Download Solutions to Exercises, Section 6.3

Instructor’s Solutions Manual, Section 6.3
Exercise 1
Solutions to Exercises, Section 6.3
1.
For θ = 23◦ , evaluate each of the following:
(a) cos(2θ)
(b) 2 cos θ
[This exercise and the next one emphasize that cos(2θ) does not equal
2 cos θ.]
solution
(a) Note that 2 × 23 = 46. Using a calculator working in degrees, we have
cos 46◦ ≈ 0.694658.
(b) Using a calculator working in degrees, we have
2 cos 23◦ ≈ 2 × 0.920505 = 1.841010.
Instructor’s Solutions Manual, Section 6.3
2.
Exercise 2
For θ = 7 radians, evaluate each of the following:
(a) cos(2θ)
(b) 2 cos θ
solution
(a) Note that 2 × 7 = 14. Using a calculator working in radians, we have
cos 14 ≈ 0.136737.
(b) Using a calculator working in radians, we have
2 cos 7 ≈ 2 × 0.753902 = 1.507804.
Instructor’s Solutions Manual, Section 6.3
3.
Exercise 3
For θ = −5 radians, evaluate each of the following:
(a) sin(2θ)
(b) 2 sin θ
[This exercise and the next one emphasize that sin(2θ) does not equal
2 sin θ.]
solution
(a) Note that 2 × (−5) = −10. Using a calculator working in radians, we
have
sin(−10) ≈ 0.544021.
(b) Using a calculator working in radians, we have
2 sin(−5) ≈ 2 × 0.9589 = 1.9178.
Instructor’s Solutions Manual, Section 6.3
4.
Exercise 4
For θ = 100◦ , evaluate each of the following:
(a) sin(2θ)
(b) 2 sin θ
solution
(a) Note that 2 × 100◦ = 200◦ . Using a calculator working in degrees, we
have
sin 200◦ ≈ −0.34202.
(b) Using a calculator working in degrees, we have
2 sin 100◦ ≈ 2 × 0.984808 = 1.969616.
Instructor’s Solutions Manual, Section 6.3
5.
Exercise 5
For θ = 6 radians, evaluate each of the following:
(a) cos
θ
2
(b)
cos θ
2
[This exercise and the next one emphasize that cos θ2 does not equal
cos θ
2 .]
solution
(a) Using a calculator working in radians, we have
6
cos 2 = cos 3 ≈ −0.989992.
(b) Using a calculator working in radians, we have
0.96017
cos 6
≈
= 0.480085.
2
2
Instructor’s Solutions Manual, Section 6.3
6.
Exercise 6
For θ = −80◦ , evaluate each of the following:
(a) cos
θ
2
(b)
cos θ
2
solution
(a) Using a calculator working in degrees, we have
cos(
−80◦
2 )
= cos(−40◦ ) ≈ 0.766044.
(b) Using a calculator working in degrees, we have
0.173648
cos(−80◦ )
≈
= 0.086824.
2
2
Instructor’s Solutions Manual, Section 6.3
7.
Exercise 7
For θ = 65◦ , evaluate each of the following:
(a) sin
θ
2
(b)
sin θ
2
[This exercise and the next one emphasize that sin θ2 does not equal
solution
(a) Using a calculator working in degrees, we have
sin
65◦
2
= sin 32.5◦ ≈ 0.537300.
(b) Using a calculator working in degrees, we have
0.906308
sin 65◦
≈
= 0.453154.
2
2
sin θ
2 .]
Instructor’s Solutions Manual, Section 6.3
8.
Exercise 8
For θ = 9 radians, evaluate each of the following:
(a) sin
θ
2
(b)
sin θ
2
solution
(a) Using a calculator working in radians, we have
9
sin 2 = sin 4.5 ≈ −0.97753.
(b) Using a calculator working in radians, we have
0.412118
sin 9
≈
= 0.206059.
2
2
Instructor’s Solutions Manual, Section 6.3
Exercise 9
√
5−1
9. Given that sin 18◦ = 4 , find an exact expression for cos 36◦ .
[The value used here for sin 18◦ is derived in Problem 101 in this section.]
solution To evaluate cos 36◦ , use one of the double-angle formulas
for cos(2θ) with θ = 18◦ :
cos 36◦ = 1 − 2 sin2 18◦
=1−2
√5−1 2
4
=1−2
3−√5 8
=
√
5+1
4 .
Instructor’s Solutions Manual, Section 6.3
Exercise 10
√
5+1
3π
3π
10. Given that sin 10 = 4 , find an exact expression for cos 5 .
3π
[Problem 71 asks you to explain how the value for sin 10 used here
follows from the solution to Exercise 9.]
solution To evaluate cos
3π
for cos(2θ) with θ = 10 :
cos
3π
5
3π
5 ,
use one of the double-angle formulas
= 1 − 2 sin2 3π
10
=1−2
√5+1 2
4
=1−2
3+√5 8
=
√
1− 5
4 .
Instructor’s Solutions Manual, Section 6.3
Exercise 11
For Exercises 11–26, evaluate the given quantities assuming that u and
π
ν are both in the interval (0, 2 ) and
cos u =
1
3
and
sin ν =
1
4.
11. sin u
solution Because 0 < u <
π
2,
we know that sin u > 0. Thus
sin u = 1 − cos2 u = 1 −
1
9
=
8
9
=
√
2 2
3 .
Instructor’s Solutions Manual, Section 6.3
Exercise 12
12. cos ν
solution Because 0 < ν <
π
2,
we know that cos ν > 0. Thus
cos ν = 1 − sin2 ν = 1 −
1
16
=
15
16
=
√
15
4 .
Instructor’s Solutions Manual, Section 6.3
Exercise 13
13. tan u
solution To evaluate tan u, use its definition as a ratio:
sin u
=
tan u =
cos u
√
2 2
3
1
3
√
= 2 2.
Instructor’s Solutions Manual, Section 6.3
Exercise 14
14. tan ν
solution To evaluate tan ν, use its definition as a ratio:
sin ν
=
tan ν =
cos ν
1
√4
15
4
=
√1
15
=
√
15
15 .
Instructor’s Solutions Manual, Section 6.3
Exercise 15
15. cos(2u)
solution To evaluate cos(2u), use one of the double-angle formulas
for cosine:
2
7
cos(2u) = 2 cos2 u − 1 = 9 − 1 = − 9 .
