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these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions. Cofunction Identities The cofunction identities are summarized in Table 9.6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1001 ⎛ sin θ = cos ⎝ − θ⎞ ⎠ π 2 ⎛ cos θ = sin ⎝ − θ⎞ ⎠ π 2 tan θ = cot ⎛ ⎝ π 2 − θ⎞ ⎠ ⎛ cot θ = tan ⎝ − θ⎞ ⎠ π 2 ⎛ sec θ = csc ⎝ − θ⎞ ⎠ π 2 ⎛ csc θ = sec ⎝ − θ⎞ ⎠ π 2 Table 9.6 Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using cos⎛ ⎝α − β⎞ ⎠ = cos αcos β + sin αsin β, we can write ⎛ cos ⎝ π 2 Example 9.17 − θ⎞ ⎠ = cos π 2 cos θ + sin π 2 = (0)cos θ + (1)sin θ = sin θ sin θ Finding a Cofunction with the Same Value as the Given Expression in terms of its cofunction. Write tan π 9 Solution The cofunction of tan θ = cot ⎛ ⎝ π 2 − θ⎞ ⎠. Thus, ⎛ tan ⎝ π 9 ⎞ ⎠ = cot ⎛ ⎝ = cot = cot ⎛ ⎝ ⎛ ⎝ ⎞ π − ⎠ 9 − 2π 18 ⎞ ⎠ π 2 9π 18 7π 18 ⎞ ⎠ 9.9 Write sin π 7 in terms of its cofunction. Using the Sum and Difference Formulas to Verify Identities Verifying an identity means demonstrating that the equation holds for all values
of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity. 1002 Chapter 9 Trigonometric Identities and Equations Given an identity, verify using sum and difference formulas. 1. Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient. 2. Look for opportunities to use the sum and difference formulas. 3. Rewrite sums or differences of quotients as single quotients. 4. If the process becomes cumbersome, rewrite the expression in terms of sines and cosines. Example 9.18 Verifying an Identity Involving Sine Verify the identity sin(α + β) + sin(α − β) = 2 sin α cos β. Solution We see that the left side of the equation includes the sines of the sum and the difference of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β We can rewrite each using the sum and difference formulas. sin(α + β) + sin(α − β) = sin α cos β + cos α sin β + sin α cos β − cos α sin β = 2 sin α cos β We see that the identity is verified. Example 9.19 Verifying an Identity Involving Tangent Verify the following identity. sin(α − β) cos α cos β = tan α − tan β Solution We can begin by rewriting the numerator on the left side of the equation. sin(α − β) cos α cos β = sin α cos β − cos αsin β cos αcos β = sin α cos β cos α cos β − cos α sin β cos α cos β Rewrite using a common denominator. sin β = sin α cos α − cos β = tan α − tan β Cancel. Rewrite in terms of tangent. We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1003
9.10 Verify the identity: tan(π − θ) = − tan θ. Example 9.20 Using Sum and Difference Formulas to Solve an Application Problem Let L1 and L2 denote two non-vertical intersecting lines, and let θ denote the acute angle between L1 and L2. See Figure 9.12. Show that tan θ = m2 − m1 1 + m1 m2 where m1 and m2 are the slopes of L1 and L2 respectively. (Hint: Use the fact tan θ2 = m2. ) that tan θ1 = m1 and Figure 9.12 Solution Using the difference formula for tangent, this problem does not seem as daunting as it might. tan θ = tan⎛ ⎞ ⎠ ⎝θ2 − θ1 tan θ2 − tan θ1 1 + tan θ1 tan θ2 m2 − m1 1 + m1 m2 = = Example 9.21 Investigating a Guy-wire Problem For a climbing wall, a guy-wire R is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire S attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle α between the wires. See Figure 9.13. 1004 Chapter 9 Trigonometric Identities and Equations Figure 9.13 Solution Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that tan β = 47. We can 50 ⎠ = 40 50, and tan⎛ ⎝β − α⎞ = 4 5 then use difference formula for tangent. Now, substituting the values we know into the formula, we have tan⎛ ⎝β − α⎞ ⎠ = tan β − tan α 1 + tan βtan α 4 5 = 47 50 − tan α 1 + 47 50tan α − tan α⎞ ⎛ ⎠ ⎝ 47 50 4 ⎛ ⎝1 + 47 50 tan α⎞ ⎠ = 5 Use the distributive property, and then simplify the functions. ⎛ 4(1) + 4 ⎝ 47 50 ⎞ ⎠
tan α = 5 4 + 3.76 tan α = 4.7 − 5 tan α ⎞ ⎠ − 5 tan α 47 50 ⎛ ⎝ 5 tan α + 3.76 tan α = 0.7 8.76 tan α = 0.7 tan α ≈ 0.07991 tan−1(0.07991) ≈.079741 Now we can calculate the angle in degrees. α ≈ 0.079741 ⎛ ⎝ 180 π ⎞ ⎠ ≈ 4.57° Analysis Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given. Access these online resources for additional instruction and practice with sum and difference identities. • Sum and Difference Identities for Cosine (http://openstaxcollege.org/l/sumdifcos) • Sum and Difference Identities for Sine (http://openstaxcollege.org/l/sumdifsin) • Sum and Difference Identities for Tangent (http://openstaxcollege.org/l/sumdiftan) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1005 9.2 EXERCISES Verbal Explain the basis for the cofunction identities and 43. when they apply. 44. ⎛ Is there only one way to evaluate cos ⎝ ⎞ ⎠? Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer. 5π 4 45. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for f (x) = sin(x) and g(x) = cos(x). (Hint: 0 − x = − x ) Algebraic For the following exercises, find the exact value. 46. 47. 48. 49. 50. 51. ⎛ cos ⎝ ⎞ ⎠ 7π 12 ⎛ cos ⎝ ⎞ ⎠ π 12 ⎛ sin ⎝ ⎞ ⎠ 5π 12 ⎛
sin ⎝ 11π 12 ⎞ ⎠ ⎛ tan ⎝− ⎞ ⎠ π 12 ⎛ tan ⎝ 19π 12 ⎞ ⎠ For the following exercises, rewrite in terms of sin x and cos x. 52. 53. 54. 55. ⎛ ⎝x + 11π sin 6 ⎞ ⎠ ⎛ sin ⎝x − 3π 4 ⎞ ⎠ ⎛ ⎝x − 5π cos 6 ⎞ ⎠ ⎛ ⎝x + 2π cos 3 ⎞ ⎠ For the following exercises, simplify the given expression. 56. ⎛ csc ⎝ 57. ⎛ sec ⎝ − t⎞ ⎠ − θ⎞ ⎠ π 2 π 2 58. 59. cot ⎛ ⎝ π 2 − x⎞ ⎠ ⎛ tan ⎝ π 2 − x⎞ ⎠ 60. sin(2x) cos(5x) − sin(5x) cos(2x) 61. 3 2 ⎛ tan ⎝ x⎞ ⎛ x⎞ 7 ⎠ − tan ⎝ ⎠ 5 x⎞ ⎛ x⎞ ⎛ 3 7 ⎠tan 1 + tan ⎝ ⎠ ⎝ 2 5 For the following exercises, find the requested information. 62. Given that sin a = 2 3 interval ⎡ ⎣ the in b both and cos b = − 1 4, with a and, π⎞ ⎠, find sin(a + b) and π 2 cos(a − b). 63. Given that sin a = 4 5, and cos b = 1 3, with a and b both in the cos(a + b). interval ⎡ ⎣0, π 2 ⎞ ⎠, find sin(a − b) and For the following exercises, find the exact value of each expression. 64. 65. 66. ⎛ ⎝cos−1(0) − cos−1 ⎛ sin ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 ⎛ �
��cos−1 ⎛ cos ⎝ ⎞ ⎠ + sin−sin−1 ⎛ tan ⎝ 1 2 ⎞ ⎠ − cos−1 ⎛ ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Graphical For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator. 67. ⎛ cos ⎝ π 2 − x⎞ ⎠ 68. sin(π − x) ⎛ tan ⎝ π 3 + x⎞ ⎠ ⎛ sin ⎝ π 3 + x⎞ ⎠ 69. 70. 71. 1006 Chapter 9 Trigonometric Identities and Equations Extensions For the following exercises, prove the identities provided. 89. 90. 91. 92. 93. tan(x + π 4 ) = tan x + 1 1 − tan x tan(a + b) tan(a − b) = sin a cos a + sin b cos b sin a cos a − sin b cos b cos(a + b) cos a cos b = 1 − tan a tan b cos(x + y)cos(x − y) = cos2 x − sin2 y cos(x + h) − cos x h = cos xcos h − 1 h − sin xsin h h the following exercises, prove or disprove the For statements. 94. 95. 96. tan(u + v) = tan u + tan v 1 − tan u tan v tan(u − v) = tan u − tan v 1 + tan u tan v tan(x + y) 1 + tan x tan x = tan x + tan y 1 − tan2 x tan2 y 97. prove or disprove sin⎛ If α, β, and γ are angles in the same triangle, then ⎠ = sin γ. ⎝α + β⎞ If α, β, and y are angles in the same triangle, then 98. prove or disprove tan α + tan β + tan γ = tan α tan β tan γ ⎛ tan ⎝ π 4 − x⎞ ⎠ 72. 73. 74. ⎛ cos ⎝ 7π 6 + x⎞ �
�� ⎛ sin ⎝ π 4 + x⎞ ⎠ ⎛ cos ⎝ 5π 4 + x⎞ ⎠ For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think 2x = x + x. ) 75. f (x) = sin(4x) − sin(3x)cos x, g(x) = sin x cos(3x) 76. f (x) = cos(4x) + sin x sin(3x), g(x) = − cos x cos(3x) 77. f (x) = sin(3x)cos(6x), g(x) = − sin(3x)cos(6x) 78. f (x) = sin(4x), g(x) = sin(5x)cos x − cos(5x)sin x 79. f (x) = sin(2x), g(x) = 2 sin x cos x 80. 81. f (θ) = cos(2θ), g(θ) = cos2 θ − sin2 θ f (θ) = tan(2θ), g(θ) = tan θ 1 + tan2 θ 82. f (x) = sin(3x)sin x, g(x) = sin2(2x)cos2 x − cos2(2x)sin2 x 83. f (x) = tan( − x), g(x) = tan x − tan(2x) 1 − tan x tan(2x) Technology the following exercises, For find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. 84. sin(75°) 85. sin(195°) 86. 87. 88. cos(165°) cos(345°) tan(−15°) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1007 9.3 \| Double-Angle, Half-Angle, and Reduction Formulas Learning Objectives In this section, you will: 9.3.
1 Use double-angle formulas to find exact values. 9.3.2 Use double-angle formulas to verify identities. 9.3.3 Use reduction formulas to simplify an expression. 9.3.4 Use half-angle formulas to find exact values. Figure 9.14 Bicycle ramps for advanced riders have a steeper incline than those designed for novices. Bicycle ramps made for competition (see Figure 9.14) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θ such that tan θ = 5. The angle is 3 divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one. Using Double-Angle Formulas to Find Exact Values In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α = β. Deriving the double-angle formula for sine begins with the sum formula, sin⎛ ⎝α + β⎞ ⎠ = sin α cos β + cos α sin β If we let α = β = θ, then we have sin(θ + θ) = sin θ cos θ + cos θ sin θ sin(2θ) = 2sin θ cos θ Deriving cos⎛ ⎝α + β⎞ double-angle cosine the ⎠ = cos α cos β − sin α sin β, and letting α = β = θ, we have gives three for us options. First, starting from the sum formula, cos(θ + θ) = cos θ cos θ − sin θ sin θ cos(2θ) = cos2 θ − sin2 θ Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more variations. The first variation is: cos(2θ) = cos2 θ − sin2 θ ⎝1 − sin2 θ⎞ ⎛ = = 1 − 2sin2 θ ⎠ − sin2 θ 1008 Chapter 9 Trigonometric Identities and
Equations The second variation is: cos(2θ) = cos2 θ − sin2 θ = cos2 θ − = 2 cos2 θ − 1 ⎝1 − cos2 θ⎞ ⎛ ⎠ Similarly, to derive the double-angle formula for tangent, replacing α = β = θ in the sum formula gives tan α + tan β tan(α + β) = 1 − tan α tan β tan(θ + θ) = tan θ + tan θ 1 − tan θ tan θ 2tan θ 1 − tan2 θ tan(2θ) = Double-Angle Formulas The double-angle formulas are summarized as follows: sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 2 tan θ 1 − tan2 θ tan(2θ) = (9.24) (9.25) (9.26) Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value. 1. Draw a triangle to reflect the given information. 2. Determine the correct double-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Example 9.22 Using a Double-Angle Formula to Find the Exact Value Involving Tangent Given that tan θ = − 3 4 and θ is in quadrant II, find the following: a. b. c. sin(2θ) cos(2θ) tan(2θ) Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given tan θ = − 3, such that θ is in quadrant II. The tangent of an angle is equal to the opposite 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1009 side over the adjacent side, and because θ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: (−4
)2 + (3)2 = c2 16 + 9 = c2 25 = c2 c = 5 Now we can draw a triangle similar to the one shown in Figure 9.15. Figure 9.15 a. Let’s begin by writing the double-angle formula for sine. sin(2θ) = 2 sin θ cos θ We see that we to need to find sin θ and cos θ. Based on Figure 9.15, we see that the hypotenuse equals 5, so sin θ = 3 5. Substitute these values into the equation, and simplify., and cos θ = − 4 5 Thus, sin(2θ) = 2 ⎛ ⎞ ⎛ 3 ⎝− 4 ⎝ ⎠ 5 5 = − 24 25 ⎞ ⎠ b. Write the double-angle formula for cosine. Again, substitute the values of the sine and cosine into the equation, and simplify. cos(2θ) = cos2 θ − sin2 θ 2 ⎞ ⎠ ⎛ ⎝ 3 5 cos(2θ) = 2 − ⎛ ⎞ ⎝− 4 ⎠ 5 − 9 = 16 25 25 = 7 25 c. Write the double-angle formula for tangent. tan(2θ) = 2 tan θ 1 − tan2 θ In this formula, we need the tangent, which we were given as tan θ = − 3 4. Substitute this value into the equation, and simplify. 1010 Chapter 9 Trigonometric Identities and Equations tan(2θ) = 2 ⎞ ⎠ ⎛ ⎝− 3 4 ⎛ ⎝− 16 ⎛ = − 3 16 ⎝ 7 2 = − 24 7 ⎞ ⎠ 9.11 Given sin α = 5 8, with θ in quadrant I, find cos(2α). Example 9.23 Using the Double-Angle Formula for Cosine without Exact Values Use the double-angle formula for cosine to write cos(6x) in terms of cos(3x). Solution cos(6x) = cos(3x + 3x) = cos 3x cos 3x − sin 3x sin 3x = cos2 3x − sin2 3x Analysis This example
illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function. Using Double-Angle Formulas to Verify Identities Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side. Example 9.24 Using the Double-Angle Formulas to Verify an Identity Verify the following identity using double-angle formulas: 1 + sin(2θ) = (sin θ + cos θ)2 Solution We will work on the right side of the equal sign and rewrite the expression until it matches the left side. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1011 (sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ ⎠ + 2 sin θ cos θ ⎝sin2 θ + cos2 θ⎞ ⎛ = = 1 + 2 sin θ cos θ = 1 + sin(2θ) Analysis This process is not complicated, as long as we recall the perfect square formula from algebra: (a ± b)2 = a2 ± 2ab + b2 where a = sin θ and b = cos θ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent. 9.12 Verify the identity: cos4 θ − sin4 θ = cos(2θ). Example 9.25 Verifying a Double-Angle Identity for Tangent Verify the identity: Solution tan(2θ) = 2 cot θ − tan θ In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation. tan(2θ) = = = = ⎞ ⎠ 1 tan θ 2 tan θ 1 − tan2 θ 2 tan θ⎛ 1 ⎝ tan θ ⎛ (1 − tan2 θ) ⎝ 2 tan θ − tan2 θ 2 cot θ
− tan θ tan θ 1 Double-angle formula Multiply by a term that results in desired numerator. ⎞ ⎠ Use reciprocal identity for 1 tan θ. Analysis Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show Let’s work on the right side. 2tan θ 1 − tan2 θ = 2 cot θ − tan θ 1012 Chapter 9 Trigonometric Identities and Equations 2 cot θ − tan θ = 2 1 tan θ − tan θ ⎛ ⎝ tan θ tan θ ⎞ ⎠ = 2 tan θ tan θ (tan θ ) − tan θ(tan θ) 2 tan θ 1 − tan2 θ When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier. = 1 9.13 Verify the identity: cos(2θ)cos θ = cos3 θ − cos θ sin2 θ. Use Reduction Formulas to Simplify an Expression The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the doubleangle formulas. We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ) = 1 − 2 sin2 θ. Solve for sin2 θ : cos(2θ) = 1 − 2 sin2 θ 2 sin2 θ = 1 − cos(2θ) 1 − cos(2θ) 2 sin2 θ = Next, we use the
formula cos(2θ) = 2 cos2 θ − 1. Solve for cos2 θ : cos(2θ) = 2 cos2 θ − 1 1 + cos(2θ) = 2 cos2 θ 1 + cos(2θ) 2 = cos2 θ The last reduction formula is derived by writing tangent in terms of sine and cosine: Substitute the reduction formulas. tan2 θ = sin2 θ cos2 θ 1 − cos(2θ) 2 1 + cos(2θ) 2 = ⎛ ⎞ ⎠ ⎝ 2 1 + cos(2θ) ⎞ ⎠ = = 1 − cos(2θ) ⎛ ⎝ 2 1 − cos(2θ) 1 + cos(2θ) Reduction Formulas The reduction formulas are summarized as follows: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) 1013 (9.27) (9.28) (9.29) Example 9.26 Writing an Equivalent Expression Not Containing Powers Greater Than 1 Write an equivalent expression for cos4 x that does not involve any powers of sine or cosine greater than 1. Solution We will apply the reduction formula for cosine twice. cos4 x = ⎛ ⎝cos2 x + cos(2x) 2 ⎞ ⎠ ⎛ ⎞ ⎝1 + 2cos(2x) + cos2(2x cos(2x) + 1 4 cos(2x) + 1 8 cos(2x) + 1 8 ⎞ ⎠ 1 + cos2(2x) 2 cos(4x) ⎛ ⎝ + 1 8 cos(4x) Substitute reduction formula for cos 2 x. Substitute reduction formula for cos 2 x. Analysis The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra. Example 9.27 Using the Power-Reducing Formulas to Prove an Identity Use
the power-reducing formulas to prove sin3 (2x) = ⎤ sin(2x) ⎦ ⎡ ⎣1 − cos(4x)⎤ ⎦ ⎡ ⎣ 1 2 Solution We will work on simplifying the left side of the equation: ⎤ ⎡ sin3(2x) = [sin(2x)] ⎣sin2(2x) ⎦ 1 − cos(4x) ⎡ = sin(2x) ⎣ 2 ⎤ ⎦ ⎛ = sin(2x) ⎝ 1 2 ⎞ ⎠[1 − cos(4x)] = 1 2 [sin(2x)][1 − cos(4x)] Substitute the power-reduction formula. 1014 Chapter 9 Trigonometric Identities and Equations Analysis Note that in this example, we substituted for sin2 (2x). The formula states We let θ = 2x, so 2θ = 4x. 1 − cos(4x) 2 sin2 θ = 1 − cos(2θ) 2 9.14 Use the power-reducing formulas to prove that 10 cos4 x = 15 4 + 5 cos(2x) + 5 4 cos(4x). Using Half-Angle Formulas to Find Exact Values The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θ with α the half-angle formula for sine 2 ⎞ ⎠. Note that the half-angle formulas are preceded by a ± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α 2 ⎛ is found by simplifying the equation and solving for sin ⎝ terminates. α 2, The half-angle formula for sine is derived as follows: sin2 ⎛ ⎝ α 2 ⎛ sin ⎝ α 2 To derive the half-angle formula for cosine, we have sin2 − cos(2θ) 2 ⎛ ⎝cos2 ⋅ α 2 2 = 1 − cos α 2 ⎠ = ± 1 − cos α ⎞ 2 cos2
⎛ ⎝ α 2 ⎛ cos ⎝ α 2 For the tangent identity, we have cos2 θ = ⎞ ⎠ ⎞ ⎠ = 1 + cos(2θ) 2 ⎛ ⎝2 ⋅ α 1 + cos 2 2 = 1 + cos α 2 ⎠ = ± 1 + cos α ⎞ 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1015 tan2 θ = ⎞ ⎠ = 1 − cos(2θ) 1 + cos(2θ) ⎛ ⎝2 ⋅ α 1 − cos 2 ⎛ ⎝2 ⋅ α 1 + cos 2 = 1 − cos α 1 + cos α ⎠ = ± 1 − cos α 1 + cos α ⎞ tan2⎛ ⎝ α 2 ⎛ tan ⎝ α 2 Half-Angle Formulas The half-angle formulas are as follows: ⎞ ⎞ 2 2 ⎛ sin ⎝ ⎠ = ± 1 − cos α α 2 ⎠ = ± 1 + cos α α ⎛ cos ⎝ 2 ⎠ = ± 1 − cos α α 1 + cos α 2 ⎛ tan ⎝ = sin α 1 + cos α = 1 − cos α ⎞ sin α ⎞ ⎠ ⎞ ⎠ (9.30) (9.31) (9.32) Example 9.28 Using a Half-Angle Formula to Find the Exact Value of a Sine Function Find sin(15°) using a half-angle formula. Solution Since 15° = 30° 2, we use the half-angle formula for sine: sin 30° 2 = 1 − cos30 Remember that we can check the answer with a graphing calculator. Analysis Notice that we used only the positive root because sin(15°) is positive. 1016 Chapter 9 Trigonometric Identities and Equations Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle. 1. Draw a triangle to represent the given information. 2. Determine the correct half-angle formula. 3
. Substitute values into the formula based on the triangle. 4. Simplify. Example 9.29 Finding Exact Values Using Half-Angle Identities Given that tan α = 8 15 and α lies in quadrant III, find the exact value of the following: a. b. c. ⎛ sin ⎝ ⎞ ⎠ α 2 ⎛ cos ⎝ ⎞ ⎠ α 2 ⎛ tan ⎝ ⎞ ⎠ α 2 Solution Using the given information, we can draw the triangle shown in Figure 9.16. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α = − 8 17 and cos α = − 15 17. Figure 9.16 a. Before we start, we must remember that if α is in quadrant III, α 2 180° 2 < 270° 2. This means that the terminal side of α 2, we begin by writing the half-angle formula for sine. Then we substitute the value of the < To find sin α 2 cosine we found from the triangle in Figure 9.16 and simplify. is in quadrant II, since 90° < < 135°. then 180° < α < 270°, so α 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1017 sin α 2 = ± 1 − cos α 1 − = ± 2 ⎛ ⎝− 15 17 2 ⎞ ⎠ ⋅ 1 2 = ± 32 17 2 = ± 32 17 = ± 16 17 = ± 4 17 = 4 17 17 because the angle terminates in quadrant II and sine is positive in, we will write the half-angle formula for cosine, substitute the value of the cosine we found We choose the positive value of sin α 2 quadrant II. b. To find cos α 2 from the triangle in Figure 9.16, and simplify. cos α 2 = ± 1 + cos α 1 + = ± 2 ⎛ ⎝− 15 17 2 ⎞ ⎠ ⋅ 1 2 = ± 2 17 2 = ± 2 17 = ± 1 17 = − 17 17 because the angle is in quadrant II because cosine is negative in, we write the half-angle formula for tangent. Again, we
substitute the value of the cosine We choose the negative value of cos α 2 quadrant II. c. To find tan α 2 we found from the triangle in Figure 9.16 and simplify. tan α 2 = ± 1 − cos α 1 + cos α 1 − ( − 15 17) 1 + ( − 15 17) = ± = ± 32 17 2 17 = ± 32 2 = − 16 = −4 1018 Chapter 9 Trigonometric Identities and Equations We choose the negative value of tan α 2 II. because α 2 lies in quadrant II, and tangent is negative in quadrant 9.15 Given that sin α = − 4 5 and α lies in quadrant IV, find the exact value of cos ⎛ ⎝ ⎞ ⎠. α 2 Example 9.30 Finding the Measurement of a Half Angle Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for highlevel competition with an angle of θ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tan θ = 5 3 for higher-level competition, what is the measurement of the angle for novice competition? Solution Since the angle for novice competition measures half the steepness of the angle for the high level competition, and tan θ = 5 for high competition, we can find cos θ from the right triangle and the Pythagorean theorem so 3 that we can use the half-angle identities. See Figure 9.17. 32 + 52 = 34 c = 34 Figure 9.17 We see that cos θ = 3 34 = 3 34 34. We can use the half-angle formula for tangent: tan θ 2 = 1 − cos θ 1 + cos θ. Since tan θ is in the first quadrant, so is tan θ 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1019 tan θ 2 = 1 − 3 34 34 1 + 3 34 34 = 34 − 3 34 34 34 + 3 34 34 = 34 − 3 34 34 + 3 34 ≈ 0.57 We can take the inverse tangent to find the angle: tan−1(0.57) ≈ 29.7°. So the angle of the ramp for novice competition is ≈ 29.7°. Access these
online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas. • Double-Angle Identities (http://openstaxcollege.org/l/doubleangiden) • Half-Angle Identities (http://openstaxcollege.org/l/halfangleident) 1020 Chapter 9 Trigonometric Identities and Equations 9.3 EXERCISES Verbal Explain how to determine the reduction identities from 99. the double-angle identity cos(2x) = cos2 x − sin2 x. 100. Explain how to determine the double-angle formula for tan(2x) using the double-angle formulas for cos(2x) and sin(2x). 101. ⎛ tan ⎝ ⎛ by cos ⎝ ⎛ tan ⎝ x 2 ⎛ by dividing the formula for sin ⎝ We can determine the half-angle formula for ⎠ = 1 − cos x x ⎞ ⎞ ⎠ 1 + cos x 2 x 2 ⎞ ⎠ that do not involve any square roots. ⎞ ⎠. Explain how to determine two formulas for x 2 102. ⎛ exercise for tan ⎝ For the half-angle formula given in the previous ⎞ ⎠, explain why dividing by 0 is not a concern. (Hint: examine the values of cos x necessary for the denominator to be 0.) x 2 Algebraic For the following exercises, find the exact values of a) sin(2x), b) cos(2x), and c) tan(2x) without solving for x. 103. If sin x = 1 8, and x is in quadrant I. 104. If cos x = 2 3, and x is in quadrant I. 105. If cos x = − 1 2, and x is in quadrant III. 106. If tan x = −8, and x is in quadrant IV. For the following exercises, find the values of the six trigonometric functions if the conditions provided hold. 107. 108. cos(2θ) = 3 5 and 90° ≤ θ ≤ 180° cos(2θ) = 1 2 and 180° ≤ θ ≤ 270° For the following exercises, simplify to one trigonometric expression. 109. 110. ⎛ 2 sin ⎝ π 4 �
�� ⎛ ⎠ 2 cos ⎝ ⎞ ⎠ π 4 ⎛ 4 sin ⎝ π 8 ⎛ ⎞ ⎠ cos ⎝ ⎞ ⎠ π 8 This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, find the exact value using halfangle formulas. 111. ⎛ sin ⎝ ⎞ ⎠ π 8 112. ⎛ cos ⎝− 11π 12 ⎞ ⎠ 113. ⎛ sin ⎝ 11π 12 ⎞ ⎠ 114. ⎛ cos ⎝ ⎞ ⎠ 7π 8 115. ⎛ tan ⎝ ⎞ ⎠ 5π 12 116. ⎛ tan ⎝− 3π 12 ⎞ ⎠ 117. ⎛ ⎝− 3π tan 8 ⎞ ⎠ For the following exercises, find the exact values of a) ⎛ without solving for x. sin ⎝ ⎞ ⎛ ⎠, and c) tan ⎝ ⎞ ⎛ ⎠, b) cos ⎝ ⎞ ⎠ x 2 x 2 x 2 118. 119. If tan x = − 4 3, and x is in quadrant IV. If sin x = − 12 13, and x is in quadrant III. 120. If csc x = 7, and x is in quadrant II. 121. If sec x = − 4, and x is in quadrant II. For the following exercises, use Figure 9.18 to find the requested half and double angles. Figure 9.18 122. Find sin(2θ), cos(2θ), and tan(2θ). 123. Find sin(2α), cos(2α), and tan(2α). ⎛ Find sin ⎝ θ 2 ⎛ ⎞ ⎠, cos ⎝ θ 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠. θ 2 124. 125. Chapter 9 Trigonometric Identities and Equations 1021 145. sin2 x cos2 x 146. tan2 x sin x 147.
