Solutions to Exercises, Section 5.6

Transcription

1 Instructor s Solutions Manual, Section 5.6 Exercise 1 Solutions to Exercises, Section 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 ( ) (b) Note that 7 2 = 49. Using a calculator working in degrees, we have cos

2 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 ( ) (b) Note that 5 2 = 25. Using a calculator working in radians, we have cos

3 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 ( ) (b) Note that 4 2 = 16. Using a calculator working in radians, we have sin

4 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 ( ) (b) Note that ( 8) 2 = 64. Using a calculator working in degrees, we have sin

5 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 = and sin π 2 8 = [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 =

6 Instructor s Solutions Manual, Section 5.6 Exercise 6 6. sin( π 8 ) sin( π 8 ) = sin π 8 = 2 2 2

7 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 ( = 1 2 = = Because sin π 12 > 0, taking square roots of both sides of the equation above gives sin π =. 2 ) 2

8 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 = = ) 2 Because cos π 8 above gives > 0, taking square roots of both sides of the equation cos π =. 2

9 Instructor s Solutions Manual, Section 5.6 Exercise 9 9. sin( π 12 ) sin( π 12 ) = sin π 12 = 2 3 2

10 Instructor s Solutions Manual, Section 5.6 Exercise cos( π 8 ) cos( π 8 ) = cos π 8 =

11 Instructor s Solutions Manual, Section 5.6 Exercise tan π 12 tan π 12 = sin π 12 cos π = = = = 2 3

12 Instructor s Solutions Manual, Section 5.6 Exercise tan π 8 tan π 8 = sin π 8 cos π = = = = = 2 1

13 Instructor s Solutions Manual, Section 5.6 Exercise tan( π 12 ) tan( π 12 ) = tan π 12 = (2 3) = 3 2

14 Instructor s Solutions Manual, Section 5.6 Exercise tan( π 8 ) tan( π 8 ) = tan π 8 = ( 2 1) = 1 2

15 Instructor s Solutions Manual, Section 5.6 Exercise cos 25π 12 Because 25π 12 = π π, we have cos 25π 12 = cos( π π) = cos π =. 2

16 Instructor s Solutions Manual, Section 5.6 Exercise cos 17π 8 Because 17π 8 = π 8 + 2π, we have cos 17π 8 = cos( π 8 + 2π) = cos π =. 2

17 Instructor s Solutions Manual, Section 5.6 Exercise sin 25π 12 Because 25π 12 = π π, we have sin 25π 12 = sin( π π) = sin π =. 2

18 Instructor s Solutions Manual, Section 5.6 Exercise sin 17π 8 Because 17π 8 = π 8 + 2π, we have sin 17π 8 = sin( π 8 + 2π) = sin π =. 2

19 Instructor s Solutions Manual, Section 5.6 Exercise tan 25π 12 Because 25π 12 = π π, we have tan 25π 12 = tan( π π) = tan π 12 = 2 3.

20 Instructor s Solutions Manual, Section 5.6 Exercise tan 17π 8 Because 17π 8 = π 8 + 2π, we have tan 17π 8 = tan( π 8 + 2π) = tan π 8 = 2 1.

21 Instructor s Solutions Manual, Section 5.6 Exercise cos 13π 12 Because 13π 12 = π 12 + π, we have cos 13π 12 = cos( π 12 + π) = cos π =. 2

22 Instructor s Solutions Manual, Section 5.6 Exercise cos 9π 8 Because 9π 8 = π 8 + π, we have cos 9π 8 = cos( π 8 + π) = cos π =. 2

23 Instructor s Solutions Manual, Section 5.6 Exercise sin 13π 12 Because 13π 12 = π 12 + π, we have sin 13π 12 = sin( π 12 + π) = sin π =. 2

24 Instructor s Solutions Manual, Section 5.6 Exercise sin 9π 8 Because 9π 8 = π 8 + π, we have sin 9π 8 = sin( π 8 + π) = sin π =. 2

25 Instructor s Solutions Manual, Section 5.6 Exercise tan 13π 12 Because 13π 12 = π 12 + π, we have tan 13π 12 = tan( π 12 + π) = tan π 12 = 2 3.

