6, Im(z) = 0, \|z\| = 6 arg(z) = {2πk \| k is an integer} and Arg(z) = 0. √ Im(z) = 3 , Re(z) = 0, √ 12. z = i 3 7, 7cis π 2 √ \|z\| = 3 7 2 + 2πk \| k is an integer and Arg(z) = π 2 . √ 7 = 3 arg(z) = π 13. z = 3 + 4i = 5cis arctan 4 3 , Re(z) = 3, Im(z) = 4, \|z\| = 5 arg(z) = arctan 4 3 + 2πk \| k is an integer and Arg(z) = arctan 4 3 . 1008 14. z = √ 2 + i = √ arctan √ 2 2 , Re(z) = √ 2, Im(z) = 1, \|z\| = arg(z) = arctan + 2πk \| k is an integer and Arg(z) = arctan 3cis √ 2 2 √ 3 √ 2 2 . Applications of Trigonometry 15. z = −7 + 24i = 25cis π − arctan 24 7 , Re(z) = −7, Im(z) = 24, \|z\| = 25 arg(z) = π − arctan 24 7 + 2πk \| k is an integer and Arg(z) = π − arctan 24 7 . 16. z = −2 + 6i = 2 √ 10cis (π − arctan (3)), Re(z) = −2, Im(z) = 6, √ \|z\| = 2 10 arg(z) = {π − arctan (3) + 2πk \| k is an integer} and Arg(z) = π − arctan (3). 17. z = −12 − 5i = 13cis π + arctan 5 12 , Re(z) = −12, Im(z) = −5, \|z\| = 13 arg(z) = π + arctan 5 12 + 2πk \| k is an integer and Arg(z) = arctan 5 12 18. z = −5 − 2i = √ 29cis π + arctan 2 5 , Re(z) = −5, Im(z) = −2, arg(z) = π + arctan 2 5 5cis arctan − 1 2 19. z = 4 − 2i = 2 √ + 2πk \| k is an integer and Arg(z) = arctan 2 5 √ , Re(z) = 4, Im(z) = −2, \|z\| = 2 arg(z) = arctan − 1 2 + 2πk \| k is an integer and Arg(z) = arctan − 1 2 − π. √ \|z\| = 29 − π. 5 = − arctan 1 2 . 20. z = 1 − 3i = √ 10cis (arctan (−3)), Re(z) = 1, Im(z) = −3, \|z\| = √ 10 arg(z) = {arctan (−3) + 2πk \| k is an integer} and Arg(z) = arctan (−3) = − arctan(3). 21. z = 6cis(0) = 6 √ 23. z = 7 25. z = 4cis 2π 3 2cis π 4 = −2 + 2i = 7 + 7i √ 3 27. z = 9cis (π) = −9 √ 3 + i = 22. z = 2cis π 6 24. z = 3cis π 2 6cis 3π 4 26. z = √ = 3i √ = − √ 3 + i 3 28. z = 3cis 4π 3 √ 30. z = = − 3 2 − 3i √ √ 3 2 13 13cis 3π 2 = −i 32. z = 12cis − π 3 = 6 − 6i √ 3 29. z = 7cis − 3π 4 31. z = 1 2 cis 7π 33. z = 8cis π 12 35. z = 5cis arctan + 4i 2 − √ 3 = 3 + 4i √ 37. z = 15cis (arctan (−2)) = 3 5 − 6i √ 5 √ 38. z = 34. z = 2cis 7π 8 √ 36 10cis arctan 1 3 √ 3cis arctan − 2 = 1 − i √ 2 39. z = 50cis π − arctan 7 24 = −48 + 14i 40. z = 1 2 cis π + arctan 5 12 = − 6 13 − 5i 26 11.7 Polar Form of Complex Numbers 1009 In Exercises 41 - 52, we have that z = − 3 we get the following. √ 3 2 + 3 2 i = 3cis 5π 6 41. zw = 18cis 7π 12 44. z4 = 81cis − 2π 3 47. z3w2 = 972cis(0) 50. z3 w2 = 3 4 cis(π) √ 53. −2 + 2i 33 = 64 and w = 3 √ √ 2 − 3i 2 = 6cis − π 4 so 43. w z = 2cis 11π 46. z5w2 = 8748cis − π 3 12 42. z w = 1 2 cis − 11π 12 45. w3 = 216cis − 3π 4 48. z2 w = 3 2 cis − π 12 49. w z2 = 2 3 cis π 12 51. w2 z3 = 4 √ 3 cis(π) 54. (− 3 − i)3 = −8i 52. w z 6 = 64cis − π 2 55. (−3 + 3i)4 = −324 √ 56. ( 3 + i)4 = −8 + 8i √ 3 57. 5 2 + 5 2 i3 = − 125 4 + 125 58. 59. 3 2 − 3 2 i3 = − 27 4 − 27 4 i 60. 62. (2 + 2i)5 = −128 − 128i 63. ( 3 − i)5 = −16 3 − 16i 81 − 8i √ √ 3 81 √ 2 2 + 4 √ 2 2 i 61. = −1 64. (1 − i)8 = 16 65. Since z = 4i = 4cis π 2 w0 = 2cis π 4 = √ 2 + i 2 we
have √ 66. Since z = −25i = 25cis 3π 2 we have w0 = 5cis 3π w1 = 2cis 5π 4 √ = − √ 2 − i 2 w1 = 5cis 7π 67. Since z = 1 + i √ 3 = 2cis π 3 we have w0 = √ 2cis w1 = √ 2cis 7π 68. Since z = 5 2 − 5 3 2 i = 5cis 5π 3 we have √ w0 = √ 5cis 5π 6 = − √ 15 2 + √ 5 2 i w1 = √ 5cis 11π 6 = √ 15 2 − √ 5 2 i 69. Since z = 64 = 64cis (0) we have w0 = 4cis (0) = 4 w1 = 4cis 2π 3 = −2 + 2i √ 3 w2 = 4cis 4π 3 = −2 − 2i √ 3 1010 Applications of Trigonometry 70. Since z = −125 = 125cis (π) we have w0 = 5cis w1 = 5cis (π) = −5 w2 = 5cis 5π 71. Since z = i = cis π 2 we have w0 = cis w1 = cis 5π w2 = cis 3π 2 = −i 72. Since z = −8i = 8cis 3π 2 we have w0 = 2cis π 2 = 2i w1 = 2cis 7π 6 √ = − 3 − i w2 = cis 11π 6 = √ 3 − i 73. Since z = 16 = 16cis (0) we have w0 = 2cis (0) = 2 w2 = 2cis (π) = −2 74. Since z = −81 = 81cis (π) we have w1 = 2cis π w3 = 2cis 3π = 2i = −2i 2 2 w0 = 3cis π 4 = 3 2 2 + 3 √ w2 = 3cis 5π w1 = 3cis 3π 4 w3 = 3cis 7π 75. Since z = 64 = 64cis(0) we have w0 = 2cis(0) = 2 w3 = 2cis (π) = −2 √ w1 = 2cis π w4 = 2cis − 2π 3 3 = 1 + 3i = −1 − √ w2 = 2cis 2π 3i w5 = 2cis − π = −1 + √ = 1 − 3 3i 3 √ 3i 76. Since z = −729 = 729cis(π) we have w0 = 3cis w1 = 3cis π 2 = 3i w2 = 3cis 5π w3 = 3cis 7π w4 = 3cis − 3π 2 = −3i w5 = 3cis − 11π 77. Note: In the answers for w0 and w2 the ﬁrst rectangular form comes from applying the 3 − π appropriate Sum or Diﬀerence Identity ( π 4 , respectively) and the second comes from using the Half-Angle Identities. √ √ = 3 2 √ w0 = 3 4 and 17π 12 = 2π 3 + 3π 12 = 2cis π 12 2+ 2 2− 2 6− 6 4 + i − √ 2 √ w1 = 3 2cis 3π 4 √ w2 = 3 2cis 17π 12 √ − √ 6 √ = 3 2 2− 4 √ √ 3 2− 2 + i √ 3 √ 2+ 2 11.7 Polar Form of Complex Numbers 1011 5 78. w0 = cis(0) = 1 w1 = cis 2π w2 = cis 4π w3 = cis 6π w4 = cis 8π ≈ 0.309 + 0.951i ≈ −0.809 + 0.588i ≈ −0.809 − 0.588i ≈ 0.309 − 0.951i 5 5 5 79. p(x) = x12 −4096 = (x−2)(x+2)(x2 +4)(x2 −2x+4)(x2 +2x+4)(x2 −2 √ 3x+4)(x2 +2 √ 3+4) 1012 Applications of Trigonometry 11.8 Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For many applications, real numbers suﬃce; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suﬃce. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. (Foreshadowing the use of bearings in the exercises, perhaps?) To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors.1 A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this text, we shall adopt2 the ‘arrow’ notation, so the symbol v is read as ‘the vector v’. Below is a typical vector v with endpoints P (1, 2) and Q (4, 6). The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct −−→ v completely from P and Q, we write v = P Q, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) Q (4, 6) P (1, 2) −−→ P Q v = While it is true that P and Q completely determine v, it is important to note that since vectors are deﬁned in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as v is considered to be the same vector as v, regardless of its initial point. In the case of our vector v above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v = 3, 4, where the ﬁrst number, 3, is called the x-component of v and the second number, 4, is called the y-component of v. If we wanted to reconstruct v = 3, 4 with initial point P (−2, 3), then we would ﬁnd the terminal point of v by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q(1, 7), as seen below. 1The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’ 2Other textbook authors use bold vectors such as v. We ﬁnd that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the ‘arrow’ notation. 3If this idea of ‘over’ and ‘up’ seems familiar, it should. The slope of the line segment containing v is 4 3 . 11.8 Vectors 1013 Q (1, 7) up 4 P (−2, 3) over 3 v = 3, 4 with initial point P (−2, 3). The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in our deﬁnition below. Deﬁnition 11.5. Suppose v is represented by a directed line segment with initial point P (x0, y0) and terminal point Q (x1, y1). The component form of v is given by −−→ P Q = x1 − x0, y1 − y0 v = Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, v1, v2 = v 1 and v2 = v 2. (Again, think about this before reading on.) We now set about deﬁning operations on vectors. Suppose we are given two vectors v and w. The sum, or resultant vector v + w is obtained as follows. First, plot v. Next, plot w so that its initial point is the terminal point of v. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. It is helpful to think of the vector v + w as the ‘net result’ of moving along v then moving along w. 2 if and only if v1 = v 1, v v + w w v v, w, and v + w Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 11.8.1. A plane leaves an airport with an airspeed5 of 175 miles per hour at a bearing of N40◦E. A 35 mile per hour wind is blowing at a bearing of S60◦E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. 4If necessary, review page 905 and Section 11.3. 5That is, the speed of the plane relative to the air around it. If there were no wind, plane’s airspeed would be the same as its speed as observed from the ground. How does wind aﬀect this? Keep reading! 1014 Applications of Trigonometry Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook.6 We let v denote the plane’s velocity and w denote the wind’s velocity in the diagram below. The ‘true’ speed and bearing is found by analyzing the resultant vector, v + w. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 175, the magnitude of w, which is 35, and the magnitude of v + w, which we’ll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 and 175 measures 100◦. N 40◦ w 60◦ v v + w N 35 100◦ 175 α c 40◦ E E 60◦ 31850 − 12250 cos(100◦) ≈ 184, which means the From the Law of Cosines, we determine c = true speed of the plane is (approximately) 184 miles per hour. To determine the true bearing of the plane, we need to determine the angle α. Using the Law of Cosines once more,7 we ﬁnd cos(α) = c2+29400 so that α ≈ 11◦. Given the geometry of the situation, we add α to the given 40◦ and ﬁnd the true bearing of the plane to be (approximately) N51◦E. 350c Our next step is to deﬁne addition of vectors component-wise to match the geometric action.8 Deﬁnition 11.6. Suppose v = v1, v2 and w = w1, w2. The vector v + w is deﬁned by v + w = v1 + w1, v2 + w2 Example 11.8.2. Let v = 3, 4 and suppose w = and interpret this sum geometrically. −−→ P Q where P (−3, 7) and Q(−2, 5). Find v + w Solution. Before can add the vectors using Deﬁnition 11.6, we need to write w in component form. Using Deﬁnition 11.5, we get w = −2 − (−3), 5 − 7 = 1, −2. Thus 6See Section 10.1.1, for instance. 7Or, since our given angle, 100◦, is obtuse, we could use the Law of Sines without any ambiguity here. 8Adding vectors ‘component-wise’ should seem hauntingly familiar. Compare this with how matrix addition was deﬁned in section 8.3. In fact, in more advanced courses such as Linear Algebra, vectors are deﬁned as 1 × n or n × 1 matrices, depending on the situation. 11.8 Vectors 1015 v + w = 3, 4 + 1, −2 = 3 + 1, 4 + (−2) = 4, 2 To visualize this sum, we draw v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w with its initial point at (3, 4). Moving one to the right and two down, we ﬁnd the terminal point of w to be (4, 2). We see that the vector v + w has initial point (0, 0) and terminal point (4, 2) so its component form is 4, 2, as required In order for vector addition to enjoy the same kinds of properties as real number addition, it is necessary to extend our deﬁnition of vectors to include a ‘zero vector’, 0 = 0, 0. Geometrically, 0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader may well object to the inclusion of 0, since after all, vectors are supposed to have both a magnitude (length) and a direction. While it seems clear that the magnitude of 0 should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undeﬁned, but this minor hiccup in the natural ﬂow of things is worth the beneﬁts we reap by including 0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition Commutative Property: For all vectors v and w, v + w = w + v. Associative Property: For all v
ectors u, v and w, (u + v) + w = u + (v + w). Identity Property: The vector 0 acts as the additive identity for vector addition. That is, for all vectors v. Inverse Property: Every vector v has a unique additive inverse, denoted −v. That is, for every vector v, there is a vector −v so that v + (−v) = (−v) + v = 0. 1016 Applications of Trigonometry The properties in Theorem 11.18 are easily veriﬁed using the deﬁnition of vector addition.9 For the commutative property, we note that if v = v1, v2 and w = w1, w2 then v + w = v1, v2 + w1, w2 = v1 + w1, v2 + w2 = w1 + v1, w2 + v2 = w + v Geometrically, we can ‘see’ the commutative property by realizing that the sums v + w and w + v are the same directed diagonal determined by the parallelogram below Demonstrating the commutative property of vector addition. The proofs of the associative and identity properties proceed similarly, and the reader is encouraged to verify them and provide accompanying diagrams. The existence and uniqueness of the additive inverse is yet another property inherited from the real numbers. Given a vector v = v1, v2, suppose we wish to ﬁnd a vector w = w1, w2 so that v + w = 0. By the deﬁnition of vector addition, we have v1 + w1, v2 + w2 = 0, 0, and hence, v1 + w1 = 0 and v2 + w2 = 0. We get w1 = −v1 and w2 = −v2 so that w = −v1, −v2. Hence, v has an additive inverse, and moreover, it is unique and can be obtained by the formula −v = −v1, −v2. Geometrically, the vectors v = v1, v2 and −v = −v1, −v2 have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum v + (−v) results in starting at the initial point of v and ending back at the initial point of v, or in other words, the net result of moving v then −v is not moving at all. v −v Using the additive inverse of a vector, we can deﬁne the diﬀerence of two vectors, v − w = v + (− w). If v = v1, v2 and w = w1, w2 then 9The interested reader is encouraged to compare Theorem 11.18 and the ensuing discussion with Theorem 8.3 in Section 8.3 and the discussion there. 11.8 Vectors 1017 v − w = v + (− w) = v1, v2 + −w1, −w2 = v1 + (−w1) , v2 + (−w2) = v1 − w1, v2 − w2 In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v − w geometrically, we note w + (v − w) = w + (v + (− w)) Deﬁnition of Vector Subtraction = w + ((− w) + v) Commutativity of Vector Addition = ( w + (− w)) + v Associativity of Vector Addition = 0 + v = v Deﬁnition of Additive Inverse Deﬁnition of Additive Identity This means that the ‘net result’ of moving along w then moving along v − w is just v itself. From the diagram below, we see that v − w may be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. It is also worth mentioning that in the parallelogram determined by the vectors v and w, the vector v − w is one of the diagonals – the other being v + w Next, we discuss scalar multiplication – that is, taking a real number times a vector. We deﬁne scalar multiplication for vectors in the same way we deﬁned it for matrices in Section 8.3. Deﬁnition 11.7. If k is a real number and v = v1, v2, we deﬁne kv by kv = k v1, v2 = kv1, kv2 Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector and reversing its direction (if k < 0) as demonstrated below. 1018 Applications of Trigonometry v 2v 1 2 v −2v Note that, by deﬁnition 11.7, (−1)v = (−1) v1, v2 = (−1)v1, (−1)v2 = −v1, −v2 = −v. This, and other properties of scalar multiplication are summarized below. Theorem 11.19. Properties of Scalar Multiplication Associative Property: For every vector v and scalars k and r, (kr)v = k(rv). Identity Property: For all vectors v, 1v = v. Additive Inverse Property: For all vectors v, −v = (−1)v. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k + r)v = kv + rv Distributive Property of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k, k(v + w) = kv + k w Zero Product Property: If v is vector and k is a scalar, then kv = 0 if and only if k = 0 or v = 0 The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the deﬁnition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let v = v1, v2. If k and r are scalars then (kr)v = (kr) v1, v2 = (kr)v1, (kr)v2 Deﬁnition of Scalar Multiplication = k(rv1), k(rv2) Associative Property of Real Number Multiplication = k rv1, rv2 = k (r v1, v2) = k(rv) Deﬁnition of Scalar Multiplication Deﬁnition of Scalar Multiplication 11.8 Vectors 1019 The remaining properties are proved similarly and are left as exercises. Our next example demonstrates how Theorem 11.19 allows us to do the same kind of algebraic manipulations with vectors as we do with variables – multiplication and division of vectors notwithstanding. If the pedantry seems familiar, it should. This is the same treatment we gave Example 8.3.1 in Section 8.3. As in that example, we spell out the solution in excruciating detail to encourage the reader to think carefully about why each step is justiﬁed. Example 11.8.3. Solve 5v − 2 (v + 1, −2) = 0 for v. Solution. 5v − 2 (v + 1, −2) = 0 5v + (−1) [2 (v + 1, −2)] = 0 5v + [(−1)(2)] (v + 1, −2) = 0 5v + (−2) (v + 1, −2) = 0 5v + [(−2)v + (−2) 1, −2] = 0 5v + [(−2)v + (−2)(1), (−2)(−2)] = 0 [5v + (−2)v] + −2, 4 = 0 (5 + (−2))v + −2, 4 = 0 3v + −2, 4 = 0 (3v + −2, 4) + (− −2, 4) = 0 + (− −2, 4) 3v + [−2, 4 + (− −2, 4)] = 0 + (−1) −2, 4 3v + 0 = 0 + (−1)(−2), (−1)(4) 3v = 2, −4 1 3 1 3 (3v) = 1 (3) v = 1 3 1v = 2 v = 2 3 (2, −4) (2), (−4) A vector whose initial point is (0, 0) is said to be in standard position. If v = v1, v2 is plotted in standard position, then its terminal point is necessarily (v1, v2). (Once more, think about this before reading on.) y (v1, v2) v = v1, v2 in standard position. x 1020 Applications of Trigonometry Plotting a vector in standard position enables us to more easily quantify the concepts of magnitude and direction of the vector. We can convert the point (v1, v2) in rectangular coordinates to a pair (r, θ) in polar coordinates where r ≥ 0. The magnitude of v, which we said earlier was length 2 and is denoted by v. From Section 11.4, we of the directed line segment, is r = know v1 = r cos(θ) = v cos(θ) and v2 = r sin(θ) = v sin(θ). From the deﬁnition of scalar multiplication and vector equality, we get 1 + v2 v2 v = v1, v2 = v cos(θ), v sin(θ) = v cos(θ), sin(θ) This motivates the following deﬁnition. Deﬁnition 11.8. Suppose v is a vector with component form v = v1, v2. Let (r, θ) be a polar representation of the point with rectangular coordinates (v1, v2) with r ≥ 0. The magnitude of v, denoted v, is given by v = r = 1 + v2 v2 2 If v = 0, the (vector) direction of v, denoted ˆv is given by ˆv = cos(θ), sin(θ) Taken together, we get v = v cos(θ), v sin(θ). A few remarks are in order. First, we note that if v = 0 then even though there are inﬁnitely many angles θ which satisfy Deﬁnition 11.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if θ and θ both satisfy the conditions of Deﬁnition 11.8, then cos(θ) = cos(θ) and sin(θ) = sin(θ), and as such, cos(θ), sin(θ) = cos(θ), sin(θ) making ˆv is well-deﬁned.10 If v = 0, then v = 0, 0, and we know from Section 11.4 that (0, θ) is a polar representation for the origin for any angle θ. For this reason, ˆ0 is undeﬁned. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem 11.20. Properties of Magnitude and Direction: Suppose v is a vector. v ≥ 0 and v = 0 if and only if v = 0 For all scalars k, k v = \|k\|v. If v = 0 then v = vˆv, so that ˆv = 1 v v. 1 + v2 v2 v2 1 + v2 1 + v2 2 = 0 if and only of v2 The proof of the ﬁrst property in Theorem 11.20 is a direct consequence of the deﬁnition of v. If v = v1, v2, then v = 2 which is by deﬁnition greater than or equal to 0. Moreover, 2 = 0 if and only if v1 = v2 = 0. Hence, v = 0 if and only if v = 0, 0 = 0, as required. The second property is a result of the deﬁnition of magnitude and scalar multiplication along with a propery of radicals. If v = v1, v2 and k is a scalar then 10If this all looks familiar, it should. The interested reader is invited to compare Deﬁnition 11.8 to Deﬁnition 11.2 in Section 11.7. 11.8 Vectors 1021 k v = k v1, v2 = kv1, kv2 Deﬁnition of scalar multiplication 2 = = = √ (kv1)2 + (kv2)2 Deﬁnition of magnitude 1 + k2v2 k2v2 1 + v2 k2(v2 2 ) k2 v2 1 + v2 = = \|k\| 1 + v2 v2 = \|k\|v Product Rule for Radicals k2 = \|k\| Since √ 2 2 The equation v = vˆv in Theorem 11.20 is a consequence of the deﬁnitions of v and ˆv and was worked out in the discussion just prior to Deﬁnition 11.8 on page 1020. In words, the equation v = vˆv says that any given vector is the product of its magnitude and its direction – an important v is a result of concept to keep in mind when studying and using vectors. The equation ˆv = solving v = vˆv for ˆv by multiplying11 both sides of the equation by 1 v and using the properties of Theorem 11.19. We are overdue for an example. 1 v Example 11.8.4. 1. Find the component form of the vector v with v = 5 so that when v is plotted in standard position, it lies in Quadrant II and makes a 60◦ angle12 with the negative x-axis. 2. For v = 3, −3 √ 3, ﬁnd v and θ, 0 ≤ θ < 2π so that v = v cos(θ), sin(θ). 3. For the vectors v = 3, 4 and w = 1, −2, ﬁnd the following. (a) ˆv (b) v − 2 w (c) v − 2 w (d) ˆw Solution. 1. We are told that v = 5 and are given information about its direction, so we can use the formula v = vˆv to get the component form of v. To determine ˆv, we appeal to Deﬁnition 11.8. We are told that v lies in Quadrant II and makes a 60◦ angle with the negative x-axis, so the polar form of the terminal point of v, when plotted in standard position is (5, 120◦
). (See the diagram below.) Thus ˆv = cos (120◦) , sin (120◦) = , so v = vˆv = √ − . 11Of course, to go from v = vˆv to ˆv = 1 v v, we are essentially ‘dividing both sides’ of the equation by the scalar v. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the properties in play. 12Due to the utility of vectors in ‘real-world’ applications, we will usually use degree measure for the angle when giving the vector’s direction. However, since Carl doesn’t want you to forget about radians, he’s made sure there are examples and exercises which use them. 1022 Applications of Trigonometry y 5 4 3 2 1 v 60◦ θ = 120◦ −3 −2 −1 1 2 3 x 2. For v = 3, −3 √ 3, we get v = (3)2 + (−3 √ ﬁnd the θ we’re after by converting the point with rectangular coordinates (3, −3 form (r, θ) where r = v > 0. From Section 11.4, we have tan(θ) = −3 (3, −3 may check our answer by verifying v = 3, −3 3) is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick θ = 5π √ 3)2 = 6. In light of Deﬁnition 11.8, we can 3) to polar √ 3. Since 3 . We √ 3 = − . 3 = 6 cos 5π 3 , sin 5π 3 √ √ 3 3. (a) Since we are given the component form of v, we’ll use the formula ˆv = 5 3, 4, we have v = 25 = 5. Hence, ˆv = 1 32 + 42 = √ √ 5 1 v . v. For (b) We know from our work above that v = 5, so to ﬁnd v−2 w, we need only ﬁnd w. 12 + (−2)2 = Since w = 1, −2, we get w = 5. Hence. (c) In the expression v −2 w, notice that the arithmetic on the vectors comes ﬁrst, then the magnitude. Hence, our ﬁrst step is to ﬁnd the component form of the vector v − 2 w. We get v − 2 w = 3, 4 − 2 1, −2 = 1, 8. Hence, v − 2 w = 1, 8 = 65. √ 12 + 82 = √ √ (d) To ﬁnd ˆw, we ﬁrst need ˆw. Using the formula ˆw = 1 w which we found the in the previous problem, we get ˆw = 1√ 5 √ √ 5 5 . Hence, ˆw = 25 + 20 . 25 = w along with w = 5, 1, −2 = 1√ 5 , − 2√ 5 = The process exempliﬁed by number 1 in Example 11.8.4 above by which we take information about the magnitude and direction of a vector and ﬁnd the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Example 11.8.1 below. Example 11.8.5. A plane leaves an airport with an airspeed of 175 miles per hour with bearing N40◦E. A 35 mile per hour wind is blowing at a bearing of S60◦E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution: We proceed as we did in Example 11.8.1 and let v denote the plane’s velocity and w denote the wind’s velocity, and set about determining v + w. If we regard the airport as being 11.8 Vectors 1023 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50◦ and −30◦, respectively. Hence, the component form of v = 175 cos(50◦), sin(50◦) = 175 cos(50◦), 175 sin(50◦) and the component form of w = 35 cos(−30◦), 35 sin(−30◦). Since we have no convenient way to express the exact values of cosine and sine of 50◦, we leave both vectors in terms of cosines and sines.13 Adding corresponding components, we ﬁnd the resultant vector v + w = 175 cos(50◦) + 35 cos(−30◦), 175 sin(50◦) + 35 sin(−30◦). To ﬁnd the ‘true’ speed of the plane, we compute the magnitude of this resultant vector v + w = (175 cos(50◦) + 35 cos(−30◦))2 + (175 sin(50◦) + 35 sin(−30◦))2 ≈ 184 Hence, the ‘true’ speed of the plane is approximately 184 miles per hour. To ﬁnd the true bearing, we need to ﬁnd the angle θ which corresponds to the polar form (r, θ), r > 0, of the point (x, y) = (175 cos(50◦) + 35 cos(−30◦), 175 sin(50◦) + 35 sin(−30◦)). Since both of these coordinates are positive,14 we know θ is a Quadrant I angle, as depicted below. Furthermore, tan(θ) = y x = 175 sin(50◦) + 35 sin(−30◦) 175 cos(50◦) + 35 cos(−30◦) , so using the arctangent function, we get θ ≈ 39◦. Since, for the purposes of bearing, we need the angle between v + w and the positive y-axis, we take the complement of θ and ﬁnd the ‘true’ bearing of the plane to be approximately N51◦E. y (N) y (N) v v v + w 40◦ 50◦ −30◦ w 60◦ x (E) θ w x (E) In part 3d of Example 11.8.4, we saw that ˆw = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Deﬁnition 11.9. Unit Vectors: Let v be a vector. If v = 1, we say that v is a unit vector. 13Keeping things ‘calculator’ friendly, for once! 14Yes, a calculator approximation is the quickest way to see this, but you can also use good old-fashioned inequalities and the fact that 45◦ ≤ 50◦ ≤ 60◦. 1024 Applications of Trigonometry 1 v If v is a unit vector, then necessarily, v = vˆv = 1 · ˆv = ˆv. Conversely, we leave it as an exercise15 In practice, if v is a unit v is a unit vector for any nonzero vector v. to show that ˆv = vector we write it as ˆv as opposed to v because we have reserved the ‘ˆ’ notation for unit vectors. 1 The process of multiplying a nonzero vector by the factor v to produce a unit vector is called ‘normalizing the vector,’ and the resulting vector ˆv is called the ‘unit vector in the direction of v’. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. (You should take the time to show this.) As a result, we visualize normalizing a nonzero vector v as shrinking16 its terminal point, when plotted in standard position, back to the Unit Circle. y v 1 ˆv −1 1 x −1 Visualizing vector normalization ˆv = 1 v v Of all of the unit vectors, two deserve special mention. Deﬁnition 11.10. The Principal Unit Vectors: The vector ˆı is deﬁned by ˆı = 1, 0 The vector ˆ is deﬁned by ˆı = 0, 1 We can think of the vector ˆı as representing the positive x-direction, while ˆ represents the positive y-direction. We have the following ‘decomposition’ theorem.17 Theorem 11.21. Principal Vector Decomposition Theorem: Let v be a vector with component form v = v1, v2. Then v = v1ˆı + v2ˆ. The proof of Theorem 11.21 is straightforward. Since ˆı = 1, 0 and ˆ = 0, 1, we have from the deﬁnition of scalar multiplication and vector addition that v1ˆı + v2ˆ = v1 1, 0 + v2 0, 1 = v1, 0 + 0, v2 = v1, v2 = v 15One proof uses the properties of scalar multiplication and magnitude. If v = 0, consider ˆv = 1 v . Use v the fact that v ≥ 0 is a scalar and consider factoring. 16. . . if v > 1 . . . 17We will see a generalization of Theorem 11.21 in Section 11.9. Stay tuned! 11.8 Vectors 1025 Geometrically, the situation looks like this: y v2ˆ v = v1, v2 ˆ ˆı v1ˆı x v = v1, v2 = v1ˆı + v2ˆ. We conclude this section with a classic example which demonstrates how vectors are used to model forces. A ‘force’ is deﬁned as a ‘push’ or a ‘pull.’ The intensity of the push or pull is the magnitude of the force, and is measured in Netwons (N) in the SI system or pounds (lbs.) in the English system.18 The following example uses all of the concepts in this section, and should be studied in great detail. Example 11.8.6. A 50 pound speaker is suspended from the ceiling by two support braces. If one of them makes a 60◦ angle with the ceiling and the other makes a 30◦ angle with the ceiling, what are the tensions on each of the supports? Solution. We represent the problem schematically below and then provide the corresponding vector diagram. 30◦ 60◦ 30◦ 60◦ T1 T2 30◦ 60◦ w 50 lbs. We have three forces acting on the speaker: the weight of the speaker, which we’ll call w, pulling the speaker directly downward, and the forces on the support rods, which we’ll call T1 and T2 (for ‘tensions’) acting upward at angles 60◦ and 30◦, respectively. We are looking for the tensions on the support, which are the magnitudes T1 and T2. In order for the speaker to remain stationary,19 we require w + T1 + T2 = 0. Viewing the common initial point of these vectors as the 18See also Section 11.1.1. 19This is the criteria for ‘static equilbrium’. 1026 Applications of Trigonometry origin and the dashed line as the x-axis, we use Theorem 11.20 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directly downwards with a magnitude of 50 pounds. That is, w = 50 and ˆw = −ˆ = 0, −1. Hence, w = 50 0, −1 = 0, −50. For the force in the ﬁrst support, we get T1 = T1 cos (60◦) , sin (60◦) T1 2 , √ T1 2 3 = For the second support, we note that the angle 30◦ is measured from the negative x-axis, so the angle needed to write T2 in component form is 150◦. Hence T2 = T2 cos (150◦) , sin (150◦) √ = − T2 2 3 , T2 2 The requirement w + T1 + T2 = 0 gives us this vector equation. 0, −50 + , T1 2 T1 2 − T1 2 T2 2 √ 3 + √ 3 , T1 2 − √ , 3 T2 2 T2 2 + 3 w + T1 + T2 = 0 √ T2 2 = 0, 0 − 50 = 0, 0 Equating the corresponding components of the vectors on each side, we get a system of linear equations in the variables T1 and T2.    (E1) (E2) T1 2 3 √ + T1 2 √ T2 2 3 − = 0 T2 2 − 50 = 0 From (E1), we get T1 = T2 which yields 2 T2 − 50 = 0. Hence, T2 = 25 pounds and T1 = T2 3. Substituting that into (E2) gives ( T2 √ 2 3 = 25 + T2 3 √ 3 pounds. 2 − 50 = 0 √ √ √ 3) 11.8 Vectors 11.8.1 Exercises 1027 In Exercises 1 - 10, use the given pair of vectors v and w to ﬁnd the following quantities. State whether the result is a vector or a scalar. • v + w • w − 2v • − wv • wˆv Finally, verify that the vectors satisfy the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 1. v = 12, −5, w = 3, 4 2. v = −7, 24, w = −5, −12 3. v = 2, −1, w = −2, 4 √ √ 3, 1, w = 2 5. v = − √ 7. v = 3. v = 4. v = 10, 4, w = −2, 5 61, − 3 2 √ 3 9. v = 3ˆı + 4ˆ, w = −2ˆ 10. v = 1 2 (ˆı + ˆ), w = 1 2 (ˆı − ˆ) In Exercises 11 - 25, ﬁnd the component form of the vector v using the information given about its magnitude and direction. Give exact values. 11. v = 6; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive x-axis 12. v = 3; when drawn in standard position v lies in Quadrant I and makes a 45◦
