Download Solutions to Exercises, Section 5.6

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) cos2 θ
(b) cos(θ 2 )
[Exercises 1 and 2 emphasize that cos2 θ does not equal cos(θ 2 ).]
solution
(a) Using a calculator working in degrees, we have
cos2 7◦ = (cos 7◦ )2 ≈ (0.992546)2 ≈ 0.985148.
(b) Note that 72 = 49. Using a calculator working in degrees, we have
cos 49◦ ≈ 0.656059.
Instructor’s Solutions Manual, Section 5.6
2.
Exercise 2
For θ = 5 radians, evaluate each of the following:
(a) cos2 θ
(b) cos(θ 2 )
solution
(a) Using a calculator working in radians, we have
cos2 5 = (cos 5)2 ≈ (0.283662)2 ≈ 0.0804642.
(b) Note that 52 = 25. Using a calculator working in radians, we have
cos 25 ≈ 0.991203.
Instructor’s Solutions Manual, Section 5.6
3.
Exercise 3
For θ = 4 radians, evaluate each of the following:
(a) sin2 θ
(b) sin(θ 2 )
[Exercises 3 and 4 emphasize that sin2 θ does not equal sin(θ 2 ).]
solution
(a) Using a calculator working in radians, we have
sin2 4 = (sin 4)2 ≈ (−0.756802)2 ≈ 0.57275.
(b) Note that 42 = 16. Using a calculator working in radians, we have
sin 16 ≈ −0.287903.
Instructor’s Solutions Manual, Section 5.6
4.
Exercise 4
For θ = −8◦ , evaluate each of the following:
(a) sin2 θ
(b) sin(θ 2 )
solution
(a) Using a calculator working in degrees, we have
2
sin2(−8)◦ = sin(−8)◦ ≈ (−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
√
√
2
+
3
2− 2
π
π
and sin 8 =
.
cos 12 =
2
2
[These values for cos
Section 6.3.]
π
12
and sin
π
8
will be derived in Examples 4 and 5 in
π
5. cos(− 12 )
solution
π
cos(− 12 )
= cos
π
12
=
2+
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 6
6. sin(− π8 )
solution
π
sin(− 8 )
= − sin
π
8
2−
=−
2
√
2
Instructor’s Solutions Manual, Section 5.6
Exercise 7
π
7. sin 12
solution We know that
π
π
cos2 12 + sin2 12 = 1.
Thus
π
π
sin2 12 = 1 − cos2 12
2 + √ 3 2
=1−
2
√
2+ 3
=1−
4
√
2− 3
.
=
4
π
Because sin 12 > 0, taking square roots of both sides of the equation
above gives
√
2− 3
π
.
sin 12 =
2
Instructor’s Solutions Manual, Section 5.6
Exercise 8
8. cos π8
solution We know that
π
π
cos2 8 + sin2 8 = 1.
Thus
π
π
cos2 8 = 1 − sin2 8
2 − √ 2 2
=1−
2
√
2− 2
=1−
4
√
2+ 2
.
=
4
Because cos
above gives
π
8
> 0, taking square roots of both sides of the equation
cos
π
8
=
2+
2
√
2
.
Instructor’s Solutions Manual, Section 5.6
Exercise 9
π
9. sin(− 12
)
solution
π
sin(− 12 )
= − sin
π
12
2−
=−
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 10
10. cos(− π8 )
solution
π
cos(− 8 )
= cos
π
8
=
2+
2
√
2
Instructor’s Solutions Manual, Section 5.6
π
11. tan 12
solution
π
tan 12 =
π
sin 12
π
cos 12
√
2− 3
=
√
2+ 3
√
√
2− 3
2− 3
= √ ·
√
2+ 3
2− 3
√
2− 3
= √
4−3
√
=2− 3
Exercise 11
Instructor’s Solutions Manual, Section 5.6
12. tan π8
solution
tan
π
8
=
sin π8
cos π8
√
2− 2
=
√
2+ 2
√
√
2− 2
2− 2
= √ ·
√
2+ 2
2− 2
√
2− 2
= √
4−2
√
√
2
2− 2
·√
= √
2
2
√
= 2−1
Exercise 12
Instructor’s Solutions Manual, Section 5.6
Exercise 13
π
13. tan(− 12
)
solution
π
π
tan(− 12 ) = − tan 12 = −(2 −
√
3) =
√
3−2
Instructor’s Solutions Manual, Section 5.6
Exercise 14
14. tan(− π8 )
solution
π
tan(− 8 ) = − tan
π
8
√
√
= −( 2 − 1) = 1 − 2
Instructor’s Solutions Manual, Section 5.6
Exercise 15
15. cos 25π
12
solution Because
25π
12
=
π
12
cos
+ 2π , we have
25π
12
π
= cos( 12 + 2π )
π
= cos 12
√
2+ 3
.