Instructor’s Solutions Manual, Section 6.3
Exercise 16
16. cos(2ν)
solution To evaluate cos(2ν), use one of the double-angle formulas
for cosine:
2
7
cos(2ν) = 1 − 2 sin2 ν = 1 − 16 = 8 .
Instructor’s Solutions Manual, Section 6.3
Exercise 17
17. sin(2u)
solution To evaluate sin(2u), use the double-angle formula for sine:
sin(2u) = 2 cos u sin u = 2 ·
1
3
·
√
2 2
3
=
√
4 2
9 .
Instructor’s Solutions Manual, Section 6.3
Exercise 18
18. sin(2ν)
solution To evaluate sin(2ν), use the double-angle formula for sine:
sin(2ν) = 2 cos ν sin ν = 2 ·
√
15
4
·
1
4
=
√
15
8 .
Instructor’s Solutions Manual, Section 6.3
Exercise 19
19. tan(2u)
solution To evaluate tan(2u), use its definition as a ratio:
√
4 2
√
sin(2u)
= 97 = − 4 7 2 .
tan(2u) =
cos(2u)
−9
Alternatively, we could have used the double-angle formula for tangent,
which will produce the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 20
20. tan(2ν)
solution To evaluate tan(2ν), use its definition as a ratio:
sin(2ν)
=
tan(2ν) =
cos(2ν)
√
15
8
7
8
=
√
15
7 .
Alternatively, we could have used the double-angle formula for tangent,
which will produce the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 21
21. cos u
2
solution Because 0 <
u
2
<
π
4,
we know that cos u
2 > 0. Thus
cos
u
2
1 + cos u
2
$
$
%4
%
&
& 1 + 13
= 3 = 23 =
=
2
2
=
√
6
3 .
Instructor’s Solutions Manual, Section 6.3
Exercise 22
22. cos ν2
solution Because 0 <
cos ν2 =
ν
2
<
π
4,
we know that cos ν2 > 0. Thus
$
√
%
√
% 1 + 15
&
1 + cos ν
4 + 15
4
=
=
.
2
2
8
Instructor’s Solutions Manual, Section 6.3
Exercise 23
23. sin u
2
solution Because 0 <
u
2
<
π
4,
we know that sin u
2 > 0. Thus
sin
u
2
1 − cos u
2
$
$
%2
%
&
& 1 − 13
= 3 = 13 =
=
2
2
=
√1
3
=
√
3
3 .
Instructor’s Solutions Manual, Section 6.3
24. sin ν2
solution Because 0 <
sin ν2 =
ν
2
<
π
4,
we know that sin ν2 > 0. Thus
$
√
%
√
% 1 − 15
&
1 − cos ν
4 − 15
4
=
=
.
2
2
8
Exercise 24
Instructor’s Solutions Manual, Section 6.3
Exercise 25
25. tan u
2
solution To evaluate tan u
2 , use its definition as a ratio:
tan
u
2
=
sin u
2
cos
u
2
√
=
3
√3
6
3
=
√
√3
6
=
√1
2
=
√
2
2 .
Alternatively, we could have used the half-angle formula for tangent,
which will produce the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 26
26. tan ν2
solution
√
tan
ν
2
1−
1 − cos ν
=
=
1
sin ν
4
15
4
=4−
√
15
Instructor’s Solutions Manual, Section 6.3
Exercise 27
For Exercises 27–42, evaluate the given quantities assuming that u and
π
ν are both in the interval ( 2 , π) and
sin u =
1
5
and
sin ν =
1
6.
27. cos u
solution Because
π
2
< u < π , we know that cos u < 0. Thus
cos u = − 1 − sin2 u = − 1 −
1
25
24
25
√
2 6
− 5 .
=−
=
Instructor’s Solutions Manual, Section 6.3
Exercise 28
28. cos ν
solution Because
π
2
< ν < π , we know that cos ν < 0. Thus
cos ν = − 1 − sin2 ν = − 1 −
1
36
=−
=−
35
36
√
35
6 .
Instructor’s Solutions Manual, Section 6.3
Exercise 29
29. tan u
solution To evaluate tan u, use its definition as a ratio:
1
√
sin u
1
6
= 25√6 = − 2√6 = − 12 .
tan u =
cos u
− 5
Instructor’s Solutions Manual, Section 6.3
Exercise 30
30. tan ν
solution To evaluate tan ν, use its definition as a ratio:
1
√
sin ν
1
35
= √635 = − √35 = − 35 .
tan ν =
cos ν
− 6
Instructor’s Solutions Manual, Section 6.3
Exercise 31
31. cos(2u)
solution To evaluate cos(2u), use one of the double-angle formulas
for cosine:
2
23
cos(2u) = 1 − 2 sin2 u = 1 − 25 = 25 .
Instructor’s Solutions Manual, Section 6.3
Exercise 32
32. cos(2ν)
solution To evaluate cos(2ν), use one of the double-angle formulas
for cosine:
2
17
cos(2ν) = 1 − 2 sin2 ν = 1 − 36 = 18 .
Instructor’s Solutions Manual, Section 6.3
Exercise 33
33. sin(2u)
solution To evaluate sin(2u), use the double-angle formula for sine:
2√6 sin(2u) = 2 cos u sin u = 2 · − 5 ·
1
5
√
= − 4256 .
Instructor’s Solutions Manual, Section 6.3
Exercise 34
34. sin(2ν)
solution To evaluate sin(2ν), use the double-angle formula for sine:
√35 sin(2ν) = 2 cos ν sin ν = 2 · − 6 ·
1
6
=−
√
35
18 .
Instructor’s Solutions Manual, Section 6.3
Exercise 35
35. tan(2u)
solution To evaluate tan(2u), use its definition as a ratio:
√
√
−4 6
sin(2u)
= 2325 = − 4236 .
tan(2u) =
cos(2u)
25
Alternatively, we could have used the double-angle formula for tangent,
which will produce the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 36
36. tan(2ν)
solution To evaluate tan(2ν), use its definition as a ratio:
√
√
− 35
sin(2ν)
35
= 1718 = − 17
.
tan(2ν) =
cos(2ν)
18
Alternatively, we could have used the double-angle formula for tangent,
which will produce the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 37
37. cos u
2
solution Because
π
4
<
cos
u
2
=
u
2
<
π
2,
we know that cos u
2 > 0. Thus
1 + cos u
2
$
$
√
√
%
%
√
% 5−2 6
%1 − 2 6
&
&
5−2 6
5
5
.