26 Instructor s Solutions Manual, Section 5.6 Exercise tan 9π 8 Because 9π 8 = π 8 + π, we have tan 9π 8 = tan( π 8 + π) = tan π 8 = 2 1.

27 Instructor s Solutions Manual, Section 5.6 Exercise cos 5π 12 cos 5π 12 = sin( π 2 5π 12 ) = sin π = 2

28 Instructor s Solutions Manual, Section 5.6 Exercise cos 3π 8 cos 3π 8 = sin( π 2 3π 8 ) = sin π 8 = 2 2 2

29 Instructor s Solutions Manual, Section 5.6 Exercise cos( 5π 12 ) cos( 5π 5π ) = cos 12 = 2

30 Instructor s Solutions Manual, Section 5.6 Exercise cos( 3π 8 ) cos( 3π 8 ) = cos 3π 8 = 2 2 2

31 Instructor s Solutions Manual, Section 5.6 Exercise sin 5π 12 sin 5π 12 = cos( π 2 5π 12 ) = cos π = 2

32 Instructor s Solutions Manual, Section 5.6 Exercise sin 3π 8 sin 3π 8 = cos( π 2 3π 8 ) = cos π 8 =

33 Instructor s Solutions Manual, Section 5.6 Exercise sin( 5π 12 ) sin( 5π 12 ) = sin 5π 12 =

34 Instructor s Solutions Manual, Section 5.6 Exercise sin( 3π 8 ) sin( 3π 8 ) = sin 3π 8 =

35 Instructor s Solutions Manual, Section 5.6 Exercise tan 5π 12 tan 5π 12 = 1 tan( π 2 5π 12 ) = 1 tan π 12 1 = = = = 2 + 3

36 Instructor s Solutions Manual, Section 5.6 Exercise tan 3π 8 tan 3π 8 = 1 tan( π 2 3π 8 ) = 1 tan π 8 1 = = = 2 1 = 2 + 1

37 Instructor s Solutions Manual, Section 5.6 Exercise tan( 5π 12 ) tan( 5π 5π 12 ) = tan 12 = 2 3

38 Instructor s Solutions Manual, Section 5.6 Exercise tan( 3π 8 ) tan( 3π 8 ) = tan 3π 8 = 2 1

39 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

40 Instructor s Solutions Manual, Section 5.6 Exercise tan( ν) tan( ν) = tan ν = ( 3) = 3

41 Instructor s Solutions Manual, Section 5.6 Exercise 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.

42 Instructor s Solutions Manual, Section 5.6 Exercise 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 =

43 Instructor s Solutions Manual, Section 5.6 Exercise cos( u) 5 cos( u) = cos u = 5

44 Instructor s Solutions Manual, Section 5.6 Exercise cos( ν) 10 cos( ν) = cos ν = 10

45 Instructor s Solutions Manual, Section 5.6 Exercise sin u sin u = 1 cos 2 u = = = 2 5 = 2 5 5

46 Instructor s Solutions Manual, Section 5.6 Exercise sin ν Because π 2 <ν<π, we know that sin ν>0. Thus sin ν = 1 cos 2 ν = = 10 = 3 10 =