angle with the positive x-axis 13. v = 2 3 ; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive y-axis 14. v = 12; when drawn in standard position v lies along the positive y-axis 15. v = 4; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the negative x-axis √ 16. v = 2 3; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the positive y-axis 17. v = 7 2 ; when drawn in standard position v lies along the negative x-axis √ 18. v = 5 6; when drawn in standard position v lies in Quadrant III and makes a 45◦ angle with the negative x-axis 19. v = 6.25; when drawn in standard position v lies along the negative y-axis 1028 20. v = 4 √ 3; when drawn in standard position v lies in Quadrant IV and makes a 30◦ angle Applications of Trigonometry with the positive x-axis 21. v = 5 √ 2; when drawn in standard position v lies in Quadrant IV and makes a 45◦ angle with the negative y-axis 22. v = 2 √ 5; when drawn in standard position v lies in Quadrant I and makes an angle measuring arctan(2) with the positive x-axis √ 23. v = 10; when drawn in standard position v lies in Quadrant II and makes an angle measuring arctan(3) with the negative x-axis 24. v = 5; when drawn in standard position v lies in Quadrant III and makes an angle measuring arctan 4 3 with the negative x-axis 25. v = 26; when drawn in standard position v lies in Quadrant IV and makes an angle measuring arctan 5 12 with the positive x-axis In Exercises 26 - 31, approximate the component form of the vector v using the information given about its magnitude and direction. Round your approximations to two decimal places. 26. v = 392; when drawn in standard position v makes a 117◦ angle with the positive x-axis 27. v = 63.92; when drawn in standard position v makes a 78.3◦ angle with the positive x-axis 28. v = 5280; when drawn in standard position v makes a 12◦ angle with the positive x-axis 29. v = 450; when drawn in standard position v makes a 210.75◦ angle with the positive x-axis 30. v = 168.7; when drawn in standard position v makes a 252◦ angle with the positive x-axis 31. v = 26; when drawn in standard position v makes a 304.5◦ angle with the positive x-axis In Exercises 32 - 52, for the given vector v, ﬁnd the magnitude v and an angle θ with 0 ≤ θ < 360◦ so that v = v cos(θ), sin(θ) (See Deﬁnition 11.8.) Round approximations to two decimal places. 32. v = 1, √ 35. v = − √ 2, 3 √ 2 33. v = 5, 5 36 38. v = 6, 0 41. v = −10ˆ 39. v = −2.5, 0 40. v = 0, 42. v = 3, 4 43. v = 12, 5 34. v = −2 √ 37. v = − 1 3 11.8 Vectors 1029 44. v = −4, 3 45. v = −7, 24 46. v = −2, −1 47. v = −2, −6 48. v = ˆı + ˆ 49. v = ˆı − 4ˆ 50. v = 123.4, −77.05 51. v = 965.15, 831.6 52. v = −114.1, 42.3 53. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S68◦W. The river is ﬂowing due east at 8 miles per hour. What is the boat’s true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 54. The HMS Sasquatch leaves port with bearing S20◦E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N60◦E, ﬁnd the HMS Sasquatch’s true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 55. If the captain of the HMS Sasquatch in Exercise 54 wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S20◦E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current, what does v + w need to be to reach Chupacabra Cove in three hours? 56. In calm air, a plane ﬂying from the Pedimaxus International Airport can reach Cliﬀs of Insanity Point in two hours by following a bearing of N8.2◦E at 96 miles an hour. (The distance between the airport and the cliﬀs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree. 57. The SS Bigfoot leaves Yeti Bay on a course of N37◦W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S40◦E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. 58. A 600 pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a 60◦ angle with the ceiling, ﬁnd the tension on each cable. Round your answer to the nearest pound. 59. Two cables are to support an object hanging from a ceiling. If the cables are each to make a 42◦ angle with the ceiling, and each cable is rated to withstand a maximum tension of 100 pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound. 1030 Applications of Trigonometry 60. A 300 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 72◦ angle with the ceiling while the right hand cable makes a 18◦ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places. 61. Two drunken college students have ﬁlled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77◦E and the other pulls at a heading of S68◦E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound. 62. Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The ﬁrst rope points N80◦W, the second points due west and the third points S80◦W. The force applied to the ﬁrst rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Does it move? If so, is it heading due west? 63. Let v = v1, v2 be any non-zero vector. Show that 1 v v has length 1. 64. We say that two non-zero vectors v and w are parallel if they have same or opposite directions. That is, v = 0 and w = 0 are parallel if either ˆv = ˆw or ˆv = − ˆw. Show that this means v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if they point in opposite directions. 65. The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line y = 2x − 4. Let v0 = 0, −4 and let s = 1, 2. Let t be any real number. Show that the vector deﬁned by v = v0 + ts, when drawn in standard position, has its terminal point on the line y = 2x − 4. (Hint: Show that v0 + ts = t, 2t − 4 for any real number t.) Now consider the non-vertical line y = mx + b. Repeat the previous analysis with v0 = 0, b and let s = 1, m. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of 0, b (the position vector of the y-intercept) and a scalar multiple of the slope vector s = 1, m. 66. Prove the associative and identity properties of vector addition in Theorem 11.18. 67. Prove the properties of scalar multiplication in Theorem 11.19. 11.8 Vectors 11.8.2 Answers 1. v + w = 15, −1, vector √ v + w = 226, scalar 1031 w − 2v = −21, 14, vector v + w = 18, scalar v w − wv = −21, 77, vector wˆv = 60 13 , − 25 13 , vector 2. v + w = −12, 12, vector w − 2v = 9, −60, vector v + w = 12 2, scalar √ v w − wv = −34, −612, vector v + w = 38, scalar wˆv = − 91 25 , 312 25 , vector 3. v + w = 0, 3, vector v + w = 3, scalar w − 2v = −6, 6, vector v + w = 3 √ 5, scalar v w − wv = −6 √ √ 5, 6 5, vector wˆv = 4, −2, vector 4. v + w = 8, 9, vector 5. 6. v + w = √ v w − wv = −14 v + w = √ 145, scalar √ 3, 3, vector √ 3, scalar v + w = 2 √ 29, vector 29, 6 v w − wv = 8 √ 3, 0, vector , scalar , vector v w − wv = − 7 5 , − 1 5 , vector w − 2v = −22, −3, vector v + w = 3 √ 29, scalar wˆv = 5, 2, vector w − 2v = 4 √ 3, 0, vector v + w = 6, scalar wˆv = −2 √ 3, 2, vector w − 2v = −2, −1, vector v + w = 2, scalar wˆv = 3 5 5 , 4 , vector √ √ − 3 2 2 , 3 2 2 7. v + w = 0, 0, vector w − 2v = , vector v + w = 0, scalar v w − wv = − √ √ 2, 2, vector v + w = 2, scalar wˆv = √ 2 2 , − √ 2 2 , vector 1032 Applications of Trigonometry 8 , vector w − 2v = −2, −2 √ 3, vector v + w = 1, scalar v w − wv = −2, −2 √ 3, vector v + w = 3, scalar wˆv = 1, √ 3, vector 9. v + w = 3, 2, vector v + w = √ 13, scalar w − 2v = −6, −10, vector v + w = 7, scalar v w − wv = −6, −18, vector 10. v + w = 1, 0, vector v + w = 1, scalar v w − wv = 0, − √ 2 2 , vector , vector wˆv = 6 5 , 8 5 w − 2v = − , scalar , vector wˆv = 1 2 , 1 2 , vector 11. v = 3, 3 √ 3 14. v = 0, 12 17. v = − 7 2 , 0 √ 20. v = 6, −2 3 12. v = √ 3 2 2 , 3 √ 15. v = −2 3, 2 √ 2 2 13. v = √ 3 3 3 , 1 √ 16. v = − 3, 3 18. v = −5 √ √ 3 3, −5 19. v = 0, −6.25 21. v = 5, −5 22. v = 2, 4 23. v = −1, 3 24. v = −3, −4 25. v = 24, −10 26. v ≈ −177.96, 349.27 27. v ≈ 12.96,
62.59 28. v ≈ 5164.62, 1097.77 29. v ≈ −386.73, −230.08 32. v = 2, θ = 60◦ 30. v ≈ −52.13, −160.44 √ 33. v = 5 2, θ = 45◦ 31. v ≈ 14.73, −21.43 34. v = 4, θ = 150◦ 35. v = 2, θ = 135◦ 36. v = 1, θ = 225◦ 38. v = 6, θ = 0◦ 39. v = 2.5, θ = 180◦ 41. v = 10, θ = 270◦ 42. v = 5, θ ≈ 53.13◦ 44. v = 5, θ ≈ 143.13◦ 45. v = 25, θ ≈ 106.26◦ 37. v = 1, θ = 240◦ √ 40. v = 7, θ = 90◦ 43. v = 13, θ ≈ 22.62◦ √ 46. v = 5, θ ≈ 206.57◦ 11.8 Vectors 1033 47. v = 2 √ 10, θ ≈ 251.57◦ 48. v = √ 2, θ ≈ 45◦ √ 49. v = 17, θ ≈ 284.04◦ 50. v ≈ 145.48, θ ≈ 328.02◦ 51. v ≈ 1274.00, θ ≈ 40.75◦ 52. v ≈ 121.69, θ ≈ 159.66◦ 53. The boat’s true speed is about 10 miles per hour at a heading of S50.6◦W. 54. The HMS Sasquatch’s true speed is about 41 miles per hour at a heading of S26.8◦E. 55. She should maintain a speed of about 35 miles per hour at a heading of S11.8◦E. 56. She should ﬂy at 83.46 miles per hour with a heading of N22.1◦E 57. The current is moving at about 10 miles per hour bearing N54.6◦W. 58. The tension on each of the cables is about 346 pounds. 59. The maximum weight that can be held by the cables in that conﬁguration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 pounds so the couch doesn’t budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street. 1034 Applications of Trigonometry 11.9 The Dot Product and Projection In Section 11.8, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we deﬁne a product of vectors. We begin with the following deﬁnition. Deﬁnition 11.11. Suppose v and w are vectors whose component forms are v = v1, v2 and w = w1, w2. The dot product of v and w is given by v · w = v1, v2 · w1, w2 = v1w1 + v2w2 For example, let v = 3, 4 and w = 1, −2. Then v · w = 3, 4 · 1, −2 = (3)(1) + (4)(−2) = −5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity v· w is often called the scalar product of v and w. The dot product enjoys the following properties. Theorem 11.22. Properties of the Dot Product Commutative Property: For all vectors v and w, v · w = w · v. Distributive Property: For all vectors u, v and w, u · (v + w) = u · v + u · w. Scalar Property: For all vectors v and w and scalars k, (kv) · w = k(v · w) = v · (k w). Relation to Magnitude: For all vectors v, v · v = v2. Like most of the theorems involving vectors, the proof of Theorem 11.22 amounts to using the deﬁnition of the dot product and properties of real number arithmetic. To show the commutative property for instance, let v = v1, v2 and w = w1, w2. Then v · w = v1, v2 · w1, w2 = v1w1 + v2w2 Deﬁnition of Dot Product = w1v1 + w2v2 = w1, w2 · v1, v2 Deﬁnition of Dot Product = w · v Commutativity of Real Number Multiplication The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that v = v1, v2 and w = w1, w2 and k is a scalar. Then (kv) · w = (k v1, v2) · w1, w2 = kv1, kv2 · w1, w2 = (kv1)(w1) + (kv2)(w2) Deﬁnition of Dot Product Deﬁnition of Scalar Multiplication = k(v1w1) + k(v2w2) Associativity of Real Number Multiplication = k(v1w1 + v2w2) = k v1, v2 · w1, w2 = k(v · w) Distributive Law of Real Numbers Deﬁnition of Dot Product We leave the proof of k(v · w) = v · (k w) as an exercise. 11.9 The Dot Product and Projection 1035 For the last property, we note that if v = v1, v2, then v · v = v1, v2 · v1, v2 = v2 where the last equality comes courtesy of Deﬁnition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justiﬁcation for each step. 2 = v2, 1 + v2 Example 11.9.1. Prove the identity: v − w2 = v2 − 2(v · w) + w2. Solution. We begin by rewriting v − w2 in terms of the dot product using Theorem 11.22. v − w2 = (v − w) · (v − w) = (v + [− w]) · (v + [− w]) = (v + [− w]) · v + (v + [− w]) · [− w] = v · (v + [− w]) + [− w] · (v + [− w]) = v · v + v · [− w] + [− w] · v + [− w] · [− w] = v · v + v · [(−1) w] + [(−1) w] · v + [(−1) w] · [(−1) w] = v · v + (−1)(v · w) + (−1)( w · v) + [(−1)(−1)]( w · w) = v · v + (−1)(v · w) + (−1)(v · w(v · w) + w · w = v2 − 2(v · w) + w2 Hence, v − w2 = v2 − 2(v · w) + w2 as required. If we take a step back from the pedantry in Example 11.9.1, we see that the bulk of the work is needed to show that (v − w)·(v − w) = v ·v −2(v · w)+ w· w. If this looks familiar, it should. Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand (v − w) · (v − w) for vectors v and w match those required to expand (v − w)(v − w) for real numbers v and w, and hence we get similar looking results. The identity veriﬁed in Example 11.9.1 plays a large role in the development of the geometric properties of the dot product, which we now explore. Suppose v and w are two nonzero vectors. If we draw v and w with the same initial point, we deﬁne the angle between v and w to be the angle θ determined by the rays containing the vectors v and w, as illustrated below. We require 0 ≤ θ ≤ π. (Think about why this is needed in the deﬁnition.) The following theorem gives us some insight into the geometric role the dot product plays. Theorem 11.23. Geometric Interpretation of Dot Product: vectors then v · w = v w cos(θ), where θ is the angle between v and w. If v and w are nonzero 1036 Applications of Trigonometry We prove Theorem 11.23 in cases. If θ = 0, then v and w have the same direction. It follows1 that there is a real number k > 0 so that w = kv. Hence, v · w = v · (kv) = k(v · v) = kv2 = kvv. Since k > 0, k = \|k\|, so kv = \|k\|v = kv by Theorem 11.20. Hence, kvv = v(kv) = vkv = v w. Since cos(0) = 1, we get v · w = kvv = v w = v w cos(0), proving that the formula holds for θ = 0. If θ = π, we repeat the argument with the diﬀerence being w = kv where k < 0. In this case, \|k\| = −k, so kv = −\|k\|v = −kv = − w. Since cos(π) = −1, we get v · w = −v w = v w cos(π), as required. Next, if 0 < θ < π, the vectors v, w and v − w determine a triangle with side lengths v, w and v − w, respectively, as seen below The Law of Cosines yields v − w2 = v2 + w2 − 2v w cos(θ). From Example 11.9.1, we know v − w2 = v2 − 2(v · w) + w2. Equating these two expressions for v − w2 gives v2 + w2 −2v w cos(θ) = v2 −2(v· w)+ w2 which reduces to −2v w cos(θ) = −2(v· w), or v · w = v w cos(θ), as required. An immediate consequence of Theorem 11.23 is the following. Theorem 11.24. Let v and w be nonzero vectors and let θ the angle between v and w. Then θ = arccos v · w v w = arccos(ˆv · ˆw) We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for θ. Since v and w are nonzero, so are v and w. Hence, we may divide both sides of v · w = v w cos(θ) by v w to get cos(θ) = v· w v w . Since 0 ≤ θ ≤ π by deﬁnition, the values of θ exactly match the . Using Theorem 11.22, we can rewrite v· w range of the arccosine function. Hence, θ = arccos v w = ˆv · ˆw, giving us the alternative formula θ = arccos(ˆv · ˆw). v· w v w = 1 vv 1 w w · We are overdue for an example. Example 11.9.2. Find the angle between the following pairs of vectors. 1. v = 3, −3 √ 3, and w = − 3, 1 √ 2. v = 2, 2, and w = 5, −5 3. v = 3, −4, and w = 2, 1 Solution. We use the formula θ = arccos v· w v w from Theorem 11.24 in each case below. 1Since v = vˆv and w = w ˆw, if ˆv = ˆw then w = wˆv = w v (vˆv) = w v v. In this case, k = w v > 0. 1037 √ 3)2 = 11.9 The Dot Product and Projection 1. We have v· w = 3, −3 √ 3·− √ 3, 1 = −3 √ 36 = 6 and w = (− 3)2 + 12 = √ √ √ √ 3−3 3 = −6 √ 4 = 2, θ = arccos 3. Since v = −6 12 = arccos √ 3 32 + (−3 √ − 3 2 = 5π 6 . 2. For v = 2, 2 and w = 5, −5, we ﬁnd v · w = 2, 2 · 5, −5 = 10 − 10 = 0. Hence, it doesn’t v· w matter what v and w are,2 θ = arccos = arccos(0) = π 2 . v w 3. We ﬁnd v · w = 3, −4 · 2, 1 = 6 − 4 = 2. Also v = √ √ w = 22 + 12 = 5, so θ = arccos = arccos isn’t the cosine of one 32 + (−4)2 = √ 25 = 5 and 2 √ 5 5 of the common angles, we leave our answer as θ = arccos 2 √ 5 25 √ 5 25 . Since 2 2 √ 5 25 . The vectors v = 2, 2, and w = 5, −5 in Example 11.9.2 are called orthogonal and we write v ⊥ w, because the angle between them is π 2 radians = 90◦. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. w v v and w are orthogonal, v ⊥ w We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem 11.25. The Dot Product Detects Orthogonality: Let v and w be nonzero vectors. Then v ⊥ w if and only if v · w = 0. To prove Theorem 11.25, we ﬁrst assume v and w are nonzero vectors with v ⊥ w. By deﬁnition, = 0. Conversely, the angle between v and w is π v· w if v and w are nonzero vectors and v · w = 0, then Theorem 11.24 gives θ = arccos = 2 . By Theorem 11.23, v · w = v w cos π 2 v w arccos to provide a diﬀerent proof about the relationship between the slopes of perpendicular lines.3 2 , so v ⊥ w. We can use Theorem 11.25 in the following example v w = arccos(0) = π 0 Example 11.9.3. Let L1 be the line y = m1x + b1 and let L2 be the line y = m2x + b2. Prove that L1 is perpendicular to L2 if and only if m1 · m2 = −1. Solution. Our strategy is to ﬁnd two vectors: v1, which has the same direction as L1, and v2, which has the same direction as L2 and show v1 ⊥ v2 if and only if m1m2 = −1. To that end, we substitute x = 0 and x = 1 into y = m1x + b1 to ﬁnd two points which lie on L1, namely P (0, b1) 2Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0, yet v · w = 0. 3See Exercise 2.1.1 in Section 2.1. 1038 Applications of Trigonometry −−→ P Q = 1 − 0, (m1 + b1) − b1 = 1, m1, an
d note that since v1 is and Q(1, m1 + b1). We let v1 = determined by two points on L1, it may be viewed as lying on L1. Hence it has the same direction as L1. Similarly, we get the vector v2 = 1, m2 which has the same direction as the line L2. Hence, L1 and L2 are perpendicular if and only if v1 ⊥ v2. According to Theorem 11.25, v1 ⊥ v2 if and only if v1 · v2 = 0. Notice that v1 · v2 = 1, m1 · 1, m2 = 1 + m1m2. Hence, v1 · v2 = 0 if and only if 1 + m1m2 = 0, which is true if and only if m1m2 = −1, as required. While Theorem 11.25 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors v and w drawn with a common initial point O below. For the moment, assume that the angle between v and w, which we’ll denote θ, is acute. We wish to develop a formula for the vector p, indicated below, which is called the orthogonal projection of v onto w. The vector p is obtained geometrically as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point −−→ of intersection R. The vector p is then deﬁned as p = OR. Like any vector, p is determined by its magnitude p and its direction ˆp according to the formula p = pˆp. Since we want ˆp to have the same direction as w, we have ˆp = ˆw. To determine p, we make use of Theorem 10.4 as applied to the right triangle ORT . We ﬁnd cos(θ) = p v , or p = v cos(θ). To get things in terms of just v and w, we use Theorem 11.23 to get p = v cos(θ) = v w cos(θ) w . Using Theorem 11.22, we rewrite v· w = v · ˆw. Hence, p = v · ˆw, and since ˆp = ˆw, we now have a formula for p completely in terms of v and w, namely p = pˆp = (v · ˆw) ˆw. w = v · = v −−→ OR p = v w θ R p O Now suppose that the angle θ between v and w is obtuse, and consider the diagram below. In this case, we see that ˆp = − ˆw and using the triangle ORT , we ﬁnd p = v cos(θ). Since θ +θ = π, it follows that cos(θ) = − cos(θ), which means p = v cos(θ) = −v cos(θ). Rewriting this last equation in terms of v and w as before, we get p = −(v · ˆw). Putting this together with ˆp = − ˆw, we get p = pˆp = −(v · ˆw)(− ˆw) = (v · ˆw) ˆw in this case as well. 11.9 The Dot Product and Projection 1039 T v w θ O θ −−→ OR p = R If the angle between v and w is π v · w = 0. It follows that v · ˆw = 0 and p = 0 = 0 ˆw = (v · ˆw) ˆw in this case, too. This gives us 2 then it is easy to show4 that p = 0. Since v ⊥ w in this case, Deﬁnition 11.12. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w(v) is given by proj w(v) = (v · ˆw) ˆw. Deﬁnition 11.12 gives us a good idea what the dot product does. The scalar v · ˆw is a measure of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. While the formula given in Deﬁnition 11.12 is theoretically appealing, because of the presence of the normalized unit vector ˆw, computing the projection using the formula proj w(v) = (v · ˆw) ˆw can be messy. We present two other formulas that are often used in practice. Theorem 11.26. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then proj w(v) = (v · ˆw) ˆw = v · w w2 w = v · w w · w w The proof of Theorem 11.26, which we leave to the reader as an exercise, amounts to using the formula ˆw = w and properties of the dot product. It is time for an example. 1 w Example 11.9.4. Let v = 1, 8 and w = −1, 2. Find p = proj w(v), and plot v, w and p in standard position. Solution. We ﬁnd v · w = 1, 8 · −1, 2 = (−1) + 16 = 15 and w · w = −1, 2 · −1, 2 = 1 + 4 = 5. Hence, p = v· w 5 −1, 2 = −3, 6. We plot v, w and p below. w· w w = 15 4In this case, the point R coincides with the point O, so p = −→ OR = −−→ OO = 0. 1040 Applications of Trigonometry 3 −2 −1 1 Suppose we wanted to verify that our answer p in Example 11.9.4 is indeed the orthogonal projection of v onto w. We ﬁrst note that since p is a scalar multiple of w, it has the correct direction, so what remains to check is the orthogonality condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v3 −2 −1 1 From the deﬁnition of vector arithmetic, p + q = v, so that q = v − p. In the case of Example 11.9.4, v = 1, 8 and p = −3, 6, so q = 1, 8−−3, 6 = 4, 2. Then q· w = 4, 2·−1, 2 = (−4)+4 = 0, which shows q ⊥ w, as required. This result is generalized in the following theorem. Theorem 11.27. Generalized Decomposition Theorem: Let v and w be nonzero vectors. There are unique vectors p and q such that v = p + q where p = k w for some scalar k, and q · w = 0. Note that if the vectors p and q in Theorem 11.27 are nonzero, then we can say p is parallel 5 to w and q is orthogonal to w. In this case, the vector p is sometimes called the ‘vector component of v parallel to w’ and q is called the ‘vector component of v orthogonal to w.’ To prove Theorem 11.27, we take p = proj w(v) and q = v − p. Then p is, by deﬁnition, a scalar multiple of w. Next, we compute q · w. 5See Exercise 64 in Section 11.8. 11.9 The Dot Product and Projection 1041 q · w = (v − p) · w Deﬁnition of q. = v · w − p · w Properties of Dot Product · w Since p = proj w(v). ( w · w) Properties of Dot Product Hence, q · w = 0, as required. At this point, we have shown that the vectors p and q guaranteed by Theorem 11.27 exist. Now we need to show that they are unique. Suppose v = p + q = p + q where the vectors p and q satisfy the same properties described in Theorem 11.27 as p and q. Then p − p = q − q, so w · (p − p ) = w · (q − q. Hence, w · (p − p ) = 0. Now there are scalars k and k so that p = k w and p = k w. This means w · (p − p ) = w · (k w − k w) = w · ([k − k ] w) = (k − k )( w · w) = (k − k ) w2. Since w = 0, w2 = 0, which means the only way w · (p − p ) = (k − k ) w2 = 0 is for k − k = 0, or k = k . This means p = k w = k w = p . With , it must be that q = q as well. Hence, we have shown there is only one way to write v as a sum of vectors as described in Theorem 11.27. We close this section with an application of the dot product. In Physics, if a constant force F is exerted over a distance d, the work W done by the force is given by W = F d. Here, we assume the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to ﬁnd the work done. Consider the scenario below where the constant force F is applied to move an object from the point P to the point Q. F F θ θ P Q To ﬁnd the work W done in this scenario, we need to ﬁnd how much of the force F is in the −−→ P Q. This is precisely what the dot product F · P Q represents. Since direction of the motion −−→ the distance the object travels is P QP Q, −−→ −−→ W = ( F · P Q) P Q = F · ( P Q cos(θ), where θ is the angle between the applied force F and the trajectory of the motion −−→ P Q, we get W = ( F · P Q) −−→ P Q = F −−→ P Q. We have proved the following. −−→ P QP Q) = F · −−→ P Q. Since −−→ P Q = 1042 Applications of Trigonometry Theorem 11.28. Work as a Dot Product: Suppose a constant force F is applied along the vector −−→ P Q. The work W done by F is given by where θ is the angle between F and −−→ P Q = F −−→ P Q cos(θ), W = F · −−→ P Q. Example 11.9.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30◦ angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a 30◦ angle for the duration of the 50 feet. 30◦ −−→ Solution. There are two ways to attack this problem. One way is to ﬁnd the vectors F and P Q −−→ mentioned in Theorem 11.28 and compute W = F · P Q. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive x-axis lies along the dashed line in the ﬁgure above. Since the force applied is a constant 10 pounds, we have F = 10. Since it is being applied at a constant angle of θ = 30◦ with respect to the positive x-axis, Deﬁnition √ 11.8 gives us F = 10 cos(30◦, sin(30◦) = 5 3, 5. Since the wagon is being pulled along 50 −−→ P Q = 50ˆı = 50 1, 0 = 50, 0. We get feet in the positive direction, the displacement vector is W = F · 3. Since force is measured in pounds and distance is measured in feet, we get W = 250 3 foot-pounds. Alternatively, we can use the formulation W = F −−→ P Q cos(θ) to get W = (10 pounds)(50 feet) cos (30◦) = 250 3, 5 · 50, 0 = 250 3 foot-pounds of work. −−→ P Q = 5 √ √ √ √ 11.9 The Dot Product and Projection 1043 11.9.1 Exercises In Exercises 1 - 20, use the pair of vectors v and w to ﬁnd the following quantities. v · w The angle θ (in degrees) between v and w proj w(v) q = v − proj w(v) (Show that q · w = 0.) 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 3. v = 1, √ 3 and w = 1, − √ 3 5. v = −2, 1 and w = 3, 6 4. v = 3, 4 and w = −6, −8 6. v = −3 √ 3, 3 and w = − 3, −1 √ 7. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı 15. v = 3 17. v = 19. v = 2ˆı + 3 1 2 , 2 ˆ and w = ˆı − ˆ √ √ 2 2 , and and 16. v = 5ˆı + 12ˆ and w = −3ˆı + 4ˆ 18. v = 20 and w = 1 2 , − √ 3 2 and 21. A force of 1500 pounds is required to tow a trailer. Find the work done towing the trailer along a ﬂat stretch of road 300 feet. Assume the force is applied in the direction of the motion. 22. Find the work done lifting a 10 pound book 3 feet straight up into the air. Assume the force of gravity is acting straight downwards. 23. Suppose Taylor ﬁlls her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a 15◦ angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two dec