=
=
=
10
2
2
Instructor’s Solutions Manual, Section 6.3
Exercise 38
38. cos ν2
solution Because
π
4
<
cos ν2 =
ν
2
<
π
2,
we know that cos ν2 > 0. Thus
$
√
%
√
% 1 − 35
&
1 + cos ν
6 − 35
6
=
=
.
2
2
12
Instructor’s Solutions Manual, Section 6.3
Exercise 39
39. sin u
2
solution Because
π
4
<
sin
u
2
=
u
2
<
π
2,
we know that sin u
2 > 0. Thus
1 − cos u
2
$
$
√
√
%
%
√
% 5+2 6
%1 + 2 6
&
&
5+2 6
5
5
.
=
=
=
10
2
2
Instructor’s Solutions Manual, Section 6.3
Exercise 40
40. sin ν2
solution Because
π
4
<
sin ν2 =
ν
2
<
π
2,
we know that sin ν2 > 0. Thus
$
√
%
√
% 1 + 35
&
1 − cos ν
6 + 35
6
=
=
.
2
2
12
Instructor’s Solutions Manual, Section 6.3
Exercise 41
41. tan u
2
solution To evaluate tan u
2 , use one of the half-angle formulas for
tangent:
√
√
1 + 256
1 − cos u
u
=
= 5 + 2 6.
tan 2 =
1
sin u
5
We could also have evaluated tan u
2 by using its definition as the ratio
u
u
of sin 2 and cos 2 , but in this case that procedure would lead to a more
complicated algebraic expression.
Instructor’s Solutions Manual, Section 6.3
Exercise 42
42. tan ν2
solution To evaluate tan ν2 , use one of the half-angle formulas for
tangent:
√
√
1 + 635
1 − cos ν
ν
=
= 6 + 35.
tan 2 =
1
sin ν
6
We could also have evaluated tan ν2 by using its definition as the ratio of
ν
ν
sin 2 and cos 2 , but in this case that procedure would lead to a more
complicated algebraic expression.
Instructor’s Solutions Manual, Section 6.3
Exercise 43
For Exercises 43–58, evaluate the given quantities assuming that u and
π
ν are both in the interval (− 2 , 0) and
1
tan u = − 7
1
tan ν = − 8 .
and
43. cos u
π
solution Because − 2 < u < 0, we know that cos u > 0 and sin u < 0.
Thus
√
− 1 − cos2 u
sin u
1
=
.
− 7 = tan u =
cos u
cos u
Squaring the first and last entries above gives
1
49
=
1 − cos2 u
.
cos2 u
Multiplying both sides by cos2 u and then by 49 gives
cos2 u = 49 − 49 cos2 u.
Thus 50 cos2 u = 49, which implies that
cos u =
49
50
=
7
√
5 2
=
√
7 2
10 .
Instructor’s Solutions Manual, Section 6.3
Exercise 44
44. cos ν
π
solution Because − 2 < ν < 0, we know that cos ν > 0 and sin ν < 0.
Thus
√
− 1 − cos2 ν
sin ν
1
=
.
− 8 = tan ν =
cos ν
cos ν
Squaring the first and last entries above gives
1
64
=
1 − cos2 ν
.
cos2 ν
Multiplying both sides by cos2 ν and then by 64 gives
cos2 ν = 64 − 64 cos2 ν.
Thus 65 cos2 ν = 64, which implies that
cos ν =
64
65
=
√8
65
=
√
8 65
65 .
Instructor’s Solutions Manual, Section 6.3
Exercise 45
45. sin u
solution Solve the equation tan u =
sin u = cos u tan u =
sin u
cos u
√
7 2
10
for sin u:
√
· (− 17 ) = − 102 .
Instructor’s Solutions Manual, Section 6.3
Exercise 46
46. sin ν
solution Solve the equation tan ν =
sin ν = cos ν tan ν =
sin ν
cos ν
√
8 65
65
for sin ν:
· (− 18 ) = −
√
65
65 .
Instructor’s Solutions Manual, Section 6.3
Exercise 47
47. cos(2u)
solution To evaluate cos(2u), use one of the double-angle formulas
for cosine:
49
24
cos(2u) = 2 cos2 u − 1 = 2 · 50 − 1 = 25 .
Instructor’s Solutions Manual, Section 6.3
Exercise 48
48. cos(2ν)
solution To evaluate cos(2ν), use one of the double-angle formulas
for cosine:
64
63
cos(2ν) = 2 cos2 ν − 1 = 2 · 65 − 1 = 65 .
Instructor’s Solutions Manual, Section 6.3
Exercise 49
49. sin(2u)
solution To evaluate sin(2u), use the double-angle formula for sine:
sin(2u) = 2 cos u sin u = 2 ·
√
7 2
10
√
2
7
· (− 10 ) = − 25 .
Instructor’s Solutions Manual, Section 6.3
Exercise 50
50. sin(2ν)
solution To evaluate sin(2ν), use the double-angle formula for sine:
sin(2ν) = 2 cos ν sin ν = 2 ·
√
8 65
65
· (−
√
65
65 )
16
= − 65 .
Instructor’s Solutions Manual, Section 6.3
Exercise 51
51. tan(2u)
solution To evaluate tan(2u), use the double-angle formula for
tangent:
− 27
2 tan u
7
=
tan(2u) =
48 = − 24 .
1 − tan2 u
49
Alternatively, we could have evaluated tan(2u) by using its definition
as a ratio of sin(2u) and cos(2u), producing the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 52
52. tan(2ν)
solution To evaluate tan(2ν), use the double-angle formula for
tangent:
− 28
2 tan ν
16
=
tan(2ν) =
63 = − 63 .
1 − tan2 ν
64
Alternatively, we could have evaluated tan(2ν) by using its definition as
a ratio of sin(2ν) and cos(2ν), producing the same answer.
Instructor’s Solutions Manual, Section 6.3
Exercise 53
53. cos u
2
solution Because − π4 <
cos
u
2
=
u
2
< 0, we know that cos u
2 > 0. Thus
1 + cos u
2
$
$
√
√
%
%
√
% 10+7 2
%1 + 7 2
&
&
10 + 7 2
10
10
.
