Instructor s Solutions Manual, Section 5.6 Exercise 1 Solutions to Exercises, Section 5.6 1. For θ = 7, evaluate each of the following: (a) cos 2 θ (b) cos(θ 2 ) [Exercises 1 and 2 emphasize that cos 2 θ does not equal cos(θ 2 ).] (a) Using a calculator working in degrees, we have cos 2 7 = (cos 7 ) 2 (0.992546) 2 0.985148. (b) Note that 7 2 = 49. Using a calculator working in degrees, we have cos 49 0.656059.
Instructor s Solutions Manual, Section 5.6 Exercise 2 2. For θ = 5 radians, evaluate each of the following: (a) cos 2 θ (b) cos(θ 2 ) (a) Using a calculator working in radians, we have cos 2 5 = (cos 5) 2 (0.283662) 2 0.0804642. (b) Note that 5 2 = 25. Using a calculator working in radians, we have cos 25 0.991203.
Instructor s Solutions Manual, Section 5.6 Exercise 3 3. For θ = 4 radians, evaluate each of the following: (a) sin 2 θ (b) sin(θ 2 ) [Exercises 3 and 4 emphasize that sin 2 θ does not equal sin(θ 2 ).] (a) Using a calculator working in radians, we have sin 2 4 = (sin 4) 2 ( 0.756802) 2 0.57275. (b) Note that 4 2 = 16. Using a calculator working in radians, we have sin 16 0.287903.
Instructor s Solutions Manual, Section 5.6 Exercise 4 4. For θ = 8, evaluate each of the following: (a) sin 2 θ (b) sin(θ 2 ) (a) Using a calculator working in degrees, we have sin 2 ( 8) = ( sin( 8) ) 2 ( 0.139173) 2 0.0193692. (b) Note that ( 8) 2 = 64. Using a calculator working in degrees, we have sin 64 0.898794.
Instructor s Solutions Manual, Section 5.6 Exercise 5 In Exercises 5 38, find exact expressions for the indicated quantities, given that cos π 12 = 2 + 3 and sin π 2 8 = 2 2. 2 [These values for cos π 12 and sin π 8 Section 6.3.] will be derived in Examples 4 and 5 in 5. cos( π 12 ) cos( π 12 ) = cos π 12 = 2 + 3 2
Instructor s Solutions Manual, Section 5.6 Exercise 6 6. sin( π 8 ) sin( π 8 ) = sin π 8 = 2 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 7 7. sin π 12 We know that cos 2 π 12 + sin2 π 12 = 1. Thus sin 2 π 12 = 1 cos2 π 12 ( 2 + 3 = 1 2 = 1 2 + 3 4 = 2 3. 4 Because sin π 12 > 0, taking square roots of both sides of the equation above gives sin π 2 3 12 =. 2 ) 2
Instructor s Solutions Manual, Section 5.6 Exercise 8 8. cos π 8 We know that cos 2 π 8 + sin2 π 8 = 1. Thus cos 2 π 8 = 1 sin2 π 8 ( 2 2 = 1 2 = 1 2 2 4 = 2 + 2. 4 ) 2 Because cos π 8 above gives > 0, taking square roots of both sides of the equation cos π 2 + 2 8 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 9 9. sin( π 12 ) sin( π 12 ) = sin π 12 = 2 3 2
Instructor s Solutions Manual, Section 5.6 Exercise 10 10. cos( π 8 ) cos( π 8 ) = cos π 8 = 2 + 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 11 11. tan π 12 tan π 12 = sin π 12 cos π 12 2 3 = 2 + 3 = 2 3 2 3 2 + 3 2 3 = 2 3 4 3 = 2 3
Instructor s Solutions Manual, Section 5.6 Exercise 12 12. tan π 8 tan π 8 = sin π 8 cos π 8 2 2 = 2 + 2 = 2 2 2 2 2 + 2 2 2 = 2 2 4 2 = 2 2 2 2 2 = 2 1
Instructor s Solutions Manual, Section 5.6 Exercise 13 13. tan( π 12 ) tan( π 12 ) = tan π 12 = (2 3) = 3 2
Instructor s Solutions Manual, Section 5.6 Exercise 14 14. tan( π 8 ) tan( π 8 ) = tan π 8 = ( 2 1) = 1 2
Instructor s Solutions Manual, Section 5.6 Exercise 15 15. cos 25π 12 Because 25π 12 = π 12 + 2π, we have cos 25π 12 = cos( π 12 + 2π) = cos π 12 2 + 3 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 16 16. cos 17π 8 Because 17π 8 = π 8 + 2π, we have cos 17π 8 = cos( π 8 + 2π) = cos π 8 2 + 2 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 17 17. sin 25π 12 Because 25π 12 = π 12 + 2π, we have sin 25π 12 = sin( π 12 + 2π) = sin π 12 2 3 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 18 18. sin 17π 8 Because 17π 8 = π 8 + 2π, we have sin 17π 8 = sin( π 8 + 2π) = sin π 8 2 2 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 19 19. tan 25π 12 Because 25π 12 = π 12 + 2π, we have tan 25π 12 = tan( π 12 + 2π) = tan π 12 = 2 3.