imal places. 24. In Exercise 61 in Section 11.8, two drunken college students have ﬁlled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13◦ angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77◦E.) Find the work done by this student if the keg is dragged 42 feet. 1044 Applications of Trigonometry 25. Find the work done pushing a 200 pound barrel 10 feet up a 12.5◦ incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places. HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the 12.5◦ of the incline! 26. Prove the distributive property of the dot product in Theorem 11.22. 27. Finish the proof of the scalar property of the dot product in Theorem 11.22. 28. Use the identity in Example 11.9.1 to prove the Parallelogram Law v2 + w2 = 1 2 v + w2 + v − w2 29. We know that \|x + y\| ≤ \|x\| + \|y\| for all real numbers x and y by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that u + v ≤ u + v for all pairs of vectors u and v. (a) (Step 1) Show that u + v2 = u2 + 2u · v + v2. (b) (Step 2) Show that \|u · v\| ≤ uv. This is the celebrated Cauchy-Schwarz Inequality.6 (Hint: To show this inequality, start with the fact that \|u · v\| = \| uv cos(θ) \| and use the fact that \| cos(θ)\| ≤ 1 for all θ.) (c) (Step 3) Show that u + v2 = u2 + 2u · v + v2 ≤ u2 + 2\|u · v\| + v2 ≤ u2 + 2uv + v2 = (u + v)2. (d) (Step 4) Use Step 3 to show that u + v ≤ u + v for all pairs of vectors u and v. (e) As an added bonus, we can now show that the Triangle Inequality \|z + w\| ≤ \|z\| + \|w\| holds for all complex numbers z and w as well. Identify the complex number z = a + bi with the vector u = a, b and identify the complex number w = c + di with the vector v = c, d and just follow your nose! 6It is also known by other names. Check out this site for details. 11.9 The Dot Product and Projection 1045 11.9.2 Answers 1. v = −2, −7 and w = 5, −9 2. v = −6, −5 and w = 10, −12 v · w = 53 θ = 45◦ proj w(v = 90◦ proj w(v) = 0, 0 q = −6, −5 3. v = 1, √ 3 and w = 1, − √ 3 4. v = 3, 4 and w = −6, −8 v · w = −2 θ = 120◦ proj w(v50 θ = 180◦ proj w(v) = 3, 4 q = 0, 0 5. v = −2, 1 and w = 3, 6 v · w = 0 θ = 90◦ proj w(v) = 0, 0 q = −2, 1 3, 3 and w = − 3, −1 √ √ 6. v = −3 v · w = 6 θ = 60◦ proj w(v. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 v · w = −1 θ ≈ 93.37◦ proj w(v) = 1, 0 q = 0, 17 v · w = 63 θ ≈ 14.25◦ proj w(v) = 315 169 , 756 q = 192 169 , − 80 169 169 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 v · w = 6 θ ≈ 74.74◦ proj w(v) = 3 q = − 55 13 , − 11 13 13 , − 15 13 v · w = −62 θ ≈ 169.94◦ proj w(v) = − 248 q = − 77 65 , − 44 65 65 , 434 65 1046 Applications of Trigonometry 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 v · w = 2 θ ≈ 87.88◦ proj w(v) = 1 10 , 3 q = − 81 10 , 27 10 10 v · w = −91 θ ≈ 159.51◦ proj w(v) = 0, −91 q = 34, 0 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı v · w = −4 θ ≈ 108.43◦ proj w(v) = 0, −1 q = 3, 0 15. v = 3 2 ˆ and w = ˆı − ˆ 2ˆı + 3 v · w = 0 θ = 90◦ proj w(v) = 048 θ ≈ 163.74◦ proj w(v) = −24, 0 q = 0, 7 16. v = 5ˆı + 12ˆ and w = −3ˆı + 4ˆ v · w = 33 θ ≈ 59.49◦ proj w(v) = − 99 q = 224 25 , 168 25 25 , 132 25 17 = 75◦ and − 4 √ √ 3−1 4 3 , proj w(v) = 1+ 4 q = √ 3 1− 4 √ , 1+ 4 3 18 = 105◦ and − 4 6 proj w(v) = √ 3 √ q = 2+ 8 √ √ 2− 8 6 √ √ 2+ 8 6 , √ 6 2 20. v = 1 and 19. v = √ 3 2 , 1 2 √ and w = √ v · w = − θ = 165◦ 2 6+ 4 √ proj w(v) = √ q = 3−1 4 , 1− 4 3+1 4 √ 3 √ , 3++ 4 v · w = θ = 15◦ √ proj w(v) = 1− 4 q = √ 3 , 1− 4 3+1 4 √ 3 √ 3+1 4 , − 21. (1500 pounds)(300 feet) cos (0◦) = 450, 000 foot-pounds 22. (10 pounds)(3 feet) cos (0◦) = 30 foot-pounds 11.9 The Dot Product and Projection 1047 23. (13 pounds)(25 feet) cos (15◦) ≈ 313.92 foot-pounds 24. (100 pounds)(42 feet) cos (13◦) ≈ 4092.35 foot-pounds 25. (200 pounds)(10 feet) cos (77.5◦) ≈ 432.88 foot-pounds 1048 Applications of Trigonometry 11.10 Parametric Equations As we have seen in Exercises 53 - 56 in Section 1.2, Chapter 7 and most recently in Section 11.5, there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a function of x nor x as a function of y. In this section, we present a new concept which allows us to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. y 5 4 3 2 1 P (x, y) = (f (t), g(t)) Q O x 1 2 3 4 5 The curve C does not represent y as a function of x because it fails the Vertical Line Test and it does not represent x as a function of y because it fails the Horizontal Line Test. However, since the bug can be in only one place P (x, y) at any given time t, we can deﬁne the x-coordinate of P as a function of t and the y-coordinate of P as a (usually, but not necessarily) diﬀerent function of t. (Traditionally, f (t) is used for x and g(t) is used for y.) The independent variable t in this case is called a parameter and the system of equations x = f (t) y = g(t) is called a system of parametric equations or a parametrization of the curve C.1 The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path2 at the point Q and ﬁnally heads oﬀ into the ﬁrst quadrant. It is important to note that the curve itself is a set of points and as such is devoid of any orientation. The parametrization determines the orientation and as we shall see, diﬀerent parametrizations can determine diﬀerent orientations. If all of this seems hauntingly familiar, it should. By deﬁnition, the system of equations {x = cos(t), y = sin(t) parametrizes the Unit Circle, giving it a counter-clockwise orientation. More generally, the equations of circular motion {x = r cos(ωt), y = r sin(ωt) developed on page 732 in Section 10.2.1 are parametric equations which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counterclockwise; if ω < 0, the orientation is clockwise. The angular frequency ω determines ‘how fast’ the 1Note the use of the indeﬁnite article ‘a’. As we shall see, there are inﬁnitely many diﬀerent parametric represen- tations for any given curve. 2Here, the bug reaches the point Q at two diﬀerent times. While this does not contradict our claim that f (t) and g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on.) 11.10 Parametric Equations 1049 object moves around the circle. In particular, the equations x = 2960 cos π 12 t that model the motion of Lakeland Community College as the earth rotates (see Example 10.2.7 in Section 10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0, 24). It is time for another example. 12 t , y = 2960 sin π Example 11.10.1. Sketch the curve described by x = t2 − 3 y = 2t − 1 for t ≥ −2. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t ≥ −2, we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. t −2 −1 0 1 2 3 x(t) y(t) 1 −5 −2 −3 −3 −1 −2 1 3 1 5 6 (x(t), y(t)) (1, −5) (−2, −3) (−3, −1) (−2, 1) (1, 3) (6, 5) y 5 4 3 2 1 −2−1 −1 1 2 3 4 5 6 x −2 −3 −5 2 . Substituting this into the equation x = t2 − 3 yields x = The curve sketched out in Example 11.10.1 certainly looks like a parabola, and the presence of the t2 term in the equation x = t2 − 3 reinforces this hunch. Since the parametric equations x = t2 − 3, y = 2t − 1 given to describe this curve are a system of equations, we can use the technique of substitution as described in Section 8.7 to eliminate the parameter t and get an equation involving just x and y. To do so, we choose to solve the equation y = 2t − 1 for t to get t = y+1 − 3 or, after some rearrangement, (y + 1)2 = 4(x + 3). Thinking back to Section 7.3, we see that the graph of this equation is a parabola with vertex (−3, −1) which opens to the right, as required. Technically speaking, the equation (y + 1)2 = 4(x + 3) describes the entire parabola, while the parametric equations x = t2 − 3, y = 2t − 1 for t ≥ −2 describe only a portion of the parabola. In this case,3 we can remedy this situation by restricting the bounds on y. Since the portion of the parabola we want is exactly the part where y ≥ −5, the equation (y + 1)2 = 4(x + 3) coupled with the restriction y ≥ −5 describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. y+1 2 2 Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible, can be a great help in graphing curves determined by parametric equations. If the system of parametric equations contains algebraic functions, as was the case in Example 11.10.1, then the usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the 3We will have an example shortly where no matter how we restrict x and y, we can never accurately describe the curve once we’ve eliminated the parameter. 1050 Applications of Trigonometry system {x = f (t), y = g(t) to eliminate the parameter. If, on the other
hand, the parametrization In this case, it is often best involves the trigonometric functions, the strategy changes slightly. to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the following parametric equations. x = t3 y = 2t2 x = e−t y = e−2t 1. 2. for −1 ≤ t ≤ 1 for t ≥ 0 Solution. 3. 4. for 0 < t < π x = sin(t) y = csc(t) x = 1 + 3 cos(t) y = 2 sin(t) for 0 ≤ t ≤ 3π 2 1. To get a feel for the curve described by the system x = t3, y = 2t2 we ﬁrst sketch the graphs of x = t3 and y = 2t2 over the interval [−1, 1]. We note that as t takes on values in the interval [−1, 1], x = t3 ranges between −1 and 1, and y = 2t2 ranges between 0 and 2. This means that all of the action is happening on a portion of the plane, namely {(x, y) \| − 1 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Next, we plot a few points to get a sense of the position and orientation of the curve. Certainly, t = −1 and t = 1 are good values to pick since these are the extreme values of t. We also choose t = 0, since that corresponds to a relative minimum4 on the graph of y = 2t2. Plugging in t = −1 gives the point (−1, 2), t = 0 gives (0, 0) and t = 1 gives (1, 2). More generally, we see that x = t3 is increasing over the entire interval [−1, 1] whereas y = 2t2 is decreasing over the interval [−1, 0] and then increasing over [0, 1]. Geometrically, this means that in order to trace out the path described by the parametric equations, we start at (−1, 2) (where t = −1), then move to the right (since x is increasing) and down (since y is decreasing) to (0, 0) (where t = 0). We continue to move to the right (since x is still increasing) but now move upwards (since y is now increasing) until we reach (1, 2) (where t = 1). Finally, to get a good sense of the shape of the curve, we √ eliminate the parameter. Solving x = t3 for t, we get t = 3 x. Substituting this into y = 2t2 √ x)2 = 2x2/3. Our experience in Section 5.3 yields the graph of our ﬁnal answer gives y = 2( 3 below. x 1 −1 1 t −1 y 2 1 y 2 1 −1 1 t −1 1 x x = t3, −1 ≤ t ≤ 1 y = 2t2, −1 ≤ t ≤ 1 x = t3, y = 2t2 , −1 ≤ t ≤ 1 4You should review Section 1.6.1 if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘relative minimum’ mean. 11.10 Parametric Equations 1051 9 3 , 1 and for t = ln(3) we get 2 2. For the system x = 2e−t, y = e−2t for t ≥ 0, we proceed as in the previous example and graph x = 2e−t and y = e−2t over the interval [0, ∞). We ﬁnd that the range of x in this case is (0, 2] and the range of y is (0, 1]. Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t involve natural logarithms. Starting with t = ln(1) = 0 we get5 (2, 1), for t = ln(2) we get 1, 1 . Since t is ranging over the unbounded interval [0, ∞), 4 we take the time to analyze the end behavior of both x and y. As t → ∞, x = 2e−t → 0+ and y = e−2t → 0+ as well. This means the graph of x = 2e−t, y = e−2t approaches the point (0, 0). Since both x = 2e−t and y = e−2t are always decreasing for t ≥ 0, we know that our ﬁnal graph will start at (2, 1) (where t = 0), and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. To eliminate the parameter, one way to proceed is to solve x = 2e−t for t to get t = − ln x . Substituting 2 = x2 this for t in y = e−2t gives y = e−2(− ln(x/2)) = e2 ln(x/2) = eln(x/2)2 4 . Or, we could recognize that y = e−2t = e−t2 2 2 , we get y = x = x2 4 this way as well. Either way, the graph of x = 2e−t, y = e−2t for t ≥ 0 is a portion of the parabola y = x2 4 which starts at the point (2, 1) and heads towards, but never reaches,6 (0, 0). = x 2 , and since x = 2e−t means e− = 2e−t, t ≥ 0 y = e−2t, t ≥ 0 x = 2e−t, y = e−2t , t ≥ 0 6 gives the point 1 3. For the system {x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and y = csc(t) over the interval (0, π). We ﬁnd that the range of x is (0, 1] while the range of 2 , 2, t = π y is [1, ∞). Plotting a few friendly points, we see that t = π 2 , 2. Since t = 0 and t = π aren’t included in the 6 returns us to 1 gives (1, 1) and t = 5π domain for t, (because y = csc(t) is undeﬁned at these t-values), we analyze the behavior of the system as t approaches 0 and π. We ﬁnd that as t → 0+ as well as when t → π−, we get x = sin(t) → 0+ and y = csc(t) → ∞. Piecing all of this information together, we get that for t near 0, we have points with very small positive x-values, but very large positive y-values. As t ranges through the interval 0, π , x = sin(t) is increasing and y = csc(t) is decreasing. 2 This means that we are moving to the right and downwards, through 1 6 to (1, 1) when t = π 2 , the orientation reverses, and we start to head to the left, since x = sin(t) is now decreasing, and up, since y = csc(t) is now increasing. We pass back through 1 6 back to the points with small positive x-coordinates and large 2 , 2 when t = 5π 2 , 2 when t = π 2 . Once t = π 2 5The reader is encouraged to review Sections 6.1 and 6.2 as needed. 6Note the open circle at the origin. See the solution to part 3 in Example 1.2.1 on page 22 and Theorem 4.1 in Section 4.1 for a review of this concept. 1052 Applications of Trigonometry positive y-coordinates. To better explain this behavior, we eliminate the parameter. Using a reciprocal identity, we write y = csc(t) = 1 sin(t) . Since x = sin(t), the curve traced out by this parametrization is a portion of the graph of y = 1 x . We now can explain the unusual behavior as t → 0+ and t → π− – for these values of t, we are hugging the vertical asymptote x = 0 of the graph of y = 1 x . We see that the parametrization given above traces out the portion of y = 1 x for 0 < x ≤ 1 twice as t runs through the interval (0, π). = sin(t), 0 < t < π y = csc(t), 0 < t < π {x = sin(t), y = csc(t) , 0 < t < π 4. Proceeding as above, we set about graphing {x = 1 + 3 cos(t), y = 2 sin(t) for 0 ≤ t ≤ 3π 2 by . We see that x ranges ﬁrst graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval 0, 3π 2 from −2 to 4 and y ranges from −2 to 2. Plugging in t = 0, π 2 , π and 3π 2 gives the points (4, 0), (1, 2), (−2, 0) and (1, −2), respectively. As t ranges from 0 to π 2 , x = 1 + 3 cos(t) is decreasing, while y = 2 sin(t) is increasing. This means that we start tracing out our answer at (4, 0) and continue moving to the left and upwards towards (1, 2). For π 2 ≤ t ≤ π, x is decreasing, as is y, so the motion is still right to left, but now is downwards from (1, 2) to (−2, 0). On the interval π, 3π , x begins to increase, while y continues to decrease. Hence, the motion becomes left 2 to right but continues downwards, connecting (−2, 0) to (1, −2). To eliminate the parameter here, we note that the trigonometric functions involved, namely cos(t) and sin(t), are related by the Pythagorean Identity cos2(t) + sin2(t) = 1. Hence, we solve x = 1 + 3 cos(t) for cos(t) 3 , and we solve y = 2 sin(t) for sin(t) to get sin(t) = y to get cos(t) = x−1 2 . Substituting these expressions into cos2(t)+sin2(t) = 1 gives x−1 9 + y2 4 = 1. From Section 3 7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at (−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. = 1, or (x−1)1 π 2 π t 3π 2 −2 x = 1 + 3 cos(t), 0 ≤ t ≤ 3π 2 y 2 1 π 2 π t 3π 2 −1 1 2 3 x 4 −1 −2 y 2 1 −1 −2 y = 2 sin(t), 0 ≤ t ≤ 3π 2 {x = 1 + 3 cos(t), y = 2 sin(t) , 0 ≤ t ≤ 3π 2 11.10 Parametric Equations 1053 Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of ﬁnding parametric representations of curves. We start with the following. Parametrizations of Common Curves To parametrize y = f (x) as x runs through some interval I, let x = t and y = f (t) and let t run through I. To parametrize x = g(y) as y runs through some interval I, let x = g(t) and y = t and let t run through I. To parametrize a directed line segment with initial point (x0, y0) and terminal point (x1, y1), let x = x0 + (x1 − x0)t and y = y0 + (y1 − y0)t for 0 ≤ t ≤ 1. To parametrize (x−h)2 a2 + (y−k)2 b2 = 1 where a, b > 0, let x = h + a cos(t) and y = k + b sin(t) for 0 ≤ t < 2π. (This will impart a counter-clockwise orientation.) The reader is encouraged to verify the above formulas by eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following example. Example 11.10.3. Find a parametrization for each of the following curves and check your answers. 1. y = x2 from x = −3 to x = 2 2. y = f −1(x) where f (x) = x5 + 2x + 1 3. The line segment which starts at (2, −3) and ends at (1, 5) 4. The circle x2 + 2x + y2 − 4y = 4 4 + y2 5. The left half of the ellipse x2 9 = 1 Solution. 1. Since y = x2 is written in the form y = f (x), we let x = t and y = f (t) = t2. Since x = t, the bounds on t match precisely the bounds on x so we get x = t, y = t2 for −3 ≤ t ≤ 2. The check is almost trivial; with x = t we have y = t2 = x2 as t = x runs from −3 to 2. 2. We are told to parametrize y = f −1(x) for f (x) = x5 + 2x + 1 so it is safe to assume that f is one-to-one. (Otherwise, f −1 would not exist.) To ﬁnd a formula y = f −1(x), we follow the procedure outlined on page 384 – we start with the equation y = f (x), interchange x and y and solve for y. Doing so gives us the equation x = y5 + 2y + 1. While we could attempt to solve this equation for y, we don’t need to. We can parametrize x = f (y) = y5 + 2y + 1 by setting y = t so that x = t5 + 2t + 1. We know from our work in Section 3.1 that since f (x) = x5 + 2x + 1 is an odd-degree polynomial, the range of y = f (x) = x5 + 2x + 1 is (−∞, ∞). Hence, in order to trace out the entire graph of x = f (y) = y5 + 2y + 1, we need to let y run through all real number
s. Our ﬁnal answer to this problem is x = t5 + 2t + 1, y = t for −∞ < t < ∞. As in the previous problem, our solution is trivial to check.7 7Provided you followed the inverse function theory, of course. 1054 Applications of Trigonometry 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x0 +(x1 −x0)t and y = y0 +(y1 −y0)t for 0 ≤ t ≤ 1. While these equations at ﬁrst glance are quite a handful,8 they can be summarized as ‘starting point + (displacement)t’. To ﬁnd the equation for x, we have that the line segment starts at x = 2 and ends at x = 1. This means the displacement in the x-direction is (1 − 2) = −1. Hence, the equation for x is x = 2 + (−1)t = 2 − t. For y, we note that the line segment starts at y = −3 and ends at y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3 + 8t. Our ﬁnal answer is {x = 2 − t, y = −3 + 8t for 0 ≤ t ≤ 1. To check, we can solve x = 2 − t for t to get t = 2 − x. Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 − x), or y = −8x + 13. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t = 0, x = 2 − t = 2, and when t = 1, x = 2 − t = 1. Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required. 9 + (y−2)2 9 + (y−2)2 4. In order to use the formulas above to parametrize the circle x2 +2x+y2 −4y = 4, we ﬁrst need to put it into the correct form. After completing the squares, we get (x + 1)2 + (y − 2)2 = 9, or (x+1)2 9 = 1. Once again, the formulas x = h + a cos(t) and y = k + b sin(t) can be a challenge to memorize, but they come from the Pythagorean Identity cos2(t) + sin2(t) = 1. In the equation (x+1)2 3 . Rearranging these last two equations, we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). In order to complete one revolution around the circle, we let t range through the interval [0, 2π). We get as our ﬁnal answer {x = −1 + 3 cos(t), y = 2 + 3 sin(t) for 0 ≤ t < 2π. To check our answer, we could eliminate the parameter by solving x = −1 + 3 cos(t) for cos(t) and y = 2 + 3 sin(t) for sin(t), invoking a Pythagorean Identity, and then manipulating the resulting equation in x and y into the original equation x2 + 2x + y2 − 4y = 4. Instead, we opt for a more direct approach. We substitute x = −1 + 3 cos(t) and y = 2 + 3 sin(t) into the equation x2 + 2x + y2 − 4y = 4 and show that the latter is satisﬁed for all t such that 0 ≤ t < 2π. 9 = 1, we identify cos(t) = x+1 3 and sin(t) = y−2 (−1 + 3 cos(t))2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t))2 − 4(2 + 3 sin(t)) 1 − 6 cos(t) + 9 cos2(t) − 2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin2(t) − 8 − 12 sin(t) 9 cos2(t) + 9 sin2(t) − 5 9 cos2(t) + sin2(t) − 5 9 (1) − 5 x2 + 2x + y2 − 4y = Now that we know the parametric equations give us points on the circle, we can go through the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered as t ranges through the interval [0, 2π). 8Compare and contrast this with Exercise 65 in Section 11.8. 11.10 Parametric Equations 1055 5. In the equation x2 4 + y2 9 = 1, we can either use the formulas above or think back to the Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). The normal range on the parameter in this case is 0 ≤ t < 2π, but since we are interested in only the left half of the ellipse, we restrict t to the values which correspond to Quadrant II and Quadrant III angles, namely 2 ≤ t ≤ 3π π 2 . Substituting x = 2 cos(t) and y = 3 sin(t) into x2 = 1, which reduces to the Pythagorean Identity cos2(t) + sin2(t) = 1. This proves that the points generated by the parametric equations {x = 2 cos(t), y = 3 sin(t) lie on the ellipse x2 9 = 1. Employing 2 ≤ t ≤ 3π the techniques demonstrated in Example 11.10.2, we ﬁnd that the restriction π 2 generates the left half of the ellipse, as required. 2 . Our ﬁnal answer is {x = 2 cos(t), y = 3 sin(t) for π 4 + y2 9 = 1 gives 4 cos2(t) 2 ≤ t ≤ 3π + 9 sin2(t) 9 4 + y2 4 We note that the formulas given on page 1053 oﬀer only one of literally inﬁnitely many ways to parametrize the common curves listed there. At times, the formulas oﬀered there need to be altered to suit the situation. Two easy ways to alter parametrizations are given below. Adjusting Parametric Equations Reversing Orientation: Replacing every occurrence of t with −t in a parametric description for a curve (including any inequalities which describe the bounds on t) reverses the orientation of the curve. Shift of Parameter: Replacing every occurrence of t with (t − c) in a parametric description for a curve (including any inequalities which describe the bounds on t) shifts the start of the parameter t ahead by c units. We demonstrate these techniques in the following example. Example 11.10.4. Find a parametrization for the following curves. 1. The curve which starts at (2, 4) and follows the parabola y = x2 to end at (−1, 1). Shift the parameter so that the path starts at t = 0. 2. The two part path which starts at (0, 0), travels along a line to (3, 4), then travels along a line to (5, 0). 3. The Unit Circle, oriented clockwise, with t = 0 corresponding to (0, −1). Solution. 1. We can parametrize y = x2 from x = −1 to x = 2 using the formula given on Page 1053 as x = t, y = t2 for −1 ≤ t ≤ 2. This parametrization, however, starts at (−1, 1) and ends at (2, 4). Hence, we need to reverse the orientation. To do so, we replace every occurrence of t with −t to get x = −t, y = (−t)2 for −1 ≤ −t ≤ 2. After simplifying, we get x = −t, y = t2 for −2 ≤ t ≤ 1. We would like t to begin at t = 0 instead of t = −2. The problem here is that the parametrization we have starts 2 units ‘too soon’, so we need to introduce a ‘time delay’ of 2. Replacing every occurrence of t with (t − 2) gives x = −(t − 2), y = (t − 2)2 for −2 ≤ t − 2 ≤ 1. Simplifying yields x = 2 − t, y = t2 − 4t + 4 for 0 ≤ t ≤ 3. 1056 Applications of Trigonometry 2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the ﬁrst part of the path, we get {x = 3t, y = 4t for 0 ≤ t ≤ 1, and for the second part we get {x = 3 + 2t, y = 4 − 4t for 0 ≤ t ≤ 1. Since the ﬁrst parametrization leaves oﬀ at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Replacing t with (t − 1) in the second set of parametric equations gives {x = 3 + 2(t − 1), y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying yields {x = 1 + 2t, y = 8 − 4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as {x = f (t), y = g(t) for 0 ≤ t ≤ 2 where f (t) = 3t, 1 + 2t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 and g(t) = 4t, 8 − 4t, for 0 ≤ t ≤ 1 for 1 ≤ t ≤ 2 3. We know that {x = cos(t), y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to (1, 0), so the ﬁrst order of business is to reverse the orientation. Replacing t with −t gives {x = cos(−t), y = sin(−t) for 0 ≤ −t < 2π, which simpliﬁes9 to {x = cos(t), y = − sin(t) for −2π < t ≤ 0. This parametrization gives a clockwise orientation, but t = 0 still corresponds to the point (1, 0); the point (0, −1) is reached when t = − 3π 2 . Our strategy is to ﬁrst get the parametrization to ‘start’ at the point (0, −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0. We know that any interval of length 2π will parametrize the entire circle, so we keep the equations {x = cos(t), y = − sin(t) , but start the parameter t at − 3π 2 , and ﬁnd the upper bound by adding 2π so − 3π 2 . The reader can verify that {x = cos(t), y = − sin(t) for − 3π 2 traces out the Unit Circle clockwise starting at the point (0, −1). We now shift the parameter by introducing a ‘time delay’ of 3π 2 units by replacing every occurrence of t with t − 3π 2 . This 2 simpliﬁes10 to {x = − sin(t), y = − cos(t) for 0 ≤ t < 2π, as required. . We get x = cos t − 3π 2 , y = − sin t − 3π 2 2 ≤ t < π 2 ≤ t − 3π 2 ≤ t < π for − 3π 2 < π We put our answer to Example 11.10.4 number 3 to good use to derive the equation of a cycloid. Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below. Let θ be the angle in radians which measures the amount of clockwise rotation experienced by the radius highlighted in the ﬁgure. y r P (x, y) θ 9courtesy of the Even/Odd Identities 10courtesy of the Sum/Diﬀerence Formulas x 11.10 Parametric Equations 1057 Our goal is to ﬁnd parametric equations for the coordinates of the point P (x, y) in terms of θ. From our work in Example 11.10.4 number 3, we know that clockwise motion along the Unit Circle starting at the point (0, −1) can be modeled by the equations {x = − sin(θ), y = − cos(θ) for 0 ≤ θ < 2π. (We have renamed the parameter ‘θ’ to match the context of this problem.) To model this motion on a circle of radius r, all we need to do11 is multiply both x and y by the factor r which yields {x = −r sin(θ), y = −r cos(θ) . We now need to adjust for the fact that the circle isn’t stationary with center (0, 0), but rather, is rolling along the positive x-axis. Since the velocity v is constant, we know that at time t, the center of the circle has traveled a distance vt down the positive x-axis. Furthermore, since the radius of the circle is r and the circle isn’t moving vertically, we know that the center of the circle is always r units above the x-axis. Putting these two facts together, we have that at time t, the center of the circle is at the point (vt, r). From Section 10.1.1, we know v = rθ t , or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is (rθ, r). As a result, we need to modify the equations {x = −r sin(θ), y = −r cos(θ) by shifting the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y
-coordinate up r units12 (by adding r to the expression for y). We get {x = −r sin(θ) + rθ, y = −r cos(θ) + r , which can be written as {x = r(θ − sin(θ)), y = r(1 − cos(θ)) . Since the motion starts at θ = 0 and proceeds indeﬁnitely, we set θ ≥ 0. We end the section with a demonstration of the graphing calculator. Example 11.10.5. Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive x-axis as described above. Graph your answer using a calculator. Solution. We have r = 3 which gives the equations {x = 3(t − sin(t)), y = 3(1 − cos(t)) for t ≥ 0. (Here we have returned to the convention of using t as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility. Using a calculator to graph parametric equations is very similar to graphing polar equations on a calculator.13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the equations and advance to the ‘Window’ screen. As always, the challenge is to determine appropriate bounds on the parameter, t, as well as for x and y. We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π, so 11If we replace x with x 2 = 1 which reduces to x2 + y2 = r2. In the language of Section 1.7, we are stretching the graph by a factor of r in both the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r. r in the equation for the Unit Circle x2 + y2 = 1, we obtain x r and y with y 2 + y r r 12Does this seem familiar? See Example 11.1.1 in Section 11.1. 13See page 959 in Section 11.5. 1058 Applications of Trigonometry it seems reasonable to keep these as our bounds on t. The ‘Tstep’ seems reasonably small – too large a value here can lead to incorrect graphs.14 We know from our derivation of the equations of the cycloid that the center of the generating circle has coordinates (rθ, r), or in this case, (3t, 3). Since t ranges between 0 and 2π, we set x to range between 0 and 6π. The values of y go from the bottom of the circle to the top, so y ranges between 0 and 6. Below we graph the cycloid with these settings, and then extend t to range from 0 to 6π which forces x to range from 0 to 18π yielding three arches of the cycloid. (It is instructive to note that keeping the y settings between 0 and 6 messes up the geometry of the cycloid. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches.) 14Again, see page 959 in Section 11.5. 11.10 Parametric Equations 1059 11.10.1 Exercises In Exercises 1 - 20, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. x = 4t − 3 y = 6t − 2 for 0 ≤ t ≤ 1 x = 2t y = t2 for − 1 ≤ t ≤ 2 x = t2 + 2t + 1 y = t + 1 for t ≤ 1 x = t y = t3 for − ∞ < t < ∞ x = cos(t) y = sin(t) for − 1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π x = 2 cos(t) y = sec(t) for 0 ≤ t < π 2 x = sec(t) y = tan(t) for − π 2 < t < π 2 x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 x = cos(t) y = t for 0 ≤ t ≤ π 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. x = 4t − 1 y = 3 − 4t for + 2t − t2 18 − t2 x = 1 9 y = 1 3 t for 0 ≤ t ≤ 3 for t ≥ −3 x = t3 y = t for − ∞ < t < ∞ x = 3 cos(t) y = 3 sin(t) for 0 ≤ t ≤ π x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π x = 2 tan(t) y = cot(t) for 0 < t < π 2 x = sec(t) y = tan(t) for π 2 < t < 3π 2 x = tan(t) y = 2 sec(t) for π 2 < t < 3π 2 x = sin(t) y = t for − π 2 ≤ t ≤ π 2 In Exercises 21 - 24, plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 21. 23. x = t3 − 3t y = t2 − 4 x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 2 22. 24. x = 4 cos3(t) y = 4 sin3(t) x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π for 0 ≤ t ≤ 2π 1060 Applications of Trigonometry In Exercises 25 - 39, ﬁnd a parametric description for the given oriented curve. 25. the directed line segment from (3, −5) to (−2, 2) 26. the directed line segment from (−2, −1) to (3, −4) 27. the curve y = 4 − x2 from (−2, 0) to (2, 0). 28. the curve y = 4 − x2 from (−2, 0) to (2, 0) (Shift the parameter so t = 0 corresponds to (−2, 0).) 29. the curve x = y2 − 9 from (−5, −2) to (0, 3). 30. the curve x = y2 − 9 from (0, 3) to (−5, −2). (Shift the parameter so t = 0 corresponds to (0, 3).) 31. the circle x2 + y2 = 25, oriented counter-clockwise 32. the circle (x − 1)2 + y2 = 4, oriented counter-clockwise 33. the circle x2 + y2 − 6y = 0, oriented counter-clockwise 34. the circle x2 + y2 − 6y = 0, oriented clockwise (Shift the parameter so t begins at 0.) 35. the circle (x − 3)2 + (y + 1)2 = 117, oriented counter-clockwise 36. the ellipse (x − 1)2 + 9y2 = 9, oriented counter-clockwise 37. the ellipse 9x2 + 4y2 + 24y = 0, oriented counter-clockwise 38. the ellipse 9x2 + 4y2 + 24y = 0, oriented clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 39. the triangle with vertices (0, 0), (3, 0), (0, 4), oriented counter-clockwise (Shift the parameter so t = 0 corresponds to (0, 0).) 40. Use parametric equations and a graphing utility to graph the inverse of f (x) = x3 + 3x − 4. 41. Every polar curve r = f (θ) can be translated to a system of parametric equations with parameter θ by {x = r cos(θ) = f (θ) cos(θ), y = r sin(θ) = f (θ) sin(θ) . Convert r = 6 cos(2θ) to a system of parametric equations. Check your answer by graphing r = 6 cos(2θ) by hand using the techniques presented in Section 11.5 and then graphing the parametric equations you found using a graphing utility. 42. Use your results from Exercises 3 and 4 in Section 11.1 to ﬁnd the parametric equations which model a passenger’s position as they ride the London Eye. 11.10 Parametric Equations 1061 Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the projectile is given by15    x = v0 cos(θ) t 1 2 y = − gt2 + v0 sin(θ) t + s0 for 0 ≤ t ≤ T where v0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile is launched,16 g is the acceleration due to gravity, s0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground. (See the ﬁgure below.) y θ s0 (x(T ), 0) x 43. Carl’s friend Jason competes in Highland Games Competitions across the country. In one event, the ‘hammer throw’, he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of 42◦ with respect to the horizontal with an initial speed of 33 feet per second, ﬁnd the parametric equations for the ﬂight of the hammer. (Here, use g = 32 ft. s2 .) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility. 44. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve y = − g sec2(θ) 2v2 0 x2 + tan(θ)x + s0 Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is y = 0 sin2(θ) v2 2g + s0 when x = v2 0 sin(2θ) 2g 15A nice mix of vectors and Calculus are needed to derive this. 16We’ve seen this before. It’s the angle of elevation which was deﬁned on page 753. 1062 Applications of Trigonometry 45. In another event, the ‘sheaf toss’, Jason throws a 20 pound weight for height. If the weight is released 5 feet above the ground at an angle of 85◦ with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from part 44 to determine how fast the sheaf was launched into the air. (Once again, use g = 32 ft. s2 .) 46. Suppose θ = π formula given for y(t) above using g = 9.8 m in Exercise 25 in Section 2.3. What is x(t) in this case? 2 . (The projectile was launched vertically.) Simplify the general parametric s2 and compare that to the formula for s(t) given In Exercises 47 - 52, we explore the hyperbolic cosine function, denoted cosh(t), and the hyperbolic sine function, denoted sinh(t), deﬁned below: cosh(t) = et + e−t 2 and sinh(t) = et − e−t 2 47. Using a graphing utility as needed, verify that the domain of cosh(t) is (−∞, ∞) and the range of cosh(t) is [1, ∞). 48. Using a graphing utility as needed, verify that the domain and range of sinh(t) are both (−∞, ∞). 49. Show that {x(t) = cosh(t), y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola x2 − y2 = 1. (Hence the use of the adjective ‘hyperbolic.’) 50. Compare the deﬁnitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in Exercise 83f in Section 11.7. 51. Four other hyperbolic functions are waiting to be deﬁned: the hyperbolic secant sech(t), the hyperbolic cosecant csch(t), the hyperbolic tangent tanh(t) and the hyperbolic cotangent coth(t). Deﬁne these functions in terms of cosh(t) and sinh(t), then convert them to formulas involving et and e−t. Consult a suitable reference (a Calculus book, or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. 52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4. 11.10 Parametric Equations 1063 11.10.2 Answers x = 4t − 3 y = 6t − 2 1. for 3 −2 −1 −1 −2 2. x = 4t − 1 y = 3 − 4t y for 0 ≤ t ≤ 1 3 2 1 −1 −1 1 2 3 x x = 2t y = t2 3. for − + 2t − t2 4. for 3 −2 −1 1 2 3 4 x −1 1 2 x x = t2 + 2t + 1 y = t + 1 5. for t ≤ 1 6. 18 − t2 x = 1 9 y = 1 3 t for t ≥ −3 1 2 3 4 5 x −3 −2 −1 −1 −2 Applications of Trigonometry x = t3 y = t 8. for − ∞ < t < ∞ y 1 −4 −3 −2 −1 −1 1 2 3 4 x 1064 7. x = t y = t3 y for − ∞ < t < ∞ 4 3 2 1 −1 −1 −2 −3 −4 1 x 9. x = cos(t) y = sin(t) y 1 for − π 2 ≤ t ≤ π 2 −1 x 1 −1 10. x = 3 cos(t) y = 3 sin(t) y for 0 ≤ t