=
=
=
20
2
2
Instructor’s Solutions Manual, Section 6.3
Exercise 54
54. cos ν2
solution Because − π4 <
cos
ν
2
=
=
ν
2
< 0, we know that cos ν2 > 0. Thus
1 + cos ν
2
$
%
%1 +
&
√
8 65
65
2
$
√
%
√
% 65+8 65
&
65 + 8 65
65
.
=
=
130
2
Instructor’s Solutions Manual, Section 6.3
Exercise 55
55. sin u
2
solution Because − π4 <
sin
u
2
=−
u
2
< 0, we know that sin u
2 < 0. Thus
1 − cos u
2
$
$
√
√
%
%
√
% 10−7 2
%1 − 7 2
&
&
10 − 7 2
10
10
.
=−
=−
=−
20
2
2
Instructor’s Solutions Manual, Section 6.3
Exercise 56
56. sin ν2
solution Because − π4 <
sin
ν
2
=−
< 0, we know that sin ν2 < 0. Thus
1 − cos ν
2
$
%
%1 −
&
=−
ν
2
√
8 65
65
2
$
√
%
√
% 65−8 65
&
65 − 8 65
65
.
=−
=−
130
2
Instructor’s Solutions Manual, Section 6.3
Exercise 57
57. tan u
2
solution To evaluate tan u
2 , use one of the half-angle formulas for
tangent:
√
tan
u
2
1 − 7102
1 − cos u
√
=
=
=7−
2
sin u
− 10
10
√
2
√
= 7 − 5 2.
u
We could also have evaluated tan 2 by using its definition as the ratio
u
u
of sin 2 and cos 2 , but in this case that procedure would lead to a more
complicated algebraic expression.
Instructor’s Solutions Manual, Section 6.3
Exercise 58
58. tan ν2
solution To evaluate tan ν2 , use one of the half-angle formulas for
tangent:
√
tan
ν
2
1 − 8 6565
1 − cos ν
√
=
=
=8−
65
sin ν
− 65
=8−
ν
√65
65
√
65.
We could also have evaluated tan 2 by using its definition as the ratio of
ν
ν
sin 2 and cos 2 , but in this case that procedure would lead to a more
complicated algebraic expression.
Instructor’s Solutions Manual, Section 6.3
59.
Suppose 0 < θ <
π
2
Exercise 59
and sin θ = 0.4.
(a) Without using a double-angle formula, evaluate sin(2θ).
(b) Without using an inverse trigonometric function, evaluate sin(2θ)
again.
[Your solutions to (a) and (b), which are obtained through different
methods, should be the same, although they might differ by a tiny
amount due to using approximations rather than exact amounts.]
solution
(a) Because 0 < θ <
π
2
and sin θ = 0.4, we see that
θ = sin−1 0.4 ≈ 0.411517 radians.
Thus
2θ ≈ 0.823034 radians.
Hence
sin(2θ) ≈ sin(0.823034) ≈ 0.733212.
(b) To use the double-angle formula to evaluate sin(2θ), we must first
π
evaluate cos θ. Because 0 < θ < 2 , we know that cos θ > 0. Thus
√
√
cos θ = 1 − sin2 θ = 1 − 0.16 = 0.84
≈ 0.916515.
Instructor’s Solutions Manual, Section 6.3
Exercise 59
Now
sin(2θ) = 2 cos θ sin θ ≈ 2(0.916515)(0.4)
= 0.733212.
Instructor’s Solutions Manual, Section 6.3
60.
Suppose 0 < θ <
π
2
Exercise 60
and sin θ = 0.2.
(a) Without using a double-angle formula, evaluate sin(2θ).
(b) Without using an inverse trigonometric function, evaluate sin(2θ)
again.
solution
(a) Because 0 < θ <
π
2
and sin θ = 0.2, we see that
θ = sin−1 0.2 ≈ 0.201358 radians.
Thus
2θ ≈ 0.402716 radians.
Hence
sin(2θ) ≈ sin(0.402716) ≈ 0.391918.
(b) To use the double-angle formula to evaluate sin(2θ), we must first
π
evaluate cos θ. Because 0 < θ < 2 , we know that cos θ > 0. Thus
√
√
cos θ = 1 − sin2 θ = 1 − 0.04 = 0.96
≈ 0.979796.
Now
sin(2θ) = 2 cos θ sin θ ≈ 2(0.979796)(0.2)
= 0.391918.
Instructor’s Solutions Manual, Section 6.3
61.
Exercise 61
Suppose − π2 < θ < 0 and cos θ = 0.3.
(a) Without using a double-angle formula, evaluate cos(2θ).
(b) Without using an inverse trigonometric function, evaluate cos(2θ)
again.
solution
π
(a) Because − 2 < θ < 0 and cos θ = 0.3, we see that
θ = − cos−1 0.3 ≈ −1.2661 radians.
Thus
2θ ≈ −2.5322 radians.
Hence
cos(2θ) ≈ cos(−2.5322) ≈ −0.82.
(b) Using a double-angle formula, we have
cos(2θ) = 2 cos2 θ − 1 ≈ 2(0.3)2 − 1
= −0.82.
Instructor’s Solutions Manual, Section 6.3
62.
Exercise 62
Suppose − π2 < θ < 0 and cos θ = 0.8.
(a) Without using a double-angle formula, evaluate cos(2θ).
(b) Without using an inverse trigonometric function, evaluate cos(2θ)
again.
solution
π
(a) Because − 2 < θ < 0 and cos θ = 0.8, we see that
θ = − cos−1 0.8 ≈ −0.6435 radians.
Thus
2θ ≈ −1.287 radians.
Hence
cos(2θ) ≈ cos(−1.287) ≈ 0.28.
(b) Using a double-angle formula, we have
cos(2θ) = 2 cos2 θ − 1 ≈ 2(0.8)2 − 1
= 0.28.
Instructor’s Solutions Manual, Section 6.3
Exercise 63
63. Find an exact expression for sin 15◦ .
θ
solution Use the half-angle formula for sin 2 with θ = 30◦ (choose
the plus sign associated with the square root because sin 15◦ is
positive), getting
◦
1 − cos 30◦
2
$
$
√
√
%
%
√
% (1 − 3 ) · 2
%1 − 3
&
&
2− 3
2
2
=
=
.
=
2
2·2
2
sin 15 =
Instructor’s Solutions Manual, Section 6.3
Exercise 64
64. Find an exact expression for cos 22.5◦ .
θ
solution Use the half-angle formula for cos 2 with θ = 45◦ (choose
the plus sign associated with the square root because cos 22.5◦ is
positive), getting
◦
1 + cos 45◦
2
$
$
√
√
%
%
√
% (1 + 2 ) · 2
%1 + 2
&
&
2+ 2
2
2
=
=
.
