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2
4
= −
= −
2
5
3
= −
= −
= −
C H A P T E R 2
Analytic Trigonometry
Section 2.1 Using Fundamental Identities 1. tan u
2. csc u
9. sin θ 3
= − , cos θ 4
> 0 θ is in Quadrant IV.
3 9 7
3. cot u cos θ = 1 − − =
− 3
1 − = 16 4
4. csc u tan θ =
sin θ = 4 = − 3 3 7
5. 1 cos θ 7 7 7
4
6. − sin u
sec θ 1 1 4 = = = = 4 7
7. sec x
5
, tan x < 0 x is in Quadrant II.
cos θ 7 7 7
4
2 cot θ 1 1 7
= = = −
cos x = 1
= 1
= − 2
tan θ −
3 3 7
sec x −
5 5 2
csc θ
1 1 4 = = = −
sin x =
1 − −
2 =
1 − 4
= 21
25 5
sin θ −
3 3 4
2
tan x =
sin x
21
= 5 = − 21
10. cos θ = , sin θ 3
< 0 θ is in Quadrant IV.
2
cos x − 2 2
sin θ
= − 1 − 2
= − 1 − 4
= − 5
5
csc x = 1
= 5
= 5 21
9 3
− 5
sin x 21 21 tan θ = sin θ = 2 = −
5
cot x = 1
= − 2
= − 2 21
cos θ 2 2
3
tan x 21 21 1 1 3
8. csc x
7
, tan x > 0 x is in Quadrant III.
sec θ = cos θ
= 2
= 2
3
6 cot θ =
1 =
1 = −
2 2 5
sin x = 1
= 1
= − 6 tan θ
− 5 5 5
csc x −
7 7 6
csc θ
2
= 1
= 1
= − 3 3 5
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2
7 cos x = − 1 −
−
6
− 6
= − 1 − 36
= − 13
49 7
sin θ −
5 5 5 3
tan x = sin x
cos x = 7 =
6 =
− 13 13
6 13
13
7
sec x = 1
= 1
= − 7
= − 7 13
cos x −
13 13 13 7
cot x = 1
= 1
= 13
tan x 6 6
13
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2
2
2
csc x = − 1 + 7
= −
= −
cos x
2
214 Chapte r 2 A nalytic Trigo nomet ry
11.
tan x = 2, cos x > 0 x is in Quadrant I. 15. cos x(1 + tan 2 x) = cos x(sec2 x)
3 1
cot x = 1
= 1
= 3
= cos x cos2 x
tan x 2 2
3
2
4 13
= 1
cos x
= sec x
sec x = 1 + = 3
3
1 + = 9 3
9 13
Matches (f).
cos x 1 1
csc x = 1 + = 1 + = 16. cot x sec x = ⋅ = = csc x
2 4 2 sin x cos x sin x
sin x = 1
= 1
= 2
= 2 13 Matches (a).
csc x 13 13 13
sec2 x − 1 tan 2 x
sin 2 x 1
2 17. = = ⋅ = sec2 x
cos x = 1
= 1
= 3
= 3 13
sin 2 x sin 2 x cos2 x sin 2 x
sec x 13 13 13
3
Matches (e).
cos2 (π 2) − x
sin 2 x
sin x
cot x = 1
= 1
= 3 18.
cos x =
cos x = sin x = tan x sin x cos x
tan x 2 2
3
Matches (d).
12.
cot x = 7 , sin x < 0 x is in Quadrant III.
4
19.
tan θ cot θ
1 tan θ
= tan θ
tan x = 1
= 1
= 4 sec θ 1
cot x 7 7
4
4
16 65
cos θ
= 1 1
sec x = − 1 + = − 7
2
= −
1 + = − 49 7
1 + 49
= − 65
π
cos θ = cos θ
4 16 4 20. cos − x sec x = sin x sec x
sin x = 1
= 1
= − 4 4 65 2
1
csc x −
65 65 65 4
= sin x cos x
cos x = 1
= 1
= − 7 7 65
= tan x
sec x −
65 65 65 7
21. tan 2 x − tan 2 x sin 2 x = tan 2 x(1 − sin 2 x) = tan 2 x cos2 x
2
13. sec x cos x = 1
cos x
= sin x
⋅ cos2 x cos2 x
= 1
Matches (c).
22.
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= si
n 2
x
sin 2 x sec2 x − sin 2 x = sin 2 x(sec2 x − 1)
14. cot 2 x − csc2 x = (csc2 x − 1) − csc2 x
= −1 = sin 2 x tan 2 x
Matches (b).
23. sec x − 1
= ( sec x + 1)( sec x − 1)
sec x − 1 sec x − 1 = sec x + 1
24. cos x − 2
= cos x − 2
cos2 x − 4 (cos x + 2)(cos x − 2)
= 1 cos x + 2
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= −
2
Section 2.1 U s i ng Fundam ental Ident i ties 215
25. 1 − 2 cos
x + cos x = (1 − cos x)
26. sec4 x − tan 4 x = (sec2 x + tan 2 x)(sec2 x − tan 2 x)2 4 2
2 ( 2 2
)( ) = (sin 2 x) = sec x + tan x 1
27.