≤ π 3 2 1 −3 −2 −1 1 2 3 x 11. x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π 12. x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π y 4 3 2 1 1 2 x −4 −3 −2 −1 −1 −2 −3 −4 y 3 2 1 −3 −1 −1 1 3 x 11.10 Parametric Equations 1065 13. x = 2 cos(t) y = sec(t) y for 0 ≤ t < π 2 14. x = 2 tan(t) y = cot(t) y for 15. x = sec(t) y = tan(t) for − π 2 < t < π 2 y 16. x = sec(t) y = tan(t) < t < 3π 2 π 2 for y 4 3 2 1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1066 Applications of Trigonometry 17. x = tan(t) y = 2 sec(t) π 2 for − y < t < π 2 18. x = tan(t) y = 2 sec(t) < t < 3π 2 π 2 for y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −3 −4 19. x = cos(t) y = t for 0 < t < π 20. x = sin(t) y = t for − 1 1 x y π 2 −1 1 x − π 2 21. x = t3 − 3t y = t2 − 4 for − 2 ≤ t ≤ 2 22. x = 4 cos3(t) y = 4 sin3(t) y for 0 ≤ t ≤ 2π y 1 2 x −2 −1 −1 −2 −3 −4 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 11.10 Parametric Equations 1067 23. x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 24. x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π 1 1 x y 7 5 3 1 −1 −3 −5 −7 26. 28. 30. 32. 34. −1 x = 5t − 2 y = −1 − 3t x = t − 2 y = 4t − t2 x = t2 − 6t y = 3 − t for 0 ≤ t ≤ 1 for 0 ≤ t ≤ 4 for cos(t) y = 2 sin(t) x = 3 cos(t) y = 3 − 3 sin(t) x = 1 + 3 cos(t) y = sin(t) for 0 ≤ t < 2π for 0 ≤ t < 2π for 0 ≤ t < 2π x = 3 − 5t y = −5 + 7t for 0 ≤ t ≤ 1 for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 3 for 0 ≤ t < 2π x = t y = 4 − t2 x = t2 − 9 y = t x = 5 cos(t) y = 5 sin(t) x = 3 cos(t) x = 3 + √ y = −1 + x = 2 cos(t) 25. 27. 29. 31. 33. 35. 37. 38. y = 3 + 3 sin(t) for 0 ≤ t < 2π 117 cos(t) √ 117 sin(t) for 0 ≤ t < 2π 36. for 0 ≤ t < 2π y = 3 sin(t) − 3    x = 2 cos t − y = −3 − 3 sin π 2 = 2 sin(t) π 2 t − = −3 + 3 cos(t) for 0 ≤ t < 2π 39. {x(t), y(t) where: x(t) =    3t, 0 ≤ t ≤ 1 6 − 3t, 1 ≤ t ≤ 2 0, 2 ≤ t ≤ 3 y(t) =    0, 0 ≤ t ≤ 1 4t − 4, 1 ≤ t ≤ 2 12 − 4t, 2 ≤ t ≤ 3 1068 Applications of Trigonometry 40. The parametric equations for the inverse are x = t3 + 3t − 4 y = t for − ∞ < t < ∞ 41. r = 6 cos(2θ) translates to x = 6 cos(2θ) cos(θ) y = 6 cos(2θ) sin(θ) for 0 ≤ θ < 2π. 42. The parametric equations which describe the locations of passengers on the London Eye are x = 67.5 cos π y = 67.5 sin π 15 t − π 15 t − π 2 = 67.5 sin π 15 t 2 + 67.5 = 67.5 − 67.5 cos π 15 t for − ∞ < t < ∞ x = 33 cos(42◦)t 43. The parametric equations for the hammer throw are y = −16t2 + 33 sin(42◦)t + 6 for t ≥ 0. To ﬁnd when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or 1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after it was launched into the air. To ﬁnd how far away the hammer hits the ground, we ﬁnd x(1.61) ≈ 39.48 feet from where it was thrown into the air. 45. We solve y = 0 sin2(θ) v2 2g + s0 = 0 sin2(85◦) v2 2(32) + 5 = 31.5 to get v0 = ±41.34. The initial speed of the sheaf was approximately 41.34 feet per second. Index nth root of a complex number, 1000, 1001 principal, 397 nth Roots of Unity, 1006 u-substitution, 273 x-axis, 6 x-coordinate, 6 x-intercept, 25 y-axis, 6 y-coordinate, 6 y-intercept, 25 abscissa, 6 absolute value deﬁnition of, 173 inequality, 211 properties of, 173 acidity of a solution pH, 432 acute angle, 694 adjoint of a matrix, 622 alkalinity of a solution pH, 432 amplitude, 794, 881 angle acute, 694 between two vectors, 1035, 1036 central angle, 701 complementary, 696 coterminal, 698 decimal degrees, 695 deﬁnition, 693 degree, 694 DMS, 695 initial side, 698 measurement, 693 negative, 698 obtuse, 694 of declination, 761 of depression, 761 of elevation, 753 of inclination, 753 oriented, 697 positive, 698 quadrantal, 698 radian measure, 701 reference, 721 right, 694 standard position, 698 straight, 693 supplementary, 696 terminal side, 698 vertex, 693 angle side opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary deﬁnition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 1069 Index 1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arcsine deﬁnition of, 820 graph of, 820 properties of, 820 arctangent deﬁnition of, 824 graph of, 823 properties of, 824 argument of a complex number deﬁnition of, 991 properties of, 995 of a function, 55 of a logarithm, 425 of a trigonometric function, 793 arithmetic sequence, 654 associative property for function composition, 366 matrix addition, 579 matrix multiplication, 585 scalar multiplication, 581 vector addition, 1015 scalar multiplication, 1018 asymptote horizontal formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 of a hyperbola, 531 slant determination of, 312 formal deﬁnition of, 311 slant (oblique), 311 vertical formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 augmented matrix, 568 average angular velocity, 707 average cost, 346 average cost function, 82 average rate of change, 160 average velocity, 706 axis of symmetry, 191 back substitution, 560 bearings, 905 binomial coeﬃcient, 683 Binomial Theorem, 684 Bisection Method, 277 BMI, body mass index, 355 Boyle’s Law, 350 buﬀer solution, 478 cardioid, 951 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauchy’s Bound, 269 center of a circle, 498 of a hyperbola, 531 of an ellipse, 516 Index 1071 central angle, 701 change of base formulas, 442 characteristic polynomial, 626 Charles’s Law, 355 circle center of, 498 deﬁnition of, 498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix addition, 579 vector addition, 1015 dot product, 1034 complementary angles, 696 Complex Factorization Theorem, 290 complex number nth root, 1000, 1001 nth Roots of Unity, 1006 argument deﬁnition of, 991 properties of, 995 conjugate deﬁnition of, 288 properties of, 289 deﬁnition of, 2, 287, 991 imaginary part, 991 imaginary unit, i, 287 modulus deﬁnition of, 991 properties of, 993 polar form cis-notation, 995 principal argument, 991 real part, 991 rectangular form, 991 set of, 2 complex plane, 991 component form of a vector, 1013 composite function deﬁnition of, 360 properties of, 367 compound interest, 470 conic sections deﬁnition, 495 conjugate axis of a hyperbola, 532 conjugate of a complex number deﬁnition of, 288 properties of, 289 Conjugate Pairs Theorem, 291 consistent system, 553 constant function as a horizontal line, 156 formal deﬁnition of, 101 intuitive deﬁnition of, 100 constant of proportionality, 350 constant term of a polynomial, 236 continuous, 241 continuously compounded interest, 472 contradiction, 549 coordinates Cartesian, 6 polar, 919 rectangular, 919 correlation coeﬃcient, 226 cosecant graph of, 801 of an angle, 744, 752 properties of, 802 cosine graph of, 791 of an angle, 717, 730, 744 properties of, 791 1072 cost average, 82, 346 ﬁxed, start-up, 82 variable, 159 cost function, 82 cotangent graph of, 805 of an angle, 744, 752 properties of, 806 coterminal angle, 698 Coulomb’s Law, 355 Cramer’s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273 determinant of a matrix deﬁnition of, 614 properties of, 616 Diﬀerence Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 diﬀerence quotient, 79 dimension of a matrix, 567 direct variation, 350 directrix of a conic section in polar form, 981 of a parabola, 505 discriminant Index of a conic, 979 of a quadratic equation, 195 trichotomy, 195 distance deﬁnition, 10 distance formula, 11 distributive property matrix matrix multiplication, 585 scalar multiplication, 581 vector dot product, 1034 scalar multiplication, 1018 DMS, 695 domain applied, 60 deﬁnition of, 45 implied, 58 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to orthogonality, 1037 relation to vector magnitude, 1034 work, 1042 Double Angle Identities, 776 earthquake Richter Scale, 431 eccentricity, 522, 981 eigenvalue, 626 eigenvector, 626 ellipse center, 516 deﬁnition of, 516 eccentricity, 522 foci, 516 from slicing a cone, 496 guide rectangle, 519 major axis, 516 minor axis, 516 Index 1073 reﬂective property, 523 standard equation, 519 vertices, 516 ellipsis (. . . ), 31, 651 empty set, 2 end behavior of f (x) = axn, n even, 240 of f (x) = axn, n odd, 240 of a function graph, 239 polynomial, 243 entry in a matrix, 567 equation contradiction, 549 graph of, 23 identity, 549 linear of n variables, 554 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 extended interval notation, 756 Factor Theorem, 258 factorial, 654, 681 ﬁxed cost, 82 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 981 focus (foci) of a hyperbola, 531 of a parabola, 505 of an ellipse, 516 free variable, 552 frequency angular, 708, 881 of a sinusoid, 795 ordinary, 708, 881 function (absolute) maximum, 101 (absolute, global) minimum, 101 absolute value, 173 algebraic, 399 argument, 55 arithmetic, 76 as a process, 55, 378 average cost, 82 circular, 744 composite deﬁni
tion of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 diﬀerence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 linear, 156 local (relative) maximum, 101 local (relative) minimum, 101 logarithmic, 422 1074 Index notation, 55 odd, 95 one-to-one, 381 periodic, 790 piecewise-deﬁned, 62 polynomial, 235 price-demand, 82 product, 76 proﬁt, 82 quadratic, 188 quotient, 76 range, 45 rational, 301 revenue, 82 smooth, 241 sum, 76 transformation of graphs, 120, 135 zero, 95 fundamental cycle of y = cos(x), 791 Fundamental Graphing Principle for equations, 23 for functions, 93 for polar equations, 938 Fundamental Theorem of Algebra, 290 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 geometric sequence, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of, 2 greatest integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regression, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 deﬁnition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior, 239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quadrantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974
row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 1018 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of, 795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁnition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zero
1 2 3 4 x −1 −2 −3 −4 The horizontal number line is usually called the x-axis while the vertical number line is usually called the y-axis.6 As with the usual number line, we imagine these axes extending oﬀ indeﬁnitely in both directions.7 Having two number lines allows us to locate the positions of points oﬀ of the number lines as well as points on the lines themselves. For example, consider the point P on the next page. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line from the y-axis to P . This process is sometimes called ‘projecting’ the point P to the x- (respectively y-) axis. We then describe the point P using the ordered pair (2, −4). The ﬁrst number in the ordered pair is called the abscissa or x-coordinate and the second is called the ordinate or y-coordinate.8 Taken together, the ordered pair (2, −4) comprise the Cartesian coordinates9 of the point P . In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of ‘the point (2, −4).’ We can think of (2, −4) as instructions on how to 5So named in honor of Ren´e Descartes. 6The labels can vary depending on the context of application. 7Usually extending oﬀ towards inﬁnity is indicated by arrows, but here, the arrows are used to indicate the direction of increasing values of x and y. 8Again, the names of the coordinates can vary depending on the context of the application. If, for example, the horizontal axis represented time we might choose to call it the t-axis. The ﬁrst number in the ordered pair would then be the t-coordinate. 9Also called the ‘rectangular coordinates’ of P – see Section 11.4 for more details. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 7 reach P from the origin (0, 0) by moving 2 units to the right and 4 units downwards. Notice that the order in the ordered pair is important − if we wish to plot the point (−4, 2), we would move to the left 4 units from the origin and then move upwards 2 units, as below on the right. y 4 3 2 1 (−4, 2) y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 P −1 −2 −3 −4 P (2, −4) When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs (x, y) as x and y take values from the real numbers. Below is a summary of important facts about Cartesian coordinates. Important Facts about the Cartesian Coordinate Plane (a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d. (x, y) lies on the x-axis if and only if y = 0. (x, y) lies on the y-axis if and only if x = 0. The origin is the point (0, 0). It is the only point common to both axes. Example 1.1.2. Plot the following points: A(5, 8), B − 5 F (0, 5), G(−7, 0), H(0, −9), O(0, 0).10 2 , 3, C(−5.8, −3), D(4.5, −1), E(5, 0), Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the y-coordinate is 0 we move along the x-axis only. 10The letter O is almost always reserved for the origin. 8 Relations and Functions y F (0, 5) A(5, 8) O(0, 0) E(5, 0(−7, 0) −9 −8 −7 −6 −5 −4 −3 −2 −(−5.8, −3) D(4.5, −1) −1 −2 −3 −4 −5 −6 −7 −8 −9 H(0, −9) The axes divide the plane into four regions called quadrants. They are labeled with Roman numerals and proceed counterclockwise around the plane: Quadrant II x < 0, y > 0 y 4 3 2 1 Quadrant I x > 0, y > 0 −4 −3 −2 −1 1 2 3 4 x Quadrant III x < 0, y < 0 −1 −2 −3 −4 Quadrant IV x > 0, y < 0 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 9 For example, (1, 2) lies in Quadrant I, (−1, 2) in Quadrant II, (−1, −2) in Quadrant III and (1, −2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to that point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative y-axis (if x = 0). For example, (0, 4) lies on the positive y-axis whereas (−117, 0) lies on the negative x-axis. Such points do not belong to any of the four quadrants. One of the most important concepts in all of Mathematics is symmetry.11 There are many types of symmetry in Mathematics, but three of them can be discussed easily using Cartesian Coordinates. Deﬁnition 1.3. Two points (a, b) and (c, d) in the plane are said to be symmetric about the x-axis if a = c and b = −d symmetric about the y-axis if a = −c and b = d symmetric about the origin if a = −c and b = −d Schematically, Q(−x, y) y 0 P (x, y) x R(−x, −y) S(x, −y) In the above ﬁgure, P and S are symmetric about the x-axis, as are Q and R; P and Q are symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are Q and S. Example 1.1.3. Let P be the point (−2, 3). Find the points which are symmetric to P about the: 1. x-axis 2. y-axis 3. origin Check your answer by plotting the points. Solution. The ﬁgure after Deﬁnition 1.3 gives us a good way to think about ﬁnding symmetric points in terms of taking the opposites of the x- and/or y-coordinates of P (−2, 3). 11According to Carl. Jeﬀ thinks symmetry is overrated. 10 Relations and Functions 1. To ﬁnd the point symmetric about the x-axis, we replace the y-coordinate with its opposite to get (−2, −3). 2. To ﬁnd the point symmetric about the y-axis, we replace the x-coordinate with its opposite to get (2, 3). 3. To ﬁnd the point symmetric about the origin, we replace the x- and y-coordinates with their opposites to get (2, −3). y 3 2 1 (2, 3) P (−2, 3) −3 −2 −1 1 2 3 x −1 −2 −3 (−2, −3) (2, −3) One way to visualize the processes in the previous example is with the concept of a reﬂection. If we start with our point (−2, 3) and pretend that the x-axis is a mirror, then the reﬂection of (−2, 3) across the x-axis would lie at (−2, −3). If we pretend that the y-axis is a mirror, the reﬂection of (−2, 3) across that axis would be (2, 3). If we reﬂect across the x-axis and then the y-axis, we would go from (−2, 3) to (−2, −3) then to (2, −3), and so we would end up at the point symmetric to (−2, 3) about the origin. We summarize and generalize this process below. Reﬂections To reﬂect a point (x, y) about the: x-axis, replace y with −y. y-axis, replace x with −x. origin, replace x with −x and y with −y. 1.1.3 Distance in the Plane Another important concept in Geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Suppose we have two points, P (x0, y0) and Q (x1, y1) , in the plane. By the distance d between P and Q, we mean the length of the line segment joining P with Q. (Remember, given any two distinct points in the plane, there is a unique line containing both 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 11 points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation below on the left. Q (x1, y1) Q (x1, y1) d d P (x0, y0) P (x0, y0) (x1, y0) With a little more imagination, we can envision a right triangle whose hypotenuse has length d as drawn above on the right. From the latter ﬁgure, we see that the lengths of the legs of the triangle are \|x1 − x0\| and \|y1 − y0\| so the Pythagorean Theorem gives us \|x1 − x0\|2 + \|y1 − y0\|2 = d2 (x1 − x0)2 + (y1 − y0)2 = d2 (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get Equation 1.1. The Distance Formula: The distance d between the points P (x0, y0) and Q (x1, y1) is: d = (x1 − x0)2 + (y1 − y0)2 It is not always the case that the points P and Q lend themselves to constructing such a triangle. If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader in Exercise 35 to verify Equation 1.1 for these cases. Example 1.1.4. Find and simplify the distance between P (−2, 3) and Q(1, −3). Solution. d = = = So the distance is 3 √ 5. = 3 √ (x1 − x0)2 + (y1 − y0)2 (1 − (−2))2 + (−3 − 3)2 9 + 36 √ 5 12 Relations and Functions Example 1.1.5. Find all of the points with x-coordinate 1 which are 4 units from the point (3, 2). Solution. We shall soon see that the points we wish to ﬁnd are on the line x = 1, but for now we’ll just view them as points of the form (1, y). Visually, y 3 2 1 −1 −2 −3 (3, 2) distance is 4 units 2 3 x (1, y) We require that the distance from (3, 2) to (1, y) be 4. The Distance Formula, Equation 1.1, yields d = (x1 − x0)2 + (y1 − y0)2 (1 − 3)2 + (y − 2)2 4 + (y − 2)2 4 + (y − 2)22 4 = 4 = 42 = 16 = 4 + (y − 2)2 12 = (y − 2)2 √ (y − 2)2 = 12 2 12 √ 3 √ y = 2 ± 2 3 squaring both sides extracting the square root We obtain two answers: (1, 2 + 2 why there are two answers. √ 3) and (1, 2 − 2 √ 3). The reader is encouraged to think about Related to ﬁnding the distance between two points is the problem of ﬁnding the midpoint of the line segment connecting two points. Given two points, P (x0, y0) and Q (x1, y1), the midpoint M of P and Q is deﬁned to be the point on the line segment connecting P and Q whose distance from P is equal to its distance from Q. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 13 Q (x1, y1) M P (x0, y0) If we think of reaching M by going ‘halfway over’ and ‘halfway up’ we get the following formula. Equation 1.2. The Midpoint Formula: The midpoint M of the line segment connecting P (x0, y0) and Q (x1, y1) is: M = x0 + x1 2 , y0 + y1 2 If we let d denote the distance between P and Q, we leave it as Exercise 36 to show that the distance between P and M is d/2 which is the same as the distance between
M and Q. This suﬃces to show that Equation 1.2 gives the coordinates of the midpoint. Example 1.1.6. Find the midpoint of the line segment connecting P (−2, 3) and Q(1, −3). Solution. M = = = x0 + x1 , 2 (−2) + 1 2 y0 + y1 2 , 3 + (−3 The midpoint is − 1 2 , 0. We close with a more abstract application of the Midpoint Formula. We will revisit the following example in Exercise 72 in Section 2.1. Example 1.1.7. If a = b, prove that the line y = x equally divides the line segment with endpoints (a, b) and (b, a). Solution. To prove the claim, we use Equation 1.2 to ﬁnd the midpoint Since the x and y coordinates of this point are the same, we ﬁnd that the midpoint lies on the line y = x, as required. 14 Relations and Functions 1.1.4 Exercises 1. Fill in the chart below: Set of Real Numbers Interval Notation Region on the Real Number Line {x \| − 1 ≤ x < 5} {x \| − 5 < x ≤ 0} {x \| x ≤ 3} {x \| x ≥ −3} [0, 3) (−3, 3) (−∞, 9) 7 7 2 5 4 In Exercises 2 - 7, ﬁnd the indicated intersection or union and simplify if possible. Express your answers in interval notation. 2. (−1, 5] ∩ [0, 8) 3. (−1, 1) ∪ [0, 6] 4. (−∞, 4] ∩ (0, ∞) 5. (−∞, 0) ∩ [1, 5] 6. (−∞, 0) ∪ [1, 5] 7. (−∞, 5] ∩ [5, 8) In Exercises 8 - 19, write the set using interval notation. 8. {x \| x = 5} 9. {x \| x = −1} 10. {x \| x = −3, 4} 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 15 11. {x \| x = 0, 2} 12. {x \| x = 2, −2} 13. {x \| x = 0, ±4} 14. {x \| x ≤ −1 or x ≥ 1} 15. {x \| x < 3 or x ≥ 2} 16. {x \| x ≤ −3 or x > 0} 17. {x \| x ≤ 5 or x = 6} 18. {x \| x > 2 or x = ±1} 19. {x \| − 3 < x < 3 or x = 4} 20. Plot and label the points A(−3, −7), B(1.3, −2), C(π, √ 10), D(0, 8), E(−5.5, 0), F (−8, 4), G(9.2, −7.8) and H(7, 5) in the Cartesian Coordinate Plane given below9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −3 −4 −5 −6 −7 −8 −9 21. For each point given in Exercise 20 above Identify the quadrant or axis in/on which the point lies. Find the point symmetric to the given point about the x-axis. Find the point symmetric to the given point about the y-axis. Find the point symmetric to the given point about the origin. 16 Relations and Functions In Exercises 22 - 29, ﬁnd the distance d between the points and the midpoint M of the line segment which connects them. 22. (1, 2), (−3, 5) 23. (3, −10), (−1, 2) 24. 26. 1 2 24 5 √ , − 45, 11 − , 5 12, √ √ 28. 2 . 19 5 √ 25. − 2 3 27. √ 2, − √ 8, − 12 20, 27. 29. (0, 0), (x, y) 30. Find all of the points of the form (x, −1) which are 4 units from the point (3, 2). 31. Find all of the points on the y-axis which are 5 units from the point (−5, 3). 32. Find all of the points on the x-axis which are 2 units from the point (−1, 1). 33. Find all of the points of the form (x, −x) which are 1 unit from the origin. 34. Let’s assume for a moment that we are standing at the origin and the positive y-axis points due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be? 35. Verify the Distance Formula 1.1 for the cases when: (a) The points are arranged vertically. (Hint: Use P (a, y0) and Q(a, y1).) (b) The points are arranged horizontally. (Hint: Use P (x0, b) and Q(x1, b).) (c) The points are actually the same point. (You shouldn’t need a hint for this one.) 36. Verify the Midpoint Formula by showing the distance between P (x1, y1) and M and the distance between M and Q(x2, y2) are both half of the distance between P and Q. 37. Show that the points A, B and C below are the vertices of a right triangle. (a) A(−3, 2), B(−6, 4), and C(1, 8) (b) A(−3, 1), B(4, 0) and C(0, −3) 38. Find a point D(x, y) such that the points A(−3, 1), B(4, 0), C(0, −3) and D are the corners of a square. Justify your answer. 39. Discuss with your classmates how many numbers are in the interval (0, 1). 40. The world is not ﬂat.12 Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all? 12There are those who disagree with this statement. Look them up on the Internet some time when you’re bored. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 17 1.1.5 Answers 1. Set of Real Numbers Interval Notation Region on the Real Number Line {x \| − 1 ≤ x < 5} [−1, 5) {x \| 0 ≤ x < 3} {x \| 2 < x ≤ 7} [0, 3) (2, 7] {x \| − 5 < x ≤ 0} (−5, 0] {x \| − 3 < x < 3} (−3, 3) {x \| 5 ≤ x ≤ 7} [5, 7] {x \| x ≤ 3} (−∞, 3] {x \| x < 9} (−∞, 9) {x \| x > 4} (4, ∞) {x \| x ≥ −3} [−3, ∞) 5 3 7 0 3 7 3 9 −1 0 2 −5 −3 5 4 −3 2. (−1, 5] ∩ [0, 8) = [0, 5] 3. (−1, 1) ∪ [0, 6] = (−1, 6] 4. (−∞, 4] ∩ (0, ∞) = (0, 4] 5. (−∞, 0) ∩ [1, 5] = ∅ 6. (−∞, 0) ∪ [1, 5] = (−∞, 0) ∪ [1, 5] 7. (−∞, 5] ∩ [5, 8) = {5} 8. (−∞, 5) ∪ (5, ∞) 9. (−∞, −1) ∪ (−1, ∞) 10. (−∞, −3) ∪ (−3, 4) ∪ (4, ∞) 11. (−∞, 0) ∪ (0, 2) ∪ (2, ∞) 12. (−∞, −2) ∪ (−2, 2) ∪ (2, ∞) 13. (−∞, −4) ∪ (−4, 0) ∪ (0, 4) ∪ (4, ∞) 18 Relations and Functions 14. (−∞, −1] ∪ [1, ∞) 15. (−∞, ∞) 16. (−∞, −3] ∪ (0, ∞) 17. (−∞, 5] ∪ {6} 18. {−1} ∪ {1} ∪ (2, ∞) 19. (−3, 3) ∪ {4} 20. The required points A(−3, −7), B(1.3, −2), C(π, √ 10), D(0, 8), E(−5.5, 0), F (−8, 4), G(9.2, −7.8), and H(7, 5) are plotted in the Cartesian Coordinate Plane below. y D(0, 8) H(7, 5) √ 10) C(π (−8, 4) E(−5.5, 0) −9 −8 −7 −6 −5 −4 −3 −2 −(1.3, −2) −1 −2 −3 −4 −5 −6 −7 −8 −9 A(−3, −7) G(9.2, −7.8) 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 19 21. (a) The point A(−3, −7) is (b) The point B(1.3, −2) is in Quadrant III symmetric about x-axis with (−3, 7) symmetric about y-axis with (3, −7) symmetric about origin with (3, 7) in Quadrant IV symmetric about x-axis with (1.3, 2) symmetric about y-axis with (−1.3, −2) symmetric about origin with (−1.3, 2) (c) The point C(π, √ 10) is (d) The point D(0, 8) is in Quadrant I symmetric about x-axis with (π, − symmetric about y-axis with (−π, symmetric about origin with (−π, − √ √ √ 10) 10) 10) on the positive y-axis symmetric about x-axis with (0, −8) symmetric about y-axis with (0, 8) symmetric about origin with (0, −8) (e) The point E(−5.5, 0) is (f) The point F (−8, 4) is on the negative x-axis symmetric about x-axis with (−5.5, 0) symmetric about y-axis with (5.5, 0) symmetric about origin with (5.5, 0) in Quadrant II symmetric about x-axis with (−8, −4) symmetric about y-axis with (8, 4) symmetric about origin with (8, −4) (g) The point G(9.2, −7.8) is in Quadrant IV symmetric about x-axis with (9.2, 7.8) symmetric about y-axis with (−9.2, −7.8) symmetric about origin with (−9.2, 7.8) (h) The point H(7, 5) is in Quadrant I symmetric about x-axis with (7, −5) symmetric about y-axis with (−7, 5) symmetric about origin with (−7, −5) 22. d = 5, M = −1, 7 2 26, M = 1, 3 2 24. d = √ √ √ 26. d = 28. d = 30. (3 + 74, M = 13 10 , − 13 √ 10 83, M = 4 √ 2 5, 5 √ √ 7, −1), (3 − √ 32. (−1 + 3, 0), (−1 − 7, −1) √ 3, 0) 34. (−3, −4), 5 miles, (4, −4) 3 29. d = √ 23. d = 4 √ 25. d = 27. d = 3 37 10, M = (1, −4 , − x2 + y2, M = x 5 31. (0, 3) √ 33 37. (a) The distance from A to B is \|AB\| = 52, and the distance from B to C is \|BC\| = , we are guaranteed by the converse of the Pythagorean Theorem that the triangle is a right triangle. 13, the distance from A to C is \|AC\| = 652 132 65. Since √ √ = √ + √ 522 √ √ (b) Show that \|AC\|2 + \|BC\|2 = \|AB\|2 20 1.2 Relations Relations and Functions From one point of view,1 all of Precalculus can be thought of as studying sets of points in the plane. With the Cartesian Plane now fresh in our memory we can discuss those sets in more detail and as usual, we begin with a deﬁnition. Deﬁnition 1.4. A relation is a set of points in the plane. Since relations are sets, we can describe them using the techniques presented in Section 1.1.1. That is, we can describe a relation verbally, using the roster method, or using set-builder notation. Since the elements in a relation are points in the plane, we often try to describe the relation graphically or algebraically as well. Depending on the situation, one method may be easier or more convenient to use than another. As an example, consider the relation R = {(−2, 1), (4, 3), (0, −3)}. As written, R is described using the roster method. Since R consists of points in the plane, we follow our instinct and plot the points. Doing so produces the graph of R. y 4 3 2 1 (−2, 1) (4, 3) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 (0, −3) The graph of R. In the following example, we graph a variety of relations. Example 1.2.1. Graph the following relations. 1. A = {(0, 0), (−3, 1), (4, 2), (−3, 2)} 2. HLS1 = {(x, 3) \| − 2 ≤ x ≤ 4} 3. HLS2 = {(x, 3) \| − 2 ≤ x < 4} 4. V = {(3, y) \| y is a real number} 5. H = {(x, y) \| y = −2} 6. R = {(x, y) \| 1 < y ≤ 3} 1Carl’s, of course. 1.2 Relations Solution. 21 1. To graph A, we simply plot all of the points which belong to A, as shown below on the left. 2. Don’t let the notation in this part fool you. The name of this relation is HLS1, just like the name of the relation in number 1 was A. The letters and numbers are just part of its name, just like the numbers and letters of the phrase ‘King George III’ were part of George’s name. In words, {(x, 3) \| − 2 ≤ x ≤ 4} reads ‘the set of points (x, 3) such that −2 ≤ x ≤ 4.’ All of these points have the same y-coordinate, 3, but the x-coordinate is allowed to vary between −2 and 4, inclusive. Some of the points which belong to HLS1 include some friendly points like: (−2, 3), (−1, 3), (0, 3), (1, 3), (2, 3), (3, 3), and (4, 3). However, HLS1 also contains the points (0.829, 3), − 5 It is impossible2 to list all of these points, which is why the variable x is used. Plotting several friendly representative points should convince you that HLS1 describes the horizontal line segment from the point (−2, 3) up to and including the point (4, 3). π, 3), and so on. 6 , 34 −3 −2 −1 1 2