=
2
2·2
2
cos 22.5 =
Instructor’s Solutions Manual, Section 6.3
Exercise 65
π
65. Find an exact expression for sin 24
.
θ
π
solution Using the half-angle formula for sin 2 with θ = 12 (and
π
choosing the plus sign associated with the square root because sin 24 is
positive), we have
π
1 − cos 12
π
.
sin 24 =
2
π
π
Note that 12
radians equals 15◦ . Substituting for cos 12
the value for
cos 15◦ from Example 4 gives
π
sin 24 =
$
√ √
%
%
3
& 1 − 2+
2
2
√
2− 2+ 3
=
.
2
Instructor’s Solutions Manual, Section 6.3
Exercise 66
π
66. Find an exact expression for cos 16
.
θ
π
solution Using the half-angle formula for cos 2 with θ = 8 (and
π
choosing the plus sign associated with the square root because cos 16 is
positive), we have
π
1 + cos 8
π
.
cos 16 =
2
Note that π8 radians equals 22.5◦ . Substituting for cos π8 the value for
cos 22.5◦ found in Exercise 64 gives
π
cos 16 =
$
√ √
%
%
2
& 1 + 2+
2
2
√
2+ 2+ 2
=
.
2
Instructor’s Solutions Manual, Section 6.3
Exercise 67
67. Find a formula for sin(4θ) in terms of cos θ and sin θ.
solution Use the double-angle formula for sine, with θ replaced by
2θ, getting
sin(4θ) = 2 cos(2θ) sin(2θ).
Now use the double-angle formulas for the expressions on the right
side, getting
sin(4θ) = 2(2 cos2 θ − 1)(2 cos θ sin θ)
= 4(2 cos2 θ − 1) cos θ sin θ.
Instructor’s Solutions Manual, Section 6.3
Exercise 68
68. Find a formula for cos(4θ) in terms of cos θ.
solution Use one of the double-angle formulas for cosine, with θ
replaced by 2θ, getting
cos(4θ) = 2 cos2(2θ) − 1.
Now use the same double-angle formula for cosine again, getting
cos(4θ) = 2(2 cos2 θ − 1)2 − 1
= 8 cos4 θ − 8 cos2 θ + 1.
Instructor’s Solutions Manual, Section 6.3
Exercise 69
69. Find constants a, b, and c such that
cos4 θ = a + b cos(2θ) + c cos(4θ)
for all θ.
solution One of the double-angle formulas for cos(2θ) can be
written in the form
1 + cos(2θ)
.
cos2 θ =
2
Squaring both sides, we get
cos4 θ =
1 + 2 cos(2θ) + cos2(2θ)
.
4
We now see that we need an expression for cos2(2θ), which we can
obtain by replacing θ by 2θ in the formula above for cos2 θ:
cos2(2θ) =
1 + cos(4θ)
.
2
Substituting this expression into the expression above for cos4 θ gives
cos4 θ =
=
1 + 2 cos(2θ) +
1+cos(4θ)
2
4
3
8
+
Thus a = 38 , b = 12 , and c = 18 .
1
2
cos(2θ) +
1
8
cos(4θ).
Instructor’s Solutions Manual, Section 6.3
Exercise 70
70. Find constants a, b, and c such that
sin4 θ = a + b cos(2θ) + c cos(4θ)
for all θ.
solution One of the double-angle formulas for cos(2θ) can be
written in the form
1 − cos(2θ)
.
sin2 θ =
2
Squaring both sides, we get
sin4 θ =
1 − 2 cos(2θ) + cos2(2θ)
.
4
In the solution to the previous exercise we obtained the following
formula for cos2(2θ):
cos2(2θ) =
1 + cos(4θ)
.
2
Substituting this expression into the expression above for sin4 θ gives
sin4 θ =
=
1 − 2 cos(2θ) +
1+cos(4θ)
2
4
3
8
−
1
2
Thus a = 38 , b = − 12 , and c = 18 .
cos(2θ) +
1
8
cos(4θ).
Instructor’s Solutions Manual, Section 6.3
Problem 71
Solutions to Problems, Section 6.3
71. Explain how the equation sin 3π
10 =
Exercise 9.
solution Note that
3π
10
√
5+1
4
follows from the solution to
radians = 54◦ . Thus
sin
3π
10
= sin 54◦
= cos 36◦
=
√
5+1
4 ,
where the second equality comes from the identity sin θ ◦ = cos(90 − θ)◦
and the third equality comes from the solution to Exercise 9.
Instructor’s Solutions Manual, Section 6.3
72. Show that
(cos x + sin x)2 = 1 + sin(2x)
for every number x.
solution
(cos x + sin x)2 = cos2 x + 2 cos x sin x + sin2 x
= 1 + 2 cos x sin x
= 1 + sin(2x)
Problem 72
Instructor’s Solutions Manual, Section 6.3
73. Show that
cos(2θ) ≤ cos2 θ
for every angle θ.
solution
cos(2θ) = cos2 θ − sin2 θ
≤ cos2 θ
Problem 73
Instructor’s Solutions Manual, Section 6.3
74. Show that
| sin(2θ)| ≤ 2| sin θ|
for every angle θ.
solution For every angle θ we have
| sin(2θ)| = |2 cos θ sin θ|
= 2| cos θ| · | sin θ|
≤ 2| sin θ|,
where the inequality above holds because | cos θ| ≤ 1.
Problem 74
Instructor’s Solutions Manual, Section 6.3
Problem 75
75. Do not ever make the mistake of thinking that
sin(2θ)
= sin θ
2
is a valid identity. Although the equation above is false in general, it is
true for some special values of θ. Find all values of θ that satisfy the
equation above.
solution Note that
2 cos θ sin θ
sin(2θ)
=
= cos θ sin θ.
2
2
Thus sin(2θ)
= sin θ if and only if cos θ sin θ = sin θ, which is equivalent
2
to the equation
sin θ(1 − cos θ) = 0.
This last equation holds if and only if either sin θ = 0 or 1 − cos θ = 0.
Now sin θ = 0 if and only if θ is an integer multiple of π , and
1 − cos θ = 0 if and only if θ is an even multiple of π . Hence each
solution to the equation 1 − cos θ = 0 is already a solution to the
sin(2θ)
= sin θ if and only
equation sin θ = 0. We can thus conclude that
2
if θ is an integer multiple of π .