=
2
Instructor’s Solutions Manual, Section 5.6
Exercise 16
16. cos 17π
8
solution Because
17π
8
=
π
8
cos
+ 2π , we have
17π
8
π
= cos( 8 + 2π )
π
= cos 8
√
2+ 2
.
=
2
Instructor’s Solutions Manual, Section 5.6
Exercise 17
17. sin 25π
12
solution Because
25π
12
=
π
12
sin
+ 2π , we have
25π
12
π
= sin( 12 + 2π )
π
= sin 12
√
2− 3
.
=
2
Instructor’s Solutions Manual, Section 5.6
Exercise 18
18. sin 17π
8
solution Because
17π
8
=
π
8
sin
+ 2π , we have
17π
8
π
= sin( 8 + 2π )
π
= sin 8
√
2− 2
.
=
2
Instructor’s Solutions Manual, Section 5.6
Exercise 19
19. tan 25π
12
solution Because
25π
12
=
π
12
tan
+ 2π , we have
25π
12
π
= tan( 12 + 2π )
π
= tan 12
√
= 2 − 3.
Instructor’s Solutions Manual, Section 5.6
Exercise 20
20. tan 17π
8
solution Because
17π
8
=
π
8
tan
+ 2π , we have
17π
8
π
= tan( 8 + 2π )
π
= tan 8
√
= 2 − 1.
Instructor’s Solutions Manual, Section 5.6
Exercise 21
21. cos 13π
12
solution Because
13π
12
=
π
12
cos
+ π , we have
13π
12
π
= cos( 12 + π )
π
= − cos 12
√
2+ 3
.
=−
2
Instructor’s Solutions Manual, Section 5.6
Exercise 22
22. cos 9π
8
solution Because
9π
8
=
π
8
+ π , we have
cos
9π
8
π
= cos( 8 + π )
π
= − cos 8
√
2+ 2
.
=−
2
Instructor’s Solutions Manual, Section 5.6
Exercise 23
23. sin 13π
12
solution Because
13π
12
=
π
12
sin
+ π , we have
13π
12
π
= sin( 12 + π )
π
= − sin 12
√
2− 3
.
=−
2
Instructor’s Solutions Manual, Section 5.6
Exercise 24
24. sin 9π
8
solution Because
9π
8
=
π
8
+ π , we have
sin
9π
8
π
= sin( 8 + π )
π
= − sin 8
√
2− 2
.
=−
2
Instructor’s Solutions Manual, Section 5.6
Exercise 25
25. tan 13π
12
solution Because
13π
12
=
π
12
tan
+ π , we have
13π
12
π
= tan( 12 + π )
π
= tan 12
√
= 2 − 3.
Instructor’s Solutions Manual, Section 5.6
Exercise 26
26. tan 9π
8
solution Because
9π
8
=
π
8
+ π , we have
tan
9π
8
π
= tan( 8 + π )
π
= tan 8
√
= 2 − 1.