3 4 x −4 −3 −2 −1 1 2 3 4 x The graph of A The graph of HLS1 3. HLS2 is hauntingly similar to HLS1. In fact, the only diﬀerence between the two is that instead of ‘−2 ≤ x ≤ 4’ we have ‘−2 ≤ x < 4’. This means that we still get a horizontal line segment which includes (−2, 3) and extends to (4, 3), but we do not include (4, 3) because of the strict inequality x < 4. How do we denote this on our graph? It is a common mistake to make the graph start at (−2, 3) end at (3, 3) as pictured below on the left. The problem with this graph is that we are forgetting about the points like (3.1, 3), (3.5, 3), (3.9, 3), (3.99, 3), and so forth. There is no real number that comes ‘immediately before’ 4, so to describe the set of points we want, we draw the horizontal line segment starting at (−2, 3) and draw an open circle at (4, 3) as depicted below on the right. 2Really impossible. The interested reader is encouraged to research countable versus uncountable sets. 22 Relations and Functions 4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x This is NOT the correct graph of HLS2 The graph of HLS2 4. Next, we come to the relation V , described as the set of points (3, y) such that y is a real number. All of these points have an x-coordinate of 3, but the y-coordinate is free to be whatever it wants to be, without restriction.3 Plotting a few ‘friendly’ points of V should convince you that all the points of V lie on the vertical line4 x = 3. Since there is no restriction on the y-coordinate, we put arrows on the end of the portion of the line we draw to indicate it extends indeﬁnitely in both directions. The graph of V is below on the left. 5. Though written slightly diﬀerently, the relation H = {(x, y) \| y = −2} is similar to the relation V above in that only one of the coordinates, in this case the y-coordinate, is speciﬁed, leaving x to be ‘free’. Plotting some representative points gives us the horizontal line y = −2. y 4 3 2 1 −1 −2 −3 −4 1 2 3 4 x y −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 The graph of H The graph of V 6. For our last example, we turn to R = {(x, y) \| 1 < y ≤ 3}. As in the previous example, x is free to be whatever it likes. The value of y, on the other hand, while not completely free, is permitted to roam between 1 and 3 excluding 1, but including 3. After plotting some5 friendly elements of R, it should become clear that R consists of the region between the horizontal 3We’ll revisit the concept of a ‘free variable’ in Section 8.1. 4Don’t worry, we’ll be refreshing your memory about vertical and horizontal lines in just a moment! 5The word ‘some’ is a relative term. It may take 5, 10, or 50 points until you see the pattern. 1.2 Relations 23 lines y = 1 and y = 3. Since R requires that the y-coordinates be greater than 1, but not equal to 1, we dash the line y = 1 to indicate that those points do not belong to R. y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x The graph of R The relations V and H in the previous example lead us to our ﬁnal way to describe relations: algebraically. We can more succinctly describe the points in V as those points which satisfy the equation ‘x = 3’. Most likely, you have seen equations like this before. Depending on the context, ‘x = 3’ could mean we have solved an equation for x and arrived at the solution x = 3. In this case, however, ‘x = 3’ describes a set of points in the plane whose x-coordinate is 3. Similarly, the relation H above can be described by the equation ‘y = −2’. At some point in your mathematical upbringing, you probably learned the following. Equations of Vertical and Horizontal Lines The graph of the equation x = a is a vertical line through (a, 0). The graph of the equation y = b is a horizontal line through (0, b). Given that the very simple equations x = a and y = b produced lines, it’s natural to wonder what shapes other equations might yield. Thus our next objective is to study the graphs of equations in a more general setting as we continue to unite Algebra and Geometry. 1.2.1 Graphs of Equations In this section, we delve more deeply into the connection between Algebra and Geometry by focusing on graphing relations described by equations. The main idea of this section is the following. The Fundamental Graphing Principle The graph of an equation is the set of points which satisfy the equation. That is, a point (x, y) is on the graph of an equation if and only if x and y satisfy the equation. Here, ‘x and y satisfy the equation’ means ‘x and y make the equation true’. It is at this point that we gain some insight into the word ‘relation’. If the equation to be graphed contains both x and y, then the equation itself is what is relating the two variables. More speciﬁcally, in the next two examples, we consider the graph of the equation x2 + y3 = 1. Even though it is not speciﬁcally 24 Relations and Functions spelled out, what we are doing is graphing the relation R = {(x, y) \| x2 + y3 = 1}. The points (x, y) we graph belong to the relation R and are necessarily related by the equation x2 + y3 = 1, since it is those pairs of x and y which make the equation true. Example 1.2.2. Determine whether or not (2, −1) is on the graph of x2 + y3 = 1. Solution. We substitute x = 2 and y = −1 into the equation to see if the equation is satisﬁed. Hence, (2, −1) is not on the graph of x2 + y3 = 1. (2)2 + (−1)3 ? = 1 3 = 1 We could spend hours randomly guessing and checking to see if points are on the graph of the equation. A more systematic approach is outlined in the following example. Example 1.2.3. Graph x2 + y3 = 1. Solution. To eﬃciently generate points on the graph of this equation, we ﬁrst solve for y x2 + y3 = 1 y3 = 1 − x2 y3 = 3√ y = 3√ We now substitute a value in for x, determine the corresponding value y, and plot the resulting point (x, y). For example, substituting x = −3 into the equation yields 1 − x2 1 − x2 3 y = 3 1 − x2 = 3 1 − (−3)2 = 3√ −8 = −2, so the point (−3, −2) is on the graph. Continuing in this manner, we generate a table of points which are on the graph of the equation. These points are then plotted in the plane as shown below. y y x −2 −3 √ −2 − 3 −2 3 (x, y) (−3, −2) √ 3 (−2, − 3 3) (−1, 0) (0, 1) (1, 0) √ (2, − 3 3) (3, −2) 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 Remember, these points constitute only a small sampling of the points on the graph of this equation. To get a better idea of the shape of the graph, we could plot more points until we feel comfortable 1.2 Relations 25 ‘connecting the dots’. Doing so would result in a curve similar to the one pictured below on the far left. y 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 Don’t worry if you don’t get all of the little bends and curves just right − Calculus is where the art of precise graphing takes center stage. For now, we will settle with our naive ‘plug and plot’ approach to graphing. If you feel like all of this tedious computation and plotting is beneath you, then you can reach for a graphing calculator, input the formula as shown above, and graph. Of all of the points on the graph of an equation, the places where the graph crosses or touches the axes hold special signiﬁcance. These are called the intercepts of the graph. Intercepts come in two distinct varieties: x-intercepts and y-intercepts. They are deﬁned below. Deﬁnition 1.5. Suppose the graph of an equation is given. A point on a graph which is also on the x-axis is called an x-intercept of the graph. A point on a graph which is also on the y-axis is called an y-intercept of the graph. In our previous example the graph had two x-intercepts, (−1, 0) and (1, 0), and one y-intercept, (0, 1). The graph of an equation can have any number of intercepts, including none at all! Since x-intercepts lie on the x-axis, we can ﬁnd them by setting y = 0 in the equation. Similarly, since y-intercepts lie on the y-axis, we can ﬁnd them by setting x = 0 in the equation. Keep in mind, intercepts are points and therefore must be written as ordered pairs. To summarize, Finding the Intercepts of the Graph of an Equation Given an equation involving x and y, we ﬁnd the intercepts of the graph as follows: x-intercepts have the form (x, 0); set y = 0 in the equation and solve for x. y-intercepts have the form (0, y); set x = 0 in the equation and solve for y. Another fact which you may have noticed about the graph in the previous example is that it seems to be symmetric about the y-axis. To actually prove this analytically, we assume (x, y) is a generic point on the graph of the equation. That is, we assume x2 + y3 = 1 is true. As we learned in Section 1.1, the point symmetric to (x, y) about the y-axis is (−x, y). To show that the graph is 26 Relations and Functions symmetric about the y-axis, we need to show that (−x, y) satisﬁes the equation x2 + y3 = 1, too. Substituting (−x, y) into the equation gives (−x)2 + (y)3 ? = 1 x2 + y3 = 1 Since we are assuming the original equation x2 + y3 = 1 is true, we have shown that (−x, y) satisﬁes the equation (since it leads to a true result) and hence is on the graph. In this way, we can check whether the graph of a given equation possesses any of the symmetries discussed in Section 1.1. We summarize the procedure in the following result. Testing the Graph of an Equation for Symmetry To test the graph of an equation for symmetry about the y-axis − substitute (−x, y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the y-axis. about the x-axis – substitute (x, −y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the x-axis. about the origin - substitute (−x, −y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the origin. Intercepts and symmetry are two tools which can help us sketch the graph of an equation analytically, as demonstrated in the next example. Example 1.2.4. Find the x- and y-intercepts (if any) of the graph of (x − 2)2 + y2 = 1. Tes
t for symmetry. Plot additional points as needed to complete the graph. Solution. To look for x-intercepts, we set y = 0 and solve (x − 2)2 + y2 = 1 (x − 2)2 + 02 = 1 (x − 2)2 = 1 √ (x − 2)2 = 1 x − 2 = ±1 extract square roots x = 2 ± 1 x = 3, 1 We get two answers for x which correspond to two x-intercepts: (1, 0) and (3, 0). Turning our attention to y-intercepts, we set x = 0 and solve 1.2 Relations 27 (x − 2)2 + y2 = 1 (0 − 2)2 + y2 = 1 4 + y2 = 1 y2 = −3 Since there is no real number which squares to a negative number (Do you remember why?), we are forced to conclude that the graph has no y-intercepts. Plotting the data we have so far, we get y 1 −1 (1, 0) (3, 0) 1 2 3 4 x Moving along to symmetry, we can immediately dismiss the possibility that the graph is symmetric about the y-axis or the origin. If the graph possessed either of these symmetries, then the fact that (1, 0) is on the graph would mean (−1, 0) would have to be on the graph. (Why?) Since (−1, 0) would be another x-intercept (and we’ve found all of these), the graph can’t have y-axis or origin symmetry. The only symmetry left to test is symmetry about the x-axis. To that end, we substitute (x, −y) into the equation and simplify (x − 2)2 + y2 = 1 ? = 1 (x − 2)2 + (−y)2 (x − 2)2 + y2 = 1 Since we have obtained our original equation, we know the graph is symmetric about the x-axis. This means we can cut our ‘plug and plot’ time in half: whatever happens below the x-axis is reﬂected above the x-axis, and vice-versa. Proceeding as we did in the previous example, we obtain y 1 −1 1 2 3 4 x 28 Relations and Functions A couple of remarks are in order. First, it is entirely possible to choose a value for x which does not correspond to a point on the graph. For example, in the previous example, if we solve for y as is our custom, we get y = ± 1 − (x − 2)2. Upon substituting x = 0 into the equation, we would obtain y = ± 1 − (0 − 2)3, which is not a real number. This means there are no points on the graph with an x-coordinate of 0. When this happens, we move on and try another point. This is another drawback of the ‘plug-and-plot’ approach to graphing equations. Luckily, we will devote much of the remainder of this book to developing techniques which allow us to graph entire families of equations quickly.6 Second, it is instructive to show what would have happened had we tested the equation in the last example for symmetry about the y-axis. Substituting (−x, y) into the equation yields (x − 2)2 + y2 = . (−x − 2)2 + y2 ((−1)(x + 2))2 + y2 (x + 2)2 + y2 This last equation does not appear to be equivalent to our original equation. However, to actually prove that the graph is not symmetric about the y-axis, we need to ﬁnd a point (x, y) on the graph whose reﬂection (−x, y) is not. Our x-intercept (1, 0) ﬁts this bill nicely, since if we substitute (−1, 0) into the equation we get (x − 2)2 + y2 (−1 − 2)2 + 02 ? = 1 = 1 9 = 1. This proves that (−1, 0) is not on the graph. 6Without the use of a calculator, if you can believe it! 1.2 Relations 1.2.2 Exercises In Exercises 1 - 20, graph the given relation. 1. {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 2. {(−2, 0), (−1, 1), (−1, −1), (0, 2), (0, −2), (1, 3), (1, −3)} 29 3. {(m, 2m) \| m = 0, ±1, ±2} 4. 6 k , k \| k = ±1, ±2, ±3, ±4, ±5, ±6 5. n, 4 − n2 \| n = 0, ±1, ±2 6. √ j, j \| j = 0, 1, 4, 9 7. {(x, −2) \| x > −4} 8. {(x, 3) \| x ≤ 4} 9. {(−1, y) \| y > 1} 10. {(2, y) \| y ≤ 5} 11. {(−2, y) \| − 3 < y ≤ 4} 12. {(3, y) \| − 4 ≤ y < 3} 13. {(x, 2) \| − 2 ≤ x < 3} 14. {(x, −3) \| − 4 < x ≤ 4} 15. {(x, y) \| x > −2} 16. {(x, y) \| x ≤ 3} 17. {(x, y) \| y < 4} 18. {(x, y) \| x ≤ 3, y < 2} 19. {(x, y) \| x > 0, y < 4} 20. {(x, y } In Exercises 21 - 30, describe the given relation using either the roster or set-builder method. 21. y 4 3 2 1 −4 −3 −2 −1 −1 1 x Relation A 22. y 3 2 1 −4 −3 −2 −1 1 2 3 4 x Relation B 30 23. y 5 4 3 2 1 −1 −2 −3 1 2 3 x Relations and Functions 24. y 3 2 1 −3 −2 −1 −1 x −2 −3 −4 Relation C Relation D 25. 27. y 3 2 1 −4 −3 −2 −1 1 2 3 4 x Relation E y 3 2 1 26. 28. y 4 3 2 1 −3 −2 −1 1 2 3 x Relation F y 3 2 1 −3 −2 −1 −1 1 2 3 x −4 −3 −2 −1 −1 1 2 3 x −2 −3 Relation G −2 −3 Relation H 1.2 Relations 29. y 5 4 3 2 1 −1 −1 1 2 3 4 5 x Relation I In Exercises 31 - 36, graph the given line. 31. x = −2 33. y = 3 35. x = 0 31 30. y 2 1 −4 −3 −2 −1 −1 1 2 3 4 5 x −2 −3 Relation J 32. x = 3 34. y = −2 36. y = 0 Some relations are fairly easy to describe in words or with the roster method but are rather diﬃcult, if not impossible, to graph. Discuss with your classmates how you might graph the relations given in Exercises 37 - 40. Please note that in the notation below we are using the ellipsis, . . . , to denote that the list does not end, but rather, continues to follow the established pattern indeﬁnitely. For the relations in Exercises 37 and 38, give two examples of points which belong to the relation and two points which do not belong to the relation. 37. {(x, y) \| x is an odd integer, and y is an even integer.} 38. {(x, 1) \| x is an irrational number } 39. {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5), . . .} 40. {. . . , (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9), . . .} For each equation given in Exercises 41 - 52: Find the x- and y-intercept(s) of the graph, if any exist. Follow the procedure in Example 1.2.3 to create a table of sample points on the graph of the equation. Plot the sample points and create a rough sketch of the graph of the equation. Test for symmetry. If the equation appears to fail any of the symmetry tests, ﬁnd a point on the graph of the equation whose reﬂection fails to be on the graph as was done at the end of Example 1.2.4 32 Relations and Functions 41. y = x2 + 1 43. y = x3 − x 45. y = √ x − 2 47. 3x − y = 7 49. (x + 2)2 + y2 = 16 51. 4y2 − 9x2 = 36 42. y = x2 − 2x − 8 44. y = x3 4 − 3x √ 46. y = 2 x + 4 − 2 48. 3x − 2y = 10 50. x2 − y2 = 1 52. x3y = −4 The procedures which we have outlined in the Examples of this section and used in Exercises 41 - 52 all rely on the fact that the equations were “well-behaved”. Not everything in Mathematics is quite so tame, as the following equations will show you. Discuss with your classmates how you might approach graphing the equations given in Exercises 53 - 56. What diﬃculties arise when trying to apply the various tests and procedures given in this section? For more information, including pictures of the curves, each curve name is a link to its page at www.wikipedia.org. For a much longer list of fascinating curves, click here. 53. x3 + y3 − 3xy = 0 Folium of Descartes 54. x4 = x2 + y2 Kampyle of Eudoxus 55. y2 = x3 + 3x2 Tschirnhausen cubic 56. (x2 + y2)2 = x3 + y3 Crooked egg 57. With the help of your classmates, ﬁnd examples of equations whose graphs possess symmetry about the x-axis only symmetry about the y-axis only symmetry about the origin only symmetry about the x-axis, y-axis, and origin Can you ﬁnd an example of an equation whose graph possesses exactly two of the symmetries listed above? Why or why not? 1.2 Relations 1.2.3 Answers 1. 5. −3 −2 −2 −1 −1 1 2 x −2 −3 −4 y 4 3 2 1 −2 −1 1 2 x 33 2. 4. y 3 2 1 −2 −1 −1 1 2 x −2 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 6 34 7. Relations and Functions y 8. −4 −3 −2 −1 −3 −4 −3 −2 −1 1 2 3 4 x 9. y 101 1 2 x 11. y 4 3 2 1 −3 −2 −1 −1 x −2 −3 5 4 3 2 1 −1 −2 −3 1 2 3 x 12. y 3 2 1 −1 −2 −3 −4 1 2 3 x 1.2 Relations 35 14. 16. 18. 13. 15. 17. y 3 2 1 −4 −3 −2 −2 −1 −1 1 2 3 x −2 −3 y 4 3 2 1 −3 −2 −1 1 2 3 x y −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 y y 3 2 1 −1 −2 −3 3 2 1 −1 −2 −3 1 2 3 x 1 2 3 x 19. y 20. y 4 3 2 1 −2 −1 1 x 36 Relations and Functions 21. A = {(−4, −1), (−2, 1), (0, 3), (1, 4)} 22. B = {(x, 3) \| x ≥ −3} 23. C = {(2, y) \| y > −3} 24. D = {(−2, y) \| − 4 ≤ y < 3} 25. E = {(x, 2) \| − 4 ≤ x < 3} 26. F = {(x, y) \| y ≥ 0} 27. G = {(x, y) \| x > −2} 28. H = {(x, y) \| − 3 < x ≤ 2} 29. I = {(x, y) \| x ≥ 0,y ≥ 0} 30. J = {(x, y) \| − 4 < x < 5, −3 < y < 2} 31. 33. 35. y 3 2 1 −3 −2 −1 −1 x −2 −3 The line x = −2 y 3 2 1 −3 −2 −1 1 2 3 x The line y = 3 y 3 2 1 32. y 3 2 1 −1 −2 −3 1 2 3 x The line x = 3 34. y −3 −2 −1 −1 1 2 3 x −2 −3 The line y = −2 36. y 3 2 1 −3 −2 −1 −1 1 2 3 x −3 −2 −1 −1 1 2 3 x −2 −3 −2 −3 The line x = 0 is the y-axis The line y = 0 is the x-axis 1.2 Relations 41. y = x2 + 1 37 42. y = x2 − 2x − 8 The graph has no x-intercepts x-intercepts: (4, 0), (−2, 0) y-intercept: (0, 1) x −2 −1 0 1 2 y (x, y) 5 (−2, 5) 2 (−1, 2) (0, 1) 1 (1, 2) 2 (2, 5) 5 y 5 4 3 2 1 −2−1 1 2 x The graph is not symmetric about the x-axis (e.g. (2, 5) is on the graph but (2, −5) is not) The graph is symmetric about the y-axis The graph is not symmetric about the origin (e.g. (2, 5) is on the graph but (−2, −5) is not) y-intercept: (0, −8) y 7 0 (x, y) x (−3, 7) −3 −2 (−2, 0) −1 −5 (−1, −5) (0, −8) (1, −9) (2, −8) (3, −5) (4, 0) (5, 7) 0 −8 1 −9 2 −8 3 −3−2−1 −2 −3 −4 −5 −6 −7 −8 −9 The graph is not symmetric about the x-axis (e.g. (−3, 7) is on the graph but (−3, −7) is not) The graph is not symmetric about the y-axis (e.g. (−3, 7) is on the graph but (3, 7) is not) The graph is not symmetric about the origin (e.g. (−3, 7) is on the graph but (3, −7) is not) 38 43. y = x3 − x x-intercepts: (−1, 0), (0, 0), (1, 0) y-intercept: (0, 0) x y (x, y) −2 −6 (−2, −6) (−1, 0) −1 (0, 0) 0 (1, 0) 1 (2, 62−1 −1 1 2 x −2 −3 −4 −5 −6 The graph is not symmetric about the x-axis. (e.g. (2, 6) is on the graph but (2, −6) is not) The graph is not symmetric about the y-axis. (e.g. (2, 6) is on the graph but (−2, 6) is not) The graph is symmetric about the origin. Relations and Functions 44. y = x3 4 − 3x x-intercepts: ±2 √ 3, 0 , (0, 0) y-intercept: (0, 0) x y −4 −4 9 −3 4 −2 4 11 −1 4 0 0 1 − 11 4 2 −4 3 − 9 4 4 4 (x, y) (−4, −4) −3, 9 4 (−2, 4) −1, 11 4 (0, 0) 1, − 11 4 (2, −4) 3, − 9 4 (4, 4) y 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −2 −3 −4 The graph is not symmetric about the x-axis (e.g. (−4, −4) is on the graph but (−4, 4) is not) The graph is not symmetric about the y-axis (e.g. (−4, −4) i
s on the graph but (4, −4) is not) The graph is symmetric about the origin 1.2 Relations 45. y = √ x − 2 x-intercept: (2, 0) The graph has no y-intercepts √ 46-intercept: (−3, 0) y-intercept: (0, 2) 39 y (x, y) 0 (2, 0) 1 (3, 1) (6, 2) 2 3 (11, 3) x 2 3 6 11 10 11 x The graph is not symmetric about the x-axis (e.g. (3, 1) is on the graph but (3, −1) is not) The graph is not symmetric about the y-axis (e.g. (3, 1) is on the graph but (−3, 1) is not) The graph is not symmetric about the origin (e.g. (3, 1) is on the graph but (−3, −1) is not) x −4 −3 −2 −1 0 1 y −2 0 2 − 2 −2, 3 − 2 −2, 2 5 − 2 −2, y (x, y) (−4, −2) (−3, 00, 24−3−2−1 −1 1 2 x −2 −3 The graph is not symmetric about the x-axis (e.g. (−4, −2) is on the graph but (−4, 2) is not) The graph is not symmetric about the y-axis (e.g. (−4, −2) is on the graph but (4, −2) is not) The graph is not symmetric about the origin (e.g. (−4, −2) is on the graph but (4, 2) is not) 40 Relations and Functions 47. 3x − y = 7 Re-write as: y = 3x − 7. 48. 3x − 2y = 10 Re-write as: y = 3x−10 2 . x-intercept: ( 7 3 , 0) y-intercept: (0, −7) x y (x, y) −2 −13 (−2, −13) −1 −10 (−1, −10) (0, −7) (1, −4) (2, −1) (3, 2) 0 −7 1 −4 2 −1 2 3 y 3 2 1 −2−1 −1 1 2 3 x −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 x-intercepts: 10 3 , 0 y-intercept: (0, −5) x y −2 −8 −1 − 13 2 0 −5 1 − 7 2 2 −2 (x, y) (−2, −8) −1, − 13 2 (0, −5) 1, − 7 2 (2, −2) y 2 1 −3−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 −8 −9 The graph is not symmetric about the x-axis (e.g. (3, 2) is on the graph but (3, −2) is not) The graph is not symmetric about the y-axis (e.g. (3, 2) is on the graph but (−3, 2) is not) The graph is not symmetric about the origin (e.g. (3, 2) is on the graph but (−3, −2) is not) The graph is not symmetric about the x-axis (e.g. (2, −2) is on the graph but (2, 2) is not) The graph is not symmetric about the y-axis (e.g. (2, −2) is on the graph but (−2, −2) is not) The graph is not symmetric about the origin (e.g. (2, −2) is on the graph but (−2, 2) is not) 1.2 Relations 41 49. (x + 2)2 + y2 = 16 Re-write as y = ± 16 − (x + 2)2. x-intercepts: (−6, 0), (2, 0) y-intercepts: 0, ±2 3 √ y x −6 0 √ −4 ±2 −2 ±4 √ 0 ±2 0 2 (x, y) (−6, 0) √ 3 −4, ±2 3 3 √ (−2, ±4) 3 0, ±2 (2, 0) y 5 4 3 2 1 50. x2 − y2 = 1 √ Re-write as: y = ± x2 − 1. x-intercepts: (−1, 0), (1, 0) The graph has no y-intercepts x −3 ± −2 ± −x, y) √ √ 8 (−3, ± 3 (−2, ± 8) 3) (−1, 0) (1, 0) √ √ (2, ± (3, ± 3) 8) 3 8 −7−6−5−4−3−2−1 −1 1 2 3 x −2 −3 −4 −5 The graph is symmetric about the x-axis The graph is not symmetric about the y-axis (e.g. (−6, 0) is on the graph but (6, 0) is not) The graph is not symmetric about the origin (e.g. (−6, 0) is on the graph but (6, 0) is not) y 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 The graph is symmetric about the x-axis The graph is symmetric about the y-axis The graph is symmetric about the origin 42 51. 4y2 − 9x2 = 36 Re-write as: y = ± √ 9x2+36 2 . The graph has no x-intercepts y-intercepts: (0, ±3) Relations and Functions 52. x3y = −4 Re-write as: y = − 4 x3 . The graph has no x-intercepts The graph has no y-intercepts x y √ −4 ±3 √ −2 ±3 ±3 √ √ 0 2 ±3 4 ±3 √ √ (x, y) 5 −4, ±3 2 −2, ±3 (0, ±3) √ √ 2, ±3 4, ±3 2 5 2 5 5 2 x −2 −1 − 1 2 1 y 1 2 4 32 2 −32 ( 1 1 −4 2 − 1 2 (x, y) (−2, 1 2 ) (−1, 4) (− 1 2 , 32) 2 , −32) (1, −4) (24−3−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 y 32 4 −4 −2 −1 1 x 2 The graph is symmetric about the x-axis −32 The graph is symmetric about the y-axis The graph is symmetric about the origin The graph is not symmetric about the x-axis (e.g. (1, −4) is on the graph but (1, 4) is not) The graph is not symmetric about the y-axis (e.g. (1, −4) is on the graph but (−1, −4) is not) The graph is symmetric about the origin 1.3 Introduction to Functions 43 1.3 Introduction to Functions One of the core concepts in College Algebra is the function. There are many ways to describe a function and we begin by deﬁning a function as a special kind of relation. Deﬁnition 1.6. A relation in which each x-coordinate is matched with only one y-coordinate is said to describe y as a function of x. Example 1.3.1. Which of the following relations describe y as a function of x? 1. R1 = {(−2, 1), (1, 3), (1, 4), (3, −1)} 2. R2 = {(−2, 1), (1, 3), (2, 3), (3, −1)} Solution. A quick scan of the points in R1 reveals that the x-coordinate 1 is matched with two diﬀerent y-coordinates: namely 3 and 4. Hence in R1, y is not a function of x. On the other hand, every x-coordinate in R2 occurs only once which means each x-coordinate has only one corresponding y-coordinate. So, R2 does represent y as a function of x. Note that in the previous example, the relation R2 contained two diﬀerent points with the same y-coordinates, namely (1, 3) and (2, 3). Remember, in order to say y is a function of x, we just need to ensure the same x-coordinate isn’t used in more than one point.1 To see what the function concept means geometrically, we graph R1 and R2 in the plane2 −1 −1 1 2 3 x −2 −1 −1 1 2 3 x The graph of R1 The graph of R2 The fact that the x-coordinate 1 is matched with two diﬀerent y-coordinates in R1 presents itself graphically as the points (1, 3) and (1, 4) lying on the same vertical line, x = 1. If we turn our attention to the graph of R2, we see that no two points of the relation lie on the same vertical line. We can generalize this idea as follows Theorem 1.1. The Vertical Line Test: A set of points in the plane represents y as a function of x if and only if no two points lie on the same vertical line. 1We will have occasion later in the text to concern ourselves with the concept of x being a function of y. In this case, R1 represents x as a function of y; R2 does not. 44 Relations and Functions It is worth taking some time to meditate on the Vertical Line Test; it will check to see how well you understand the concept of ‘function’ as well as the concept of ‘graph’. Example 1.3.2. Use the Vertical Line Test to determine which of the following relations describes y as a function of x. y 4 3 2 1 −1 1 x −1 The graph of R The graph of S Solution. Looking at the graph of R, we can easily imagine a vertical line crossing the graph more than once. Hence, R does not represent y as a function of x. However, in the graph of S, every vertical line crosses the graph at most once, so S does represent y as a function of x. In the previous test, we say that the graph of the relation R fails the Vertical Line Test, whereas the graph of S passes the Vertical Line Test. Note that in the graph of R there are inﬁnitely many vertical lines which cross the graph more than once. However, to fail the Vertical Line Test, all you need is one vertical line that ﬁts the bill, as the next example illustrates. Example 1.3.3. Use the Vertical Line Test to determine which of the following relations describes y as a function of x1 1 x −1 −1 1 x −1 The graph of S1 The graph of S2 1.3 Introduction to Functions 45 Solution. Both S1 and S2 are slight modiﬁcations to the relation S in the previous example whose graph we determined passed the Vertical Line Test. In both S1 and S2, it is the addition of the point (1, 2) which threatens to cause trouble. In S1, there is a point on the curve with x-coordinate 1 just below (1, 2), which means that both (1, 2) and this point on the curve lie on the vertical line x = 1. (See the picture below and the left.) Hence, the graph of S1 fails the Vertical Line Test, so y is not a function of x here. However, in S2 notice that the point with x-coordinate 1 on the curve has been omitted, leaving an ‘open circle’ there. Hence, the vertical line x = 1 crosses the graph of S2 only at the point (1, 2). Indeed, any vertical line will cross the graph at most once, so we have that the graph of S2 passes the Vertical Line Test. Thus it describes y as a function of x1 −1 x −1 1 x −1 S1 and the line x = 1 The graph of G for Ex. 1.3.4 Suppose a relation F describes y as a function of x. The sets of x- and y-coordinates are given special names which we deﬁne below. Deﬁnition 1.7. Suppose F is a relation which describes y as a function of x. The set of the x-coordinates of the points in F is called the domain of F . The set of the y-coordinates of the points in F is called the range of F . We demonstrate ﬁnding the domain and range of functions given to us either graphically or via the roster method in the following example. Example 1.3.4. Find the domain and range of the function F = {(−3, 2), (0, 1), (4, 2), (5, 2)} and of the function G whose graph is given above on the right. Solution. The domain of F is the set of the x-coordinates of the points in F , namely {−3, 0, 4, 5} and the range of F is the set of the y-coordinates, namely {1, 2}. To determine the domain and range of G, we need to determine which x and y values occur as coordinates of points on the given graph. To ﬁnd the domain, it may be helpful to imagine collapsing the curve to the x-axis and determining the portion of the x-axis that gets covered. This is called projecting the curve to the x-axis. Before we start projecting, we need to pay attention to two 46 Relations and Functions subtle notations on the graph: the arrowhead on the lower left corner of the graph indicates that the graph continues to curve downwards to the left forever more; and the open circle at (1, 3) indicates that the point (1, 3) isn’t on the graph, but all points on the curve leading up to that point are. y project down 1 1 x −1 −1 1 x −1 project up The graph of G The graph of G We see from the ﬁgure that if we project the graph of G to the x-axis, we get all real numbers less than 1. Using interval notation, we write the domain of G as (−∞, 1). To determine the range of G, we project the curve to the y-axis as follows: y 4 3 2 1 project left project right y 4 3 2 1 −1 1 x −1 −1 1 x −1 The graph of G The graph of G Note that even though there is an open circle at (1, 3), we still include the y value of 3 in our range, since the point (−1, 3) is on the graph of G. We see that the range of G is all real n