Instructor s Solutions Manual, Section 5.6 Exercise 20 20. tan 17π 8 Because 17π 8 = π 8 + 2π, we have tan 17π 8 = tan( π 8 + 2π) = tan π 8 = 2 1.
Instructor s Solutions Manual, Section 5.6 Exercise 21 21. cos 13π 12 Because 13π 12 = π 12 + π, we have cos 13π 12 = cos( π 12 + π) = cos π 12 2 + 3 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 22 22. cos 9π 8 Because 9π 8 = π 8 + π, we have cos 9π 8 = cos( π 8 + π) = cos π 8 2 + 2 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 23 23. sin 13π 12 Because 13π 12 = π 12 + π, we have sin 13π 12 = sin( π 12 + π) = sin π 12 2 3 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 24 24. sin 9π 8 Because 9π 8 = π 8 + π, we have sin 9π 8 = sin( π 8 + π) = sin π 8 2 2 =. 2
Instructor s Solutions Manual, Section 5.6 Exercise 25 25. tan 13π 12 Because 13π 12 = π 12 + π, we have tan 13π 12 = tan( π 12 + π) = tan π 12 = 2 3.
Instructor s Solutions Manual, Section 5.6 Exercise 26 26. tan 9π 8 Because 9π 8 = π 8 + π, we have tan 9π 8 = tan( π 8 + π) = tan π 8 = 2 1.
Instructor s Solutions Manual, Section 5.6 Exercise 27 27. cos 5π 12 cos 5π 12 = sin( π 2 5π 12 ) = sin π 2 3 12 = 2
Instructor s Solutions Manual, Section 5.6 Exercise 28 28. cos 3π 8 cos 3π 8 = sin( π 2 3π 8 ) = sin π 8 = 2 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 29 29. cos( 5π 12 ) cos( 5π 5π 2 3 12 ) = cos 12 = 2
Instructor s Solutions Manual, Section 5.6 Exercise 30 30. cos( 3π 8 ) cos( 3π 8 ) = cos 3π 8 = 2 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 31 31. sin 5π 12 sin 5π 12 = cos( π 2 5π 12 ) = cos π 2 + 3 12 = 2
Instructor s Solutions Manual, Section 5.6 Exercise 32 32. sin 3π 8 sin 3π 8 = cos( π 2 3π 8 ) = cos π 8 = 2 + 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 33 33. sin( 5π 12 ) sin( 5π 12 ) = sin 5π 12 = 2 + 3 2
Instructor s Solutions Manual, Section 5.6 Exercise 34 34. sin( 3π 8 ) sin( 3π 8 ) = sin 3π 8 = 2 + 2 2
Instructor s Solutions Manual, Section 5.6 Exercise 35 35. tan 5π 12 tan 5π 12 = 1 tan( π 2 5π 12 ) = 1 tan π 12 1 = 2 3 1 = 2 3 2 + 3 2 + 3 = 2 + 3 4 3 = 2 + 3
Instructor s Solutions Manual, Section 5.6 Exercise 36 36. tan 3π 8 tan 3π 8 = 1 tan( π 2 3π 8 ) = 1 tan π 8 1 = 2 1 1 2 + 1 = 2 1 2 1 2 + 1 = 2 1 = 2 + 1
Instructor s Solutions Manual, Section 5.6 Exercise 37 37. tan( 5π 12 ) tan( 5π 5π 12 ) = tan 12 = 2 3
Instructor s Solutions Manual, Section 5.6 Exercise 38 38. tan( 3π 8 ) tan( 3π 8 ) = tan 3π 8 = 2 1
Instructor s Solutions Manual, Section 5.6 Exercise 39 Suppose u and ν are in the interval ( π 2,π), with tan u = 2 and tan ν = 3. In Exercises 39 66, find exact expressions for the indicated quantities. 39. tan( u) tan( u) = tan u = ( 2) = 2
Instructor s Solutions Manual, Section 5.6 Exercise 40 40. tan( ν) tan( ν) = tan ν = ( 3) = 3
Instructor s Solutions Manual, Section 5.6 Exercise 41 41. cos u We know that 2 = tan u = sin u cos u. To find cos u, make the substitution sin u = 1 cos 2 u in the equation above (this substitution is valid because π 2 <u<π, which implies that sin u>0), getting 1 cos 2 = 2 u. cos u Now square both sides of the equation above, then multiply both sides by cos 2 u and rearrange to get the equation 5 cos 2 u = 1. Thus cos u = 1 5 (the possibility that cos u equals 1 5 is eliminated because π 2 <u<π, which implies that cos u<0). This can be written as cos u = 5 5.