Instructor’s Solutions Manual, Section 5.6
Exercise 27
27. cos 5π
12
solution
cos
5π
12
=
π
sin( 2
−
5π
12 )
= sin
π
12
=
2−
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 28
28. cos 3π
8
solution
cos
3π
8
=
π
sin( 2
−
3π
8 )
= sin
π
8
=
2−
2
√
2
Instructor’s Solutions Manual, Section 5.6
Exercise 29
29. cos(− 5π
12 )
solution
5π
cos(− 12 )
= cos
5π
12
=
2−
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 30
30. cos(− 3π
8 )
solution
3π
cos(− 8 )
= cos
3π
8
=
2−
2
√
2
Instructor’s Solutions Manual, Section 5.6
Exercise 31
31. sin 5π
12
solution
sin
5π
12
=
π
cos( 2
−
5π
12 )
= cos
π
12
=
2+
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 32
32. sin 3π
8
solution
sin
3π
8
=
π
cos( 2
−
3π
8 )
= cos
π
8
=
2+
2
√
2
Instructor’s Solutions Manual, Section 5.6
Exercise 33
33. sin(− 5π
12 )
solution
5π
sin(− 12 )
= − sin
5π
12
2+
=−
2
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 34
34. sin(− 3π
8 )
solution
3π
sin(− 8 )
= − sin
3π
8
2+
=−
2
√
2
Instructor’s Solutions Manual, Section 5.6
Exercise 35
35. tan 5π
12
solution
tan
5π
12
=
1
tan( π2
−
=
1
π
tan 12
=
1
√
2− 3
5π
12 )
√
2+ 3
1
√ ·
√
=
2− 3 2+ 3
√
2+ 3
=
4−3
√
=2+ 3
Instructor’s Solutions Manual, Section 5.6
Exercise 36
36. tan 3π
8
solution
tan
3π
8
=
=
1
tan( π2
−
3π
8 )
1
tan π8
1
= √
2−1
√
2+1
1
·√
= √
2−1
2−1
√
2+1
=
2−1
√
= 2+1
Instructor’s Solutions Manual, Section 5.6
Exercise 37
37. tan(− 5π
12 )
solution
5π
tan(− 12 ) = − tan
5π
12
= −2 −
√
3
Instructor’s Solutions Manual, Section 5.6
Exercise 38
38. tan(− 3π
8 )
solution
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)
solution tan(−u) = − tan u = −(−2) = 2
Instructor’s Solutions Manual, Section 5.6
40. tan(−ν)
solution
tan(−ν) = − tan ν = −(−3) = 3
Exercise 40
Instructor’s Solutions Manual, Section 5.6
Exercise 41
41. cos u
solution We know that
−2 = tan u
=
sin u
.
cos u
√
To find cos u, make the substitution sin u = 1 − cos2 u in the equation
π
above (this substitution is valid because 2 < u < π , which implies that
sin u > 0), getting
√
1 − cos2 u
.
−2 =
cos u
Now square both sides of the equation above, then multiply both sides
by cos2 u and rearrange to get the equation
5 cos2 u = 1.
1
1
Thus cos u = − √5 (the possibility that cos u equals √5 is eliminated
π
because 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
solution We know that
−3 = tan ν
=
sin ν
.
cos ν
√
To find cos ν, make the substitution sin ν = 1 − cos2 ν in the equation
π
above (this substitution is valid because 2 < ν < π , which implies that
sin ν > 0), getting
√
1 − cos2 ν
.
−3 =
cos ν
Now square both sides of the equation above, then multiply both sides
by cos2 ν and rearrange to get the equation
10 cos2 ν = 1.
1
1
Thus cos ν = − √10 (the possibility that cos ν equals √10 is eliminated
π
because 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
43. cos(−u)
√
solution cos(−u) = cos u = −
5
5
Exercise 43
Instructor’s Solutions Manual, Section 5.6
Exercise 44
44. cos(−ν)
solution
√
cos(−ν) = cos ν = −
10
10
Instructor’s Solutions Manual, Section 5.6
45. sin u
solution
sin u = 1 − cos2 u
1
= 1−
5
4
=
5
2
= √
5
√
2 5
=
5
Exercise 45
Instructor’s Solutions Manual, Section 5.6
Exercise 46
46. sin ν
solution Because
π
2
< ν < π , we know that sin ν > 0. Thus
sin ν = 1 − cos2 ν
1
= 1−
10
9
=
10
3
= √
10
√
3 10
.