umbers less than or equal to 4, or, in interval notation, (−∞, 4]. 1.3 Introduction to Functions 47 All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section1.2 may or may not describe y as a function of x. The algebraic representation of functions is possibly the most important way to view them so we need a process for determining whether or not an equation of a relation represents a function. (We delay the discussion of ﬁnding the domain of a function given algebraically until Section 1.4.) Example 1.3.5. Determine which equations represent y as a function of x. 1. x3 + y2 = 1 2. x2 + y3 = 1 3. x2y = 1 − 3y Solution. For each of these equations, we solve for y and determine whether each choice of x will determine only one corresponding value of y. 1. 2. 3. x3 + y2 = 1 √ y2 = 1 − x3 y2 = y = ± 1 − x3 √ 1 − x3 extract square roots 1 − 03 = ±1, so that (0, 1) If we substitute x = 0 into our equation for y, we get y = ± and (0, −1) are on the graph of this equation. Hence, this equation does not represent y as a function of x. √ x2 + y3 = 1 3 y3 = 1 − x2 y3 = 3√ y = 3√ 1 − x2 1 − x2 For every choice of x, the equation y = 3√ equation describes y as a function of x. 1 − x2 returns only one value of y. Hence, this x2y = 1 − 3y x2y + 3y = 1 y x2 + 3 = 1 y = 1 x2 + 3 factor For each choice of x, there is only one value for y, so this equation describes y as a function of x. We could try to use our graphing calculator to verify our responses to the previous example, but we immediately run into trouble. The calculator’s “Y=” menu requires that the equation be of the form ‘y = some expression of x’. If we wanted to verify that the ﬁrst equation in Example 1.3.5 48 Relations and Functions does not represent y as a function of x, we would need to enter two separate expressions into the calculator: one for the positive square root and one for the negative square root we found when solving the equation for y. As predicted, the resulting graph shown below clearly fails the Vertical Line Test, so the equation does not represent y as a function of x. Thus in order to use the calculator to show that x3 + y2 = 1 does not represent y as a function of x we needed to know analytically that y was not a function of x so that we could use the calculator properly. There are more advanced graphing utilities out there which can do implicit function plots, but you need to know even more Algebra to make them work properly. Do you get the point we’re trying to make here? We believe it is in your best interest to learn the analytic way of doing things so that you are always smarter than your calculator. 1.3 Introduction to Functions 49 1.3.1 Exercises In Exercises 1 - 12, determine whether or not the relation represents y as a function of x. Find the domain and range of those relations which are functions. 1. {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 2. {(−3, 0), (1, 6), (2, −3), (4, 2), (−5, 6), (4, −9), (6, 2)} 3. {(−3, 0), (−7, 6), (5, 5), (6, 4), (4, 9), (3, 0)} 4. {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12), . . .} 5. {(x, y) \| x is an odd integer, and y is an even integer} 6. {(x, 1) \| x is an irrational number} 7. {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5), . . . } 8. {. . . , (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9), . . . } 9. {(−2, y) \| − 3 < y < 4} 10. {(x, 3) \| − 2 ≤ x < 4} 11. {x, x2 \| x is a real number} 12. {x2, x \| x is a real number} In Exercises 13 - 32, determine whether or not the relation represents y as a function of x. Find the domain and range of those relations which are functions. 13. y 4 3 2 1 14. y 4 3 2 1 −4 −3 −2 −1 1 x −1 −4 −3 −2 −1 1 x −1 50 15. 17. 19. 21. y 5 4 3 2 1 −2 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3−2−1 −1 −2 −3 −4 −5 Relations and Functions 16. 18. 20. 22. y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 y 4 3 2 1 −4 −3 −2 −5 −4 −3 −2 −1 1 2 3 x −1 −2 y 5 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 1.3 Introduction to Functions 51 23. 25. 27. 29. y 5 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 y 4 3 2 1 24. y 5 4 3 2 1 −1 −1 −2 −3 −4 −5 262 −1 1 2 x −2 −1 1 2 x y 4 3 2 1 28. y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x y 2 1 30. y 2 1 −2 −1 1 2 x −3 −2 −1 1 2 3 x −1 −2 −1 −2 52 31. y 2 1 Relations and Functions 32. y 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −1 −2 In Exercises 33 - 47, determine whether or not the equation represents y as a function of x. 33. y = x3 − x 36. x2 − y2 = 1 39. x = y2 + 4 42. y = √ 4 − x2 34. y = 37. y = √ x − 2 x x2 − 9 40. y = x2 + 4 43. x2 − y2 = 4 45. 2x + 3y = 4 46. 2xy = 4 35. x3y = −4 38. x = −6 41. x2 + y2 = 4 44. x3 + y3 = 4 47. x2 = y2 48. Explain why the population P of Sasquatch in a given area is a function of time t. What would be the range of this function? 49. Explain why the relation between your classmates and their email addresses may not be a function. What about phone numbers and Social Security Numbers? The process given in Example 1.3.5 for determining whether an equation of a relation represents y as a function of x breaks down if we cannot solve the equation for y in terms of x. However, that does not prevent us from proving that an equation fails to represent y as a function of x. What we really need is two points with the same x-coordinate and diﬀerent y-coordinates which both satisfy the equation so that the graph of the relation would fail the Vertical Line Test 1.1. Discuss with your classmates how you might ﬁnd such points for the relations given in Exercises 50 - 53. 50. x3 + y3 − 3xy = 0 52. y2 = x3 + 3x2 51. x4 = x2 + y2 53. (x2 + y2)2 = x3 + y3 1.3 Introduction to Functions 53 1.3.2 Answers 1. Function 2. Not a function domain = {−3, −2, −1, 0, 1, 2 ,3} range = {0, 1, 4, 9} 3. Function domain = {−7, −3, 3, 4, 5, 6} range = {0, 4, 5, 6, 9} 4. Function domain = {1, 4, 9, 16, 25, 36, . . .} = {x \| x is a perfect square} range = {2, 4, 6, 8, 10, 12, . . .} = {y \| y is a positive even integer} 5. Not a function 6. Function domain = {x \| x is irrational} range = {1} 7. Function 8. Function domain = {x\|x = 2n for some whole number n} range = {y \| y is any whole number} domain = {x \| x is any integer} range = y \| y = n2 for some integer n 9. Not a function 10. Function 11. Function domain = (−∞, ∞) range = [0, ∞) domain = [−2, 4), range = {3} 12. Not a function 13. Function 14. Not a function domain = {−4, −3, −2, −1, 0, 1} range = {−1, 0, 1, 2, 3, 4} 15. Function domain = (−∞, ∞) range = [1, ∞) 17. Function domain = [2, ∞) range = [0, ∞) 19. Not a function 16. Not a function 18. Function domain = (−∞, ∞) range = (0, 4] 20. Function domain = [−5, −3) ∪ (−3, 3) range = (−2, −1) ∪ [0, 4) 54 Relations and Functions 21. Function domain = [−2, ∞) range = [−3, ∞) 23. Function domain = [−5, 4) range = [−4, 4) 25. Function domain = (−∞, ∞) range = (−∞, 4] 27. Function domain = [−2, ∞) range = (−∞, 3] 29. Function domain = (−∞, 0] ∪ (1, ∞) range = (−∞, 1] ∪ {2} 31. Not a function 22. Not a function 24. Function domain = [0, 3) ∪ (3, 6] range = (−4, −1] ∪ [0, 4] 26. Function domain = (−∞, ∞) range = (−∞, 4] 28. Function domain = (−∞, ∞) range = (−∞, ∞) 30. Function domain = [−3, 3] range = [−2, 2] 32. Function domain = (−∞, ∞) range = {2} 33. Function 34. Function 35. Function 36. Not a function 37. Function 38. Not a function 39. Not a function 40. Function 41. Not a function 42. Function 45. Function 43. Not a function 44. Function 46. Function 47. Not a function 1.4 Function Notation 55 1.4 Function Notation In Deﬁnition 1.6, we described a function as a special kind of relation − one in which each xcoordinate is matched with only one y-coordinate. In this section, we focus more on the process by which the x is matched with the y. If we think of the domain of a function as a set of inputs and the range as a set of outputs, we can think of a function f as a process by which each input x is matched with only one output y. Since the output is completely determined by the input x and the process f , we symbolize the output with function notation: ‘f (x)’, read ‘f of x.’ In other words, f (x) is the output which results by applying the process f to the input x. In this case, the parentheses here do not indicate multiplication, as they do elsewhere in Algebra. This can cause confusion if the context is not clear, so you must read carefully. This relationship is typically visualized using a diagram similar to the one below. f x Domain (Inputs) y = f (x) Range (Outputs) The value of y is completely dependent on the choice of x. For this reason, x is often called the independent variable, or argument of f , whereas y is often called the dependent variable. As we shall see, the process of a function f is usually described using an algebraic formula. For example, suppose a function f takes a real number and performs the following two steps, in sequence 1. multiply by 3 2. add 4 If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to our result from step 1 which yields 15 + 4 = 19. Using function notation, we would write f (5) = 19 to indicate that the result of applying the process f to the input 5 gives the output 19. In general, if we use x for the input, applying step 1 produces 3x. Following with step 2 produces 3x + 4 as our ﬁnal output. Hence for an input x, we get the output f (x) = 3x + 4. Notice that to check our formula for the case x = 5, we replace the occurrence of x in the formula for f (x) with 5 to get f (5) = 3(5) + 4 = 15 + 4 = 19, as required. 56 Relations and Functions Example 1.4.1. Suppose a function g is described by applying the following steps, in sequence 1. add 4 2. multiply by 3 Determine g(5) and ﬁnd an expression for g(x). Solution. Starting with 5, step 1 gives 5 + 4 = 9. Continuing with step 2, we get (3)(9) = 27. To ﬁnd a formula for g(x), we start with our input x. Step 1 produces x + 4. We now wish to multiply this entire quantity by 3, so we use a parentheses: 3(x + 4) = 3x + 12. Hence, g(x) = 3x +
12. We can check our formula by replacing x with 5 to get g(5) = 3(5) + 12 = 15 + 12 = 27 . Most of the functions we will encounter in College Algebra will be described using formulas like the ones we developed for f (x) and g(x) above. Evaluating formulas using this function notation is a key skill for success in this and many other Math courses. Example 1.4.2. Let f (x) = −x2 + 3x + 4 1. Find and simplify the following. (a) f (−1), f (0), f (2) (b) f (2x), 2f (x) (c) f (x + 2), f (x) + 2, f (x) + f (2) 2. Solve f (x) = 4. Solution. 1. (a) To ﬁnd f (−1), we replace every occurrence of x in the expression f (x) with −1 f (−1) = −(−1)2 + 3(−1) + 4 = −(1) + (−3) + 4 = 0 Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. (b) To ﬁnd f (2x), we replace every occurrence of x with the quantity 2x f (2x) = −(2x)2 + 3(2x) + 4 = −(4x2) + (6x) + 4 = −4x2 + 6x + 4 The expression 2f (x) means we multiply the expression f (x) by 2 2f (x) = 2 −x2 + 3x + 4 = −2x2 + 6x + 8 1.4 Function Notation 57 (c) To ﬁnd f (x + 2), we replace every occurrence of x with the quantity x + 2 f (x + 2) = −(x + 2)2 + 3(x + 2) + 4 = − x2 + 4x + 4 + (3x + 6) + 4 = −x2 − 4x − 4 + 3x + 6 + 4 = −x2 − x + 6 To ﬁnd f (x) + 2, we add 2 to the expression for f (x) f (x) + 2 = −x2 + 3x + 4 + 2 = −x2 + 3x + 6 From our work above, we see f (2) = 6 so that f (x) + f (2) = −x2 + 3x + 4 + 6 = −x2 + 3x + 10 2. Since f (x) = −x2 + 3x + 4, the equation f (x) = 4 is equivalent to −x2 + 3x + 4 = 4. Solving we get −x2 + 3x = 0, or x(−x + 3) = 0. We get x = 0 or x = 3, and we can verify these answers by checking that f (0) = 4 and f (3) = 4. A few notes about Example 1.4.2 are in order. First note the diﬀerence between the answers for f (2x) and 2f (x). For f (2x), we are multiplying the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely diﬀerent results. Along these lines, note that f (x + 2), f (x) + 2 and f (x) + f (2) are three diﬀerent expressions as well. Even though function notation uses parentheses, as does multiplication, there is no general ‘distributive property’ of function notation. Finally, note the practice of using parentheses when substituting one algebraic expression into another; we highly recommend this practice as it will reduce careless errors. Suppose now we wish to ﬁnd r(3) for r(x) = 2x x2−9 . Substitution gives r(3) = 2(3) (3)2 − 9 = 6 0 , which is undeﬁned. (Why is this, again?) The number 3 is not an allowable input to the function r; in other words, 3 is not in the domain of r. Which other real numbers are forbidden in this formula? We think back to arithmetic. The reason r(3) is undeﬁned is because substitution results in a division by 0. To determine which other numbers result in such a transgression, we set the denominator equal to 0 and solve x2 − 9 = 0 x2 = 9 √ x2 = x = ±3 √ 9 extract square roots 58 Relations and Functions As long as we substitute numbers other than 3 and −3, the expression r(x) is a real number. Hence, we write our domain in interval notation1 as (−∞, −3) ∪ (−3, 3) ∪ (3, ∞). When a formula for a function is given, we assume that the function is valid for all real numbers which make arithmetic sense when substituted into the formula. This set of numbers is often called the implied domain2 of the function. At this stage, there are only two mathematical sins we need to avoid: division by 0 and extracting even roots of negative numbers. The following example illustrates these concepts. Example 1.4.3. Find the domain3 of the following functions. 1. g(x) = √ 4 − 3x 3. f (x) = 2 4x x − 3 1 − 5. r(t) = 4 √ t + 3 6 − Solution. √ 2. h(x) = 5 4 − 3x 4. F (x) = √ 4 2x + 1 x2 − 1 6. I(x) = 3x2 x 1. The potential disaster for g is if the radicand4 is negative. To avoid this, we set 4 − 3x ≥ 0. 3 , the expression From this, we get 3x ≤ 4 or x ≤ 4 4 − 3x ≥ 0, and the formula g(x) returns a real number. Our domain is −∞, 4 3 3 . What this shows is that as long as x ≤ 4 . 2. The formula for h(x) is hauntingly close to that of g(x) with one key diﬀerence − whereas the expression for g(x) includes an even indexed root (namely a square root), the formula for h(x) involves an odd indexed root (the ﬁfth root). Since odd roots of real numbers (even negative real numbers) are real numbers, there is no restriction on the inputs to h. Hence, the domain is (−∞, ∞). 3. In the expression for f , there are two denominators. We need to make sure neither of them is 0. To that end, we set each denominator equal to 0 and solve. For the ‘small’ denominator, we get x − 3 = 0 or x = 3. For the ‘large’ denominator 1See the Exercises for Section 1.1. 2or, ‘implicit domain’ 3The word ‘implied’ is, well, implied. 4The ‘radicand’ is the expression ‘inside’ the radical. 1.4 Function Notation 59 1 − 4x x − 3 = 0 1 = (1)(x − 3) = 4x x − 3 4x x − 3 x − 3 = 4x −3 = 3x −1 = x (x − 3) clear denominators So we get two real numbers which make denominators 0, namely x = −1 and x = 3. Our domain is all real numbers except −1 and 3: (−∞, −1) ∪ (−1, 3) ∪ (3, ∞). 4. In ﬁnding the domain of F , we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a fourth (even-indexed) root. Our strategy is to determine the restrictions imposed by each part and select the real numbers which satisfy both conditions. To satisfy the fourth root, we require 2x + 1 ≥ 0. From this we get 2x ≥ −1 or x ≥ − 1 2 . Next, we round up the values of x which could cause trouble in the denominator by setting the denominator equal to 0. We get x2 − 1 = 0, or x = ±1. Hence, in order for a real number x to be in the domain of F , x ≥ − 1 2 but x = ±1. In interval notation, this set is − 1 2 , 1 ∪ (1, ∞). 5. Don’t be put oﬀ by the ‘t’ here. It is an independent variable representing a real number, just like x does, and is subject to the same restrictions. As in the previous problem, we have double danger here: we have a square root and a denominator. To satisfy the square root, we need a non-negative radicand so we set t + 3 ≥ 0 to get t ≥ −3. Setting the denominator equal to zero gives 6 − t + 3 = 6. Squaring both sides gives t + 3 = 36, or t = 33. Since we squared both sides in the course of solving this equation, we need to check our answer.5 Sure enough, when t = 33, 6 − 36 = 0, so t = 33 will cause problems in the denominator. At last we can ﬁnd the domain of r: we need t ≥ −3, but t = 33. Our ﬁnal answer is [−3, 33) ∪ (33, ∞). t + 3 = 0, or . It’s tempting to simplify I(x) = 3x2 x = 3x, and, since there are no longer any denominators, claim that there are no longer any restrictions. However, in simplifying I(x), we are assuming x = 0, since 0 0 is undeﬁned.6 Proceeding as before, we ﬁnd the domain of I to be all real numbers except 0: (−∞, 0) ∪ (0, ∞). It is worth reiterating the importance of ﬁnding the domain of a function before simplifying, as evidenced by the function I in the previous example. Even though the formula I(x) simpliﬁes to 5Do you remember why? Consider squaring both sides to ‘solve’ 6More precisely, the fraction 0 0 is an ‘indeterminant form’. Calculus is required tame such beasts. √ t + 1 = −2. 60 Relations and Functions 3x, it would be inaccurate to write I(x) = 3x without adding the stipulation that x = 0. It would be analogous to not reporting taxable income or some other sin of omission. 1.4.1 Modeling with Functions The importance of Mathematics to our society lies in its value to approximate, or model real-world phenomenon. Whether it be used to predict the high temperature on a given day, determine the hours of daylight on a given day, or predict population trends of various and sundry real and mythical beasts,7 Mathematics is second only to literacy in the importance humanity’s development.8 It is important to keep in mind that anytime Mathematics is used to approximate reality, there are always limitations to the model. For example, suppose grapes are on sale at the local market for $1.50 per pound. Then one pound of grapes costs $1.50, two pounds of grapes cost $3.00, and so forth. Suppose we want to develop a formula which relates the cost of buying grapes to the amount of grapes being purchased. Since these two quantities vary from situation to situation, we assign them variables. Let c denote the cost of the grapes and let g denote the amount of grapes purchased. To ﬁnd the cost c of the grapes, we multiply the amount of grapes g by the price $1.50 dollars per pound to get c = 1.5g In order for the units to be correct in the formula, g must be measured in pounds of grapes in which case the computed value of c is measured in dollars. Since we’re interested in ﬁnding the cost c given an amount g, we think of g as the independent variable and c as the dependent variable. Using the language of function notation, we write c(g) = 1.5g where g is the amount of grapes purchased (in pounds) and c(g) is the cost (in dollars). For example, c(5) represents the cost, in dollars, to purchase 5 pounds of grapes. In this case, c(5) = 1.5(5) = 7.5, so it would cost $7.50. If, on the other hand, we wanted to ﬁnd the amount of grapes we can purchase for $5, we would need to set c(g) = 5 and solve for g. In this case, c(g) = 1.5g, so solving c(g) = 5 is equivalent to solving 1.5g = 5 Doing so gives g = 5 1.5 = 3.3. This means we can purchase exactly 3.3 pounds of grapes for $5. Of course, you would be hard-pressed to buy exactly 3.3 pounds of grapes,9 and this leads us to our next topic of discussion, the applied domain10 of a function. Even though, mathematically, c(g) = 1.5g has no domain restrictions (there are no denominators and no even-indexed radicals), there are certain values of g that don’t make any physical sense. For example, g = −1 corresponds to ‘purchasing’ −1 pounds of grapes.11 Also, unless the ‘local market’ mentioned is the State of California (or some other exporter of grapes), it also doesn’t make much sense for g = 500,000,000, either. So the real
ity of the situation limits what g can be, and 7See Sections 2.5, 11.1, and 6.5, respectively. 8In Carl’s humble opinion, of course . . . 9You could get close... within a certain speciﬁed margin of error, perhaps. 10or, ‘explicit domain’ 11Maybe this means returning a pound of grapes? 1.4 Function Notation 61 these limits determine the applied domain of g. Typically, an applied domain is stated explicitly. In this case, it would be common to see something like c(g) = 1.5g, 0 ≤ g ≤ 100, meaning the number of pounds of grapes purchased is limited from 0 up to 100. The upper bound here, 100 may represent the inventory of the market, or some other limit as set by local policy or law. Even with this restriction, our model has its limitations. As we saw above, it is virtually impossible to buy exactly 3.3 pounds of grapes so that our cost is exactly $5. In this case, being sensible shoppers, we would most likely ‘round down’ and purchase 3 pounds of grapes or however close the market scale can read to 3.3 without being over. It is time for a more sophisticated example. Example 1.4.4. The height h in feet of a model rocket above the ground t seconds after lift-oﬀ is given by h(t) = −5t2 + 100t, 0, if 0 ≤ t ≤ 20 t > 20 if 1. Find and interpret h(10) and h(60). 2. Solve h(t) = 375 and interpret your answers. Solution. 1. We ﬁrst note that the independent variable here is t, chosen because it represents time. Secondly, the function is broken up into two rules: one formula for values of t between 0 and 20 inclusive, and another for values of t greater than 20. Since t = 10 satisﬁes the inequality 0 ≤ t ≤ 20, we use the ﬁrst formula listed, h(t) = −5t2 + 100t, to ﬁnd h(10). We get h(10) = −5(10)2 + 100(10) = 500. Since t represents the number of seconds since lift-oﬀ and h(t) is the height above the ground in feet, the equation h(10) = 500 means that 10 seconds after lift-oﬀ, the model rocket is 500 feet above the ground. To ﬁnd h(60), we note that t = 60 satisﬁes t > 20, so we use the rule h(t) = 0. This function returns a value of 0 regardless of what value is substituted in for t, so h(60) = 0. This means that 60 seconds after lift-oﬀ, the rocket is 0 feet above the ground; in other words, a minute after lift-oﬀ, the rocket has already returned to Earth. 2. Since the function h is deﬁned in pieces, we need to solve h(t) = 375 in pieces. For 0 ≤ t ≤ 20, h(t) = −5t2 + 100t, so for these values of t, we solve −5t2 + 100t = 375. Rearranging terms, we get 5t2 − 100t + 375 = 0, and factoring gives 5(t − 5)(t − 15) = 0. Our answers are t = 5 and t = 15, and since both of these values of t lie between 0 and 20, we keep both solutions. For t > 20, h(t) = 0, and in this case, there are no solutions to 0 = 375. In terms of the model rocket, solving h(t) = 375 corresponds to ﬁnding when, if ever, the rocket reaches 375 feet above the ground. Our two answers, t = 5 and t = 15 correspond to the rocket reaching this altitude twice – once 5 seconds after launch, and again 15 seconds after launch.12 12What goes up . . . 62 Relations and Functions The type of function in the previous example is called a piecewise-deﬁned function, or ‘piecewise’ function for short. Many real-world phenomena, income tax formulas13 for example, are modeled by such functions. By the way, if we wanted to avoid using a piecewise function in Example 1.4.4, we could have used h(t) = −5t2 + 100t on the explicit domain 0 ≤ t ≤ 20 because after 20 seconds, the rocket is on the ground and stops moving. In many cases, though, piecewise functions are your only choice, so it’s best to understand them well. Mathematical modeling is not a one-section topic. It’s not even a one-course topic as is evidenced by undergraduate and graduate courses in mathematical modeling being oﬀered at many universities. Thus our goal in this section cannot possibly be to tell you the whole story. What we can do is get you started. As we study new classes of functions, we will see what phenomena they can be used to model. In that respect, mathematical modeling cannot be a topic in a book, but rather, must be a theme of the book. For now, we have you explore some very basic models in the Exercises because you need to crawl to walk to run. As we learn more about functions, we’ll help you build your own models and get you on your way to applying Mathematics to your world. 13See the Internal Revenue Service’s website 1.4 Function Notation 1.4.2 Exercises 63 In Exercises 1 - 10, ﬁnd an expression for f (x) and state its domain. 1. f is a function that takes a real number x and performs the following three steps in the order given: (1) multiply by 2; (2) add 3; (3) divide by 4. 2. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) multiply by 2; (3) divide by 4. 3. f is a function that takes a real number x and performs the following three steps in the order given: (1) divide by 4; (2) add 3; (3) multiply by 2. 4. f is a function that takes a real number x and performs the following three steps in the order given: (1) multiply by 2; (2) add 3; (3) take the square root. 5. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) multiply by 2; (3) take the square root. 6. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) take the square root; (3) multiply by 2. 7. f is a function that takes a real number x and performs the following three steps in the order given: (1) take the square root; (2) subtract 13; (3) make the quantity the denominator of a fraction with numerator 4. 8. f is a function that takes a real number x and performs the following three steps in the order given: (1) subtract 13; (2) take the square root; (3) make the quantity the denominator of a fraction with numerator 4. 9. f is a function that takes a real number x and performs the following three steps in the order given: (1) take the square root; (2) make the quantity the denominator of a fraction with numerator 4; (3) subtract 13. 10. f is a function that takes a real number x and performs the following three steps in the order given: (1) make the quantity the denominator of a fraction with numerator 4; (2) take the square root; (3) subtract 13. In Exercises 11 - 18, use the given function f to ﬁnd and simplify the following: f (3) f (4x) f (x − 4) f (−1) 4f (x) f (x) − 4 f 3 2 f (−x) f x2 64 Relations and Functions 11. f (x) = 2x + 1 13. f (x) = 2 − x2 15. f (x) = x x − 1 17. f (x) = 6 12. f (x) = 3 − 4x 14. f (x) = x2 − 3x + 2 16. f (x) = 2 x3 18. f (x) = 0 In Exercises 19 - 26, use the given function f to ﬁnd and simplify the following: f (2) 2f (a) f 2 a 19. f (x) = 2x − 5 21. f (x) = 2x2 − 1 2x + 1 23. f (x) = 25. f (x) = √ x 2 f (−2) f (a + 2) f (a) 2 f (2a) f (a) + f (2) f (a + h) 20. f (x) = 5 − 2x 22. f (x) = 3x2 + 3x − 2 24. f (x) = 117 26. f (x) = 2 x In Exercises 27 - 34, use the given function f to ﬁnd f (0) and solve f (x) = 0 27. f (x) = 2x − 1 29. f (x) = 2x2 − 6 31. f (x) = √ x + 4 28. f (x) = 3 − 2 5 x 30. f (x) = x2 − x − 12 √ 32. f (x) = 1 − 2x 33. f (x) = 3 4 − x   35. Let f (x) =  (a) f (−4) (d) f (3.001) 34. f (x) = 3x2 − 12x 4 − x2 √ x + 5 9 − x2 −x + 5 x ≤ −3 if if −3 < x ≤ 3 x > 3 if Compute the following function values. (b) f (−3) (e) f (−3.001) (c) f (3) (f) f (2) 1.4 Function Notation 65 36. Let f (x) =    √ x2 1 − x2 x x ≤ −1 if if −1 < x ≤ 1 x > 1 if Compute the following function values. (a) f (4) (d) f (0) (b) f (−3) (e) f (−1) (c) f (1) (f) f (−0.999) In Exercises 37 - 62, ﬁnd the (implied) domain of the function. 37. f (x) = x4 − 13x3 + 56x2 − 19 38. f (x) = x2 + 4 39. f (x) = 41. f (x) = 43. f (x) = 45. f (x) = x − 2 x + 1 2x x2 + 3 x + 4 x2 − 36 √ 3 − x 47. f (x) = 9x √ x + 3 49. f (x) = √ 6x − 2 √ 51. f (x) = 3 6x − 2 53. f (x) = √ 6x − 2 x2 − 36 55. s(t) = t t − 8 57. b(θ) = 59. α(y 61. T (t) = √ t − 8 5 − t 40. f (x) = 3x x2 + x − 2 42. f (x) = 44. f (x) = 46. f (x) = 48. f (x) = 2x x2 − 3 x − 2 x − 2 √ 2x + 5 √ 7 − x x2 + 1 50. f (x) = √ 6 6x − 2 6 √ 6x − 2 52. f (x) = 54. f (x) = 56. Q(r) = 58. A(x) = 60. g(v) = 62. u(w) = 4 − √ 3 6x − 2 x2 + 36 √ r r − 8 √ 1 4 − 1 v2 66 Relations and Functions 63. The area A enclosed by a square, in square inches, is a function of the length of one of its sides x, when measured in inches. This relation is expressed by the formula A(x) = x2 for x > 0. Find A(3) and solve A(x) = 36. Interpret your answers to each. Why is x restricted to x > 0? 64. The area A enclosed by a circle, in square meters, is a function of its radius r, when measured in meters. This relation is expressed by the formula A(r) = πr2 for r > 0. Find A(2) and solve A(r) = 16π. Interpret your answers to each. Why is r restricted to r > 0? 65. The volume V enclosed by a cube, in cubic centimeters, is a function of the length of one of its sides x, when measured in centimeters. This relation is expressed by the formula V (x) = x3 for x > 0. Find V (5) and solve V (x) = 27. Interpret your answers to each. Why is x restricted to x > 0? 66. The volume V enclosed by a sphere, in cubic feet, is a function of the radius of the sphere r, 3 r3 for r > 0. Find when measured in feet. This relation is expressed by the formula V (r) = 4π V (3) and solve V (r) = 32π 3 . Interpret your answers to each. Why is r restricted to r > 0? 67. The height of an object dropped from the roof of an eight story building is modeled by: h(t) = −16t2 + 64, 0 ≤ t ≤ 2. Here, h is the height of the object oﬀ the ground, in feet, t seconds after the object is dropped. Find h(0) and solve h(t) = 0. Interpret your answers to each. Why is t restricted to 0 ≤ t ≤ 2? 68. The temperature T in degrees Fahrenheit t hours after 6 AM is given by T (t) = − 1 2 t2 + 8t + 3 for 0 ≤ t ≤ 12. Find and interpret T (0), T (6) and T (12). 69. The function C(x) = x2 − 10x + 27 models the cost, in hundreds of dollars, to produce x thousa