Instructor s Solutions Manual, Section 5.6 Exercise 42 42. cos ν We know that 3 = tan ν = sin ν cos ν. To find cos ν, make the substitution sin ν = 1 cos 2 ν in the equation above (this substitution is valid because π 2 <ν<π, which implies that sin ν>0), getting 1 cos 3 = 2 ν. cos ν Now square both sides of the equation above, then multiply both sides by cos 2 ν and rearrange to get the equation 10 cos 2 ν = 1. Thus cos ν = 1 10 (the possibility that cos ν equals 1 10 is eliminated because π 2 <ν<π, which implies that cos ν<0). This can be written as cos ν = 10 10.
Instructor s Solutions Manual, Section 5.6 Exercise 43 43. cos( u) 5 cos( u) = cos u = 5
Instructor s Solutions Manual, Section 5.6 Exercise 44 44. cos( ν) 10 cos( ν) = cos ν = 10
Instructor s Solutions Manual, Section 5.6 Exercise 45 45. sin u sin u = 1 cos 2 u = = 4 5 1 1 5 = 2 5 = 2 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 46 46. sin ν Because π 2 <ν<π, we know that sin ν>0. Thus sin ν = 1 cos 2 ν = 1 1 10 9 = 10 = 3 10 = 3 10 10.
Instructor s Solutions Manual, Section 5.6 Exercise 47 47. sin( u) sin( u) = sin u = 2 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 48 48. sin( ν) sin( ν) = sin ν = 3 10 10
Instructor s Solutions Manual, Section 5.6 Exercise 49 49. cos(u + 4π) 5 cos(u + 4π) = cos u = 5
Instructor s Solutions Manual, Section 5.6 Exercise 50 50. cos(ν 6π) 10 cos(ν 6π) = cos ν = 10
Instructor s Solutions Manual, Section 5.6 Exercise 51 51. sin(u 6π) sin(u 6π) = sin u = 2 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 52 52. sin(ν + 10π) sin(ν + 10π) = sin ν = 3 10 10
Instructor s Solutions Manual, Section 5.6 Exercise 53 53. tan(u + 8π) tan(u + 8π) = tan u = 2
Instructor s Solutions Manual, Section 5.6 Exercise 54 54. tan(ν 4π) tan(ν 4π) = tan ν = 3
Instructor s Solutions Manual, Section 5.6 Exercise 55 55. cos(u 3π) 5 cos(u 3π) = cos u = 5
Instructor s Solutions Manual, Section 5.6 Exercise 56 56. cos(ν + 5π) 10 cos(ν + 5π) = cos ν = 10
Instructor s Solutions Manual, Section 5.6 Exercise 57 57. sin(u + 5π) sin(u + 5π) = sin u = 2 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 58 58. sin(ν 7π) sin(ν 7π) = sin ν = 3 10 10
Instructor s Solutions Manual, Section 5.6 Exercise 59 59. tan(u 9π) tan(u 9π) = tan u = 2
Instructor s Solutions Manual, Section 5.6 Exercise 60 60. tan(ν + 3π) tan(ν + 3π) = tan ν = 3
Instructor s Solutions Manual, Section 5.6 Exercise 61 61. cos( π 2 u) cos( π 2 u) = sin u = 2 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 62 62. cos( π 2 ν) cos( π 2 ν) = sin ν = 3 10 10
Instructor s Solutions Manual, Section 5.6 Exercise 63 63. sin( π 2 u) sin ( π 2 u) = cos u = 5 5
Instructor s Solutions Manual, Section 5.6 Exercise 64 64. sin( π 2 ν) sin ( π 2 ν) = cos ν = 10 10
Instructor s Solutions Manual, Section 5.6 Exercise 65 65. tan( π 2 u) tan( π 2 u) = 1 tan u = 1 2
Instructor s Solutions Manual, Section 5.6 Exercise 66 66. tan( π 2 ν) tan( π 2 ν) = 1 tan ν = 1 3
Instructor s Solutions Manual, Section 5.6 Problem 67 Solutions to Problems, Section 5.6 67. Show that (cos θ + sin θ) 2 = 1 + 2 cos θ sin θ for every number θ. [Expressions such as cos θ sin θ mean (cos θ)(sin θ), not cos(θ sin θ).] (cos θ + sin θ) 2 = cos 2 θ + sin 2 θ + 2 cos θ sin θ = 1 + 2 cos θ sin θ
Instructor s Solutions Manual, Section 5.6 Problem 68 68. Show that sin x 1 cos x = 1 + cos x sin x for every number x that is not an integer multiple of π. If x is any number, then sin 2 x = 1 cos 2 x = (1 + cos x)(1 cos x). If x is not an integer multiple of π, then neither sin x nor 1 cos x equals 0, which means that the equation above can be divided by sin x(1 cos x), giving sin x 1 cos x = 1 + cos x sin x.