Instructor’s Solutions Manual, Section 6.3
Problem 76
76. Explain why there does not exist an angle θ such that cos θ sin θ = 23 .
2
solution Suppose there is an angle θ such that cos θ sin θ = 3 .
Multiplying both sides of this equation by 2, we have
4
3
4
= 2 cos θ sin θ = sin(2θ).
4
However, 3 > 1, and thus there is no angle whose sine equals 3 . In
other words, the equation above cannot hold, which means that our
2
assumption that there is an angle θ such that cos θ sin θ = 3 must be
2
false. Hence there does not exist an angle θ such that cos θ sin θ = 3 .
Instructor’s Solutions Manual, Section 6.3
Problem 77
77. Show that
| cos θ sin θ| ≤
1
2
for every angle θ.
solution For every angle θ we have
2 cos θ sin θ | cos θ sin θ| = 2
sin(2θ) =
2
=
| sin(2θ)|
2
≤
1
2
where the inequality above holds because the sine of every angle is in
the interval [−1, 1].
Instructor’s Solutions Manual, Section 6.3
Problem 78
78. Do not ever make the mistake of thinking that
cos(2θ)
= cos θ
2
is a valid identity.
(a) Show that the equation above is false whenever 0 < θ <
π
2.
(b) Show that there exists an angle θ in the interval ( π2 , π ) satisfying
the equation above.
solution
(a) Note that
2 cos2 θ − 1
1
cos(2θ)
=
= cos2 θ − .
2
2
2
Thus the equation
cos(2θ)
2
= cos θ is equivalent to the equation
cos2 θ − cos θ −
1
= 0.
2
Think of the equation above as a quadratic equation with cos θ as the
variable. By the quadratic formula, we have
√
1 ± 1 − 4(− 21 )
1± 3
cos θ =
=
.
2
2
√
√
Because 3 > 1, we have 1+2 3 > 1. Thus choosing the plus sign above
would give cos θ > 1, which is impossible.
Instructor’s Solutions Manual, Section 6.3
Problem 78
Choosing the minus sign above gives cos θ =
then cos θ > 0. Thus we conclude that
π
0<θ < 2.
cos(2θ)
2
√
1− 3
2
< 0. If 0 < θ <
π
2,
= cos θ whenever
(b) As we saw in the solution to part (a), we will have
cos(2θ)
= cos θ
2
if θ satisfies the equation cos θ =
that 0 >
√
1− 3
2
√
1− 3
2 .
Because 1 <
√
3 < 3, we know
> −1. Hence there is a number θ in the interval ( π2 , π )
such that cos θ =
√
1− 3
2
√
1− 3
2 ). Thus
cos(2θ)
= cos θ.
2
(namely, take θ = cos−1
number θ in the interval
( π2 , π )
such that
there is a
Instructor’s Solutions Manual, Section 6.3
Problem 79
79. Without doing any algebraic manipulations, explain why
(2 cos2 θ − 1)2 + (2 cos θ sin θ)2 = 1
for every angle θ.
solution The first quantity in parentheses above equals cos(2θ). The
second quantity in parentheses above equals sin(2θ). Because the
cosine squared plus sine squared of every angle equals 1, the equation
above holds.
Instructor’s Solutions Manual, Section 6.3
Problem 80
80. Find angles u and ν such that cos(2u) = cos(2ν) but cos u = cos ν.
solution The equation cos(2u) = cos(2ν) can be rewritten as
2 cos2 u − 1 = 2 cos2 ν − 1.
Thus if we find numbers u and ν such that cos u = 1 and cos ν = −1,
then we will have cos(2u) = cos(2ν) but cos u = cos ν. One easy choice
is to take u = 0 and ν = π .
Instructor’s Solutions Manual, Section 6.3
Problem 81
81. Show that if cos(2u) = cos(2ν), then | cos u| = | cos ν|.
solution Suppose cos(2u) = cos(2ν). Then the double angle formula
for cosine implies that
2 cos2 u − 1 = 2 cos2 ν − 1,
which implies that
cos2 u = cos2 ν.
The equation above implies that
cos u = ± cos ν,
which implies that
| cos u| = | cos ν|.
Instructor’s Solutions Manual, Section 6.3
Problem 82
82. Find angles u and ν such that sin(2u) = sin(2ν) but | sin u| = | sin ν|.
solution One possibility is to take u = 0 and ν =
π
2.
Then we have
π
sin(2u) = sin 0 = 0 = sin π = sin(2 2 ) = sin(2ν)
but
| sin u| = | sin 0| = |0| = 0 = 1 = |1| = | sin
π
2|
= | sin ν|.
Instructor’s Solutions Manual, Section 6.3
83. Show that
sin2 (2θ) = 4(sin2 θ − sin4 θ)
for all θ.
solution
sin2 (2θ) = (2 cos θ sin θ)2
= 4 cos2 θ sin2 θ
= 4(1 − sin2 θ) sin2 θ
= 4(sin2 θ − sin4 θ)
Problem 83
Instructor’s Solutions Manual, Section 6.3
Problem 84
84. Find a formula that expresses sin2 (2θ) only in terms of cos θ.
solution
sin2 (2θ) = (2 cos θ sin θ)2
= 4 cos2 θ sin2 θ
= 4 cos2 θ(1 − cos2 θ)
= 4(cos2 θ − cos4 θ)
Instructor’s Solutions Manual, Section 6.3
Problem 85
85. Show that
(cos θ + sin θ)2 (cos θ − sin θ)2 + sin2(2θ) = 1
for all angles θ.
solution
(cos θ + sin θ)2 (cos θ − sin θ)2 + sin2(2θ)
2
= (cos θ + sin θ)(cos θ − sin θ) + sin2(2θ)
= (cos2 θ − sin2 θ)2 + sin2(2θ)
= cos2(2θ) + sin2(2θ)
=1
Instructor’s Solutions Manual, Section 6.3
Problem 86
86. Suppose θ is not an integer multiple of π . Explain why the point
(1, 2 cos θ) is on the line containing the point sin θ, sin(2θ) and the
origin.
solution The slope of the line containing the points sin θ, sin(2θ)
sin(2θ)−0
2 cos θ sin θ
, which equals 2 cos θ.
and (0, 0) is (sin θ)−0 , which equals
sin θ
Thus the equation of the line in the xy-plane that contains the points
sin θ, sin(2θ) and (0, 0) is
y = 2(cos θ)x.