=
10
Instructor’s Solutions Manual, Section 5.6
47. sin(−u)
√
2 5
solution sin(−u) = − sin u = −
5
Exercise 47
Instructor’s Solutions Manual, Section 5.6
48. sin(−ν)
√
3 10
solution sin(−ν) = − sin ν = −
10
Exercise 48
Instructor’s Solutions Manual, Section 5.6
49. cos(u + 4π )
solution cos(u + 4π ) = cos u = −
Exercise 49
√
5
5
Instructor’s Solutions Manual, Section 5.6
50. cos(ν − 6π )
solution cos(ν − 6π ) = cos ν = −
Exercise 50
√
10
10
Instructor’s Solutions Manual, Section 5.6
51. sin(u − 6π )
√
2 5
solution sin(u − 6π ) = sin u =
5
Exercise 51
Instructor’s Solutions Manual, Section 5.6
52. sin(ν + 10π )
√
3 10
solution sin(ν + 10π ) = sin ν =
10
Exercise 52
Instructor’s Solutions Manual, Section 5.6
53. tan(u + 8π )
solution tan(u + 8π ) = tan u = −2
Exercise 53
Instructor’s Solutions Manual, Section 5.6
54. tan(ν − 4π )
solution tan(ν − 4π ) = tan ν = −3
Exercise 54
Instructor’s Solutions Manual, Section 5.6
55. cos(u − 3π )
√
5
solution cos(u − 3π ) = − cos u =
5
Exercise 55
Instructor’s Solutions Manual, Section 5.6
56. cos(ν + 5π )
solution cos(ν + 5π ) = − cos ν =
Exercise 56
√
10
10
Instructor’s Solutions Manual, Section 5.6
57. sin(u + 5π )
√
2 5
solution sin(u + 5π ) = − sin u = −
5
Exercise 57
Instructor’s Solutions Manual, Section 5.6
58. sin(ν − 7π )
√
3 10
solution sin(ν − 7π ) = − sin ν = −
10
Exercise 58
Instructor’s Solutions Manual, Section 5.6
59. tan(u − 9π )
solution tan(u − 9π ) = tan u = −2
Exercise 59
Instructor’s Solutions Manual, Section 5.6
60. tan(ν + 3π )
solution tan(ν + 3π ) = tan ν = −3
Exercise 60
Instructor’s Solutions Manual, Section 5.6
61. cos( π2 − u)
solution cos( π2 − u) = sin u =
√
2 5
5
Exercise 61
Instructor’s Solutions Manual, Section 5.6
62. cos( π2 − ν)
solution cos( π2 − ν) = sin ν =
√
3 10
10
Exercise 62
Instructor’s Solutions Manual, Section 5.6
63. sin( π2 − u)
π
solution sin
2
√
5
− u = cos u = −
5
Exercise 63
Instructor’s Solutions Manual, Section 5.6
64. sin( π2 − ν)
π
solution sin
2
√
10
− ν = cos ν = −
10
Exercise 64
Instructor’s Solutions Manual, Section 5.6
65. tan( π2 − u)
solution tan( π2 − u) =
1
1
=−
tan u
2
Exercise 65
Instructor’s Solutions Manual, Section 5.6
66. tan( π2 − ν)
solution tan( π2 − ν) =
1
1
=−
tan ν
3
Exercise 66
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).]
solution
(cos θ + sin θ)2 = cos2 θ + sin2 θ + 2 cos θ sin θ
= 1 + 2 cos θ sin θ
Instructor’s Solutions Manual, Section 5.6
68. Show that
Problem 68
1 + cos x
sin x
=
1 − cos x
sin x
for every number x that is not an integer multiple of π .
solution If x is any number, then
sin2 x = 1 − cos2 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
1 + cos x
sin x
=
.
1 − cos x
sin x
Instructor’s Solutions Manual, Section 5.6
Problem 69
69. Show that
cos3 θ + cos2 θ sin θ + cos θ sin2 θ + sin3 θ
= cos θ + sin θ
for every number θ.