Instructor s Solutions Manual, Section 5.6 Problem 69 69. Show that cos 3 θ + cos 2 θ sin θ + cos θ sin 2 θ + sin 3 θ = cos θ + sin θ for every number θ. [Hint: Try replacing the cos 2 θ term above with 1 sin 2 θ and replacing the sin 2 θ term above with 1 cos 2 θ.] cos 3 θ + cos 2 θ sin θ + cos θ sin 2 θ + sin 3 θ = cos 3 θ + (1 sin 2 θ) sin θ + cos θ(1 cos 2 θ) + sin 3 θ = cos 3 θ + sin θ sin 3 θ + cos θ cos 3 θ + sin 3 θ = cos θ + sin θ
Instructor s Solutions Manual, Section 5.6 Problem 70 70. Show that sin 2 θ = tan2 θ 1 + tan 2 θ for all θ except odd multiples of π 2. If θ is not an odd multiple of π 2, then tan θ is defined and sin 2 θ(1 + tan 2 θ) = sin 2 θ (1 + sin2 θ ) cos 2 θ ( cos = sin 2 2 θ + sin 2 θ ) θ cos 2 θ ( = sin 2 1 ) θ cos 2 θ ( sin θ ) 2 = cos θ = tan 2 θ. Dividing both sides of the equation above by 1 + tan 2 θ shows that sin 2 θ = tan2 θ 1 + tan 2 θ.
Instructor s Solutions Manual, Section 5.6 Problem 71 71. Find a formula for cos θ solely in terms of tan θ. Suppose θ is not an odd multiple of π 2. Using the result from the previous problem, we have cos 2 θ = 1 sin 2 θ = 1 tan2 θ 1 + tan 2 θ = 1 + tan2 θ 1 + tan 2 θ tan2 θ 1 + tan 2 θ = 1 1 + tan 2 θ. Thus 1 cos θ =± 1 + tan 2 θ, where the plus sign is chosen if cos θ>0 and the minus sign is chosen if cos θ<0.
Instructor s Solutions Manual, Section 5.6 Problem 72 72. Find a formula for tan θ solely in terms of sin θ. Suppose θ is not an odd multiple of π 2, which means that tan θ is defined. We have tan θ = sin θ cos θ =± 1 sin θ sin 2 θ, where the plus sign is chosen if cos θ>0 and the minus sign is chosen if cos θ<0.
Instructor s Solutions Manual, Section 5.6 Problem 73 73. Is cosine an even function, an odd function, or neither? Recall that cos( θ) = cos θ for every number θ. Thus cosine is an even function.
Instructor s Solutions Manual, Section 5.6 Problem 74 74. Is sine an even function, an odd function, or neither? Recall that sin( θ) = sin θ for every number θ. Thus sine is an odd function.
Instructor s Solutions Manual, Section 5.6 Problem 75 75. Is tangent an even function, an odd function, or neither? Recall that tan( θ) = tan θ for every number θ in the domain of tangent. Thus tangent is an odd function.
Instructor s Solutions Manual, Section 5.6 Problem 76 76. Explain why sin 3 + sin 357 = 0. The radius that makes an angle of 357 with the positive horizontal axis is the same as the radius that makes an angle of 3 with the positive horizontal axis. Thus sin 3 + sin 357 = sin 3 + sin( 3 ) = sin 3 sin 3 = 0.
Instructor s Solutions Manual, Section 5.6 Problem 77 77. Explain why cos 85 + cos 95 = 0. Using the trigonometric identity for cos(90 θ), we have cos 85 = cos(90 5) = sin 5. Using the trigonometric identity for cos(90 θ) and the trigonometric identity for sin( θ), we have cos 95 = cos ( 90 ( 5) ) = sin( 5) = sin 5. The two equations above show that cos 85 + cos 95 = 0. The following figure may also help explain the result: 1 Here the radius of the unit circle that makes an angle of 95 with the positive horizontal axis is the reflection through the vertical axis of the radius that makes an angle of 85 with the positive horizontal axis. Thus cos 95 = cos 85.