nd pens. Find and interpret C(0), C(2) and C(5). 70. Using data from the Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Use your calculator to ﬁnd F (0), F (14) and F (28). Round your answers to two decimal places and interpret your answers to each. 71. The population of Sasquatch in Portage County can be modeled by the function P (t) = 150t t+15 , where t represents the number of years since 1803. Find and interpret P (0) and P (205). Discuss with your classmates what the applied domain and range of P should be. 72. For n copies of the book Me and my Sasquatch, a print on-demand company charges C(n) dollars, where C(n) is determined by the formula C(n) =    15n 13.50n 12n if if if 1 ≤ n ≤ 25 25 < n ≤ 50 n > 50 (a) Find and interpret C(20). 1.4 Function Notation 67 (b) How much does it cost to order 50 copies of the book? What about 51 copies? (c) Your answer to 72b should get you thinking. Suppose a bookstore estimates it will sell 50 copies of the book. How many books can, in fact, be ordered for the same price as those 50 copies? (Round your answer to a whole number of books.) 73. An on-line comic book retailer charges shipping costs according to the following formula S(n) = 1.5n + 2.5 0 if if 1 ≤ n ≤ 14 n ≥ 15 where n is the number of comic books purchased and S(n) is the shipping cost in dollars. (a) What is the cost to ship 10 comic books? (b) What is the signiﬁcance of the formula S(n) = 0 for n ≥ 15? 74. The cost C (in dollars) to talk m minutes a month on a mobile phone plan is modeled by C(m) = 25 25 + 0.1(m − 1000) 0 ≤ m ≤ 1000 if if m > 1000 (a) How much does it cost to talk 750 minutes per month with this plan? (b) How much does it cost to talk 20 hours a month with this plan? (c) Explain the terms of the plan verbally. 75. In Section 1.1.1 we deﬁned the set of integers as Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.14 The greatest integer of x, denoted by x, is deﬁned to be the largest integer k with k ≤ x. (a) Find 0.785, 117, −2.001, and π + 6 (b) Discuss with your classmates how x may be described as a piecewise deﬁned function. HINT: There are inﬁnitely many pieces! (c) Is a + b = a + b always true? What if a or b is an integer? Test some values, make a conjecture, and explain your result. 76. We have through our examples tried to convince you that, in general, f (a + b) = f (a) + f (b). It has been our experience that students refuse to believe us so we’ll try again with a diﬀerent approach. With the help of your classmates, ﬁnd a function f for which the following properties are always true. (a) f (0) = f (−1 + 1) = f (−1) + f (1) 14The use of the letter Z for the integers is ostensibly because the German word zahlen means ‘to count.’ 68 Relations and Functions (b) f (5) = f (2 + 3) = f (2) + f (3) (c) f (−6) = f (0 − 6) = f (0) − f (6) (d) f (a + b) = f (a) + f (b) regardless of what two numbers we give you for a and b. How many functions did you ﬁnd that failed to satisfy the conditions above? Did f (x) = x2 work? What about f (x) = ? Did you ﬁnd an attribute x or f (x) = 3x + 7 or f (x) = common to those functions that did succeed? You should have, because there is only one extremely special family of functions that actually works here. Thus we return to our previous statement, in general, f (a + b) = f (a) + f (b). 1 x √ 1.4 Function Notation 1.4.3 Answers 1. f (x) = 2x+3 4 Domain: (−∞, ∞) 69 2. f (x) = 2(x+3) 4 = x+3 2 Domain: (−∞, ∞) 3. f (x Domain: (−∞, ∞) √ 4. f (x) = 2x + 3 Domain: − 3 2 , ∞ 5. f (x) = 2(x + 3) = √ 2x + 6 Domain: [−3, ∞) 7. f (x) = 4√ x−13 Domain: [0, 169) ∪ (169, ∞) 6. f (x) = 2 √ x + 3 Domain: [−3, ∞) 8. f (x) = 4√ x−13 Domain: (13, ∞) 9. f (x) = 4√ x − 13 Domain: (0, ∞) 11. For f (x) = 2x + 1 10. f (x) = 4 x − 13 = 2√ x − 13 Domain: (0, ∞) f (3) = 7 f (−1) = −1 f 3 2 = 4 f (4x) = 8x + 1 4f (x) = 8x + 4 f (−x) = −2x + 1 f (x − 4) = 2x − 7 f (x) − 4 = 2x − 3 f x2 = 2x2 + 1 12. For f (x) = 3 − 4x f (3) = −9 f (−1) = 7 f 3 2 = −3 f (4x) = 3 − 16x 4f (x) = 12 − 16x f (−x) = 4x + 3 f (x − 4) = 19 − 4x f (x) − 4 = −4x − 1 f x2 = 3 − 4x2 70 Relations and Functions 13. For f (x) = 2 − x2 f (3) = −7 f (−14x) = 2 − 16x2 4f (x) = 8 − 4x2 f (−x) = 2 − x2 f (x − 4) = −x2 + 8x − 14 f (x) − 4 = −x2 − 2 f x2 = 2 − x4 14. For f (x) = x2 − 3x + 2 f (3) = 2 f (−14x) = 16x2 − 12x + 2 4f (x) = 4x2 − 12x + 8 f (−x) = x2 + 3x + 2 f (x − 4) = x2 − 11x + 30 f (x) − 4 = x2 − 3x − 2 f x2 = x4 − 3x2 + 2 15. For f (x) = x x−1 f (3) = 3 2 f (4x) = 4x 4x−1 f (x − 4) = x−4 x−5 16. For f (x) = 2 x3 f (3) = 2 27 f (4x) = 1 32x3 f (x − 4) = 2 (x−4)3 = 2 x3−12x2+48x−64 17. For f (x) = 6 f (3) = 6 f (4x) = 6 f (−1) = 1 2 4f (x) = 4x x−1 f (x) − 4 = x x−1 − 4 = 4−3x x−1 f (−1) = −2 4f (x) = 8 x3 f (x) − 4 = 2 x3 − 4 = 2−4x3 x3 f (−1) = 6 4f (x) = 24 f (x − 4) = 6 f (x (−x) = x x+1 f x2 = x2 x2−1 f 3 2 = 16 27 f (−x) = − 2 x3 f x2 = 2 x6 f 3 2 = 6 f (−x) = 6 f x2 = 6 1.4 Function Notation 71 18. For f (x) = 0 f (3) = 0 f (−1) = 0 f (4x) = 0 4f (x) = 0 f (x − 4) = 0 f (x) − 4 = −4 f 3 2 = 0 f (−x) = 0 f x2 = 0 19. For f (x) = 2x − 5 f (2) = −1 f (−2) = −9 f (2a) = 4a − 5 2f (a) = 4a − 10 f (a + 2) = 2a − 1 f (a) + f (2) = 2a − −5a a 20. For f (x) = 5 − 2x f (a) 2 = 2a−5 2 f (a + h) = 2a + 2h − 5 f (2) = 1 f (−2) = 9 f (2a) = 5 − 4a 2f (a) = 10 − 4a f (a + 2) = 1 − 2a f (a) + f (2) = 6 − 2a f 2 a = 5 − 4 a = 5a−4 a 21. For f (x) = 2x2 − 1 f (a) 2 = 5−2a 2 f (a + h) = 5 − 2a − 2h f (2) = 7 f (−2) = 7 f (2a) = 8a2 − 1 2f (a) = 4a2 − 2 f (a + 2) = 2a2 + 8a + 7 f (a) + f (2) = 2a2 + 6 f 2 a = 8 a2 − 1 = 8−a2 a2 f (a) 2 = 2a2−1 2 f (a + h) = 2a2 + 4ah + 2h2 − 1 72 Relations and Functions 22. For f (x) = 3x2 + 3x − 2 f (2) = 16 f (−2) = 4 f (2a) = 12a2 + 6a − 2 2f (a) = 6a2 + 6a − 4 f (a+2) = 3a2 +15a+16 f (a) + f (2) = 3a2 + 3a + 14 f 2 a = 12 a2 + 6 a − 2 = 12+6a−2a2 a2 23. For f (x) = √ 2x + 1 f (a) 2 = 3a2+3a−2 2 f (a + h) = 3a2 + 6ah + 3h2 + 3a + 3h − 2 f (2) = √ 5 f (−2) is not real f (2a) = √ 4a + 1 2f (a) = 2 √ 2a + 1 f (a + 2) = √ 2a + 5 f (a)+f (2) = √ 2a + 1+ √ 5 f (a) 2 = √ 2a+1 2 f (a+h) = √ 2a + 2h + +4 a 24. For f (x) = 117 f (2) = 117 f (−2) = 117 f (2a) = 117 2f (a) = 234 f (a + 2) = 117 f (a) + f (2) = 234 f 2 a = 117 25. For f (x) = x 2 f (2) = 1 2f (aa) 2 = 117 2 f (a + h) = 117 f (−2) = −1 f (2a) = a f (a + 2) = a+2 2 f (a) 2 = a 4 f (a) + f (2) = a 2 + 1 = a+2 2 f (a + h) = a+h 2 1.4 Function Notation 73 26. For f (x) = 2 x f (2) = 1 2f (a (−2) = −1 f (a + 2) = 2 a+2 f (a) 2 = 1 a f (2a) = 1 a f (a) + f (2) = 2 a + 1 = a+2 2 f (a + h) = 2 a+h 27. For f (x) = 2x − 1, f (0) = −1 and f (x) = 0 when x = 1 2 28. For f (x) = 3 − 2 5 x, f (0) = 3 and f (x) = 0 when x = 15 2 29. For f (x) = 2x2 − 6, f (0) = −6 and f (x) = 0 when x = ± √ 3 30. For f (x) = x2 − x − 12, f (0) = −12 and f (x) = 0 when x = −3 or x = 4 31. For f (x) = 32. For f (x) = √ √ x + 4, f (0) = 2 and f (x) = 0 when x = −4 1 − 2x, f (0) = 1 and f (x) = 0 when x = 1 2 33. For f (x) = 3 4−x , f (0) = 3 4 and f (x) is never equal to 0 34. For f (x) = 3x2−12x 4−x2 , f (0) = 0 and f (x) = 0 when x = 0 or x = 4 35. (a) f (−4) = 1 (b) f (−3) = 2 (d) f (3.001) = 1.999 (e) f (−3.001) = 1.999 36. (a) f (4) = 4 (d) f (0) = 1 37. (−∞, ∞) 39. (−∞, −1) ∪ (−1, ∞) 41. (−∞, ∞) (b) f (−3) = 9 (e) f (−1) = 1 (c) f (3) = 0 √ (f) f (2) = 5 (c) f (1) = 0 (f) f (−0.999) ≈ 0.0447 38. (−∞, ∞) 40. (−∞, −2) ∪ (−2, 1) ∪ (1, ∞) 42. (−∞, − √ √ 3) ∪ (− √ 3, 3) ∪ ( √ 3, ∞) 43. (−∞, −6) ∪ (−6, 6) ∪ (6, ∞) 44. (−∞, 2) ∪ (2, ∞) 45. (−∞, 3] 46. − 5 2 , ∞ 74 47. [−3, ∞) 49. 1 3 , ∞ 51. (−∞, ∞) 53. 1 3 , 6 ∪ (6, ∞) 55. (−∞, 8) ∪ (8, ∞) 57. (8, ∞) 59. (−∞, 8) ∪ (8, ∞) 61. [0, 5) ∪ (5, ∞) Relations and Functions 48. (−∞, 7] 50. 1 523, ∞) 54. (−∞, ∞) 56. [0, 8) ∪ (8, ∞) 58. [7, 9] 60. −∞, − , 1 2 ∪ 1 2 , ∞ 62. [0, 25) ∪ (25, ∞) 63. A(3) = 9, so the area enclosed by a square with a side of length 3 inches is 9 square inches. The solutions to A(x) = 36 are x = ±6. Since x is restricted to x > 0, we only keep x = 6. This means for the area enclosed by the square to be 36 square inches, the length of the side needs to be 6 inches. Since x represents a length, x > 0. 64. A(2) = 4π, so the area enclosed by a circle with radius 2 meters is 4π square meters. The solutions to A(r) = 16π are r = ±4. Since r is restricted to r > 0, we only keep r = 4. This means for the area enclosed by the circle to be 16π square meters, the radius needs to be 4 meters. Since r represents a radius (length), r > 0. 65. V (5) = 125, so the volume enclosed by a cube with a side of length 5 centimeters is 125 cubic centimeters. The solution to V (x) = 27 is x = 3. This means for the volume enclosed by the cube to be 27 cubic centimeters, the length of the side needs to 3 centimeters. Since x represents a length, x > 0. 66. V (3) = 36π, so the volume enclosed by a sphere with radius 3 feet is 36π cubic feet. The is r = 2. This means for the volume enclosed by the sphere to be 32π 3 solution to V (r) = 32π 3 cubic feet, the radius needs to 2 feet. Since r represents a radius (length), r > 0. 67. h(0) = 64, so at the moment the object is dropped oﬀ the building, the object is 64 feet oﬀ of the ground. The solutions to h(t) = 0 are t = ±2. Since we restrict 0 ≤ t ≤ 2, we only keep t = 2. This means 2 seconds after the object is dropped oﬀ the building, it is 0 feet oﬀ the ground. Said diﬀerently, the object hits the ground after 2 seconds. The restriction 0 ≤ t ≤ 2 restricts the time to be between the moment the object is released and the moment it hits the ground. 68. T (0) = 3, so at 6 AM (0 hours after 6 AM), it is 3◦ Fahrenheit. T (6) = 33, so at noon (6 hours after 6 AM), the temperature is 33◦ Fahrenheit. T (12) = 27, so at 6 PM (12 hours after 6 AM), it is 27◦ Fahrenheit. 1.4 Function Notation 75 69. C(0) = 27, so to make 0 pens, it costs15 $2700. C(2) = 11, so to make 2000 pens, it costs $1100. C(5) = 2, so to make 5000
pens, it costs $2000. 70. F (0) = 16.00, so in 1980 (0 years after 1980), the average fuel economy of passenger cars in the US was 16.00 miles per gallon. F (14) = 20.81, so in 1994 (14 years after 1980), the average fuel economy of passenger cars in the US was 20.81 miles per gallon. F (28) = 22.64, so in 2008 (28 years after 1980), the average fuel economy of passenger cars in the US was 22.64 miles per gallon. 71. P (0) = 0 which means in 1803 (0 years after 1803), there are no Sasquatch in Portage County. 22 ≈ 139.77, so in 2008 (205 years after 1803), there were between 139 and 140 P (205) = 3075 Sasquatch in Portage County. 72. (a) C(20) = 300. It costs $300 for 20 copies of the book. (b) C(50) = 675, so it costs $675 for 50 copies of the book. C(51) = 612, so it costs $612 for 51 copies of the book. (c) 56 books. 73. (a) S(10) = 17.5, so it costs $17.50 to ship 10 comic books. (b) There is free shipping on orders of 15 or more comic books. 74. (a) C(750) = 25, so it costs $25 to talk 750 minutes per month with this plan. (b) Since 20 hours = 1200 minutes, we substitute m = 1200 and get C(1200) = 45. It costs $45 to talk 20 hours per month with this plan. (c) It costs $25 for up to 1000 minutes and 10 cents per minute for each minute over 1000 minutes. 75. (a) 0.785 = 0, 117 = 117, −2.001 = −3, and π + 6 = 9 15This is called the ‘ﬁxed’ or ‘start-up’ cost. We’ll revisit this concept on page 82. 76 Relations and Functions 1.5 Function Arithmetic In the previous section we used the newly deﬁned function notation to make sense of expressions such as ‘f (x) + 2’ and ‘2f (x)’ for a given function f . It would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The following deﬁnitions allow us to add, subtract, multiply and divide functions using the arithmetic we already know for real numbers. Function Arithmetic Suppose f and g are functions and x is in both the domain of f and the domain of g.a The sum of f and g, denoted f + g, is the function deﬁned by the formula (f + g)(x) = f (x) + g(x) The diﬀerence of f and g, denoted f − g, is the function deﬁned by the formula (f − g)(x) = f (x) − g(x) The product of f and g, denoted f g, is the function deﬁned by the formula (f g)(x) = f (x)g(x) The quotient of f and g, denoted , is the function deﬁned by the formula f g f g (x) = f (x) g(x) , provided g(x) = 0. aThus x is an element of the intersection of the two domains. In other words, to add two functions, we add their outputs; to subtract two functions, we subtract their outputs, and so on. Note that while the formula (f + g)(x) = f (x) + g(x) looks suspiciously like some kind of distributive property, it is nothing of the sort; the addition on the left hand side of the equation is function addition, and we are using this equation to deﬁne the output of the new function f + g as the sum of the real number outputs from f and g. Example 1.5.1. Let f (x) = 6x2 − 2x and g(x) = 3 − 1 x . 1. Find (f + g)(−1) 2. Find (f g)(2) 3. Find the domain of g − f then ﬁnd and simplify a formula for (g − f )(x). 1.5 Function Arithmetic 77 4. Find the domain of g f then ﬁnd and simplify a formula for g f (x). Solution. 1. To ﬁnd (f + g)(−1) we ﬁrst ﬁnd f (−1) = 8 and g(−1) = 4. By deﬁnition, we have that (f + g)(−1) = f (−1) + g(−1) = 8 + 4 = 12. 2. To ﬁnd (f g)(2), we ﬁrst need f (2) and g(2). Since f (2) = 20 and g(2) = 5 2 , our formula yields (f g)(2) = f (2)g(2) = (20) 5 2 = 50. 3. One method to ﬁnd the domain of g−f is to ﬁnd the domain of g and of f separately, then ﬁnd the intersection of these two sets. Owing to the denominator in the expression g(x) = 3 − 1 x , we get that the domain of g is (−∞, 0) ∪ (0, ∞). Since f (x) = 6x2 − 2x is valid for all real numbers, we have no further restrictions. Thus the domain of g − f matches the domain of g, namely, (−∞, 0) ∪ (0, ∞). A second method is to analyze the formula for (g − f )(x) before simplifying and look for the usual domain issues. In this case, (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x , so we ﬁnd, as before, the domain is (−∞, 0) ∪ (0, ∞). Moving along, we need to simplify a formula for (g − f )(x). In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x = 3 − 1 x − 6x2 + 2x get common denominators = = = + − − 1 x 6x3 x 2x2 3x x x 3x − 1 − 6x3 − 2x2 x −6x3 − 2x2 + 3x − 1 x 4. As in the previous example, we have two ways to approach ﬁnding the domain of g f . First, we can ﬁnd the domain of g and f separately, and ﬁnd the intersection of these two sets. In f (x) , we are introducing a new denominator, namely f (x), so we addition, since need to guard against this being 0 as well. Our previous work tells us that the domain of g is (−∞, 0) ∪ (0, ∞) and the domain of f is (−∞, ∞). Setting f (x) = 0 gives 6x2 − 2x = 0 (x) = g(x) g f 78 Relations and Functions 3 . As a result, the domain of g f is all real numbers except x = 0 and x = 1 3 , or or x = 0, 1 (−∞, 0) ∪ 0, 1 3 ∪ 1 3 , ∞. Alternatively, we may proceed as above and analyze the expression simplifying. In this case, g f (x) = g(x) f (x) before g f (x) = g(x) f (x) = 3 − 1 x 6x2 − 2x We see immediately from the ‘little’ denominator that x = 0. To keep the ‘big’ denominator away from 0, we solve 6x2 − 2x = 0 and get x = 0 or x = 1 3 . Hence, as before, we ﬁnd the domain of g f to be (−∞, 0) ∪ 0, 1 ∪ 1 3 Next, we ﬁnd and simplify a formula for (x). 3 , ∞. g f g f (x(x) f (x) 3 − 1 x 6x2 − 2x 3 − 1 x · x x x 3 − 6x2 − 2x 1 x (6x2 − 2x) x 3x − 1 (6x2 − 2x) x 3x − 1 2x2(3x − 1) 1 (3x − 1) 2x2 (3x − 1) 1 2x2 simplify compound fractions factor cancel Please note the importance of ﬁnding the domain of a function before simplifying its expression. In number 4 in Example 1.5.1 above, had we waited to ﬁnd the domain of g f until after simplifying, we’d just have the formula 1 2x2 to go by, and we would (incorrectly!) state the domain as (−∞, 0)∪(0, ∞), since the other troublesome number, x = 1 3 , was canceled away.1 1We’ll see what this means geometrically in Chapter 4. 1.5 Function Arithmetic 79 Next, we turn our attention to the diﬀerence quotient of a function. Deﬁnition 1.8. Given a function f , the diﬀerence quotient of f is the expression f (x + h) − f (x) h We will revisit this concept in Section 2.1, but for now, we use it as a way to practice function notation and function arithmetic. For reasons which will become clear in Calculus, ‘simplifying’ a diﬀerence quotient means rewriting it in a form where the ‘h’ in the deﬁnition of the diﬀerence quotient cancels from the denominator. Once that happens, we consider our work to be done. Example 1.5.2. Find and simplify the diﬀerence quotients for the following functions 1. f (x) = x2 − x − 2 Solution. 2. g(x) = 3 2x + 1 3. r(x) = √ x 1. To ﬁnd f (x + h), we replace every occurrence of x in the formula f (x) = x2 − x − 2 with the quantity (x + h) to get f (x + h) = (x + h)2 − (x + h) − 2 = x2 + 2xh + h2 − x − h − 2. So the diﬀerence quotient is f (x + h) − f (x) h = = = = = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2 h x2 + 2xh + h2 − x − h − 2 − x2 + x + 2 h 2xh + h2 − h h h (2x + h − 1) h h (2x + h − 1) h factor cancel = 2x + h − 1. 80 Relations and Functions 2. To ﬁnd g(x + h), we replace every occurrence of x in the formula g(x) = 3 2x+1 with the quantity (x + h) to get which yields g(x + h) − g(x) h g(x + h) = = 3 2(x + h) + 1 3 2x + 2h + 1 , 3 2x + 2h + 1 h 3 2x + 2h + 1 h − − 3 2x + 1 3 2x + 1 · (2x + 2h + 1)(2x + 1) (2x + 2h + 1)(2x + 1) 3(2x + 1) − 3(2x + 2h + 1) h(2x + 2h + 1)(2x + 1) 6x + 3 − 6x − 6h − 3 h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6 (2x + 2h + 1)(2x + 1) . = = = = = = = Since we have managed to cancel the original ‘h’ from the denominator, we are done. 3. For r(x) = √ x, we get r(x + h) = √ x + h so the diﬀerence quotient is r(x + h) − r(x In order to cancel the ‘h’ from the denominator, we rationalize the numerator by multiplying by its conjugate.2 2Rationalizing the numerator !? How’s that for a twist! 1.5 Function Arithmetic 81 r(x + h) − r(x Multiply by the conjugate. Diﬀerence of Squares. √ x + h2 )2 x (x + h Since we have removed the original ‘h’ from the denominator, we are done. As mentioned before, we will revisit diﬀerence quotients in Section 2.1 where we will explain them geometrically. For now, we want to move on to some classic applications of function arithmetic from Economics and for that, we need to think like an entrepreneur.3 Suppose you are a manufacturer making a certain product.4 Let x be the production level, that is, the number of items produced in a given time period. It is customary to let C(x) denote the function which calculates the total cost of producing the x items. The quantity C(0), which represents the cost of producing no items, is called the ﬁxed cost, and represents the amount of money required to begin production. Associated with the total cost C(x) is cost per item, or average cost, denoted C(x) and read ‘C-bar’ of x. To compute C(x), we take the total cost C(x) and divide by the number of items produced x to get C(x) = C(x) x On the retail end, we have the price p charged per item. To simplify the dialog and computations in this text, we assume that the number of items sold equals the number of items produced. From a 3Not really, but “entrepreneur” is the buzzword of the day and we’re trying to be trendy. 4Poorly designed resin Sasquatch statues, for example. Feel free to choose your own entrepreneurial fantasy. 82 Relations and Functions retail perspective, it seems natural to think of the number of items sold, x, as a function of the price charged, p. After all, the retailer can easily adjust the price to sell more product. In the language of functions, x would be the dependent variable and p would be the independent variable or, using function notation, we have a function x(p). While we will adopt thi
s convention later in the text,5 we will hold with tradition at this point and consider the price p as a function of the number of items sold, x. That is, we regard x as the independent variable and p as the dependent variable and speak of the price-demand function, p(x). Hence, p(x) returns the price charged per item when x items are produced and sold. Our next function to consider is the revenue function, R(x). The function R(x) computes the amount of money collected as a result of selling x items. Since p(x) is the price charged per item, we have R(x) = xp(x). Finally, the proﬁt function, P (x) calculates how much money is earned after the costs are paid. That is, P (x) = (R − C)(x) = R(x) − C(x). We summarize all of these functions below. Summary of Common Economic Functions Suppose x represents the quantity of items produced and sold. The price-demand function p(x) calculates the price per item. The revenue function R(x) calculates the total money collected by selling x items at a price p(x), R(x) = x p(x). The cost function C(x) calculates the cost to produce x items. The value C(0) is called the ﬁxed cost or start-up cost. The average cost function C(x) = C(x) x Here, we necessarily assume x > 0. calculates the cost per item when making x items. The proﬁt function P (x) calculates the money earned after costs are paid when x items are produced and sold, P (x) = (R − C)(x) = R(x) − C(x). It is high time for an example. Example 1.5.3. Let x represent the number of dOpi media players (‘dOpis’6) produced and sold in a typical week. Suppose the cost, in dollars, to produce x dOpis is given by C(x) = 100x + 2000, for x ≥ 0, and the price, in dollars per dOpi, is given by p(x) = 450 − 15x for 0 ≤ x ≤ 30. 1. Find and interpret C(0). 2. Find and interpret C(10). 3. Find and interpret p(0) and p(20). 4. Solve p(x) = 0 and interpret the result. 5. Find and simplify expressions for the revenue function R(x) and the proﬁt function P (x). 6. Find and interpret R(0) and P (0). 7. Solve P (x) = 0 and interpret the result. 5See Example 5.2.4 in Section 5.2. 6Pronounced ‘dopeys’ . . . 1.5 Function Arithmetic 83 Solution. 1. We substitute x = 0 into the formula for C(x) and get C(0) = 100(0) + 2000 = 2000. This means to produce 0 dOpis, it costs $2000. In other words, the ﬁxed (or start-up) costs are $2000. The reader is encouraged to contemplate what sorts of expenses these might be. 2. Since C(x) = C(x) x , C(10) = C(10) 10 = 3000 10 = 300. This means when 10 dOpis are produced, the cost to manufacture them amounts to $300 per dOpi. 3. Plugging x = 0 into the expression for p(x) gives p(0) = 450 − 15(0) = 450. This means no dOpis are sold if the price is $450 per dOpi. On the other hand, p(20) = 450 − 15(20) = 150 which means to sell 20 dOpis in a typical week, the price should be set at $150 per dOpi. 4. Setting p(x) = 0 gives 450 − 15x = 0. Solving gives x = 30. This means in order to sell 30 dOpis in a typical week, the price needs to be set to $0. What’s more, this means that even if dOpis were given away for free, the retailer would only be able to move 30 of them.7 5. To ﬁnd the revenue, we compute R(x) = xp(x) = x(450 − 15x) = 450x − 15x2. Since the formula for p(x) is valid only for 0 ≤ x ≤ 30, our formula R(x) is also restricted to 0 ≤ x ≤ 30. For the proﬁt, P (x) = (R − C)(x) = R(x) − C(x). Using the given formula for C(x) and the derived formula for R(x), we get P (x) = 450x − 15x2−(100x+2000) = −15x2+350x−2000. As before, the validity of this formula is for 0 ≤ x ≤ 30 only. 6. We ﬁnd R(0) = 0 which means if no dOpis are sold, we have no revenue, which makes sense. Turning to proﬁt, P (0) = −2000 since P (x) = R(x)−C(x) and P (0) = R(0)−C(0) = −2000. This means that if no dOpis are sold, more money ($2000 to be exact!) was put into producing the dOpis than was recouped in sales. In number 1, we found the ﬁxed costs to be $2000, so it makes sense that if we sell no dOpis, we are out those start-up costs. 7. Setting P (x) = 0 gives −15x2 + 350x − 2000 = 0. Factoring gives −5(x − 10)(3x − 40) = 0 so x = 10 or x = 40 3 . What do these values mean in the context of the problem? Since P (x) = R(x) − C(x), solving P (x) = 0 is the same as solving R(x) = C(x). This means that the solutions to P (x) = 0 are the production (and sales) ﬁgures for which the sales revenue exactly balances the total production costs. These are the so-called ‘break even’ points. The solution x = 10 means 10 dOpis should be produced (and sold) during the week to recoup the cost of production. For x = 40 3 = 13.3, things are a bit more complicated. Even though x = 13.3 satisﬁes 0 ≤ x ≤ 30, and hence is in the domain of P , it doesn’t make sense in the context of this problem to produce a fractional part of a dOpi.8 Evaluating P (13) = 15 and P (14) = −40, we see that producing and selling 13 dOpis per week makes a (slight) proﬁt, whereas producing just one more puts us back into the red. While breaking even is nice, we ultimately would like to ﬁnd what production level (and price) will result in the largest proﬁt, and we’ll do just that . . . in Section 2.3. 7Imagine that! Giving something away for free and hardly anyone taking advantage of it . . . 8We’ve seen this sort of thing before in Section 1.4.1. 84 1.5.1 Exercises Relations and Functions In Exercises 1 - 10, use the pair of functions f and g to ﬁnd the following values if they exist. (f + g)(2) (f g) 1 2 (f − g)(−1) f g (0) (g − f )(1) g f (−2) 1. f (x) = 3x + 1 and g(x) = 4 − x 2. f (x) = x2 and g(x) = −2x + 1 3. f (x) = x2 − x and g(x) = 12 − x2 4. f (x) = 2x3 and g(x) = −x2 − 2x − 3 √ 5. f (x) = x + 3 and g(x) = 2x − 1 √ 6. f (x) = 4 − x and g(x) = √ x + 2 7. f (x) = 2x and g(x) = 9. f (x) = x2 and g(x) = 1 2x + 1 1 x2 8. f (x) = x2 and g(x) = 3 2x − 3 10. f (x) = x2 + 1 and g(x) = 1 x2 + 1 In Exercises 11 - 20, use the pair of functions f and g to ﬁnd the domain of the indicated function then ﬁnd and simplify an expression for it. (f + g)(x) (f − g)(x) (f g)(x) f g (x) 11. f (x) = 2x + 1 and g(x) = x − 2 12. f (x) = 1 − 4x and g(x) = 2x − 1 13. f (x) = x2 and g(x) = 3x − 1 14. f (x) = x2 − x and g(x) = 7x 15. f (x) = x2 − 4 and g(x) = 3x + 6 16. f (x) = −x2 + x + 6 and g(x) = x2 − 9 17. f (x) = x 2 and g(x) = 19. f (x) = x and g(x) = 2 x √ x + 1 18. f (x) = x − 1 and g(x) = 1 x − 1 20. f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 In Exercises 21 - 45, ﬁnd and simplify the diﬀerence quotient f (x + h) − f (x) h for the given function. 21. f (x) = 2x − 5 23. f (x) = 6 25. f (x) = −x2 + 2x − 1 22. f (x) = −3x + 5 24. f (x) = 3x2 − x 26. f (x) = 4x2 1.5 Function Arithmetic 85 27. f (x) = x − x2 28. f (x) = x3 + 1 29. f (x) = mx + b where m = 0 30. f (x) = ax2 + bx + c where a = 0 31. f (x) = 33. f (x) = 2 x 1 x2 35. f (x) = 37. f (x) = 1 4x − 3 x x − 9 √ √ √ 39. f (x) = 41. f (x) = 43. f (x) = x − 9 −4x + 5 ax + b, where a = 0. 32. f (x) = 34. f (x) = 36. f (x 3x x + 1 38. f (x) = 40. f (x) = 42. f (x) = x2 2x + 1 √ 2x + 1 √ 4 − x 44. f (x) = x √ x √ 45. f (x) = 3 x. HINT: (a − b) a2 + ab + b2 = a3 − b3 In Exercises 46 - 50, C(x) denotes the cost to produce x items and p(x) denotes the price-demand function in the given economic scenario. In each Exercise, do the following: Find and interpret C(0). Find and interpret C(10). Find and interpret p(5) Find and simplify R(x). Find and simplify P (x). Solve P (x) = 0 and interpret. 46. The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15. 47. The cost, in dollars, to produce x bottles of 100% All-Natural Certiﬁed Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x ≥ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 − x, 0 ≤ x ≤ 35. 48. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is C(x) = 18x + 240, x ≥ 0 and the price-demand function, in cents per cup, is p(x) = 90 − 3x, 0 ≤ x ≤ 30. 49. The daily cost, in dollars, to produce x Sasquatch Berry Pies C(x) = 3x + 36, x ≥ 0 and the price-demand function, in dollars per pie, is p(x) = 12 − 0.5x, 0 ≤ x ≤ 24. 86 Relations and Functions 50. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x ≥ 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 − 2x, 0 ≤ x ≤ 70. In Exercises 51 - 62, let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)} . Compute the indicated value if it exists. 51. (f + g)(−3) 54. (g + f )(1) 57. 60. f g g f (−2) (−1) 52. (f − g)(2) 55. (g − f )(3) 58. 61. f g g f (−1) (3) 53. (f g)(−1) 56. (gf )(−3) 59. 62. f g g f (2) (−3) 1.5 Function Arithmetic 87 1.5.2 Answers 1. For f (x) = 3x + 1 and g(x) = 4 − x (f + g)(2) = 9 (f g) 1 2 = 35 4 (f − g)(−1) = −7 (g − f )(1) = −1 f g (0) = 1 4 g f (−2) = − 6 5 2. For f (x) = x2 and g(x) = −2x + 1 (f + g)(2) = 1 (f − g)(−1) = −2 (g − f )(1) = −2 (f g) 1 2 = 0 f g (0) = 0 g f (−2) = 5 4 3. For f (x) = x2 − x and g(x) = 12 − x2 (f + g)(2) = 10 (f − g)(−1) = −9 (g − f )(1) = 11 (f g) 1 2 = − 47 16 f g (0) = 0 g f (−2) = 4 3 4. For f (x) = 2x3 and g(x) = −x2 − 2x − 3 (f + g)(2) = 5 (f g) 1 2 = − 17 16 (f − g)(−1) = 0 (g − f )(1) = −8 f g (0) = 0 g f (−2) = 3 16 5. For f (x) = √ x + 3 and g(x) = 2x − 1 (f + g)(2) = 3 + √ 5 (f − g)(−1) = 3 + √ 2 (f g) 1 2 = 0 f g √ (0) = − 3 (g − f )(1) = −1 g f (−2) = −5 6. For f (x) = √ 4 − x and g(x) = √ (f + g)(2f − g)(−1) = −1 + √ 5 (f g) 1 2 = √ 35 2 f g √ 2 (0) = (g − f )(1) = 0 g f (−2) = 0 88 Relations and Functions 7. For f (x) = 2x and g(x) = 1 2x+1 (f + g)(2) = 21 5 (f g) 1 2 = 1 2 (f − g)(−1) = −1 f g (0) = 0 (g − f )(1) = − 5 3 g f (−2) = 1 12 8. For f (x) = x2 and g(x) = 3 2x−3 (f + g)(2) = 7 (f g) 1 2 = − 3 8 (f − g)(−1) = 8 5 f g (0) = 0 (g − f )(1) = −4 g f (−2) = − 3 28