Taking x = 1 in the equation above gives y = 2 cos θ. In other words,
the point (1, 2 cos θ) is on this line.
Instructor’s Solutions Manual, Section 6.3
87. Show that
tan2(2x) =
Problem 87
4(cos2 x − cos4 x)
(2 cos2 x − 1)2
for all numbers x except odd multiples of
π
4.
π
solution Suppose x is not an odd multiple of 4 , which implies that
π
2x is not an odd multiple of 2 , which implies that tan(2x) is defined.
Then
tan2(2x) =
sin2(2x)
cos2(2x)
=
(2 cos x sin x)2
(2 cos2 x − 1)2
=
4 cos2 x sin2 x
(2 cos2 x − 1)2
=
4 cos2 x(1 − cos2 x)
(2 cos2 x − 1)2
=
4(cos2 x − cos4 x)
.
(2 cos2 x − 1)2
Instructor’s Solutions Manual, Section 6.3
Problem 88
88. Find a formula that expresses tan2 (2θ) only in terms of sin θ.
π
solution Suppose θ is not an odd multiple of 4 , which implies that
π
2θ is not an odd multiple of 2 , which implies that tan(2θ) is defined.
Then
tan2(2θ) =
sin2(2θ)
cos2(2θ)
=
(2 cos θ sin θ)2
(1 − 2 sin2 θ)2
=
4 cos2 θ sin2 θ
(1 − 2 sin2 θ)2
=
4(1 − sin2 θ) sin2 θ
(1 − 2 sin2 θ)2
=
4(sin2 θ − sin4 θ)
.
(1 − 2 sin2 θ)2
Instructor’s Solutions Manual, Section 6.3
Problem 89
89. Find all numbers t such that
cos−1 t
= sin−1 t.
2
solution Suppose t is a number such that
cos−1 t
= sin−1 t.
2
Then t must be in the interval [−1, 1], because otherwise cos−1 t and
sin−1 t would not be defined.
The equation above implies that
cos−1 t t = sin
2
.
The half-angle formula for sin θ2 , with θ = cos−1 t, now gives
t=±
1 − cos(cos−1 t)
=±
2
1−t
.
2
Squaring both sides of this equation gives the equation
t2 =
which is equivalent to the equation
1−t
,
2
Instructor’s Solutions Manual, Section 6.3
Problem 89
2t 2 + t − 1 = 0.
1
The quadratic formula now shows that t = −1 or t = 2 .
To check whether each of these two possible values of t satisfies the
original equation, we begin with t = −1. We have
π
cos−1 (−1)
=
2
2
and
sin−1 (−1) = −
thus t = −1 does not satisfy the original equation
π
;
2
cos−1 t
2
= sin−1 t.
Now we check the other possibility t = 12 . We have
cos−1
2
thus t =
1
2
1
2
=
π
3
2
=
π
6
and
sin−1
does satisfy the original equation
1
2
cos−1 t
2
Conclusion: The only solution to the equation
=
π
;
6
= sin−1 t.
cos−1 t
2
= sin−1 t is t = 2 .
1
Instructor’s Solutions Manual, Section 6.3
Problem 90
90. Find all numbers t such that
cos−1 t =
sin−1 t
.
2
solution Suppose t is a number such that
cos−1 t =
sin−1 t
.
2
Then t must be in the interval [−1, 1], because otherwise cos−1 t and
sin−1 t would not be defined.
The equation above implies that
sin−1 t t = cos
2
.
The half-angle formula for cos θ2 , with θ = sin−1 t, now gives
t=±
1 + cos(sin−1 t)
=±
2
1+
√
1 − t2
,
2
where the last equality comes from Problem 27 in Section 5.8. Squaring
both sides of this equation gives the equation
2
t =
1+
which is equivalent to the equation
√
1 − t2
,
2
Instructor’s Solutions Manual, Section 6.3
Problem 90
2t 2 − 1 = 1 − t 2 .
Squaring both sides of this equation gives the equation
4t 4 − 4t 2 + 1 = 1 − t 2 ,
which is equivalent to the equation
0 = 4t 4 − 3t 2 = t 2 (4t 2 − 3).
that t = 0, or 4t 2 − 3 = 0, which
Thus either t 2 = √0, which implies
√
3
3
implies that t = 2 or t = − 2 .
To check whether each of these three possible values of t satisfies the
original equation, we begin with t = 0. We have
cos−1 0 =
π
2
and
sin−1 0
= 0;
2
thus t = 0 does not satisfy the original equation cos−1 t =
Now we check the possibility t =
√
3
cos−1 2
thus t =
√
3
2
π
=
6
√
3
2 .
We have
sin−1
and
2
√
3
2
=
π
3
2
=
π
;
6
does satisfy the original equation cos−1 t =
Now we check the possibility t = −
sin−1 t
2 .
√
3
2 .
We have
sin−1 t
2 .
Instructor’s Solutions Manual, Section 6.3
−1
cos
thus t = −
√
3
2
√
(− 23 )
5π
=
6
and
Problem 90
sin−1 (−
2
√
3
2 )
=
− π3
2
=−
π
;
6
does not satisfy the original equation cos−1 t =
Conclusion: The only solution to the equation cos−1 t =
sin−1 t
2
sin−1 t
2 .
is t =
√
3
2 .
Instructor’s Solutions Manual, Section 6.3
Problem 91
91. Show that
tan
θ
2
=±
1 − cos θ
1 + cos θ
for all θ except odd multiples of π .
solution Suppose θ is not an odd multiple of π . Thus
odd multiple of
π
2,
and hence tan
tan θ2 =
θ
2
is defined. Now
sin θ2
cos θ2
=
±
±
1−cos θ
2
1+cos θ
2
=±
1 − cos θ
.
1 + cos θ
θ
2
is not an
Instructor’s Solutions Manual, Section 6.3
Problem 92
92. Find a formula that expresses tan θ2 only in terms of tan θ.
θ
solution Suppose θ is not an odd multiple of π (so that tan 2 is
π
defined) and that θ is also not an odd multiple of 2 (so that tan θ is
defined). Then
tan
θ
2
=
1 − cos θ
sin θ
=
1
cos θ −
sin θ
cos θ
=
1
cos θ
1
−1
tan θ
,
where the second equality comes from dividing the numerator and
denominator of the previous expression by cos θ.