47 Instructor s Solutions Manual, Section 5.6 Exercise sin( u) sin( u) = sin u = 2 5 5

48 Instructor s Solutions Manual, Section 5.6 Exercise sin( ν) sin( ν) = sin ν =

49 Instructor s Solutions Manual, Section 5.6 Exercise cos(u + 4π) 5 cos(u + 4π) = cos u = 5

50 Instructor s Solutions Manual, Section 5.6 Exercise cos(ν 6π) 10 cos(ν 6π) = cos ν = 10

51 Instructor s Solutions Manual, Section 5.6 Exercise sin(u 6π) sin(u 6π) = sin u = 2 5 5

52 Instructor s Solutions Manual, Section 5.6 Exercise sin(ν + 10π) sin(ν + 10π) = sin ν =

53 Instructor s Solutions Manual, Section 5.6 Exercise tan(u + 8π) tan(u + 8π) = tan u = 2

54 Instructor s Solutions Manual, Section 5.6 Exercise tan(ν 4π) tan(ν 4π) = tan ν = 3

55 Instructor s Solutions Manual, Section 5.6 Exercise cos(u 3π) 5 cos(u 3π) = cos u = 5

56 Instructor s Solutions Manual, Section 5.6 Exercise cos(ν + 5π) 10 cos(ν + 5π) = cos ν = 10

57 Instructor s Solutions Manual, Section 5.6 Exercise sin(u + 5π) sin(u + 5π) = sin u = 2 5 5

58 Instructor s Solutions Manual, Section 5.6 Exercise sin(ν 7π) sin(ν 7π) = sin ν =

59 Instructor s Solutions Manual, Section 5.6 Exercise tan(u 9π) tan(u 9π) = tan u = 2

60 Instructor s Solutions Manual, Section 5.6 Exercise tan(ν + 3π) tan(ν + 3π) = tan ν = 3

61 Instructor s Solutions Manual, Section 5.6 Exercise cos( π 2 u) cos( π 2 u) = sin u = 2 5 5

62 Instructor s Solutions Manual, Section 5.6 Exercise cos( π 2 ν) cos( π 2 ν) = sin ν =

63 Instructor s Solutions Manual, Section 5.6 Exercise sin( π 2 u) sin ( π 2 u) = cos u = 5 5

64 Instructor s Solutions Manual, Section 5.6 Exercise sin( π 2 ν) sin ( π 2 ν) = cos ν = 10 10

65 Instructor s Solutions Manual, Section 5.6 Exercise tan( π 2 u) tan( π 2 u) = 1 tan u = 1 2

66 Instructor s Solutions Manual, Section 5.6 Exercise tan( π 2 ν) tan( π 2 ν) = 1 tan ν = 1 3

67 Instructor s Solutions Manual, Section 5.6 Problem 67 Solutions to Problems, Section Show that (cos θ + sin θ) 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 = cos θ sin θ

68 Instructor s Solutions Manual, Section 5.6 Problem 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.

69 Instructor s Solutions Manual, Section 5.6 Problem 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

70 Instructor s Solutions Manual, Section 5.6 Problem 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.

71 Instructor s Solutions Manual, Section 5.6 Problem 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 = 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.

72 Instructor s Solutions Manual, Section 5.6 Problem 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.

73 Instructor s Solutions Manual, Section 5.6 Problem 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.

74 Instructor s Solutions Manual, Section 5.6 Problem 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.

75 Instructor s Solutions Manual, Section 5.6 Problem 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.

76 Instructor s Solutions Manual, Section 5.6 Problem 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.

77 Instructor s Solutions Manual, Section 5.6 Problem 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.

78 Instructor s Solutions Manual, Section 5.6 Problem 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 = 19. Thus cos 71 = cos(90 17) = sin 19 and sin 71 = sin(90 17) = cos 19.

79 Instructor s Solutions Manual, Section 5.6 Problem 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

80 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.

81 Instructor s Solutions Manual, Section 5.6 Problem 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

82 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

83 Instructor s Solutions Manual, Section 5.6 Problem Show that cos(π θ) = cos θ for every angle θ. cos(π θ) = cos(θ π) = cos θ

84 Instructor s Solutions Manual, Section 5.6 Problem Show that sin(π θ) = sin θ for every angle θ. sin(π θ) = sin(θ π) = ( sin θ) = sin θ

85 Instructor s Solutions Manual, Section 5.6 Problem Show that cos(x + π 2 ) = sin x for every number x. cos(x + π 2 ) = cos( π 2 ( x)) = sin( x) = sin x

86 Instructor s Solutions Manual, Section 5.6 Problem Show that sin(t + π 2 ) = cos t for every number t. sin(t + π 2 ) = sin( π 2 ( t)) = cos( t) = cos t

87 Instructor s Solutions Manual, Section 5.6 Problem 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

88 Instructor s Solutions Manual, Section 5.6 Problem 85 tan(θ + π 2 ) = 1 tan θ.