tan4 x cos2 x 148. cos2 x sin(2x) 149. cos2 (2x)sin x 150. tan2 ⎛ ⎝ x 2 ⎞ ⎠ sin x the following exercises, For algebraically find an equivalent function, only in terms of sin x and/or cos x, and then check the answer by graphing both functions. 151. sin(4x) 152. cos(4x) Extensions For the following exercises, prove the identities. 153. sin(2x) = 2 tan x 1 + tan2 x 154. cos(2α) = 1 − tan2 α 1 + tan2 α 155. tan(2x) = 2 sin x cos x 2cos2 x − 1 156. ⎛ ⎞ ⎝sin2 x − 1 ⎠ 2 = cos(2x) + sin4 x 157. 158. 159. sin(3x) = 3 sin x cos2 x − sin3 x cos(3x) = cos3 x − 3sin2 x cos x 1 + cos(2t) sin(2t) − cos t = 2 cos t 2 sin t − 1 160. sin(16x) = 16 sin x cos x cos(2x)cos(4x)cos(8x) 161. cos(16x) = ⎛ ⎞ ⎛ ⎞ ⎝cos2 (4x) − sin2 (4x) − sin(8x) ⎝cos2 (4x) − sin2 (4x) + sin(8x) ⎠ ⎠ ⎛ Find sin ⎝ α 2 ⎞ ⎛ ⎠, cos ⎝ α 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠. α 2 For the following exercises, simplify each expression. Do not evaluate. 126. cos2(28°) − sin2(28°) 127. 2cos2(37°) − 1 128. 1 − 2 sin2(17°) 129. cos2(9x) − sin2(9x) 130. 4 sin(8x) cos(8x) 131. 6 sin(5x) cos(5x) For the following exercises, prove the given identity. 132. (sin t − cos t)2 = 1
− sin(2t) 133. sin(2x) = − 2 sin(−x) cos(−x) 134. cot x − tan x = 2 cot(2x) 135. sin(2θ) 1 + cos(2θ) tan2 θ = tan θ For the following exercises, rewrite the expression with an exponent no higher than 1. 136. cos2(5x) 137. cos2(6x) 138. sin4(8x) 139. sin4(3x) 140. cos2 x sin4 x 141. cos4 x sin2 x 142. tan2 x sin2 x Technology For the following exercises, reduce the equations to powers of one, and then check the answer graphically. 143. tan4 x 144. sin2(2x) 1022 Chapter 9 Trigonometric Identities and Equations 9.4 \| Sum-to-Product and Product-to-Sum Formulas Learning Objectives In this section, you will: 9.4.1 Express products as sums. 9.4.2 Express sums as products. Figure 9.19 The UCLA marching band (credit: Eric Chan, Flickr). A band marches down the field creating an amazing sound that bolsters the crowd. That sound travels as a wave that can be interpreted using trigonometric functions. For example, Figure 9.20 represents a sound wave for the musical note A. In this section, we will investigate trigonometric identities that are the foundation of everyday phenomena such as sound waves. Figure 9.20 Expressing Products as Sums We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity. Expressing Products as Sums for Cosine We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1023 cos α cos β + sin α sin β = cos(α − β) + cos α cos β − sin α sin β = cos(α
+ β) ___________________________________ 2 cos α cos β = cos(α − β) + cos(α + β) Then, we divide by 2 to isolate the product of cosines: cos α cos β = 1 2 [cos(α − β) + cos(α + β)] Given a product of cosines, express as a sum. 1. Write the formula for the product of cosines. 2. Substitute the given angles into the formula. 3. Simplify. Example 9.31 Writing the Product as a Sum Using the Product-to-Sum Formula for Cosine ⎛ Write the following product of cosines as a sum: 2 cos ⎝ 7x 2 ⎠ cos 3x ⎞ 2. Solution We begin by writing the formula for the product of cosines: cos α cos β = 1 2 ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ + cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ We can then substitute the given angles into the formula and simplify. ⎡ − 3x 7x ⎞ ⎛ 1 ⎣cos ⎠ ⎝ 2 2 2 ⎡ 10x 4x ⎞ ⎛ ⎛ ⎠ + cos = ⎣cos ⎝ ⎝ 2 2 = cos 2x + cos 5x ⎞ ⎛ ⎠ = (2) ⎝ ⎛ 2 cos ⎝ ⎞ ⎛ ⎠cos ⎝ 3x 2 7x 2 ⎞ ⎛ ⎠) + cos ⎝ ⎤ ⎞ ⎦ ⎠ 7x 2 + 3x 2 ⎤ ⎞ ⎦ ⎠ 9.16 Use the product-to-sum formula to write the product as a sum or difference: cos(2θ)cos(4θ). Expressing the Product of Sine and Cosine as a Sum Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine. If we add the sum and difference identities, we get: sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β +
_________________________________________ sin(α + β) + sin(α − β) = 2 sin α cos β Then, we divide by 2 to isolate the product of cosine and sine: sin α cos β = 1 2 ⎡ ⎣sin⎛ ⎝α + β⎞ ⎠ + sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ 1024 Chapter 9 Trigonometric Identities and Equations Example 9.32 Writing the Product as a Sum Containing only Sine or Cosine Express the following product as a sum containing only sine or cosine and no products: sin(4θ)cos(2θ). Solution Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify. sin α cos β = 1 2 sin(4θ)cos(2θ) = 1 2 = 1 2 [sin(α + β) + sin(α − β)] [sin(4θ + 2θ) + sin(4θ − 2θ)] [sin(6θ) + sin(2θ)] 9.17 Use the product-to-sum formula to write the product as a sum: sin(x + y)cos(x − y). Expressing Products of Sines in Terms of Cosine Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas: ⎠ = cos α cos β + sin α sin β cos⎛ − cos⎛ ⎝cos α cos β − sin α sin β⎞ ⎠ = − ⎛ ____________________________________________________ ⎠ = 2 sin α sin β ⎠ − cos⎛ cos⎛ ⎝α − β⎞ ⎝α − β⎞ ⎝α + β⎞ ⎝α + β⎞ ⎠ Then, we divide by 2 to isolate the product of sines: sin α sin β = 1 2 ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ − cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ Similarly we could express
the product of cosines in terms of sine or derive other product-to-sum formulas. The Product-to-Sum Formulas The product-to-sum formulas are as follows: [cos(α − β) + cos(α + β)] cos α cos β = 1 2 sin α cos β = 1 [sin(α + β) + sin(α − β)] 2 sin α sin β = 1 2 cos α sin β = 1 2 [sin(α + β) − sin(α − β)] [cos(α − β) − cos(α + β)] (9.33) (9.34) (9.35) (9.36) Example 9.33 Express the Product as a Sum or Difference This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1025 Write cos(3θ) cos(5θ) as a sum or difference. Solution We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify. cos α cos β = 1 2 cos(3θ)cos(5θ) = 1 2 = 1 2 [cos(α − β) + cos(α + β)] [cos(3θ − 5θ) + cos(3θ + 5θ)] [cos(2θ) + cos(8θ)] Use even-odd identity. 9.18 Use the product-to-sum formula to evaluate cos 11π 12 cos π 12. Expressing Sums as Products Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine. Let u + v = α and u − v = β. 2 2 Then = 2u 2 = u u + v 2 = 2v 2 = v Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have [sin(α + β) + sin(α − β)] sin α cos β = 1 2 ⎞ ⎠ = 1 2 ⎞ ⎠ = sin u + sin v �
�� ⎞ ⎠cos ⎝ ⎛ ⎞ ⎠cos ⎝ u − v 2 u − v 2 [sin u + sin v] ⎛ sin ⎝ ⎛ 2 sin ⎝ u + v 2 u + v 2 Substitute for(α + β) and (α − β) The other sum-to-product identities are derived similarly. Sum-to-Product Formulas The sum-to-product formulas are as follows: ⎛ sin α − sin β = 2sin ⎝ α + β ⎛ sin α + sin β = 2sin ⎝ 2 α − β 2 α + β 2 ⎛ cos α − cos β = −2sin ⎝ α − β ⎛ ⎞ ⎠cos ⎝ 2 α + β ⎞ ⎞ ⎛ ⎠cos ⎠ ⎝ 2 α − β ⎞ ⎛ ⎠sin ⎝ 2 ⎞ ⎠ ⎞ ⎠ (9.37) (9.38) (9.39) 1026 Chapter 9 Trigonometric Identities and Equations ⎛ cos α + cos β = 2cos ⎝ α + β 2 ⎛ ⎞ ⎠cos ⎝ α − β 2 ⎞ ⎠ (9.40) Example 9.34 Writing the Difference of Sines as a Product Write the following difference of sines expression as a product: sin(4θ) − sin(2θ). Solution We begin by writing the formula for the difference of sines. ⎛ sin α − sin β = 2sin ⎝ α − β 2 ⎛ ⎞ ⎠cos ⎝ α + β 2 ⎞ ⎠ Substitute the values into the formula, and simplify. ⎛ sin(4θ) − sin(2θ) = 2sin ⎝ 4θ − 2θ ⎛ ⎞ ⎠ cos ⎝ 2 6θ 2θ ⎞ ⎞ ⎛ ⎛ ⎠ cos = 2sin ⎠ ⎝ ⎝ 2 2 = 2 sin θ cos(3θ) 4θ + 2θ 2 ⎞ �
� 9.19 Use the sum-to-product formula to write the sum as a product: sin(3θ) + sin(θ). Example 9.35 Evaluating Using the Sum-to-Product Formula Evaluate cos(15°) − cos(75°). Check the answer with a graphing calculator. Solution We begin by writing the formula for the difference of cosines. ⎛ cos α − cos β = − 2 sin ⎝ α + β 2 ⎛ ⎞ sin ⎝ ⎠ α − β 2 ⎞ ⎠ Then we substitute the given angles and simplify. ⎛ cos(15°) − cos(75°) = −2sin ⎝ 15° + 75° 2 ⎞ ⎛ ⎠ sin ⎝ = −2sin(45°) sin(−30°) ⎞ ⎛ ⎝− 1 ⎠ 2 ⎛ = −2 ⎝ 2 2 ⎞ ⎠ 15° − 75° 2 ⎞ ⎠ = 2 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1027 Example 9.36 Proving an Identity Prove the identity: Solution cos(4t) − cos(2t) sin(4t) + sin(2t) = − tan t We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side. cos(4t) − cos(2t) sin(4t) + sin(2t) = = = ⎛ ⎞ 4t − 2t ⎠ sin ⎝ 2 ⎛ ⎞ 4t − 2t ⎠ cos ⎝ 2 ⎞ ⎠ ⎞ ⎠ 4t + 2t ⎛ −2 sin ⎝ 2 4t + 2t ⎛ 2 sin ⎝ 2 −2 sin(3t)sin t 2 sin(3t)cos t − 2 sin(3t)sin t 2 sin(3t)cos t = − sin t cos t = −tan t Analysis Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the
procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side. Example 9.37 Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities Verify the identity csc2 θ − 2 = cos(2θ) sin2 θ. Solution For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side. cos(2θ) sin2 θ = 1 − 2 sin2 θ sin2 θ 1 sin2 θ = csc2 θ − 2 = − 2 sin2 θ sin2 θ 9.20 Verify the identity tan θ cot θ − cos2 θ = sin2 θ. 1028 Chapter 9 Trigonometric Identities and Equations Access these online resources for additional instruction and practice with the product-to-sum and sum-to-product identities. • Sum to Product Identities (http://openstaxcollege.org/l/sumtoprod) • Sum to Product and Product to Sum Identities (http://openstaxcollege.org/l/sumtpptsum) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1029 9.4 EXERCISES Verbal Starting with 162. sin α cos β = 1 2 the product [sin(α + β) + sin(α − β)], explain to sum formula how to determine the formula for cos α sin β. Provide 163. calculating two different methods of cos(195°)cos(105°), one of which uses the product to sum. Which method is easier? Describe a situation where we would convert an 164. equation from a sum to a product and give an example. cos(45°)sin(15°) 180. sin(−345°)sin(−15°) 181. sin(195°)cos(15°) 182. sin(−45°)sin(−15°) For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine. 183. cos(23°)sin(17°)
Describe a situation where we would convert an 165. equation from a product to a sum, and give an example. 184. 2 sin(100°)sin(20°) Algebraic 185. 2 sin(−100°)sin(−20°) For the following exercises, rewrite the product as a sum or difference. 186. sin(213°)cos(8°) 166. 16 sin(16x)sin(11x) 167. 20 cos(36t)cos(6t) 168. 2 sin(5x)cos(3x) 169. 10 cos(5x)sin(10x) 170. sin(−x)sin(5x) 171. sin(3x)cos(5x) For the following exercises, rewrite the sum or difference as a product. 172. cos(6t) + cos(4t) 173. sin(3x) + sin(7x) 174. cos(7x) + cos(−7x) 175. sin(3x) − sin(−3x) 176. cos(3x) + cos(9x) 177. sin h − sin(3h) For the following exercises, evaluate the product for the following using a sum or difference of two functions. Evaluate exactly. 178. cos(45°)cos(15°) 179. 187. 2 cos(56°)cos(47°) For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine. 188. 189. 190. 191. 192. sin(76°) + sin(14°) cos(58°) − cos(12°) sin(101°) − sin(32°) cos(100°) + cos(200°) sin(−1°) + sin(−2°) For the following exercises, prove the identity. 193. cos(a + b) cos(a − b) = 1 − tan a tan b 1 + tan a tan b 194. 4 sin(3x)cos(4x) = 2 sin(7x) − 2 sinx 195. 6 cos(8x)sin(2x) sin(−6x) = −3 sin(10x)csc(6x) + 3 196. 197. 198. sin x + sin(3x) = 4 sin x cos2 x ⎝cos
3 x − cos x sin2 x⎞ ⎛ ⎠ = cos(3x) + cos x 2 2 tan x cos(3x) = sec x⎛ ⎝sin(4x) − sin(2x)⎞ ⎠ 199. cos(a + b) + cos(a − b) = 2 cos a cos b Chapter 9 Trigonometric Identities and Equations Extensions For the following exercises, prove the following sum-toproduct formulas. 215. ⎛ sin x − sin y = 2 sin ⎝ x − y 2 ⎛ ⎞ ⎠cos ⎝ x + y 2 ⎞ ⎠ 216. ⎛ cos x + cos y = 2 cos ⎝ x + y 2 ⎞ ⎛ ⎠cos ⎝ x − y 2 ⎞ ⎠ For the following exercises, prove the identity. 217. sin(6x) + sin(4x) sin(6x) − sin(4x) = tan (5x)cot x 218. cos(3x) + cos x cos(3x) − cos x = − cot (2x)cot x 219. cos(6y) + cos(8y) sin(6y) − sin(4y) = cot y cos (7y)sec (5y) 220. cos⎛ sin⎛ ⎝2y⎞ ⎝2y⎞ ⎠ − cos⎛ ⎠ + sin⎛ ⎝4y⎞ ⎝4y⎞ ⎠ ⎠ = tan y 221. sin(10x) − sin(2x) cos(10x) + cos(2x) = tan(4x) 222. cos x − cos(3x) = 4 sin2 xcos x 223. (cos(2x) − cos(4x))2 + (sin(4x) + sin(2x))2 = 4 sin2(3x) 224. ⎛ tan ⎝ π 4 − t⎞ ⎠ = 1 − tan t 1 + tan t 1030 Numeric For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions.
Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 200. cos(58°) + cos(12°) 201. sin(2°) − sin(3°) 202. cos(44°) − cos(22°) 203. cos(176°)sin(9°) 204. sin(−14°)sin(85°) Technology the following exercises, algebraically determine For whether each of the given equation is an identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator. 205. 2 sin(2x)sin(3x) = cos x − cos(5x) 206. cos(10θ) + cos(6θ) cos(6θ) − cos(10θ) = cot(2θ)cot(8θ) 207. sin(3x) − sin(5x) cos(3x) + cos(5x) = tan x 208. 2 cos(2x)cos x + sin(2x)sin x = 2 sin x 209. sin(2x) + sin(4x) sin(2x) − sin(4x) = − tan(3x)cot x For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical. 210. sin(9t) − sin(3t) cos(9t) + cos(3t) 211. 2 sin(8x)cos(6x) − sin(2x) 212. sin(3x) − sin x sin x 213. cos(5x) + cos(3x) sin(5x) + sin(3x) 214. sin x cos(15x) − cos x sin(15x) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1031 9.5 \| Solving Trigonometric Equations Learning Objectives In this section, you will: 9.5.1 Solve linear trigonometric equations in sine and cosine. 9.5.2 Solve equations involving a single trigonometric function. 9.5.3 Solve trig
onometric equations using a calculator. 9.5.4 Solve trigonometric equations that are quadratic in form. 9.5.5 Solve trigonometric equations using fundamental identities. 9.5.6 Solve trigonometric equations with multiple angles. 9.5.7 Solve right triangle problems. Figure 9.21 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill) Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. Solving Linear Trigonometric Equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π. In other words, every 2π units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2π: sin θ = sin(θ ± 2kπ) There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving
trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with 1032 Chapter 9 Trigonometric Identities and Equations a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. Example 9.38 Solving a Linear Trigonometric Equation Involving the Cosine Function Find all possible exact solutions for the equation cos θ = 1 2. Solution From the unit circle, we know that cos θ = 1 2 π 3 θ =, 5π 3 These are the solutions in the interval [0, 2π]. All possible solutions are given by θ = π 3 ± 2kπ and θ = 5π 3 ± 2kπ where k is an integer. Example 9.39 Solving a Linear Equation Involving the Sine Function Find all possible exact solutions for the equation sin t = 1 2. Solution Solving for all possible values of t means that solutions include angles beyond the period of 2π. From Figure. But the problem is asking for all possible values that 9.7, we can see that the solutions are t = π 6 solve the equation. Therefore, the answer is and t = 5π 6 where k is an integer. t = π 6 ± 2πk and t = 5π 6 ± 2πk Given a trigonometric equation, solve using algebra. 1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. 2. Substitute the trigonometric expression with a single variable, such as x or u. 3. Solve the equation the same way an algebraic equation would be solved. 4. Substitute the trigonometric expression back in for the variable in the resulting expressions. 5. Solve for the angle. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1033 Example 9.40 Solve the Linear Trigonometric Equation Solve the equation exactly: 2 cos θ − 3 = − 5, 0 ≤ θ < 2π. Solution Use algebraic techniques to solve the equation. 2 cos θ − 3 = −5 2 cos
θ = −2 cos θ = −1 θ = π 9.21 Solve exactly the following linear equation on the interval [0, 2π) : 2 sin x + 1 = 0. Solving Equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 9.7). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π, not 2π. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π, unless, of course, a problem places its own restrictions on the 2 domain. Example 9.41 Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: 2 sin2 θ − 1 = 0, 0 ≤ θ < 2π. Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ. Then we will find the angles. 2 sin2 θ − 1 = 0 2 sin2 θ = 1 sin2 θ = 1 2 sin2 θ = ± 1 2 sin θ = ± 1 2, 3π 4 θ = π 4 = ± 2 2, 7π 4, 5π 4 1034 Chapter 9 Trigonometric Identities and Equations Example 9.42 Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: csc θ = − 2, 0 ≤ θ < 4π. Solution We want all values of θ for which csc θ = − 2 over the interval 0 ≤ θ < 4π. csc θ = −2 1 sin θ = −2 sin θ = − 1 2, 11π θ = 7π 6 6, 19π 6, 23π 6 Analysis As sin θ = − 1 2, notice that all four solutions are in the