In Section 5.6 we derived the identity
1 + tan2 θ =
1
.
cos2 θ
Taking square roots of both sides shows that
1
= ± 1 + tan2 θ.
cos θ
Substituting this expression for
gives the formula
1
cos θ
into the expression above for tan
θ
2
Instructor’s Solutions Manual, Section 6.3
tan
θ
2
± 1 + tan2 θ − 1
.
=
tan θ
Problem 92
Instructor’s Solutions Manual, Section 6.3
Problem 93
93. Suppose θ is an angle such that cos θ is rational. Explain why cos(2θ)
is rational.
solution The square of every rational number is rational. Thus the
formula
cos(2θ) = 2 cos2 θ − 1
shows that if cos θ is rational, then so is cos(2θ).
Instructor’s Solutions Manual, Section 6.3
Problem 94
94. Give an example of an angle θ such that sin θ is rational but sin(2θ) is
irrational.
π
1
. Then sin θ = 2 , which
√ 6
3
2 , which is an irrational
solution One easy example is to take θ =
a rational number, but sin(2θ) = sin
number.
π
3
=
is
√
However, the example above requires knowing that 3 is irrational.
√
Most students will know that 3 is irrational and will be comfortable
using that fact, but strictly speaking in this book we have shown that
√
√
2 is irrational (see Section 0.1), with the irrationality of 3 never
discussed. Thus we give below another example that depends only on
√
the irrationality of 2.
For the second example, let θ = sin−1 31 . Thus sin θ = 13 , which is a
rational number. Now
√
cos θ = 1 − sin2 θ = 1 − 19 = 89 = 2 3 2 .
Thus
sin(2θ) = 2 cos θ sin θ = 2 ·
√
2 2
3
·
1
3
=
√
4 2
9 .
√
4 2
Because the product of two rational numbers is rational, if 9 were
√
√
√
9
4 2
rational, then 4 · 9 , which equals 2, would be rational. However, 2
is irrational. Thus we can conclude that
particular sin(2θ) is irrational.
√
4 2
9
is irrational, and in
Instructor’s Solutions Manual, Section 6.3
Problem 95
95. Give an example of an angle θ such that both sin θ and sin(2θ) are
rational.
solution The easiest example is to choose θ = 0. In that case, we
have sin θ = sin 0 = 0 and sin(2θ) = sin 0 = 0.
If an example is desired where sin θ and sin(2θ) are different rational
numbers, then an easy choice is to take θ = π2 . In that case, we have
π
sin θ = sin 2 = 1 and sin(2θ) = sin π = 0.
Instructor’s Solutions Manual, Section 6.3
Problem 96
Problems 96–101 will lead you to the discovery of an exact expression for
the value of sin 18◦ . For convenience, throughout these problems let
t = sin 18◦ .
96. Using a double-angle formula, show that cos 36◦ = 1 − 2t 2 .
solution
cos 36◦ = cos(2 · 18)◦
= 1 − 2 sin2 18◦
= 1 − 2t 2
Instructor’s Solutions Manual, Section 6.3
Problem 97
97. Using a double-angle formula and the previous problem, show that
cos 72◦ = 8t 4 − 8t 2 + 1.
solution We have
cos 72◦ = cos(2 · 36)◦
= 2 cos2 36◦ − 1
= 2(1 − 2t 2 )2 − 1
= 2(1 − 4t 2 + 4t 4 ) − 1
= 8t 4 − 8t 2 + 1,
where the third equality comes from the previous problem.
Instructor’s Solutions Manual, Section 6.3
Problem 98
98. Explain why sin 18◦ = cos 72◦ . Then using the previous problem,
explain why
8t 4 − 8t 2 − t + 1 = 0.
solution The equation sin 18◦ = cos 72◦ follows immediately from
the identity sin θ ◦ = cos(90 − θ)◦ with θ = 18◦ .
Because t = sin 18◦ , the result of the previous problem can be rewritten
as
t = sin 18◦ = cos 72◦ = 8t 4 − 8t 2 + 1,
which can be rewritten as
8t 4 − 8t 2 − t + 1 = 0.
Instructor’s Solutions Manual, Section 6.3
Problem 99
99. Verify that
8t 4 − 8t 2 − t + 1 = (t − 1)(2t + 1)(4t 2 + 2t − 1).
solution
(t − 1)(2t + 1)(4t 2 + 2t − 1)
= (2t 2 − t − 1)(4t 2 + 2t − 1)
= 8t 4 + 4t 3 − 2t 2 − 4t 3 − 2t 2 + t − 4t 2 − 2t + 1
= 8t 4 − 8t 2 − t + 1
Instructor’s Solutions Manual, Section 6.3
Problem 100
100. Explain why the two previous problems imply that
t = 1,
1
t=− ,
2
√
√
− 5−1
5−1
t=
, or t =
.
4
4
solution The two previous problems imply that
0 = 8t 4 − 8t 2 − t + 1 = (t − 1)(2t + 1)(4t 2 + 2t − 1).
Thus
t−1=0
or
2t + 1 = 0
or
4t 2 + 2t − 1 = 0.
The first equation above is equivalent to the equation t = 1, the second
1
equation above is equivalent to the equation t = − 2 , and the third
equation above is equivalent (by the quadratic formula) to t =
t=
√
5−1
4 .
√
− 5−1
4
or
Instructor’s Solutions Manual, Section 6.3
Problem 101
101. Explain why the first three values in the previous problem are not
possible values for sin 18◦ . Conclude that
√
◦
sin 18 =
5−1
.
4
π
[This value for sin 18◦ (or sin 10 if we work in radians) was used in
Exercise 9.]
solution Recall that t = sin 18◦ . Because 0◦ < 18◦ < 90◦ , we know
that 0 < sin 18◦ < 1. Thus the first possible value given in the previous
exercise is excluded because it equals 1, and the second and third
possible values given in the previous exercise are excluded because
they are negative. We conclude that t must equal the fourth possible
value given in the previous problem. In other words,
√
◦
sin 18 =
5−1
.
4
Instructor’s Solutions Manual, Section 6.3
Problem 102
102. Use the result from the previous problem to show that
√
cos 18◦ =
5+5
.
8
solution
cos 18◦ = 1 − sin2 18◦
=
1−
√ 5 − 1 2
4
=
√
5−2 5+1
1−
16
=
1−
√
3− 5
8
√
5+5
=
8