[Hint: Try replacing the cos2 θ term above with 1 − sin2 θ and replacing
the sin2 θ term above with 1 − cos2 θ.]
solution
cos3 θ + cos2 θ sin θ + cos θ sin2 θ + sin3 θ
= cos3 θ + (1 − sin2 θ) sin θ + cos θ(1 − cos2 θ) + sin3 θ
= cos3 θ + sin θ − sin3 θ + cos θ − cos3 θ + sin3 θ
= cos θ + sin θ
Instructor’s Solutions Manual, Section 5.6
70. Show that
sin2 θ =
for all θ except odd multiples of
Problem 70
tan2 θ
1 + tan2 θ
π
2.
solution If θ is not an odd multiple of
π
2,
then tan θ is defined and
sin2 θ sin2 θ(1 + tan2 θ) = sin2 θ 1 +
cos2 θ
cos2 θ + sin2 θ = sin2 θ
cos2 θ
1 = sin2 θ
cos2 θ
sin θ 2
=
cos θ
= tan2 θ.
Dividing both sides of the equation above by 1 + tan2 θ shows that
sin2 θ =
tan2 θ
.
1 + tan2 θ
Instructor’s Solutions Manual, Section 5.6
Problem 71
71. Find a formula for cos θ solely in terms of tan θ.
solution Suppose θ is not an odd multiple of
from the previous problem, we have
π
2.
Using the result
cos2 θ = 1 − sin2 θ
=1−
Thus
tan2 θ
1 + tan2 θ
=
tan2 θ
1 + tan2 θ
−
2
1 + tan θ
1 + tan2 θ
=
1
.
1 + tan2 θ
1
,
cos θ = ± 1 + tan2 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.
solution Suppose θ is not an odd multiple of
tan θ is defined. We have
tan θ =
π
2,
which means that
sin θ
cos θ
sin θ
,
= ±
1 − sin2 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
73. Is cosine an even function, an odd function, or neither?
solution Recall that
cos(−θ) = cos θ
for every number θ. Thus cosine is an even function.
Problem 73
Instructor’s Solutions Manual, Section 5.6
74. Is sine an even function, an odd function, or neither?
solution Recall that
sin(−θ) = − sin θ
for every number θ. Thus sine is an odd function.
Problem 74
Instructor’s Solutions Manual, Section 5.6
Problem 75
75. Is tangent an even function, an odd function, or neither?
solution 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.
solution 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.
solution 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.
solution 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.
solution Recall that sec θ =
sec(θ + nπ ) =
=
=
1
cos θ .
Thus
1
cos(θ + nπ )
⎧
1
⎪
⎪
⎨ cos θ
1
⎪
⎪
⎩− cos θ
if n is an even integer
if n is an odd integer
⎧
⎪
⎨sec θ
if n is an even integer
⎪
⎩− sec θ
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 π2 . 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
csc(θ + nπ ) =
=
=
Recall that cot θ =
1
tan θ .
Problem 79
1
sin(θ + nπ )
⎧
1
⎪
⎪
⎨ sin θ
if n is an even integer
1
⎪
⎪
⎩− sin θ
if n is an odd integer
⎧
⎪
⎨csc θ
if n is an even integer
⎪
⎩− csc θ
if n is an odd integer.
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.
solution
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)◦ =
sin(θ + 180n)◦ =
⎧
⎨cos θ
if n is an even integer
⎩− cos θ
if n is an odd integer
⎧
⎨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.
solution
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.
solution
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.
solution
cos(x +
π
2)
π
= cos
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.
solution
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
π
for every angle θ that is not an integer multiple of 2 . Interpret this
result in terms of the characterization of the slopes of perpendicular
lines.
solution Suppose θ is not an integer multiple of
tan(θ +
π
2)
π
= tan
=
2
π
2.
Then
− (−θ)
1
tan(−θ)
=−
1
.
tan θ
Note that tan θ equals the slope of the radius of the unit circle that
π
makes an angle of θ with the positive horizontal axis, and tan(θ + 2 )
equals the slope of the radius of the unit circle that makes an angle of
π
θ + 2 with the positive horizontal axis. These two radii are
π
π
perpendicular to each other (because θ + 2 is obtained by adding 2 to
θ), and thus the product of their slopes equals −1 (see Section 2.1). In
other words,
π (tan θ) tan(θ + 2 ) = −1,
which is equivalent to the equation
Instructor’s Solutions Manual, Section 5.6
tan(θ +
π
2)
Problem 85
=−
1
.
tan θ