third and fourth quadrants. Example 9.43 Solving an Equation Involving Tangent ⎛ Solve the equation exactly: tan ⎝θ − ⎞ ⎠ = 1, 0 ≤ θ < 2π. π 2 Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of π 4, the tangent has a value of 1. However, the angle we want is ⎛ ⎝θ − θ − ⎞ ⎛ ⎠. Thus, if tan ⎝ ⎞ ⎠ = 1, π 4 then π 2 π π = 2 4 θ = 3π 4 ± kπ Over the interval [0, 2π), we have two solutions: θ = 3π 4 and θ = 3π 4 + π = 7π 4 9.22 Find all solutions for tan x = 3. Example 9.44 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1035 Identify all Solutions to the Equation Involving Tangent Identify all exact solutions to the equation 2(tan x + 3) = 5 + tan x, 0 ≤ x < 2π. Solution We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign. 2(tan x) + 2(3) = 5 + tan x 2tan x + 6 = 5 + tan x 2tan x − tan x = 5 − 6 tan x = −1 There are two angles on the unit circle that have a tangent value of −1:θ = 3π 4 and θ = 7π 4. Solve Trigonometric Equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example 9.45 Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation sin θ = 0.8, where θ is in radians. Solution Make sure mode is set to radians
. To find θ, use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the sin−1 function. What is shown on the screen is sin−1(. The calculator is ready for the input within the parentheses. For this problem, we enter sin−1 (0.8), and press ENTER. Thus, to four decimals places, The solution is The angle measurement in degrees is sin−1(0.8) ≈ 0.9273 θ ≈ 0.9273 ± 2πk θ ≈ 53.1° θ ≈ 180° − 53.1° ≈ 126.9° Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ. 1036 Chapter 9 Trigonometric Identities and Equations Example 9.46 Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation sec θ = −4, giving your answer in radians. Solution We can begin with some algebra. sec θ = −4 1 cos θ = −4 cos θ = − 1 4 Check that the MODE is in radians. Now use the inverse cosine function. cos−1 ⎛ ⎞ ⎝− 1 ⎠ ≈ 1.8235 4 θ ≈ 1.8235 + 2πk Since π 2 ≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 9.22. Figure 9.22 So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is θ'≈ π − 1.8235 ≈ 1.3181. The other solution in quadrant III is θ'≈ π + 1.3181 ≈ 4.4597. The solutions are θ ≈ 1.8235
± 2πk and θ ≈ 4.4597 ± 2πk. 9.23 Solve cos θ = − 0.2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1037 Solving Trigonometric Equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example 9.47 Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: cos2 θ + 3 cos θ − 1 = 0, 0 ≤ θ < 2π. Solution We begin by using substitution and replacing cos θ with x. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x. We have The equation cannot be factored, so we will use the quadratic formula x = −b ± b2 − 4ac 2a. x2 + 3x − 1 = 0 Replace x with cos θ, and solve. x = −3 ± ( − 3)2 − 4(1)( − 1) 2 = −3 ± 13 2 cos θ = −3 ± 13 2 θ = cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ Note that only the + sign is used. This is because we get an error when we solve θ = cos−1 ⎛ ⎝ ⎞ on a ⎠ calculator, since the domain of the inverse cosine function is [−1, 1]. However, there is a second solution: −3 − 13 2 θ = cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ ≈ 1.26 This terminal side
of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is θ = 2π − cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ ≈ 5.02 Example 9.48 1038 Chapter 9 Trigonometric Identities and Equations Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: 2 sin2 θ − 5 sin θ + 3 = 0, 0 ≤ θ ≤ 2π. Solution Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u, or imagine it, as we factor: Now set each factor equal to zero. 2 sin2 θ − 5 sin θ + 3 = 0 (2 sin θ − 3)(sin θ − 1) = 0 2 sin θ − 3 = 0 2 sin θ = 3 sin θ = 3 2 sin θ − 1 = 0 sin θ = 1 Next solve for θ : sin θ ≠ 3 2 π 2 solution θ =., as the range of the sine function is [−1, 1]. However, sin θ = 1, giving the Analysis Make sure to check all solutions on the given domain as some factors have no solution. 9.24 Solve sin2 θ = 2 cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 9.49 Solving a Trigonometric Equation Using Algebra Solve exactly: 2 sin2 θ + sin θ = 0; 0 ≤ θ < 2π Solution This problem should appear familiar as it is similar to a quadratic. Let sin θ = x. The equation becomes 2x2 + x = 0. We begin by factoring: Set each factor equal to zero. 2x2 + x = 0 x(2x + 1) = 0 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1039 x = 0 (2x + 1) = 0 x = − 1 2 Then, substitute back into the equation the original expression sin θ for x. Thus, sin θ = 0 θ = 0, π sin �
� = − 1 2, 11π θ = 7π 6 6, 11π 6 The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, 7π 6. If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. 2 sin2 θ + sin θ = 0 sin θ(2sin θ + 1) = 0 sin θ = 0 θ = 0, π 2 sin θ + 1 = 0 2sin θ = −1 sin θ = − 1 2, 11π θ = 7π 6 6 Analysis We can see the solutions on the graph in Figure 9.23. On the interval 0 ≤ θ < 2π, four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. the graph crosses the x-axis Figure 9.23 We can verify the solutions on the unit circle in Figure 9.7 as well. Example 9.50 1040 Chapter 9 Trigonometric Identities and Equations Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: 2 sin2 θ − 3 sin θ + 1 = 0, 0 ≤ θ < 2π. Solution We can factor using grouping. Solution values of θ can be found on the unit circle. (2 sin θ − 1)(sin θ − 1) = 0 2 sin θ − 1 = 0 sin θ = 1 2 π 6 sin θ = 1 π θ = 2 θ =, 5π 6 9.25 Solve the quadratic equation 2 cos2 θ + cos θ = 0. Solving Trigonometric Equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify
the equation. Example 9.51 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2π. cos x cos(2x) + sin x sin(2x) = 3 2 Solution Notice that the left side of the equation is the difference formula for cosine. cos x cos(2x) + sin x sin(2x) = 3 2 cos(x − 2x) = 3 2 cos( − x) = 3 2 cos x = 3 2 Diffe ence formula for cosine Use the negative angle identity. From the unit circle in Figure 9.7, we see that cos x = 3 2 when x = π 6, 11π 6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1041 Example 9.52 Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: cos(2 θ) = cos θ. Solution We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: cos(2θ) = cos θ 2cos2 θ − 1 = cos θ 2 cos2 θ − cos θ − 1 = 0 (2 cos θ + 1)(cos θ − 1) = 0 2 cos θ + 1 = 0 cos θ = − 1 2 cos θ − 1 = 0 cos θ = 1 So, if cos θ = − 1 2, then θ = 2π 3 ± 2πk and θ = 4π 3 ± 2πk; if cos θ = 1, then θ = 0 ± 2πk. Example 9.53 Solving an Equation Using an Identity Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin2 θ, 0 ≤ θ < 2π. Solution If we rewrite the right side, we can write the equation in terms of cosine: 3 cos θ + 3 = 2sin2 θ ⎝1 − cos2 θ⎞ ⎛ 3 cos θ + 3 = 2 ⎠ 3 cos θ + 3 = 2
− 2cos2 θ 2 cos2 θ + 3 cos θ + 1 = 0 (2 cos θ + 1)(cos θ + 1) = 0 2 cos θ + 1 = 0 cos θ = − 1 2, 4π θ = 2π 3 3 cos θ + 1 = 0 cos θ = −1 θ = π Our solutions are θ = 2π 3, 4π 3, π. 1042 Chapter 9 Trigonometric Identities and Equations Solving Trigonometric Equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x) or cos(3x). When confronted with these equations, recall that y = sin(2x) is a horizontal compression by a factor of 2 of the function y = sin x. On an interval of 2π, we can graph two periods of y = sin(2x), as opposed to one cycle of y = sin x. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x) = 0 compared to sin x = 0. This information will help us solve the equation. Example 9.54 Solving a Multiple Angle Trigonometric Equation Solve exactly: cos(2x) = 1 2 on [0, 2π). Solution We can see that this equation is the standard equation with a multiple of an angle. If cos(α) = 1 2, we know α is in quadrants I and IV. While θ = cos−1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 will be in quadrants I and IV. Therefore, the possible angles are θ = π 3 x = 5π 6 ⎛. Does this make sense? Yes, because cos ⎝2 and θ = 5π 3 ⎛ π ⎝ 6. So, 2x = ⎞ ⎞ ⎛ ⎠ = cos ⎝ ⎠ π 3 or 2x = 5π 3, which means that x = or π 6 π 3 ⎞ ⎠ = 1 2. Are there any other possible answers? Let us return to our first step. π 6 In quadrant I, 2x =,
so x = π 3 as noted. Let us revolve around the circle again: 2x = + 2π + 6π 3 π 3 π = 3 = 7π 3 so x = 7π 6. One more rotation yields 2x = + 4π + 12π 3 π 3 π = 3 = 13π 3 x = 13π 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). In quadrant IV, 2x = 5π 3, so x = 5π 6 as noted. Let us revolve around the circle again: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1043 so x = 11π 6. One more rotation yields + 2π + 6π 3 2x = 5π 3 = 5π 3 = 11π 3 + 4π + 12π 3 2x = 5π 3 = 5π 3 = 17π 3 x = 17π 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). Our solutions are x =, 5π 6 sin(nx) = c, we must go around the unit circle n times., and 11π 6, 7π 6 π 6. Note that whenever we solve a problem in the form of Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2 + b2 = c2, and model an equation to fit a situation. 1044 Chapter 9 Trigonometric Identities and Equations Example 9.55 Using the Pythagorean Theorem to Model an Equation Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 9.24.
Figure 9.24 Solution Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem. a2 + b2 = c2 (23)2 + (69.5)2 ≈ 5359 5359 ≈ 73.2 m The angle of elevation is θ, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places. tan θ = 69.5 23 tan−1 ⎛ ⎝ 69.5 23 ⎞ ⎠ ≈ 1.2522 ≈ 71.69° The angle of elevation is approximately 71.7°, and the length of the cable is 73.2 meters. Example 9.56 Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1045 Solution For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet. See Figure 9.25. Figure 9.25 The side adjacent to θ is a and the hypotenuse is 4a. Thus, cos θ = a 4a = 1 4 ⎞ ⎠ ≈ 75.5° cos−1 ⎛ ⎝ 1 4 The elevation of the ladder forms an angle of 75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: a2 + b2 = (4a)2 b2 = (4a)2 − a2 b2 = 16a2 − a2 b2 = 15a2 b = a 15 Thus, the ladder touches the wall at a 15 feet from the ground. Access these online resources for additional instruction and practice with solving trigonometric equations. • Solving Trigonometric Equ
ations I (http://openstaxcollege.org/l/solvetrigeqI) • Solving Trigonometric Equations II (http://openstaxcollege.org/l/solvetrigeqII) • Solving Trigonometric Equations III (http://openstaxcollege.org/l/solvetrigeqIII) • Solving Trigonometric Equations IV (http://openstaxcollege.org/l/solvetrigeqIV) • Solving Trigonometric Equations V (http://openstaxcollege.org/l/solvetrigeqV) • Solving Trigonometric Equations VI (http://openstaxcollege.org/l/solvetrigeqVI) 1046 Chapter 9 Trigonometric Identities and Equations 9.5 EXERCISES Verbal Will 225. there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not. 226. When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not? 227. When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions? Algebraic For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2π. 228. 2 sin θ = − 2 229. 2 sin θ = 3 230. 2 cos θ = 1 231. 2 cos θ = − 2 232. tan θ = −1 233. tan x = 1 234. cot x + 1 = 0 235. 4 sin2 x − 2 = 0 236. csc2 x − 4 = 0 cos(2θ) = − 3 2 245. 2 sin(πθ) = 1 246. ⎛ 2 cos ⎝ π 5 θ⎞ ⎠ = 3 For the following exercises, find all exact solutions on [0, 2π). 247. sec(x)sin(x) − 2 sin(x) = 0 248. tan(x) − 2 sin(x)tan(x) = 0 249. 2 cos2 t + cos(t) = 1 250. 2 tan2(t) = 3 sec(t) 251.
2 sin(x)cos(x) − sin(x) + 2 cos(x) − 1 = 0 252. cos2 θ = 1 2 253. sec2 x = 1 254. 255. tan2 (x) = −1 + 2 tan(−x) 8 sin2(x) + 6 sin(x) + 1 = 0 256. tan5(x) = tan(x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π). For the following exercises, solve exactly on [0, 2π). 257. sin(3x)cos(6x) − cos(3x)sin(6x) = −0.9 237. 2 cos θ = 2 238. 2 cos θ = −1 239. 2 sin θ = −1 240. 2 sin θ = − 3 241. 2 sin(3θ) = 1 242. 2 sin(2θ) = 3 243. 2 cos(3θ) = − 2 244. This content is available for free at https://cnx.org/content/col11758/1.5 258. sin(6x)cos(11x) − cos(6x)sin(11x) = −0.1 259. cos(2x)cos x + sin(2x)sin x = 1 260. 6 sin(2t) + 9 sin t = 0 261. 9 cos(2θ) = 9 cos2 θ − 4 262. sin(2t) = cos t 263. cos(2t) = sin t 264. cos(6x) − cos(3x) = 0 Chapter 9 Trigonometric Identities and Equations 1047 For the following exercises, solve exactly on the interval [0, 2π). Use the quadratic formula if the equations do not factor. 265. 266. 267. 268. 269. 270. 271. 272. 273. tan2 x − 3 tan x = 0 sin2 x + sin x − 2 = 0 sin2 x − 2 sin x − 4 = 0 5 cos2 x + 3 cos x − 1 = 0 3 cos2 x − 2 cos x − 2 = 0 5 sin2 x + 2 sin x − 1 = 0 tan2 x + 5tan x − 1 = 0 cot2 x = − cot x −tan2 x − tan x
− 2 = 0 For the following exercises, find exact solutions on the interval [0, 2π). Look use trigonometric identities. opportunities for to 274. 275. sin2 x − cos2 x − sin x = 0 sin2 x + cos2 x = 0 276. sin(2x) − sin x = 0 277. cos(2x) − cos x = 0 278. 2 tan x 2 − sec2 x − sin2 x = cos2 x 279. 1 − cos(2x) = 1 + cos(2x) 280. sec2 x = 7 281. 10 sin x cos x = 6 cos x 282. −3 sin t = 15 cos t sin t 4 cos2 x − 4 = 15 cos x 8 sin2 x + 6 sin x + 1 = 0 8 cos2 θ = 3 − 2 cos θ 6 cos2 x + 7 sin x − 8 = 0 283. 284. 285. 286. 287. 12 sin2 t + cos t − 6 = 0 288. tan x = 3 sin x 289. cos3 t = cos t Graphical For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros. 290. 291. 292. 293. 294. 295. 296. 6 sin2 x − 5 sin x + 1 = 0 8 cos2 x − 2 cos x − 1 = 0 100 tan2 x + 20 tan x − 3 = 0 2 cos2 x − cos x + 15 = 0 20 sin2 x − 27 sin x + 7 = 0 2 tan2 x + 7 tan x + 6 = 0 130 tan2 x + 69 tan x − 130 = 0 Technology For the following exercises, use a calculator to find all solutions to four decimal places. 297. sin x = 0.27 298. sin x = −0.55 299. tan x = −0.34 300. cos x = 0.71 the following For equations exercises, algebraically, and then use a calculator to find the values on the interval [0, 2π). Round to four decimal places. solve the 301. 302. 303. 304. 305. 306. tan2 x + 3 tan x − 3 = 0 6 tan2 x + 13 tan x = −6 tan2 x − sec x = 1 sin2 x − 2 cos2 x = 0 2 tan2 x + 9 tan x − 6 =
0 4 sin2 x + sin(2x)sec x − 3 = 0 Chapter 9 Trigonometric Identities and Equations A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him? A 20-foot tall building has a shadow that is 55 feet 324. long. What is the angle of elevation of the sun? A 90-foot tall building has a shadow that is 2 feet 325. long. What is the angle of elevation of the sun? A spotlight on the ground 3 meters from a 2-meter tall 326. man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light? A spotlight on the ground 3 feet from a 5-foot tall 327. woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light? For the following exercises, find a solution to the following word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree. A person does a handstand with his feet touching a 328. wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall? A person does a handstand with her feet touching a 329. wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall? A 23-foot ladder is positioned next to a house. If the 330. ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping? 1048 Extensions For the following exercises, find all solutions exactly to the equations on the interval [0, 2π). 307. 308. 309. 310. 311. 312. 313. 314. 315. 316. csc2 x − 3 csc x − 4 = 0 sin2 x − cos2 x − 1 = 0 sin2 x⎛ ⎝1 − sin2 x⎞ ⎠ + cos2 x⎛ ⎝1 − sin2 x⎞ ⎠ = 0 3 sec2 x + 2 + sin2 x − tan2 x + cos2 x = 0 sin2 x − 1 + 2 cos(2x) − cos2 x = 1 tan
2 x − 1 − sec3 x cos x = 0 sin(2x) sec2 x = 0 sin(2x) 2csc2 x = 0 2 cos2 x − sin2 x − cos x − 5 = 0 1 sec2 x + 2 + sin2 x + 4 cos2 x = 4 Real-World Applications An airplane has only enough gas to fly to a city 200 317. miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly? If a loading ramp is placed next to a truck, at a height 318. of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground? If a loading ramp is placed next to a truck, at a height 319. of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground? A woman is watching a launched rocket currently 11 320. miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal? An astronaut is in a launched rocket currently 15 321. miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.) A woman is standing 8 meters away from a 10-meter 322. tall building. At what angle is she looking to the top of the building? 323. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1049 CHAPTER 9 REVIEW KEY TERMS double-angle formulas equal identities derived from the sum formulas for sine, cosine, and tangent in which the angles are even-odd identities set of equations involving trigonometric functions such that if f (−x) = − f (x), the identity is odd, and if f (−x) = f (x), the identity is even half-angle formulas identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions product-to-sum formula a trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions Pythagorean identities set of equations involving trigonometric functions based on the right triangle properties quotient identities pair of identities based on the fact that tangent is