9. For f (x) = x2 and g(x) = 1 x2 (f + g)(2) = 17 4 (f g) 1 2 = 1 (f − g)(−1) = 0 f g (0) is undeﬁned. (g − f )(1) = 0 g f (−2) = 1 16 10. For f (x) = x2 + 1 and g(x) = 1 x2+1 (f + g)(2) = 26 5 (f g) 1 2 = 1 (f − g)(−1) = 3 2 f g (0) = 1 (g − f )(1) = − 3 2 g f (−2) = 1 25 11. For f (x) = 2x + 1 and g(x) = x − 2 (f + g)(x) = 3x − 1 Domain: (−∞, ∞) (f g)(x) = 2x2 − 3x − 2 Domain: (−∞, ∞) 12. For f (x) = 1 − 4x and g(x) = 2x − 1 (f + g)(x) = −2x Domain: (−∞, ∞) (f g)(x) = −8x2 + 6x − 1 Domain: (−∞, ∞) (f − g)(x) = x + 3 Domain: (−∞, ∞) f g (x) = 2x+1 x−2 Domain: (−∞, 2) ∪ (2, ∞) (f − g)(x) = 2 − 6x Domain: (−∞, ∞) f (x) = 1−4x g 2x−1 Domain: −∞, 1 2 ∪ 1 2 , ∞ 1.5 Function Arithmetic 89 13. For f (x) = x2 and g(x) = 3x − 1 (f + g)(x) = x2 + 3x − 1 Domain: (−∞, ∞) (f g)(x) = 3x3 − x2 Domain: (−∞, ∞) 14. For f (x) = x2 − x and g(x) = 7x (f + g)(x) = x2 + 6x Domain: (−∞, ∞) (f g)(x) = 7x3 − 7x2 Domain: (−∞, ∞) (f − g)(x) = x2 − 3x + 1 Domain: (−∞, ∞) f (x) = x2 3x−1 g Domain: −∞, 1 3 ∪ 1 3 , ∞ (f − g)(x) = x2 − 8x Domain: (−∞, ∞) f g (x) = x−1 7 Domain: (−∞, 0) ∪ (0, ∞) 15. For f (x) = x2 − 4 and g(x) = 3x + 6 (f + g)(x) = x2 + 3x + 2 Domain: (−∞, ∞) (f g)(x) = 3x3 + 6x2 − 12x − 24 Domain: (−∞, ∞) 16. For f (x) = −x2 + x + 6 and g(x) = x2 − 9 (f − g)(x) = x2 − 3x − 10 Domain: (−∞, ∞) f g (x) = x−2 3 Domain: (−∞, −2) ∪ (−2, ∞) (f + g)(x) = x − 3 Domain: (−∞, ∞) (f g)(x) = −x4 + x3 + 15x2 − 9x − 54 Domain: (−∞, ∞) (f − g)(x) = −2x2 + x + 15 Domain: (−∞, ∞) f g (x) = − x+2 x+3 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 17. For f (x) = x 2 and g(x) = 2 x (f + g)(x) = x2+4 2x Domain: (−∞, 0) ∪ (0, ∞) (f g)(x) = 1 Domain: (−∞, 0) ∪ (0, ∞) (f − g)(x) = x2−4 2x Domain: (−∞, 0) ∪ (0, ∞) f g (x) = x2 4 Domain: (−∞, 0) ∪ (0, ∞) 90 Relations and Functions 18. For f (x) = x − 1 and g(x) = 1 x−1 (f + g)(x) = x2−2x+2 x−1 Domain: (−∞, 1) ∪ (1, ∞) (f − g)(x) = x2−2x x−1 Domain: (−∞, 1) ∪ (1, ∞) (f g)(x) = 1 Domain: (−∞, 1) ∪ (1, ∞) f g (x) = x2 − 2x + 1 Domain: (−∞, 1) ∪ (1, ∞) √ x + 1 19. For f (x) = x and g(x) = √ (f + g)(x) = x + Domain: [−1, ∞) x + 1 (f − g)(x) = x − Domain: [−1, ∞) √ x + 1 (f g)(x) = x √ x + 1 Domain: [−1, ∞) 20. For f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 f g (x) = x√ Domain: (−1, ∞) x+1 √ (f + g)(x) = 2 Domain: [5, ∞) x − 5 (f g)(x) = x − 5 Domain: [5, ∞) 21. 2 23. 0 25. −2x − h + 2 27. −2x − h + 1 29. m 31. 33. 35. −2 x(x + h) −(2x + h) x2(x + h)2 −4 (4x − 3)(4x + 4h − 3) (f − g)(x) = 0 Domain: [5, ∞) f g (x) = 1 Domain: (5, ∞) 22. −3 24. 6x + 3h − 1 26. 8x + 4h 28. 3x2 + 3xh + h2 30. 2ax + ah + b 32. 34. 36. 3 (1 − x − h)(1 − x) −2 (x + 5)(x + h + 5) 3 (x + 1)(x + h + 1) 1.5 Function Arithmetic 91 37. −9 (x − 9)(x + h − 9) 39. √ 41. √ 434 −4x − 4h + 5 + √ −4x + 5 a ax + ah + b + √ ax + b 38. 2x2 + 2xh + 2x + h (2x + 1)(2x + 2h + 1) 40. √ 42. √ 2 2x + 2h + 1 + √ 2x + 1 − 44. 3x2 + 3xh + h2 (x + h)3/2 + x3/2 45. 46. 47. 48. 1 (x + h)2/3 + (x + h)1/3x1/3 + x2/3 C(0) = 26, so the ﬁxed costs are $26. C(10) = 4.6, so when 10 shirts are produced, the cost per shirt is $4.60. p(5) = 20, so to sell 5 shirts, set the price at $20 per shirt. R(x) = −2x2 + 30x, 0 ≤ x ≤ 15 P (x) = −2x2 + 28x − 26, 0 ≤ x ≤ 15 P (x) = 0 when x = 1 and x = 13. These are the ‘break even’ points, so selling 1 shirt or 13 shirts will guarantee the revenue earned exactly recoups the cost of production. C(0) = 100, so the ﬁxed costs are $100. C(10) = 20, so when 10 bottles of tonic are produced, the cost per bottle is $20. p(5) = 30, so to sell 5 bottles of tonic, set the price at $30 per bottle. R(x) = −x2 + 35x, 0 ≤ x ≤ 35 P (x) = −x2 + 25x − 100, 0 ≤ x ≤ 35 P (x) = 0 when x = 5 and x = 20. These are the ‘break even’ points, so selling 5 bottles of tonic or 20 bottles of tonic will guarantee the revenue earned exactly recoups the cost of production. C(0) = 240, so the ﬁxed costs are 240¢ or $2.40. C(10) = 42, so when 10 cups of lemonade are made, the cost per cup is 42¢. p(5) = 75, so to sell 5 cups of lemonade, set the price at 75¢ per cup. R(x) = −3x2 + 90x, 0 ≤ x ≤ 30 P (x) = −3x2 + 72x − 240, 0 ≤ x ≤ 30 P (x) = 0 when x = 4 and x = 20. These are the ‘break even’ points, so selling 4 cups of lemonade or 20 cups of lemonade will guarantee the revenue earned exactly recoups the cost of production. 92 49. Relations and Functions C(0) = 36, so the daily ﬁxed costs are $36. C(10) = 6.6, so when 10 pies are made, the cost per pie is $6.60. p(5) = 9.5, so to sell 5 pies a day, set the price at $9.50 per pie. R(x) = −0.5x2 + 12x, 0 ≤ x ≤ 24 P (x) = −0.5x2 + 9x − 36, 0 ≤ x ≤ 24 P (x) = 0 when x = 6 and x = 12. These are the ‘break even’ points, so selling 6 pies or 12 pies a day will guarantee the revenue earned exactly recoups the cost of production. 50. C(0) = 1000, so the monthly ﬁxed costs are 1000 hundred dollars, or $100,000. C(10) = 120, so when 10 scooters are made, the cost per scooter is 120 hundred dollars, or $12,000. p(5) = 130, so to sell 5 scooters a month, set the price at 130 hundred dollars, or $13,000 per scooter. R(x) = −2x2 + 140x, 0 ≤ x ≤ 70 P (x) = −2x2 + 120x − 1000, 0 ≤ x ≤ 70 P (x) = 0 when x = 10 and x = 50. These are the ‘break even’ points, so selling 10 scooters or 50 scooters a month will guarantee the revenue earned exactly recoups the cost of production. 51. (f + g)(−3) = 2 52. (f − g)(2) = 3 53. (f g)(−1) = 0 54. (g + f )(1) = 0 55. (g − f )(3) = 3 56. (gf )(−3) = −8 57. 60. f g g f (−2) does not exist (−1) does not exist 58. 61. f g g f (−1) = 0 (3) = −2 59. 62. f g g f (2) = 4 (−3) = − 1 2 1.6 Graphs of Functions 93 1.6 Graphs of Functions In Section 1.3 we deﬁned a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Then in Section 1.4 we described a function as a process and deﬁned the notation necessary to work with functions algebraically. So now it’s time to look at functions graphically again, only this time we’ll do so with the notation deﬁned in Section 1.4. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions The graph of a function f is the set of points which satisfy the equation y = f (x). That is, the point (x, y) is on the graph of f if and only if y = f (x). Example 1.6.1. Graph f (x) = x2 − x − 6. Solution. To graph f , we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.2.1. Speciﬁcally, we check for intercepts, test for symmetry, and plot additional points as needed. To ﬁnd the x-intercepts, we set y = 0. Since y = f (x), this means f (x) = 0. f (x) = x2 − x − 6 0 = x2 − x − 6 0 = (x − 3)(x + 2) factor x − 3 = 0 or x + 2 = 0 x = −2, 3 So we get (−2, 0) and (3, 0) as x-intercepts. To ﬁnd the y-intercept, we set x = 0. Using function notation, this is the same as ﬁnding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y-intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.2.1, plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. x −3 −2 −1 0 1 2 3 4 f (x) 6 0 (x, f (x)) (−3, 6) (−2, 0) −4 (−1, −4) (0, −6) −6 (1, −6) −6 (2, −4) −4 (3, 0) 0 (4, 63−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 94 Relations and Functions Graphing piecewise-deﬁned functions is a bit more of a challenge. Example 1.6.2. Graph: f (x) = 4 − x2 x − 3, if x < 1 if x ≥ 1 Solution. We proceed as before – ﬁnding intercepts, testing for symmetry and then plotting additional points as needed. To ﬁnd the x-intercepts, as before, we set f (x) = 0. The twist is that we have two formulas for f (x). For x < 1, we use the formula f (x) = 4 − x2. Setting f (x) = 0 gives 0 = 4 − x2, so that x = ±2. However, of these two answers, only x = −2 ﬁts in the domain x < 1 for this piece. This means the only x-intercept for the x < 1 region of the x-axis is (−2, 0). For x ≥ 1, f (x) = x − 3. Setting f (x) = 0 gives 0 = x − 3, or x = 3. Since x = 3 satisﬁes the inequality x ≥ 1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x = 0 falls in the domain x < 1. Thus f (0) = 4 − 02 = 4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 − x2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x − 3 possesses none of the symmetries discussed in the Section 1.2.1. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x = 1, since this is where the equations change. When x = 1, we use the formula f (x) = x − 3, so the point on the graph (1, f (1)) is (1, −2). However, for all values less than 1, we use the formula f (x) = 4 − x2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9, x = 0.99, x = 0.999, and so on, we ﬁnd the corresponding y values using the formula f (x) = 4 − x2. Making a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f (x) values inch closer and closer1 to 4 − 12 = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get (x, f (x)) x f (x) (0.9, 3.19) 0.9 3.19 0.99 ≈ 3.02 (0.99, 3.02) 0.999 ≈ 3.002 (0.999, 3.002) y 4 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 −4 1We’ve just stepped into Calculus here! 1.6 Graphs of Functions 95 In the previous two examples, the x-coordinates of the x-intercepts of the graph of y = f (x) were found by solving f (x) = 0. For this reason, they are called the zeros of f . Deﬁnition 1.9. The zeros of a function f are the so
lutions to the equation f (x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f (x). Of the three symmetries discussed in Section 1.2.1, only two are of signiﬁcance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether the graph of an equation is symmetric about the y-axis by replacing x with −x and checking to see if an equivalent equation results. If we are graphing the equation y = f (x), substituting −x for x results in the equation y = f (−x). In order for this equation to be equivalent to the original equation y = f (x) we need f (−x) = f (x). In a similar fashion, we recall that to test an equation’s graph for symmetry about the origin, we replace x and y with −x and −y, respectively. Doing this substitution in the equation y = f (x) results in −y = f (−x). Solving the latter equation for y gives y = −f (−x). In order for this equation to be equivalent to the original equation y = f (x) we need −f (−x) = f (x), or, equivalently, f (−x) = −f (x). These results are summarized below. Testing the Graph of a Function for Symmetry The graph of a function f is symmetric about the y-axis if and only if f (−x) = f (x) for all x in the domain of f . about the origin if and only if f (−x) = −f (x) for all x in the domain of f . For reasons which won’t become clear until we study polynomials, we call a function even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin. Apart from a very specialized family of functions which are both even and odd,3 functions fall into one of three distinct categories: even, odd, or neither even nor odd. Example 1.6.3. Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. 1. f (x) = 5 2 − x2 2. g(x) = 5x 2 − x2 3. h(x) = 5. j(x) = x2 − 5x 2 − x3 x 100 − 1 4. i(x) = 6. p(x) = 5x 2x − x3 x + 3 if x < 0 if x ≥ 0 −x + 3, Solution. The ﬁrst step in all of these problems is to replace x with −x and simplify. 2Why are we so dismissive about symmetry about the x-axis for graphs of functions? 3Any ideas? 96 1. 2. Relations and Functions f (x) = f (−x) = f (−x) = 5 2 − x2 5 2 − (−x)2 5 2 − x2 f (−x) = f (x) Hence, f is even. The graphing calculator furnishes the following. This suggests4 that the graph of f is symmetric about the y-axis, as expected. g(x) = g(−x) = g(−x) = 5x 2 − x2 5(−x) 2 − (−x)2 −5x 2 − x2 It doesn’t appear that g(−x) is equivalent to g(x). To prove this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g(−1) = −5. This proves that g is not even, but it doesn’t rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g(−x) with −g(x) −g(x) = − 5x 2 − x2 −5x 2 − x2 = −g(x) = g(−x) Hence, g is odd. Graphically, 4‘Suggests’ is about the extent of what it can do. 1.6 Graphs of Functions 97 3. 4. The calculator indicates the graph of g is symmetric about the origin, as expected. h(x) = h(−x) = h(−x) = 5x 2 − x3 5(−x) 2 − (−x)3 −5x 2 + x3 Once again, h(−x) doesn’t appear to be equivalent to h(x). We check with an x value, for example, h(1) = 5 but h(−1) = − 5 3 . This proves that h is not even and it also shows h is not odd. (Why?) Graphically, The graph of h appears to be neither symmetric about the y-axis nor the origin. i(x) = i(−x) = i(−x) = 5x 2x − x3 5(−x) 2(−x) − (−x)3 −5x −2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however, prove 98 Relations and Functions i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x) and i(−x) and show that they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3) −5x −2x + x3 Hence, i(x) = i(−x), so i is even. The calculator supports our conclusion. 5. j(x) = x2 − x 100 j(−x) = (−x)2 − j(−x) = x2 + x 100 − 1 −x 100 − 1 − 1 The expression for j(−x) doesn’t seem to be equivalent to j(x), so we check using x = 1 to get j(1) = − 1 100 . This rules out j being even. However, it doesn’t rule out j being odd. Examining −j(x) gives 100 and j(−1) = 1 j(x) = x2 − −j(x) = − x 100 x2 − −j(x) = −x2 + + 1 − 1 − 1 x 100 x 100 The expression −j(x) doesn’t seem to match j(−x) either. Testing x = 2 gives j(2) = 149 50 and j(−2) = 151 50 , so j is not odd, either. The calculator gives: 1.6 Graphs of Functions 99 The calculator suggests that the graph of j is symmetric about the y-axis which would imply that j is even. However, we have proven that is not the case. 6. Testing the graph of y = p(x) for symmetry is complicated by the fact p(x) is a piecewisedeﬁned function. As always, we handle this by checking the condition for symmetry by checking it on each piece of the domain. We ﬁrst consider the case when x < 0 and set about ﬁnding the correct expression for p(−x). Even though p(x) = x + 3 for x < 0, p(−x) = −x + 3 here. The reason for this is that since x < 0, −x > 0 which means to ﬁnd p(−x), we need to use the other formula for p(x), namely p(x) = −x+3. Hence, for x < 0, p(−x) = −(−x)+3 = x + 3 = p(x). For x ≥ 0, p(x) = −x + 3 and we have two cases. If x > 0, then −x < 0 so p(−x) = (−x) + 3 = −x + 3 = p(x). If x = 0, then p(0) = 3 = p(−0). Hence, in all cases, p(−x) = p(x), so p is even. Since p(0) = 3 but p(−0) = p(0) = 3 = −3, we also have p is not odd. While graphing y = p(x) is not onerous to do by hand, it is instructive to see how to enter this into our calculator. By using some of the logical commands,5 we have: The calculator bears shows that the graph appears to be symmetric about the y-axis. There are two lessons to be learned from the last example. The ﬁrst is that sampling function values at particular x values is not enough to prove that a function is even or odd − despite the fact that j(−1) = −j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the Algebra which proves mathematical truths.6 5Consult your owner’s manual, instructor, or favorite video site! 6Or, in other words, don’t rely too heavily on the machine! 100 Relations and Functions 1.6.1 General Function Behavior The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them all in great detail. The purpose of this section’s discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing and constant. Before deﬁning the concepts algebraically, it is instructive to ﬁrst look at them graphically. Consider the graph of the function f below. (6, 5.5) (−2, 4.54 −3 −2 −1 −1 1 2 3 4 5 6 7 x (−4, −3) −2 −3 −4 −5 −6 −7 −8 −9 (4, −6) (3, −8) (5, −6) The graph of y = f (x) Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point (6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we are walking ‘uphill’; then between (−2, 4.5) and (3, −8), we are walking ‘downhill’; and between (3, −8) and (4, −6), we are walking ‘uphill’ once more. From (4, −6) to (5, −6), we ‘level oﬀ’, and then resume walking ‘uphill’ from (5, −6) to (6, 5.5). In other words, for the x values between −4 and −2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y = f (x), the y values on the graph are the function values, and we say that the function f is increasing on the interval [−4, −2]. Analogously, we say that f is decreasing on the interval [−2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and ﬁnally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f is increasing 1.6 Graphs of Functions 101 on [−4, −2] we do not mention the actual y values that f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs.7 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run into serious trouble in our previous example if we tried to do so because x = −2 is contained in an interval on which f was increasing and one on which it is decreasing. (There’s more on this issue – and many others – in the Exercises.) We’re now ready for the more formal algebraic deﬁnitions of what it means for a function to be increasing, decreasing or constant. Deﬁnition 1.10. Suppose f is a function deﬁned on an interval I. We say f is: increasing on I if and only if f (a) < f (b) for all real numbers a, b in I with a < b. decreasing on I if and only if f (a) > f (b) for all real numbers a, b in I with a < b. constant on I if and only if f (a) = f (b) for all real numbers a, b in I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing and constant as stated in Deﬁnition 1.10 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you ﬁnd an example of a function for which none of the concepts in Deﬁnition
1.10 apply? Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5) does not have the largest y value of all of the points on the graph of f − indeed that honor goes to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’ between x = −4 and x = 3. We say that the function f has a local maximum8 at the point (−2, 4.5), because the y-coordinate 4.5 is the largest y-value (hence, function value) on the curve ‘near’9 x = −2. Similarly, we say that the function f has a local minimum10 at the point (3, −8), since the y-coordinate −8 is the smallest function value near x = 3. Although it is tempting to say that local extrema11 occur when the function changes from increasing to decreasing or vice versa, it is not a precise enough way to deﬁne the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional deﬁnitions and thoroughly vet the pathologies they induce in the Exercises. We have one last observation to make before we proceed to the algebraic deﬁnitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see that the largest y value (the largest function value) is 5.5 at x = 6. In this case, we say the maximum12 of f is 5.5; similarly, the minimum13 of f is −8. 7The notions of how quickly or how slowly a function increases or decreases are explored in Calculus. 8Also called ‘relative maximum’. 9We will make this more precise in a moment. 10Also called a ‘relative minimum’. 11‘Maxima’ is the plural of ‘maximum’ and ‘mimima’ is the plural of ‘minimum’. ‘Extrema’ is the plural of ‘extremum’ which combines maximum and minimum. 12Sometimes called the ‘absolute’ or ‘global’ maximum. 13Again, ‘absolute’ or ‘global’ minimum can be used. 102 Relations and Functions We formalize these concepts in the following deﬁnitions. Deﬁnition 1.11. Suppose f is a function with f (a) = b. We say f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f (a) ≥ f (x) for all x in I. The value f (a) = b is called ‘a local maximum value of f ’ in this case. We say f has a local minimum at the point (a, b) if and only if there is an open interval I containing a for which f (a) ≤ f (x) for all x in I. The value f (a) = b is called ‘a local minimum value of f ’ in this case. The value b is called the maximum of f if b ≥ f (x) for all x in the domain of f . The value b is called the minimum of f if b ≤ f (x) for all x in the domain of f . It’s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function’s graph would have to look like?) We shall see examples of functions in the Exercises which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic deﬁnitions above as we explore the wonderful diversity of graphs that functions provide us. Here is the ‘tame’ example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the Exercises. Example 1.6.4. Given the graph of y = f (x) below, answer all of the following questions. y (0, 3) 4 3 2 1 (−2, 0) (2, 0) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 (−4, −3) (4, −3) 1.6 Graphs of Functions 103 1. Find the domain of f . 2. Find the range of f . 3. List the x-intercepts, if any exist. 4. List the y-intercepts, if any exist. 5. Find the zeros of f . 7. Determine f (2). 6. Solve f (x) < 0. 8. Solve f (x) = −3. 9. Find the number of solutions to f (x) = 1. 10. Does f appear to be even, odd, or neither? 11. List the intervals on which f is increasing. 12. List the intervals on which f is decreasing. 13. List the local maximums, if any exist. 14. List the local minimums, if any exist. 15. Find the maximum, if it exists. 16. Find the minimum, if it exists. Solution. 1. To ﬁnd the domain of f , we proceed as in Section 1.3. By projecting the graph to the x-axis, we see that the portion of the x-axis which corresponds to a point on the graph is everything from −4 to 4, inclusive. Hence, the domain is [−4, 4]. 2. To ﬁnd the range, we project the graph to the y-axis. We see that the y values from −3 to 3, inclusive, constitute the range of f . Hence, our answer is [−3, 3]. 3. The x-intercepts are the points on the graph with y-coordinate 0, namely (−2, 0) and (2, 0). 4. The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). 5. The zeros of f are the x-coordinates of the x-intercepts of the graph of y = f (x) which are x = −2, 2. 6. To solve f (x) < 0, we look for the x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking for where the graph is below the x-axis. This happens for the x values from −4 to −2 and again from 2 to 4. So our answer is [−4, −2) ∪ (2, 4]. 7. Since the graph of f is the graph of the equation y = f (x), f (2) is the y-coordinate of the point which corresponds to x = 2. Since the point (2, 0) is on the graph, we have f (2) = 0. 8. To solve f (x) = −3, we look where y = f (x) = −3. We ﬁnd two points with a y-coordinate of −3, namely (−4, −3) and (4, −3). Hence, the solutions to f (x) = −3 are x = ±4. 9. As in the previous problem, to solve f (x) = 1, we look for points on the graph where the y-coordinate is 1. Even though these points aren’t speciﬁed, we see that the curve has two points with a y value of 1, as seen in the graph below. That means there are two solutions to f (x) = 1. 104 Relations and Functions y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 10. The graph appears to be symmetric about the y-axis. This suggests14 that f is even. 11. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f is increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the x-axis.) 12. As we move from left to right, the graph falls from (0, 3) to (4, −3). This means f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) 13. The function has its only local maximum at (0, 3) so f (0) = 3 is the local minimum value. 14. There are no local minimums. Why don’t (−4, −3) and (4, −3) count? Let’s consider the point (−4, −3) for a moment. Recall that, in the deﬁnition of local minimum, there needs to be an open interval I which contains x = −4 such that f (−4) < f (x) for all x in I diﬀerent from −4. But if we put an open interval around x = −4 a portion of that interval will lie outside of the domain of f . Because we are unable to fulﬁll the requirements of the deﬁnition for a local minimum, we cannot claim that f has one at (−4, −3). The point (4, −3) fails for the same reason − no open interval around x = 4 stays within the domain of f . 15. The maximum value of f is the largest y-coordinate which is 3. 16. The minimum value of f is the smallest y-coordinate which is −3. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to ﬁnd the local maximums and local minimums analytically; this is the business of Calculus.15 When we have need to ﬁnd such beasts, we will resort to the calculator. Most graphing calculators have ‘Minimum’ and ‘Maximum’ features which can be used to approximate these values, as we now demonstrate. 14but does not prove 15Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra. 1.6 Graphs of Functions 105 Example 1.6.5. Let f (x) = . Use a graphing calculator to approximate the intervals on which f is increasing and those on which it is decreasing. Approximate all extrema. 15x x2 + 3 Solution. Entering this function into the calculator gives Using the Minimum and Maximum features, we get To two decimal places, f appears to have its only local minimum at (−1.73, −4.33) and its only local maximum at (1.73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn’t be too surprising. The function appears to be increasing on [−1.73, 1.73] and decreasing on (−∞, −1.73]∪[1.73, ∞). This makes −4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Example 1.6.6. Find the points on the graph of y = (x − 3)2 which are closest to the origin. Round your answers to two decimal places. Solution. Suppose a point (x, y) is on the graph of y = (x − 3)2. Its distance to the origin (0, 0) is given by d = (x − 0)2 + (y − 0)2 = = = x2 + y2 x2 + [(x − 3)2]2 x2 + (x − 3)4 Since y = (x − 3)2 x2 + (x − 3)4 is the distance from (0, 0) to the point (x, y) Given a value for x, the formula d = on the curve y = (x − 3)2. What we have deﬁned, then, is a function d(x) which we wish to 106 Relations and Functions minimize over all values of x. To accomplish this task analytically would require Calculus so as we’ve mentioned before, we can use a graphing calculator to ﬁnd an approximate solution. Using the calculator, we enter the function d(x) as shown below and graph. Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x = 2.00. To ﬁnd the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation to get y = (x − 3)2 = (2.00 − 3)2 = 1.00. So, our ﬁnal answer is (2.00, 1.00).16 (What does the y value listed on the calculator screen mean in this problem?) 16It seems silly to list a ﬁnal answer as (2.00, 1.00). Indeed, Calculus conﬁrms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the authors are overly pedantic, and a

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