Instructor s Solutions Manual, Section 5.6 Problem 78 78. Pretend that you are living in the time before calculators and computers existed, and that you have a table showing the cosines and sines of 1,2,3, and so on, up to the cosine and sine of 45. Explain how you would find the cosine and sine of 71, which are beyond the range of your table. Note that 90 71 = 19. Thus cos 71 = cos(90 17) = sin 19 and sin 71 = sin(90 17) = cos 19.
Instructor s Solutions Manual, Section 5.6 Problem 79 79. Suppose n is an integer. Find formulas for sec(θ + nπ), csc(θ + nπ), and cot(θ + nπ) in terms of sec θ, csc θ, and cot θ. Recall that sec θ = 1 cos θ. Thus sec(θ + nπ) = = 1 cos(θ + nπ) 1 cos θ 1 cos θ if n is an even integer if n is an odd integer sec θ = sec θ if n is an even integer if n is an odd integer. The formula above is valid only if sec θ is defined, which means that θ must not be an odd multiple of π 2. Similarly, the following formula for csc θ is valid only if csc θ is defined, which means that θ must not be a multiple of π. Recall that csc θ = 1 sin θ. Thus
Instructor s Solutions Manual, Section 5.6 Problem 79 csc(θ + nπ) = = 1 sin(θ + nπ) 1 sin θ 1 sin θ if n is an even integer if n is an odd integer csc θ = csc θ if n is an even integer if n is an odd integer. Recall that cot θ = 1 tan θ. Thus cot(θ + nπ) = 1 tan(θ + nπ) = 1 tan θ = cot θ. The formula above is valid only if cot θ is defined, which means that θ must not be a multiple of π.
Instructor s Solutions Manual, Section 5.6 Problem 80 80. Restate all the results in boxes in the subsection on Trigonometric Identities Involving a Multiple of π in terms of degrees instead of in terms of radians. Trigonometric identities with (θ + 180) cos(θ + 180) = cos θ sin(θ + 180) = sin θ tan(θ + 180) = tan θ Trigonometric identities with (θ + 360) cos(θ + 360) = cos θ sin(θ + 360) = sin θ tan(θ + 360) = tan θ
Instructor s Solutions Manual, Section 5.6 Problem 80 Trigonometric formulas with (θ + 180n) cos(θ + 180n) cos θ if n is an even integer = cos θ if n is an odd integer sin(θ + 180n) sin θ if n is an even integer = sin θ if n is an odd integer tan(θ + 180n) = tan θ if n is an integer
Instructor s Solutions Manual, Section 5.6 Problem 81 81. Show that cos(π θ) = cos θ for every angle θ. cos(π θ) = cos(θ π) = cos θ
Instructor s Solutions Manual, Section 5.6 Problem 82 82. Show that sin(π θ) = sin θ for every angle θ. sin(π θ) = sin(θ π) = ( sin θ) = sin θ
Instructor s Solutions Manual, Section 5.6 Problem 83 83. Show that cos(x + π 2 ) = sin x for every number x. cos(x + π 2 ) = cos( π 2 ( x)) = sin( x) = sin x
Instructor s Solutions Manual, Section 5.6 Problem 84 84. Show that sin(t + π 2 ) = cos t for every number t. sin(t + π 2 ) = sin( π 2 ( t)) = cos( t) = cos t
Instructor s Solutions Manual, Section 5.6 Problem 85 85. Show that tan(θ + π 2 ) = 1 tan θ for every angle θ that is not an integer multiple of π 2. Interpret this result in terms of the characterization of the slopes of perpendicular lines. Suppose θ is not an integer multiple of π 2. Then tan(θ + π 2 ) = tan( π 2 ( θ)) = 1 tan( θ) = 1 tan θ. Note that tan θ equals the slope of the radius of the unit circle that makes an angle of θ with the positive horizontal axis, and tan(θ + π 2 ) equals the slope of the radius of the unit circle that makes an angle of θ + π 2 with the positive horizontal axis. These two radii are perpendicular to each other (because θ + π 2 is obtained by adding π 2 to θ), and thus the product of their slopes equals 1 (see Section 2.1). In other words, (tan θ) ( tan(θ + π 2 )) = 1, which is equivalent to the equation
Instructor s Solutions Manual, Section 5.6 Problem 85 tan(θ + π 2 ) = 1 tan θ.