the ratio of sine and cosine, and cotangent is the ratio of cosine and sine reciprocal identities set of equations involving the reciprocals of basic trigonometric definitions reduction formulas function identities derived from the double-angle formulas and used to reduce the power of a trigonometric sum-to-product formula a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions KEY EQUATIONS 1050 Chapter 9 Trigonometric Identities and Equations Pythagorean identities Even-odd identities Reciprocal identities Quotient identities cos2 θ + sin2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ tan( − θ) = −tan θ cot( − θ) = −cot θ sin( − θ) = −sin θ csc( − θ) = −csc θ cos( − θ) = cos θ sec( − θ) = sec θ sin θ = cos θ = tan θ = csc θ = sec θ = cot θ = 1 csc θ 1 sec θ 1 cot θ 1 sin θ 1 cos θ 1 tan θ tan θ = sin θ cos θ cot θ = cos θ sin θ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1051 Sum Formula for Cosine Difference Formula for Cosine Sum Formula for Sine Difference Formula for Sine Sum Formula for Tangent Difference Formula for Tangent Cofunction identities cos⎛ ⎝α + β⎞ ⎠ = cos α cos β − sin αsin β cos⎛ ⎝α − β⎞ ⎠ = cos α cos β + sin α sin β sin⎛ ⎝α + β⎞ ⎠ = sin α cos β + cos α sin β sin⎛ ⎝α − β⎞ ⎠ = sin α cos β − cos α sin β tan⎛ ⎝α + β⎞ ⎠ = tan⎛ ⎝α − β⎞ ⎠ = tan α +
tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β tan θ = cot ⎛ cos θ = sin ⎝ π ⎛ sin θ = cos ⎝ ⎛ sec θ = csc ⎝ 2 π 2 ⎛ csc θ = sec ⎝ ⎛ cot θ = tan ⎝ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ 1052 Chapter 9 Trigonometric Identities and Equations Double-angle formulas Reduction formulas sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2sin2 θ = 2cos2 θ − 1 2tan θ 1 − tan2 θ tan(2θ) = sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) sin α 2 cos α 2 tan α 2 Half-angle formulas 2 2 = ± 1 − cos α = ± 1 + cos α = ± 1 − cos α 1 + cos α sin α = 1 + cos α = 1 − cos α sin α Product-to-sum Formulas cos α cos β = 1 2 sin α cos β = 1 2 sin α sin β = 1 2 cos α sin β = 1 2 [cos(α − β) + cos(α + β)] [sin(α + β) + sin(α − β)] [cos(α − β) − cos(α + β)] [sin(α + β) − sin(α − β)] Sum-to-product Formulas ⎛ sin α + sin β = 2 sin ⎝ ⎛ sin α − sin β = 2 sin ⎝ ⎛ cos α − cos β = −2 sin ⎝ 2 α + β 2 ⎛ cos α + cos β = 2 cos ⎝ α − β ⎛ ⎞ ⎞ ⎠cos ⎝ ⎠ 2 α + β �
�� ⎛ ⎞ ⎠cos ⎠ ⎝ 2 α − β ⎛ ⎞ ⎠sin ⎝ 2 α − β ⎛ ⎞ ⎠cos ⎝ 2 ⎞ ⎠ ⎞ ⎠ KEY CONCEPTS 9.1 Solving Trigonometric Equations with Identities • There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem. • Graphing both sides of an identity will verify it. See Example 9.1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1053 • Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example 9.2 and Example 9.3. • The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example 9.4. • We can create an identity and then verify it. See Example 9.5. • Verifying an identity may involve algebra with the fundamental identities. See Example 9.6 and Example 9.7. • Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example 9.8, Example 9.9, and Example 9.10. 9.2 Sum and Difference Identities • The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles. • The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See Example 9.11 and Example 9.12. • The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for s
ines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See Example 9.13. • The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See Example 9.14. • The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See Example 9.15. • The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See Example 9.16. • The cofunction identities apply to complementary angles and pairs of reciprocal functions. See Example 9.17. • Sum and difference formulas are useful in verifying identities. See Example 9.18 and Example 9.19. • Application problems are often easier to solve by using sum and difference formulas. See Example 9.20 and Example 9.20. 9.3 Double-Angle, Half-Angle, and Reduction Formulas • Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example 9.22, Example 9.23, Example 9.24, and Example 9.25. • Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example 9.26 and Example 9.27. • Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example 9.28, Example 9.29, and Example 9.30. 9.4 Sum-to-Product and Product-to-Sum Formulas • From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. • We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of
sines and cosines. See Example 9.31, Example 9.32, and Example 9.33. 1054 Chapter 9 Trigonometric Identities and Equations • We can also derive the sum-to-product identities from the product-to-sum identities using substitution. • We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines. See Example 9.34. • Trigonometric expressions are often simpler to evaluate using the formulas. See Example 9.35. • The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side. See Example 9.36 and Example 9.37. 9.5 Solving Trigonometric Equations • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example 9.38, Example 9.39, and Example 9.40. • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example 9.41, Example 9.42, and Example 9.43, and Example 9.44. • We can also solve trigonometric equations using a graphing calculator. See Example 9.45 and Example 9.46. • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example 9.47, Example 9.48, Example 9.49, and Example 9.50. • We can also use the identities to solve trigonometric equation. See Example 9.51, Example 9.52, and Example 9.53. • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example 9.54. • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example 9.
55. CHAPTER 9 REVIEW EXERCISES Solving Trigonometric Equations with Identities For the following exercises, find all solutions exactly that exist on the interval [0, 2π). 331. csc2 t = 3 332. cos2 x = 1 4 333. 2 sin θ = −1 334. tan x sin x + sin(−x) = 0 335. 9 sin ω − 2 = 4 sin2 ω 336. 1 − 2 tan(ω) = tan2(ω) 338. sin3 x + cos2 x sin x the following exercises, determine if For identities are equivalent. the given 339. sin2 x + sec2 x − 1 = ⎛ ⎛ ⎝1 − cos2 x⎞ ⎝1 + cos2 x⎞ ⎠ ⎠ cos2 x 340. tan3 x csc2 x cot2 x cos x sin x = 1 Sum and Difference Identities For the following exercises, find the exact value. 341. ⎛ tan ⎝ ⎞ ⎠ 7π 12 342. ⎛ cos ⎝ 25π 12 ⎞ ⎠ For the following exercises, use basic identities to simplify the expression. 337. sec x cos x + cos x − 1 sec x 343. 344. sin(70°)cos(25°) − cos(70°)sin(25°) cos(83°)cos(23°) + sin(83°)sin(23°) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1055 For the following exercises, prove the identity. 345. cos(4x) − cos(3x)cosx = sin2 x − 4 cos2 x sin2 x For the following exercises, prove the identity. 356. 2cos(2x) sin(2x) = cot x − tan x 346. cos(3x) − cos3 x = − cos x sin2 x − sin x sin(2x) 357. cot x cos(2x) = − sin(2x) + cot x For the following exercise, simplify the expression. 347. 1 2 ⎛ tan ⎝ x⎞ ⎛ x⎞ 1 ⎠ + tan ⎝
⎠ 8 x⎞ ⎛ x⎞ ⎛ 1 1 ⎠tan 1 − tan ⎠ ⎝ ⎝ 2 8 For the following exercises, rewrite the expression with no powers. 358. cos2 x sin4(2x) For the following exercises, find the exact value. 359. tan2 x sin3 x 348. ⎝sin−1 (0) − cos−1 ⎛ ⎛ cos ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 349. ⎝sin−1 (0) + sin−1 ⎛ ⎛ tan ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Double-Angle, Half-Angle, and Reduction Formulas For the following exercises, find the exact value. 350. Find sin(2θ), cos(2θ), cos θ = − 1 3 and θ is in the interval ⎡ ⎣ and tan(2θ) given, π⎤ π ⎦. 2 351. Find sin(2θ), cos(2θ), and tan(2θ) given sec θ = − 5 3 and θ is in the interval ⎡ ⎣, π⎤ ⎦. π 2 352. ⎛ sin ⎝ ⎞ ⎠ 7π 8 353. ⎛ sec ⎝ ⎞ ⎠ 3π 8 For the following exercises, use Figure 9.26 to find the desired quantities. Figure 9.26 354. sin(2β), cos(2β), tan(2β), sin(2α), cos(2α), and tan(2α) Sum-to-Product and Product-to-Sum Formulas For the following exercises, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. 360. ⎛ cos ⎝ π 3 ⎞ ⎛ ⎠ sin ⎝ ⎞ ⎠ π 4 ⎛ 361. 2 sin ⎝ 2π 3 ⎛ ⎞ ⎠ sin ⎝ ⎞ ⎠ 5π 6 ⎛ 362. 2 cos ⎝ π 5 ⎞ ⎛ �
�� cos ⎝ ⎞ ⎠ π 3 For the following exercises, evaluate the sum by using a product formula. Write the exact answer. 363. ⎛ sin ⎝ π 12 ⎞ ⎛ ⎠ − sin ⎝ ⎞ ⎠ 7π 12 364. ⎛ cos ⎝ 5π 12 ⎞ ⎛ ⎠ + cos ⎝ ⎞ ⎠ 7π 12 For the following exercises, change the functions from a product to a sum or a sum to a product. 365. sin(9x)cos(3x) 366. cos(7x)cos(12x) 367. sin(11x) + sin(2x) 368. cos(6x) + cos(5x) β 2 ⎞ ⎛ ⎠, cos ⎝ β 2 ⎞ ⎛ ⎠, tan ⎝ β 2 ⎞ ⎛ ⎠, sin ⎝ α 2 ⎛ ⎞ ⎠, cos ⎝ α 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠ α 2 369. tan x + 1 = 0 Solving Trigonometric Equations For the following exercises, find all exact solutions on the interval [0, 2π). 355. ⎛ sin ⎝ 1056 Chapter 9 Trigonometric Identities and Equations 370. 2 sin(2x) + 2 = 0 For the following exercises, find all exact solutions on the interval [0, 2π). 371. 2 sin2 x − sin x = 0 372. cos2 x − cos x − 1 = 0 373. 2 sin2 x + 5 sin x + 3 = 0 374. cos x − 5 sin(2x) = 0 375. 1 sec2 x + 2 + sin2 x + 4 cos2 x = 0 the following exercises, simplify the equation For algebraically as much as possible. Then use a calculator to find the solutions on the interval [0, 2π). Round to four decimal places. 376. 3 cot2 x + cot x = 1 377. csc2 x − 3 csc x − 4 = 0 For the following exercises, graph each side of the equation to find the approximate solutions on the interval [0, 2π). 378. 20
cos2 x + 21 cos x + 1 = 0 sec2 x − 2 sec x = 15 379. CHAPTER 9 PRACTICE TEST For the following exercises, simplify the given expression. 380. cos(−x)sin x cot x + sin2 x 381. sin(−x)cos(−2x)−sin(−x)cos(−2x) 382. ⎞ ⎛ ⎝sec2 θ − 1 csc(θ)cot(θ) ⎠ 383. ⎞ ⎞ ⎛ ⎛ ⎝1 + tan2 (θ) ⎝1 + cot2 (θ) cos2 (θ)sin2 (θ) ⎠ ⎠ For the following exercises, find the exact value. 384. ⎛ cos ⎝ ⎞ ⎠ 7π 12 385. ⎛ tan ⎝ ⎞ ⎠ 3π 8 This content is available for free at https://cnx.org/content/col11758/1.5 386. ⎝sin−1 ⎛ ⎛ tan ⎝ ⎞ ⎠ + tan−1 3 ⎞ ⎠ 2 2 ⎛ 387. 2sin ⎝ π 4 ⎛ ⎞ ⎠sin ⎝ ⎞ ⎠ π 6 388. ⎛ cos ⎝ 4π 3 + θ⎞ ⎠ 389. ⎛ ⎝− tan π 4 + θ⎞ ⎠ For the following exercises, simplify each expression. Do not evaluate. 390. cos2(32°)tan2(32°) 391. cot ⎞ ⎠ ⎛ ⎝ θ 2 Chapter 9 Trigonometric Identities and Equations 1057 For the following exercises, find all exact solutions to the equation on [0, 2π). 406. tan3 x − tan x sec2 x = tan(−x) 392. cos2 x − sin2 x − 1 = 0 407. sin(3x) − cos x sin(2x) = cos2 x sin x − sin3 x 393. cos2 x = cos x 4 sin2 x + 2 sin x − 3 = 0 408. sin(2
x) sin x − cos(2x) cos x = sec x 394. cos(2x) + sin2 x = 0 395. 2 sin2 x − sin x = 0 396. Rewrite the expression as a product instead of a sum: cos(2x) + cos(−8x). For the following exercise, rewrite the product as a sum or difference. 397. 8cos(15x)sin(3x) For the following exercise, rewrite the sum or difference as a product. 398. 2⎛ ⎝sin(8θ) − sin(4θ)⎞ ⎠ 399. Find all solutions of tan(x) − 3 = 0. 400. Find the solutions of sec2 x − 2 sec x = 15 on the interval [0, 2π) algebraically; then graph both sides of the equation to determine the answer. For the following exercises, find all solutions exactly on the interval 0 ≤ θ ≤ π ⎛ 401. 2cos ⎝ ⎞ ⎠ = 1 θ 2 402. 3cot(y) = 1 403. Find sin(2θ), cos(2θ), and tan(2θ) given cot θ = − 3 4 and θ is on the interval ⎡ ⎣, π⎤ ⎦. π 2 404. cos θ = 7 25 ⎛ Find sin ⎝ θ 2 and θ is in quadrant IV. ⎛ ⎞ ⎠, cos ⎝ θ 2 ⎛ ⎞ ⎠, and tan ⎝ ⎞ ⎠ given θ 2 405. Rewrite the expression sin4 x with no powers greater than 1. For the following exercises, prove the identity. 409. Plot the points and find a function of the form y = Acos(Bx + C) + D that fits the given data. x y 0 −2 1 2 2 −2 3 2 4 −2 5 2 410. The displacement h(t) in centimeters of a mass suspended by a spring is modeled by the function h(t) = 1 sin(120πt), where t is measured in seconds. 4 Find the amplitude, period, and frequency of displacement. this 411. A woman is standing 300 feet away from a 2000-foot building. If
she looks to the top of the building, at what angle above horizontal is she looking? A bored worker looks down at her from the 15th floor (1500 feet above her). At what angle is he looking down at her? Round to the nearest tenth of a degree. governed 412. Two frequencies of sound are played on an equation instrument by n(t) = 8 cos(20πt)cos(1000πt). What are the period and frequency of the “fast” and “slow” oscillations? What is the amplitude? the 413. The average monthly snowfall in a small village in the Himalayas is 6 inches, with the low of 1 inch occurring in July. Construct a function that models this behavior. During what period is there more than 10 inches of snowfall? 414. A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring t seconds after being released. At what time will the spring come to rest? In this case, use 1 cm amplitude as rest. 415. Water levels near a glacier currently average 9 feet, varying seasonally by 2 inches above and below the average and reaching their highest point in January. Due 1058 Chapter 9 Trigonometric Identities and Equations to global warming, the glacier has begun melting faster than normal. Every year, the water levels rise by a steady 3 inches. Find a function modeling the depth of the water t months from now. If the docks are 2 feet above current water levels, at what point will the water first rise above the docks? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1059 10 \| FURTHER APPLICATIONS OF TRIGONOMETRY Figure 10.1 General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr) Chapter Outline 10.1 Non-right Triangles: Law of Sines 10.2 Non-right Triangles: Law of Cosines 10.3 Polar Coordinates 10.4 Polar Coordinates: Graphs 10.5 Polar Form of Complex Numbers 10.6 Parametric Equations 10.7 Parametric Equations: Graphs 10.8 Vectors 1060 Chapter 10 Further Applications of Trigonometry Introduction The world’s largest tree by volume, named General Sherman
, stands 274.9 feet tall and resides in Northern California.[1] Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, in Trigonometric Functions including finding (https://cnx.org/content/m49369/latest/) and investigate applications more deeply and meaningfully. the height of a tree. We extend topics we introduced 10.1 \| Non-right Triangles: Law of Sines Learning Objectives In this section, you will: 10.1.1 Use the Law of Sines to solve oblique triangles. 10.1.2 Find the area of an oblique triangle using the sine function. 10.1.3 Solve applied problems using the Law of Sines. Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 10.2 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. Figure 10.2 Using the Law of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 10.3. Figure 10.3 2. AAS (angle-angle-side) We know the
measurements of two angles and a side that is not between the known angles. See Figure 10.4. 1. Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1061 Figure 10.4 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 10.5. Figure 10.5 Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 10.6. Figure 10.6 Using the right triangle relationships, we know that sin α = and sin β = h a. Solving both equations for h gives two h b different expressions for h. We then set the expressions equal to each other. h = bsin α and h = asin β bsin α = asin β ⎛ ⎛ ⎞ 1 ⎠(bsin α) = (asin β) ⎝ ⎝ ab sin α a = sin β b ⎞ ⎠ 1 ab Multiply both sides by 1 ab. Similarly, we can compare the other ratios. sin α a = sin γ c and sin β b = sin γ c Collectively, these relationships are called the Law of Sines. sin α a = sin β b = sin λ c Note the standard way of labeling triangles: angle α (alpha) is opposite side a; angle β (beta) is opposite side b; and angle γ (gamma) is opposite side c. See Figure 10.7. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. 1062 Chapter 10 Further Applications of Trigonometry Figure 10.7 Law of Sines Given a triangle with angles and opposite sides labeled as in Figure 10.
7, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ (10.1) (10.2) To solve an oblique triangle, use any pair of applicable ratios. Example 10.1 Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in Figure 10.8 to the nearest tenth. Figure 10.8 Solution The three angles must add up to 180 degrees. From this, we can determine that β = 180° − 50° − 30° = 100° To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1063 sin(50°) 10 csin(50°) 10 = sin(30°) c = sin(30°) c = sin(30 ° ) c ≈ 6.5 Multiply both sides by c. 10 sin(50°) Multiply by the reciprocal to isolate c. Similarly, to solve for b, we set up another proportion. sin(50°) 10 = sin(100°) b bsin(50°) = 10sin(100°) 10sin(100°) sin(50°) b = b ≈ 12.9 Therefore, the complete set of angles and sides is Multiply both sides by b. Multiply by the reciprocal to isolate b. α = 50° a = 10 β = 100° b ≈ 12.9 γ = 30° c ≈ 6.5 10.1 Solve the triangle shown in Figure 10.9 to the nearest tenth. Figure 10.9 Using The Law of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an
ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. Possible Outcomes for SSA Triangles Oblique triangles in the category SSA may have four different outcomes. Figure 10.10 illustrates the solutions with the known sides a and b and known angle α. 1064 Chapter 10 Further Applications of Trigonometry Figure 10.10 Example 10.2 Solving an Oblique SSA Triangle Solve the triangle in Figure 10.11 for the missing side and find the missing angle measures to the nearest tenth. Figure 10.11 Solution Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we have the proportion = sin α a = sin β b sin β sin(35°) 8 6 8sin(35°) = sin β 6 0.7648 ≈ sin β sin−1(0.7648) ≈ 49.9° β ≈ 49.9° This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1065 However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have Figure 10.12. It appears that there may be a second triangle that will fit the given criteria. Figure 10.12 The angle supplementary to β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1°. (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ, we have We can then use these measurements to solve the other triangle. Since γ′ is supplementary to γ, we have γ = 180° − 35° − 130.1° ≈ 14.9° Now we need to find c and c′. We have Finally, γ′ = 180° − 35° − 49.9° ≈ 95.1° c sin(14.9°) = c = 6 sin(35°) 6sin(14.
9°) sin(35°) ≈ 2.7 c′ sin(95.1°) = c′ = 6 sin(35°) 6sin(95.1°) sin(35°) ≈ 10.4 To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 10.13. 1066 Chapter 10 Further Applications of Trigonometry Figure 10.13 However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first triangle (a) in Figure 10.13. 10.2 Given α = 80°, a = 120, and b = 121, find the missing side and angles. If there is more than one possible solution, show both. Example 10.3 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure 10.14, solve for the unknown side and angles. Round your answers to the nearest tenth. Figure 10.14 Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85°, and its corresponding side c = 12, and we know side b = 9. We will use this proportion to solve for β. sin(85°) 12 9sin(85°) 12 = sin β 9 = sin β Isolate the unknown. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1067 To find β, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. ⎞ ⎠ β = sin−1 ⎛ ⎝ 9sin(85°) 12 β ≈ sin−1(0.7471) β ≈ 48.3° In this case, there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7°. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives if we subtract β from 180°, we find that which
is impossible, and so β ≈ 48.3°. α = 180° − 85° − 131.7° ≈ − 36.7°, To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7°. Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions. = sin(46.7 ° ) a = sin(46.7 ° ) sin(85 ° ) 12 asin(85 ° ) 12 a = 12sin(46.7 ° ) sin(85 ° ) ≈ 8.8 The complete set of solutions for the given triangle is α ≈ 46.7° a ≈ 8.8 β ≈ 48.3° b = 9 γ = 85° c = 12 10.3 Given α = 80°, a = 100, b = 10, find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. Example 10.4 Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Solution Using the given information, we can solve for the angle opposite the side of length 10. See Figure 10.15. 1068 Chapter 10 Further Applications of Trigonometry = sin(50°) 4 sin α 10 10sin(50°) sin α = 4 sin α ≈ 1.915 Figure 10.15 We can stop here without finding the value of α. Because the range of the sine function is [−1, 1], it is impossible for the sine value to be 1.915. In fact, inputting sin−1 (1.915) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. 10.4 Determine the number of triangles possible given a = 31, b = 26, β = 48°. Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as Area = 1 bh, where b is base and h is 2 height. For oblique triangles, we must find h before we can use
the area formula. Observing the two triangles in Figure 10.16, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = h c or csin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α′, or 180 − α. Figure 10.16 Thus, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1069 Similarly, Area = 1 2 (base)⎛ ⎝height⎞ ⎠ = 1 2 b(csin α) Area = 1 2 ⎝bsin γ⎞ a⎛ ⎠ = 1 2 ⎝csin β⎞ a⎛ ⎠ Area of an Oblique Triangle The formula for the area of an oblique triangle is given by Area = 1 2 = 1 2 = 1 2 bcsin α acsin β absin γ (10.3) This is equivalent to one-half of the product of two sides and the sine of their included angle. Example 10.5 Finding the Area of an Oblique Triangle Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102°. Round the area to the nearest integer. Solution Using the formula, we have absin γ Area = 1 2 Area = 1 (90)(52)sin(102°) 2 Area ≈ 2289 square units 10.5 Find the area of the triangle given β = 42°, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth. Solving Applied Problems Using the Law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. Example 10.6 Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 10.17. Round the altitude to the nearest tenth of a mile. 1070 Chapter
10 Further Applications of Trigonometry Figure 10.17 Solution To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a, and then use right triangle relationships to find the height of the aircraft, h. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. sin(130°) 20 = sin(35°) a asin(130°) = 20sin(35°) 20sin(35°) sin(130°) a = a ≈ 14.98 The distance from one station to the aircraft is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for h. sin(15°) = sin(15°) = opposite hypotenuse h a sin(15°) = h 14.98 h = 14.98sin(15°) h ≈ 3.88 The aircraft is at an altitude of approximately 3.9 miles. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1071 10.6 The diagram shown in Figure 10.18 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B, is 62°, and the distance between the viewing points of the two end zones is 145 yards. Figure 10.18 Access these online resources for additional instruction and practice with trigonometric applications. • Law of Sines: The Basics (http://openstaxcollege.org/l/sinesbasic) • Law of Sines: The Ambiguous Case (http://openstaxcollege.org/l/sinesambiguous) 1072 Chapter 10 Further Applications of Trigonometry 10.1 EXERCISES Verbal 1. Describe the altitude of a triangle. 2. Compare right triangles and oblique triangles. When can you use the Law of Sines to find a missing 3. angle? 17. a = 12, c = 17, α = 35° 18. a = 20.5, b = 35.0, β = 25
° 19. a = 7, c = 9, α = 43° 20. a = 7, b = 3, β = 24° In the Law of Sines, what is the relationship between the 4. angle in the numerator and the side in the denominator? 21. b = 13, c = 5, γ = 10° 5. What type of triangle results in an ambiguous case? 22. a = 2.3, c = 1.8, γ = 28° Algebraic 23. β = 119°, b = 8.2, a = 11.3 For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 6. 7. α = 43°, γ = 69°, a = 20 α = 35°, γ = 73°, c = 20 8. α = 60°, β = 60°, γ = 60° 9. a = 4, α = 60°, β = 100° 10. b = 10, β = 95°, γ = 30° For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. 11. Find side b when A = 37°, B = 49°, c = 5. 12. Find side a when A = 132°, C = 23°, b = 10. 13. Find side c when B = 37°, C = 21, b = 23. For the following exercises, assume α is opposite side a, β is opposite side c. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. side b, and γ is opposite 14. α = 119°, a = 14, b = 26 15. γ = 113°, b = 10, c = 32 16. b = 3.5, c = 5.3, γ = 80° This content is available
for free at https://cnx.org/content/col11758/1.5 24. Find angle A when a = 24, b = 5, B = 22°. 25. Find angle A when a = 13, b = 6, B = 20°. 26. Find angle B when A = 12°, a = 2, b = 9. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a = 5, c = 6, β = 35° 28. b = 11, c = 8, α = 28° 29. a = 32, b = 24, γ = 75° 30. a = 7.2, b = 4.5, γ = 43° Graphical For the following exercises, find the length of side x. Round to the nearest tenth. 31. 32. Chapter 10 Further Applications of Trigonometry 1073 33. 34. 35. 36. 39. 40. 41. Notice that x is an obtuse angle. 42. For the following exercises, find the measure of angle x, if possible. Round to the nearest tenth. 37. 38. For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. 43. 1074 Chapter 10 Further Applications of Trigonometry 44. 45. 46. 47. 48. This content is available for free at https://cnx.org/content/col11758/1.5 49. Extensions Find the radius of the circle in Figure 10.19. Round to 50. the nearest tenth. Figure 10.19 Find the diameter of the circle in Figure 10.20. 51. Round to the nearest tenth. Figure 10.20 Find m ∠ ADC in Figure 10.21. Round to the 52. nearest tenth. Chapter 10 Further Applications of Trigonometry 1075 Figure 10.21 Figure 10.24 53. Find AD in Figure 10.22. Round to the nearest tenth. 56. Solve the triangle in Figure 10.25. (Hint: Draw a perpendicular from H to JK). Round each answer to the nearest tenth. Figure 10.25 Figure 10.22 Solve both triangles in Figure 10.23. Round each 54. answer to the nearest tenth. 57. Solve the triangle in Figure 10.26. (Hint: Draw a perpendicular from N to LM). Round each answer to the nearest tenth. Figure 10
.26 Figure 10.23 Find AB in the parallelogram shown in Figure 55. 10.24. In Figure 10.27, ABCD is not a parallelogram. 58. ∠ m is obtuse. Solve both triangles. Round each answer to the nearest tenth. 1076 Chapter 10 Further Applications of Trigonometry Figure 10.29 61. Figure 10.30 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A and B, which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 86.2° and 83.9°, respectively. How far is the satellite from station A and how high is the satellite above the ground? Round answers to the nearest whole mile. Figure 10.27 Real-World Applications A pole leans away from the sun at an angle of 7° to the 59. vertical, as shown in Figure 10.28. When the elevation of the sun is 55°, the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth. Figure 10.28 Figure 10.30 60. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 10.29. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. 62. A communications tower is located at the top of a steep hill, as shown in Figure 10.31. The angle of inclination of the hill is 67°. A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. Find the length of the cable required for the guy wire to the nearest whole meter. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1077 be 32° and 56°, as shown in Figure 10.34. Find the distance of the plane from point A to the nearest tenth of a kilometer. Figure 10.31 The roof of a house is at a 20° angle. An 8-foot solar 63. panel is to be mounted on the roof and should be angled 38° relative to the horizontal for optimal results
. (See Figure 10.32). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. Figure 10.32 64. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° and 44°, as shown in Figure 10.33. Find the distance of the plane from point A to the nearest tenth of a kilometer. Figure 10.33 A pilot is flying over a straight highway. He determines 65. the angles of depression to two mileposts, 4.3 km apart, to Figure 10.34 In order to estimate the height of a building, two 66. students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. In order to estimate the height of a building, two 67. students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. Points A and B are on opposite sides of a lake. Point 68. C is 97 meters from A. The measure of angle BAC is determined to be 101°, and the measure of angle ACB is determined to be 53°. What is the distance from A to B, rounded to the nearest whole meter? 69. A man and a woman standing 31 2 miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. Two search teams spot a stranded climber on a 70. mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of
1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. A street light is mounted on a pole. A 6-foot-tall man is 71. standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is 1078 Chapter 10 Further Applications of Trigonometry standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. Three cities, A, B, and C, are located so that city A 72. is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 miles from city B, how far is city A from city B? Round the distance to the nearest tenth of a mile. 73. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. Brian’s house is on a corner lot. Find the area of the 74. front yard if the edges measure 40 and 56 feet, as shown in Figure 10.35. Figure 10.35 75. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. A yield sign measures 30 inches on all three sides. 76. What is the area of the sign? Naomi bought a modern dining table whose top is in 77. the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as
shown in Figure 10.36. Figure 10.36 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1079 10.2 \| Non-right Triangles: Law of Cosines Learning Objectives In this section, you will: 10.2.1 Use the Law of Cosines to solve oblique triangles. 10.2.2 Solve applied problems using the Law of Cosines. 10.2.3 Use Heron’s formula to ﬁnd the area of a triangle. Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 10.37. How far from port is the boat? Figure 10.37 Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases. Using the Law of Cosines to Solve Oblique Triangles The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level. Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure 10.38. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted. 1080 Chapter 10 Further Applications of Trigonometry Figure 10.38 We can drop a perpendicular from C to the
x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that cos θ = x(adjacent) b(hypotenuse) and sin θ = y(opposite) b(hypotenuse) In terms of θ, x = bcos θ and y = bsin θ. The (x, y) point located at C has coordinates (bcos θ, bsin θ). Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagorean Theorem. Thus, a2 = (x − c)2 + y2 = (bcos θ − c)2 + (bsin θ)2 ⎝b2 cos2 θ − 2bccos θ + c2⎞ ⎛ ⎠ + b2 sin2 θ = = b2 cos2 θ + b2 sin2 θ + c2 − 2bccos θ = b2 ⎛ ⎠ + c2 − 2bccos θ ⎝cos2 θ + sin2 θ⎞ a2 = b2 + c2 − 2bccos θ Substitute (bcos θ) for x and (bsin θ) for y. Expand the perfect square. Group terms noting that cos2 θ + sin2 θ = 1. Factor out b2. The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve. Law of Cosines The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 10.39, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as
three equations. a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ (10.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1081 Figure 10.39 To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle. cos α = cos β = cos γ = b2 + c2 − a2 2bc a2 + c2 − b2 2ac a2 + b2 − c2 2ab Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle. 1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles. 2. Apply the Law of Cosines to find the length of the unknown side or angle. 3. Apply the Law of Sines or Cosines to find the measure of a second angle. 4. Compute the measure of the remaining angle. Example 10.7 Finding the Unknown Side and Angles of a SAS Triangle Find the unknown side and angles of the triangle in Figure 10.40. Figure 10.40 Solution 1082 Chapter 10 Further Applications of Trigonometry First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines. Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b, as we know the measurement of the opposite angle β. b2 = a2 + c2 − 2accos β b2 = 102 + 122 − 2(10)(12)cos(30∘) b2 = 100 + 144 − 240 ⎛ ⎝ ⎞ ⎠ 3 2 b2 = 244 − 120 3 b = 244 − 120 3 b ≈ 6.013 Substitute the measurements for the known quantities. Evaluate the cosine and begin to simplify. Use the square root property. Because we are solving for a length,
we use only the positive square root. Now that we know the length b, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α, we have sin α a = sin β b = sin α 10 sin α = sin(30°) 6.013 10sin(30°) 6.013 α = sin−1 ⎛ ⎝ α ≈ 56.3° 10sin(30°) 6.013 Multiply both sides of the equation by 10. ⎞ ⎠ Find the inverse sine of 10sin(30°) 6.013. The other possibility for α would be α = 180° – 56.3° ≈ 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0° and 180°. Proceeding with α ≈ 56.3°, we can then find the third angle of the triangle. The complete set of angles and sides is γ = 180° − 30° − 56.3° ≈ 93.7° α ≈ 56.3° β = 30° γ ≈ 93.7° a = 10 b ≈ 6.013 c = 12 10.7 Find the missing side and angles of the given triangle: α = 30°, b = 12, c = 24. Example 10.8 Solving for an Angle of a SSS Triangle Find the angle α for the given triangle if side a = 20, side b = 25, and side c = 18. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1083 Solution For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α, we have a2 = b2 + c2 −2bccos α 202 = 252 + 182 −2(25)(18)cos α 400 = 625 + 324 − 900cos α 400 = 949 − 900cos α −549 = −900cos α −549 = cos α −900 0.61 ≈ cos α cos−1
(0.61) ≈ α α ≈ 52.4° Substitute the appropriate measurements. Simplify in each step. Isolate cos α. Find the inverse cosine. See Figure 10.41. Figure 10.41 Analysis Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method. 10.8 Given a = 5, b = 7, and c = 10, find the missing angles. Solving Applied Problems Using the Law of Cosines Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few. Example 10.9 Using the Law of Cosines to Solve a Communication Problem On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart 1084 Chapter 10 Further Applications of Trigonometry along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway. Solution For simplicity, we start by drawing a diagram similar to Figure 10.42 and labeling our given information. Figure 10.42 Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let a = 2420, b = 5050, and c = 6000. Thus, θ corresponds to the opposite side a = 2420. a2 = b2 + c2 − 2bccos θ (2420)2 = (5050)2 + (6000)2 − 2(5050)(6000)cos θ (2420)2 − (5050)2 − (6000)2 = − 2(5050)(6000)cos θ (2420)2 − (
5050)2 − (6000)2 −2(5050)(6000) = cos θ cos θ ≈ 0.9183 θ ≈ cos−1(0.9183) θ ≈ 23.3° To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 10.43. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem. Figure 10.43 Using the angle θ = 23.3° and the basic trigonometric identities, we can find the solutions. Thus This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1085 cos(23.3°) = x 5050 x = 5050cos(23.3°) x ≈ 4638.15 feet sin(23.3°) = y 5050 y = 5050sin(23.3°) y ≈ 1997.5 feet The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway. Example 10.10 Calculating Distance Traveled Using a SAS Triangle Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 10.44. Figure 10.44 Solution turned 20 degrees, so the obtuse angle of the non-right The boat triangle is the supplemental angle, 180° − 20° = 160°. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port. 1086 Chapter 10 Further Applications of Trigonometry x2 = 82 + 102 − 2(8)(10)cos(160°) x2 = 314.35 x = 314.35 x ≈ 17.7 miles The boat is about 17.7 miles from port. Using Heron’s Formula to Find the Area of a Triangle We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three
sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known. Heron’s Formula Heron’s formula finds the area of oblique triangles in which sides a, b, and c are known. Area = s(s − a)(s − b)(s − c) (10.5) is one half of the perimeter of the triangle, sometimes called the semi-perimeter. where s = (a + b + c) 2 Example 10.11 Using Heron’s Formula to Find the Area of a Given Triangle Find the area of the triangle in Figure 10.45 using Heron’s formula. Figure 10.45 Solution First, we calculate s. Then we apply the formula. s = (a + b + c) 2 s = (10 + 15 + 7) 2 = 16 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1087 Area = s(s − a)(s − b)(s − c) Area = 16(16 − 10)(16 − 15)(16 − 7) Area ≈ 29.4 The area is approximately 29.4 square units. Use Heron’s formula to find the area of a triangle with sides of lengths a = 29.7 ft, b = 42.3 ft, and 10.9 c = 38.4 ft. Example 10.12 Applying Heron’s Formula to a Real-World Problem A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 10.46 for a view of the city property. Figure 10.46 Solution Find the measurement for s, which is one-half of the perimeter. (62.4 + 43.5 + 34.1) 2 s = s = 70 m 1088 Chapter 10 Further Applications of Trigonometry Apply Heron’s formula. Area
= 70(70 − 62.4)(70 − 43.5)(70 − 34.1) Area = 506,118.2 Area ≈ 711.4 The developer has about 711.4 square meters. 10.10 Find the area of a triangle given a = 4.38 ft, b = 3.79 ft, and c = 5.22 ft. Access these online resources for additional instruction and practice with the Law of Cosines. • Law of Cosines (http://openstaxcollege.org/l/lawcosines) • Law of Cosines: Applications (http://openstaxcollege.org/l/cosineapp) • Law of Cosines: Applications 2 (http://openstaxcollege.org/l/cosineapp2) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1089 10.2 EXERCISES Verbal 97. a = 108, b = 132, c = 160; find angle C. If you are looking for a missing side of a triangle, what 78. do you need to know when using the Law of Cosines? For the following exercises, solve the triangle. Round to the nearest tenth. If you are looking for a missing angle of a triangle, 79. what do you need to know when using the Law of Cosines? 98. A = 35°, b = 8, c = 11 80. Explain what s represents in Heron’s formula. Explain the relationship between the Pythagorean 81. Theorem and the Law of Cosines. When must you use the Law of Cosines instead of the 82. Pythagorean Theorem? Algebraic is opposite side b, and γ For the following exercises, assume α is opposite side a, β is opposite side c. If possible, solve each triangle for the unknown side. Round to the nearest tenth. 83. γ = 41.2°, a = 2.49, b = 3.13 84. α = 120°, b = 6, c = 7 85. β = 58.7°, a = 10.6, c = 15.7 86. γ = 115°, a = 18, b = 23 87. α = 119°, a = 26, b = 14 88. γ = 113°, b = 10, c = 32 89. β =
67°, a = 49, b = 38 90. α = 43.1°, a = 184.2, b = 242.8 91. α = 36.6°, a = 186.2, b = 242.2 92. β = 50°, a = 105, b = 45 For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth. 93. a = 42, b = 19, c = 30; find angle A. 94. a = 14, b = 13, c = 20; find angle C. 95. a = 16, b = 31, c = 20; find angle B. 96. a = 13, b = 22, c = 28; find angle A. 99. B = 88°, a = 4.4, c = 5.2 100. C = 121°, a = 21, b = 37 101. a = 13, b = 11, c = 15 102. a = 3.1, b = 3.5, c = 5 103. a = 51, b = 25, c = 29 For the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth. Find the area of a triangle with sides of length 18 in, 104. 21 in, and 32 in. Round to the nearest tenth. Find the area of a triangle with sides of length 20 cm, 105. 26 cm, and 37 cm. Round to the nearest tenth. 106. a = 1 2 m, b = 1 3 m, c = 1 4 m 107. a = 12.4 ft, b = 13.7 ft, c = 20.2 ft 108. a = 1.6 yd, b = 2.6 yd, c = 4.1 yd Graphical For the following exercises, find the length of side x. Round to the nearest tenth. 109. 110. 111. 1090 Chapter 10 Further Applications of Trigonometry 112. 113. 114. 116. 117. For the following exercises, find the measurement of angle A. 118. 115. This content is available for free at https://cnx.org/content/col11758/1.5 119. Chapter 10 Further Applications of Trigonometry 1091 Find the measure of each angle in the triangle shown in Figure 10.47. Round to the nearest tenth. 124. Figure
10.47 For the following exercises, solve for the unknown side. Round to the nearest tenth. 120. 121. 122. 123. 125. 126. 127. 128. For the following exercises, find the area of the triangle. Round to the nearest hundredth. Chapter 10 Further Applications of Trigonometry 1092 Extensions 129. A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal. 130. The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal. 131. The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal. A regular octagon is inscribed in a circle with a radius 132. of 8 inches. (See Figure 10.48.) Find the perimeter of the octagon. 137. 138. Figure 10.48 Real-World Applications A regular pentagon is inscribed in a circle of radius 12 133. cm. (See Figure 10.49.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter. A surveyor has taken the measurements shown in 139. Figure 10.50. Find the distance across the lake. Round answers to the nearest tenth. Figure 10.49 Figure 10.50 A satellite calculates the distances and angle shown in 140. Figure 10.51 (not to scale). Find the distance between the two cities. Round answers to the nearest tenth. the following that For x2 = 25 + 36 − 60cos(52) represents the relationship of three sides of a triangle and the cosine of an angle. exercises, suppose 134. Draw the triangle. 135. Find the length of the third side. For the following exercises, find the area of the triangle. 136. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1093 Figure 10.51 141. An airplane flies 220 miles with a heading of 40°, and then flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth. A 113-foot tower is located on a hill that is inclined 142. 34° to the horizontal, as shown in Figure 10.52. A guywire is to be attached to the top
of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed. Figure 10.52 Two ships left a port at the same time. One ship 143. traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Find the distance between the two ships after 10 hours of travel. 144. The graph in Figure 10.53 represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours. Figure 10.53 A triangular swimming pool measures 40 feet on one 145. side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)? A pilot flies in a straight path for 1 hour 30 min. She 146. then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position? Los Angeles is 1,744 miles from Chicago, Chicago is 147. 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle. Philadelphia is 140 miles from Washington, D.C., 148. Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle. Two planes leave the same airport at the same time. 149. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours? Two airplanes take off in different directions. One 150. travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude? A parallelogram has sides of length 15.4 units and 9.8 151. units. Its area is 72.9 square units. Find the measure of the longer diagonal. The four sequential sides of a quadrilateral
have 152. lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral? 1094 Chapter 10 Further Applications of Trigonometry 153. The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral? Find the area of a triangular piece of land that 154. measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot. Find the area of a triangular piece of land that 155. measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1095 10.3 \| Polar Coordinates Learning Objectives In this section, you will: 10.3.1 Plot points using polar coordinates. 10.3.2 Convert from polar coordinates to rectangular coordinates. 10.3.3 Convert from rectangular coordinates to polar coordinates. 10.3.4 Transform equations between polar and rectangular forms. 10.3.5 Identify and graph polar equations by converting to rectangular equations. Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind (see Figure 10.54). How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid. Figure 10.54 Plotting Points Using Polar Coordinates When we think about plotting points in the plane, we usually think of rectangular coordinates (x, y) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled (r, θ) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane. The polar grid is scaled as the unit circle with the positive x-axis now viewed as the polar axis and the origin as the pole. The first
coordinate r is the radius or length of the directed line segment from the pole. The angle θ, measured in radians, indicates the direction of r. We move counterclockwise from the polar axis by an angle of θ, and measure a directed line segment the length of r in the direction of θ. Even though we measure θ first and then r, the polar point is written with the r-coordinate first. For example, to plot the point ⎛ units in the counterclockwise direction and ⎝2, π 4 ⎞ ⎠, we would move π 4 then a length of 2 from the pole. This point is plotted on the grid in Figure 10.55. 1096 Chapter 10 Further Applications of Trigonometry Figure 10.55 Example 10.13 Plotting a Point on the Polar Grid Plot the point ⎛ ⎝3, ⎞ ⎠ on the polar grid. π 2 Solution The angle π 2 is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located at a length of 3 units from the pole in the π 2 direction, as shown in Figure 10.56. Figure 10.56 10.11 Plot the point ⎛ ⎝2, π 3 ⎞ ⎠ in the polar grid. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1097 Example 10.14 Plotting a Point in the Polar Coordinate System with a Negative Component Plot the point ⎛ ⎝−2, π 6 ⎞ ⎠ on the polar grid. Solution We know that π 6 is located in the first quadrant. However, r = −2. We can approach plotting a point with a negative r in two ways: 1. Plot the point ⎛ ⎝2, π 6 ⎞ ⎠ by moving π 6 in the counterclockwise direction and extending a directed line segment 2 units into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant; 2. Move π 6 in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative direction, into the third quadrant. See Figure 10.57(a). Compare this
to the graph of the polar coordinate ⎛ ⎝2, ⎞ ⎠ shown in Figure 10.57(b). π 6 Figure 10.57 10.12 Plot the points ⎛ ⎝3, − ⎞ ⎠ π 6 and ⎛ ⎝2, 9π 4 ⎞ ⎠ on the same polar grid. Converting from Polar Coordinates to Rectangular Coordinates When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables x, y, r, and θ. cos θ = sin θ = x r → x = rcos θ y r → y = rsin θ 1098 Chapter 10 Further Applications of Trigonometry Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 10.58. An easy way to remember the equations above is to think of cos θ as the adjacent side over the hypotenuse and sin θ as the opposite side over the hypotenuse. Figure 10.58 Converting from Polar Coordinates to Rectangular Coordinates To convert polar coordinates (r, θ) to rectangular coordinates (x, y), let cos θ = sin θ = x r → x = rcos θ y r → y = rsin θ Given polar coordinates, convert to rectangular coordinates. 1. Given the polar coordinate (r, θ), write x = rcos θ and y = rsin θ. 2. Evaluate cos θ and sin θ. 3. Multiply cos θ by r to find the x-coordinate of the rectangular form. 4. Multiply sin θ by r to find the y-coordinate of the rectangular form. Example 10.15 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates ⎛ ⎝3, ⎞ ⎠ as rectangular coordinates. π 2 Solution Use the equivalent relationships. x = rcos θ x = 3cos π 2 y = rsin θ y = 3sin π 2 = 0 = 3 The rectangular coordinates are (0, 3). See Figure 10.59. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigon
ometry 1099 Figure 10.59 Example 10.16 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates (−2, 0) as rectangular coordinates. Solution See Figure 10.60. Writing the polar coordinates as rectangular, we have x = rcos θ x = −2cos(0) = −2 y = rsin θ y = −2sin(0) = 0 The rectangular coordinates are also (−2, 0). 1100 Chapter 10 Further Applications of Trigonometry Figure 10.60 10.13 Write the polar coordinates ⎛ ⎝−1, 2π 3 ⎞ ⎠ as rectangular coordinates. Converting from Rectangular Coordinates to Polar Coordinates To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point. Converting from Rectangular Coordinates to Polar Coordinates Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 10.61. cos θ = x r or x = rcos θ y r or y = rsin θ sin θ = r 2 = x2 + y2 y x tan θ = (10.6) Figure 10.61 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1101 Example 10.17 Writing Rectangular Coordinates as Polar Coordinates Convert the rectangular coordinates (3, 3) to polar coordinates. Solution We see that the original point (3, 3) is in the first quadrant. To find θ, use the formula tan θ = y x. This gives tan θ = 3 3 tan θ = 1 π tan−1(1) = 4 To find r, we substitute the values for x and y into the formula r = x2 + y2. We know that r must be positive, as π 4 is in the first quadrant. Thus r = 32 + 32 r = 9 + 9 r = 18 = 3 2 So, r = 3 2 and θ=, giving us the polar point ⎛ ⎝3 2, π 4 ⎞ ⎠. See Figure 10.62. π 4 Figure 10.62 Analysis There are other sets of polar coordinates that will be the same as our first solution
. For example, the points ⎝−3 2, 5π ⎝−3 2, 5π ⎝3 2, π ⎞ ⎠ will coincide with the original solution of ⎛ ⎛ ⎞ ⎠ 4 4 4 indicates a move further counterclockwise by π, which is directly opposite π. The radius is expressed as 4 ⎝3 2, − 7π 4 ⎠. The point ⎛ ⎞ ⎠ and ⎛ ⎞ − 3 2. However, the angle 5π 4 is located in the third quadrant and, as r is negative, we extend the directed 1102 Chapter 10 Further Applications of Trigonometry line segment in the opposite direction, into the first quadrant. This is the same point as ⎛ ⎝3 2, π 4 ⎞ ⎠. The point ⎝3 2, − 7π ⎛ 4 ⎞ ⎠ is a move further clockwise by − 7π 4, from π 4. The radius, 3 2, is the same. Transforming Equations between Polar and Rectangular Forms We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation. Given an equation in polar form, graph it using a graphing calculator. 1. Change the MODE to POL, representing polar form. 2. Press the Y= button to bring up a screen allowing the input of six equations: r1, r2,..., r6. 3. Enter the polar equation, set equal to r. 4. Press GRAPH. Example 10.18 Writing a Cartesian Equation in Polar Form Write the Cartesian equation x2 + y2 = 9 in polar form. Solution The goal is to eliminate x and y from the equation and introduce r and θ. Ideally, we would write the equation r as a function of θ. To obtain the polar form, we will use the relationships between (x, y) and (r, θ). Since
x = rcos θ and y = rsin θ, we can substitute and solve for r. (rcos θ)2 + (rsin θ)2 = 9 r 2 cos2 θ + r 2 sin2 θ = 9 r 2(cos2 θ + sin2 θ) = 9 r 2(1) = 9 r = ± 3 Substitute cos2 θ + sin2 θ = 1. Use the square root property. Thus, x2 + y2 = 9, r = 3, and r = − 3 should generate the same graph. See Figure 10.63. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1103 Figure 10.63 (a) Cartesian form x2 + y2 = 9 (b) Polar form r = 3 To graph a circle in rectangular form, we must first solve for y. x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x2 Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form Y1 = 9 − x2 and Y2 = − 9 − x2. Press GRAPH. Example 10.19 Rewriting a Cartesian Equation as a Polar Equation Rewrite the Cartesian equation x2 + y2 = 6y as a polar equation. Solution This equation appears similar to the previous example, but it requires different steps to convert the equation. We can still follow the same procedures we have already learned and make the following substitutions: r 2 = 6y r 2 = 6rsin θ r 2 − 6rsin θ = 0 r(r − 6sin θ) = 0 r = 0 or r = 6sin θ Use x2 + y2 = r 2. Substitute y = rsin θ. Set equal to 0. Factor and solve. We reject r = 0, as it only represents one point, (0, 0). 1104 Chapter 10 Further Applications of Trigonometry Therefore, the equations x2 + y2 = 6y and r = 6sin θ should give us the same graph. See Figure 10.64. Figure 10.64 (a) Cartesian form x2 + y2 = 6y (
b) polar form r = 6sin θ The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical. Example 10.20 Rewriting a Cartesian Equation in Polar Form Rewrite the Cartesian equation y = 3x + 2 as a polar equation. Solution We will use the relationships x = rcos θ and y = rsin θ. y = 3x + 2 rsin θ = 3rcos θ + 2 rsin θ − 3rcos θ = 2 r(sin θ − 3cos θ) = 2 r = 2 sin θ − 3cos θ Isolate r. Solve for r. 10.14 Rewrite the Cartesian equation y2 = 3 − x2 in polar form. Identify and Graph Polar Equations by Converting to Rectangular Equations We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1105 Example 10.21 Graphing a Polar Equation by Converting to a Rectangular Equation Covert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding graph. Solution The conversion is r = 2sec θ r = 2 rcos θ = 2 x = 2 cos θ Notice that the equation r = 2sec θ drawn on the polar grid is clearly the same as the vertical line x = 2 drawn on the rectangular grid (see Figure 10.65). Just as x = c is the standard form for a vertical line in rectangular form, r = csec θ is the standard form for a vertical line in polar form. Figure 10.65 (a) Polar grid (b) Rectangular coordinate system A similar discussion would demonstrate that the graph of the function r = 2csc θ will be the horizontal line y = 2. In fact, r = ccsc θ is the standard form for a horizontal line in polar form, corresponding to the rectangular form y = c. Example 10.22 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation
r = 3 1 − 2cos θ as a Cartesian equation. Solution 1106 Chapter 10 Further Applications of Trigonometry The goal is to eliminate θ and r, and introduce x and y. We clear the fraction, and then use substitution. In order to replace r with x and y, we must use the expression x2 + y2 = r 2. 3 1 − 2cos θ r = r(1 − 2cos θ − 2x = 3 r⎛ ⎝1 − 2⎛ ⎝ Use cos θ = x r to eliminate θ. r = 3 + 2x r 2 = (3 + 2x)2 x2 + y2 = (3 + 2x)2 Isolate r. Square both sides. Use x2 + y2 = r 2. The Cartesian equation is x2 + y2 = (3 + 2x)2. However, to graph it, especially using a graphing calculator or computer program, we want to isolate y. x2 + y2 = (3 + 2x)2 y2 = (3 + 2x)2 − x2 y = ± (3 + 2x)2 − x2 When our entire equation has been changed from r and θ to x and y, we can stop, unless asked to solve for y or simplify. See Figure 10.66. Figure 10.66 The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1107 In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation. x2 + y2 = (3 + 2x)2 x2 + y2 − (3 + 2x)2 = 0 x2 + y2 − (9 + 12x + 4x2) = 0 x2 + y2 − 9 − 12x − 4x2 = 0 − 3x2 − 12x + y2 =
9 3x2 + 12x − y2 = − 9 3(x2 + 4x + ) − y2 = − 9 3(x2 + 4x + 4) − y2 = − 9 + 12 3(x + 2)2 − y2 = 3 (x + 2)2 − y2 3 = 1 Multiply through by −1. Organize terms to complete the square for x. 10.15 Rewrite the polar equation r = 2sin θ in Cartesian form. Example 10.23 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = sin(2θ) in Cartesian form. Solution r = sin(2θ) r = 2sin θcos θ r = 2 2xy r 2 r 3 = 2xy ⎝ x2 + y2⎞ ⎛ ⎠ = 2xy 3 Use the double angle identity for sine. x Use cos θ = r and sin θ = y r. Simplify. Multiply both sides by r 2. As x2 + y2 = r 2, r = x2 + y2. This equation can also be written as 3 2 ⎝x2 + y2⎞ ⎛ ⎠ = 2xy or x2 + y2 = ⎛ ⎝2xy⎞ ⎠ 2 3 Access these online resources for additional instruction and practice with polar coordinates. • Introduction to Polar Coordinates (http://openstaxcollege.org/l/intropolar) • Comparing Polar and Rectangular Coordinates (http://openstaxcollege.org/l/polarrect) 1108 Chapter 10 Further Applications of Trigonometry 10.3 EXERCISES Verbal How are polar coordinates different from rectangular 156. coordinates? How are the polar axes different from the x- and y- 157. axes of the Cartesian plane? 158. Explain how polar coordinates are graphed. 159. How are the points ⎛ ⎝3, ⎠ and ⎛ ⎞ ⎝−3, π 2 ⎞ ⎠ related? π 2 160. Explain why the points ⎛ ⎝−3, ⎞ ⎠ and ⎛ ⎝3, − π 2 ⎞ ⎠ are π 2 the same. Al
gebraic the following exercises, convert the given polar For to Cartesian coordinates with r > 0 and coordinates 0 ≤ θ ≤ 2π. Remember to consider the quadrant in which is located when determining θ for the the given point point. 161. ⎛ ⎝7, 7π 6 ⎞ ⎠ 162. (5, π) 163. ⎛ ⎝6, − ⎞ ⎠ π 4 164. ⎛ ⎝−3, ⎞ ⎠ π 6 165. ⎛ ⎝4, 7π 4 ⎞ ⎠ For the following exercises, convert the given Cartesian coordinates to polar coordinates with r > 0, 0 ≤ θ < 2π. Remember to consider the quadrant in which the given point is located. 166. (4, 2) 167. (−4, 6) 168. (3, −5) 169. (−10, −13) 170. (8, 8) For the following exercises, convert the given Cartesian equation to a polar equation. This content is available for free at https://cnx.org/content/col11758/1.5 171. x = 3 172. y = 4 173. y = 4x2 174. y = 2x4 175. x2 + y2 = 4y 176. x2 + y2 = 3x 177. x2 − y2 = x 178. x2 − y2 = 3y 179. x2 + y2 = 9 180. x2 = 9y 181. y2 = 9x 182. 9xy = 1 the following exercises, convert For the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented. 183. r = 3sin θ 184. r = 4cos θ 185. 186. r = 4 sin θ + 7cos θ r = 6 cos θ + 3sin θ 187. r = 2sec θ 188. r = 3csc θ 189. r = rcos θ + 2 190. r 2 = 4sec θ csc θ 191. r = 4 192. r 2 = 4 193. Chapter 10 Further Applications of Trigonometry 1109 r = 1 4cos θ − 3sin θ 194. r = 3 cos θ − 5sin θ
Graphical For the following exercises, find the polar coordinates of the point. 195. 196. 197. 198. 199. For the following exercises, plot the points. 200. ⎛ ⎝−2, ⎞ ⎠ π 3 201. ⎛ ⎝−1, − ⎞ ⎠ π 2 202. ⎛ ⎝3.5, 7π 4 ⎞ ⎠ 203. ⎛ ⎝−4, ⎞ ⎠ π 3 204. ⎛ ⎝5, ⎞ ⎠ π 2 205. ⎛ ⎝4, −5π 4 ⎞ ⎠ 206. ⎛ ⎝3, 5π 6 ⎞ ⎠ 207. ⎛ ⎝−1.5, 7π 6 ⎞ ⎠ 208. ⎛ ⎝−2, ⎞ ⎠ π 4 1110 209. ⎛ ⎝1, 3π 2 ⎞ ⎠ For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis. 210. 5x − y = 6 211. 2x + 7y = − 3 Chapter 10 Further Applications of Trigonometry Use a graphing calculator to find the polar 228. coordinates of (−2, 0) in radians. Round to the nearest hundredth. Extensions 229. Describe the graph of r = asec θ; a > 0. 230. Describe the graph of r = asec θ; a < 0. x2 + ⎛ ⎝y − 1⎞ ⎠ 2 = 1 231. Describe the graph of r = acsc θ; a > 0. 2 = 13 232. Describe the graph of r = acsc θ; a < 0. 233. What polar equations will give an oblique line? For the following exercise, graph the polar inequality. 212. 213. (x + 2)2 + ⎛ ⎝y + 3⎞ ⎠ 214. x = 2 215. x2 + y2 = 5y 216. x2 + y2 = 3x 234. r < 4 235. 0 ≤ θ ≤ π 4 236. θ = 2373 238 239. − For the following
exercises, convert the equation from polar to rectangular form and graph on the rectangular plane. 217. r = 6 218. r = − 4 219. θ = − 2π 3 220. θ = π 4 221. r = sec θ 222. r = −10sin θ 223. r = 3cos θ Technology 224. coordinates of ⎛ Use a graphing calculator to find the rectangular ⎞ ⎠. Round to the nearest thousandth. ⎝2, − π 5 225. coordinates of ⎛ Use a graphing calculator to find the rectangular ⎞ ⎠. Round to the nearest thousandth. ⎝−3, 3π 7 Use a graphing calculator to find the polar 226. coordinates of (−7, 8) in degrees. Round to the nearest thousandth. Use a graphing calculator to find the polar 227. coordinates of (3, − 4) in degrees. Round to the nearest hundredth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1111 10.4 \| Polar Coordinates: Graphs Learning Objectives In this section you will: 10.4.1 Test polar equations for symmetry. 10.4.2 Graph polar equations by plotting points. The planets move through space in elliptical, periodic orbits about the sun, as shown in Figure 10.67. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as (r, θ). We interpret r as the distance from the sun and θ as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. Figure 10.67 Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA/JPL-Caltech) Testing Polar Equations for Symmetry Just as a rectangular equation such as y = x2 describes the relationship between x and y on a Cartesian grid, a polar equation describes a relationship between r and θ on a polar grid. Recall that the coordinate pair (r, θ) indicates that we move counterclockwise from the polar
axis (positive x-axis) by an angle of θ, and extend a ray from the pole (origin) r units in the direction of θ. All points that satisfy the polar equation are on the graph. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of r) to determine the graph of a polar equation. In the first test, we consider symmetry with respect to the line θ = (y-axis). We replace (r, θ) with (−r, − θ) to π 2 determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation r = 2sin θ; 1112 Chapter 10 Further Applications of Trigonometry r = 2sin θ −r = 2sin(−θ) −r = −2sin θ r = 2sin θ Replace (r, θ) with (−r, −θ). Identity: sin(−θ) = −sin θ. Multiply both sides by−1. This equation exhibits symmetry with respect to the line θ = π 2. In the second test, we consider symmetry with respect to the polar axis ( x -axis). We replace (r, θ) with (r, − θ) or (−r, π − θ) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation r = 1 − 2cos θ. r = 1 − 2cos θ r = 1 − 2cos(−θ) r = 1 − 2cos θ Replace (r, θ) with (r, −θ). Even/Odd identity The graph of this equation exhibits symmetry with respect to the polar axis. In the third test, we consider symmetry with respect to the pole (origin). We replace (r, θ) with (−r, θ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are
given the equation r = 2sin(3θ). r = 2sin(3θ) −r = 2sin(3θ) The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests, the polar axis, or the pole. In these does not necessarily indicate that a graph will not be symmetric about the line θ = π 2 instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. Symmetry Tests A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 10.68. Figure 10.68 (a) A graph is symmetric with respect to the line θ = (y-axis) if replacing (r, θ) with ( − r, − θ) π 2 yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing (r, θ) with (r, − θ) or (−r, π−θ) yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing (r, θ) with (−r, θ) yields an equivalent equation. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1113 Given a polar equation, test for symmetry. 1. Substitute the appropriate combination of components for (r, θ) : (−r, − θ) for θ = (r, − θ) for polar axis symmetry; and (−r, θ) for symmetry with respect to the pole. symmetry; π 2 2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry. Example 10.24 Testing a Polar Equation for Symmetry Test the equation r = 2sin θ for symmetry. Solution Test for each of the three types of symmetry. 1) Replacing (r, θ) with ( − r, − θ
) yields the same result. Thus, the graph is symmetric with respect to the line θ =. π 2 2) Replacing θ with − θ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. −r = 2sin(−θ) −r = −2sin θ Even-odd identity r = 2sin θ Multiply by −1 Passed r = 2sin(−θ) r = −2sin θ r = −2sin θ ≠ 2sin θ Failed Even-odd identity 3) Replacing r with – r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. −r = 2sin θ r = −2sin θ ≠ 2sin θ Failed Table 10.1 Analysis Using a graphing calculator, we can see that the equation r = 2sin θ is a circle centered at (0, 1) with radius. We can also see that the graph is not symmetric with the polar r = 1 and is indeed symmetric to the line θ = π 2 axis or the pole. See Figure 10.69. 1114 Chapter 10 Further Applications of Trigonometry Figure 10.69 10.16 Test the equation for symmetry: r = − 2cos θ. Graphing Polar Equations by Plotting Points To graph in the rectangular coordinate system we construct a table of x and y values. To graph in the polar coordinate system we construct a table of θ and r values. We enter values of θ into a polar equation and calculate r. However, using the properties of symmetry and finding key values of θ and r means fewer calculations will be needed. Finding Zeros and Maxima To find the zeros of a polar equation, we solve for the values of θ that result in r = 0. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for x. We use the same process for polar equations. Set r = 0, and solve for θ. For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of θ into the equation that result in the maximum value of the trigonometric functions. Consider r = 5cos θ; the maximum distance between the curve and
the pole is 5 units. The maximum value of the cosine function is 1 when θ = 0, so our polar equation is 5cos θ, and the value θ = 0 will yield the maximum \|r\|. Similarly, the maximum value of the sine function is 1 when θ =, and if our polar equation is r = 5sin θ, the value π 2 will yield the maximum \|r\|. We may find additional information by calculating values of r when θ = 0. These θ = π 2 points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation. Example 10.25 Finding Zeros and Maximum Values for a Polar Equation Using the equation in Example 10.24, find the zeros and maximum \|r\| and, if necessary, the polar axis intercepts of r = 2sin θ. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1115 Solution To find the zeros, set r equal to zero and solve for θ. 2sin θ = 0 sin θ = 0 θ = sin−1 0 θ = nπ where n is an integer Substitute any one of the θ values into the equation. We will use 0. r = 2sin(0) r = 0 The points (0, 0) and (0, ± nπ) are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept. To find the maximum value of the equation, look at the maximum value of the trigonometric function sin θ, which occurs when θ = π 2 ⎛ ± 2kπ resulting in sin ⎝ π 2 ⎞ ⎠ = 1. Substitute π 2 for θ. ⎞ ⎠ π 2 ⎛ r = 2sin ⎝ r = 2(1) r = 2 Analysis The point ⎛ ⎝2, π 2 ⎞ ⎠ will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See Table 10.1 and Figure 10.70. 1116 Chapter 10 Further Applications of Trigonometry θ 0 π 6
π 3 π 2 2π 3 5π 6 π r = 2sin θ r = 2sin(0) = 0 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 π 6 ⎛ r = 2sin ⎝ π 3 ⎞ ⎠ ≈ 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 2 π 2 r 0 1 1.73 2 ⎛ r = 2sin ⎝ 2π 3 ⎞ ⎠ ≈ 1.73 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 5π 6 r = 2sin(π) = 0 1 0 Table 10.1 Figure 10.70 Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros 10.17 and maximum values of \|r\| : r = 3cos θ. Investigating Circles Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1117 There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations. Formulas for the Equation of a Circle Some of the formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ, where a is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is \|a\| ⎠. Figure 2, or one-half the diameter. For r = acos θ, the center is ⎛ ⎝ ⎞ ⎠. For r = asin θ,, 0 the center is ⎛ ⎝, π⎞ a 2 a 2 10.71 shows the graphs of these four
circles. Figure 10.71 Example 10.26 Sketching the Graph of a Polar Equation for a Circle Sketch the graph of r = 4cos θ. Solution First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the ± kπ. zeros and maximum \|r\| for r = 4cos θ. First, set r = 0, and solve for θ. Thus, a zero occurs at θ = π 2 A key point to plot is ⎛ ⎝0, ⎞ ⎠. π 2 To find the maximum value of r, note that the maximum value of the cosine function is 1 when θ = 0 ± 2kπ. Substitute θ = 0 into the equation: r = 4cos θ r = 4cos(0) r = 4(1) = 4 The maximum value of the equation is 4. A key point to plot is (4, 0). As r = 4cos θ is symmetric with respect to the polar axis, we only need to calculate r-values for θ over the interval [0, π]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to Table 10.2. The graph is shown in Figure 10.72. 1118 Chapter 10 Further Applications of Trigonometry π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π 3.46 2.83 2 0 −2 −2.83 −3.46 4 θ r 0 4 Table 10.2 Figure 10.72 Investigating Cardioids While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own. Formulas for a Cardioid The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a b = 1. The cardioid graph passes through the pole, as we can see in Figure 10.73. Figure 10.73 This content is available for free at https://

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