28.
= sin 4 x
cot3 x + cot 2 x + cot x + 1 = cot 2 x(cot x + 1) + (cot x + 1) = (cot x + 1)(cot 2 x + 1) = (cot x + 1)csc2 x
sec3 x − sec2 x − sec x + 1 = sec2 x(sec x − 1) − (sec x − 1) = (sec2 x − 1)(sec x − 1) = tan 2 x(sec x − 1)
= sec 2 x + tan 2 x
29.
3 sin 2
x − 5 sin x − 2 = (3 sin x + 1)(sin x − 2) 38.
cot u sin u + tan u cos u = cos u
(sin u) + sin u
(cos u)
30.
6 cos2 x + 5 cos x − 6 = (3 cos x − 2)(2 cos x + 3) sin u
= cos u + sin u
cos u
2 2 2
31. cot 2 x + csc x − 1 = (csc2 x − 1) + csc x − 1 39. 1 − sin x
= cos x = cos2 x tan 2 x = (cos2 x)
sin x
2 2 2
= csc2 x + csc x − 2
= (csc x − 1)(csc x + 2)
csc x − 1 cot x
= sin 2 x
cos x
32.
sin 2 x + 3 cos x + 3 = (1 − cos2 x) + 3 cos x + 3
= − cos2 x + 3 cos x + 4
= −(cos2 x − 3 cos x − 4) = −(cos x + 1)(cos x − 4)
40. cos2 y
1 − sin y
1 − sin 2 y =
1 − sin y
= (1 + sin y)(1 − sin y)
1 − sin y
= 1 + sin y
33.
tan θ csc θ
= sin θ
⋅ 1
= 1
= sec θ
41. (sin x + cos x)2 = sin 2 x + 2 sin x cos x + cos2 x
= (sin 2 x + cos2 x) + 2 sin x cos x
cos θ sin θ cos θ = 1 + 2 sin x cos x
34. tan(− x) cos x = −tan x cos x
sin x ⋅ cos x
cos x = −sin x
42. (2 csc x + 2)(2 csc x − 2) = 4 csc2 x − 4
= 4(csc2 x − 1) = 4 cot 2 x
35.
sin φ (csc φ − sin φ ) = (sin φ ) 1
− sin 2 φ
43.
1 +
1 =
1 − cos x + 1 + cos x
sin φ 1 + cos x 1 − cos x (1 + cos x)(1 − cos x)
= 1 − sin 2 φ = cos2 φ = 2 1 − cos2 x
36.
cos x(sec x − cos x) = cos x 1
− cos x
= 2
cos x
sin 2 x
= 1 − cos2 x
= sin 2 x
44.
= 2 csc2 x
1 −
1 =
sec x − 1 − (sec x + 1)
37. sin β tan β + cos β = (sin β ) sin β
+ cos β
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cos β
sec x + 1 sec x − 1
(sec x + 1)(sec x −
1)
= sec x − 1 − sec x − 1
sin 2 β = +
cos β
cos2 β
cos β
sec2 x − 1
= −2
sin 2 β + cos2 β =
cos β
tan 2 x
1 = −2 2
= 1 tan x
cos β = sec β
= −2 cot 2 x
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216 Chapte r 2 A nalytic Trigo nomet ry
45. cos x
− cos x
= cos x(1 − sin x) − cos x(1 + sin x)
1 + sin x 1 − sin x (1 + sin x)(1 − sin x)
= cos x − sin x cos x − cos x − sin x cos x
(1 + sin x)(1 − sin x)
= − 2sin x cos x
1 − sin 2 x
= − 2sin x cos x
cos2 x
= − 2sin x
cos x
= − 2 tan x
46.
sin x
1 + cos x +
sin x =
1 − cos x
sin x(1 − cos x) + sin x(1 + cos x) (1 + cos x)(1 − cos x)
= sin x − sin x cos x + sin x + sin x cos x
(1 + cos x)(1 − cos x)
= 2sin x 1 − cos2 x
= 2sin x sin 2 x
= 2 sin x
= 2 csc x
47. sec2 x
tan x − = tan 2 x − sec2 x 50.
5 ⋅
tan x − sec x = 5( tan x − sec x)
tan x tan x
= −1
= −cot x tan x
2 2
tan x + sec x tan x − sec x tan 2 x − sec2 x
5( tan x − sec x) =
−1
= 5(sec x − tan x)
48. cos x 1 + sin x cos x + (1 + sin x)
+ =
y = 1
sin x cot x + cos x1 + sin x cos x cos x(1 + sin x)
cos2 x + 1 + 2 sin x + sin 2 x 51. 1 ( )
2
= 1
cos x cos x(1 + sin x) = sin x + cos x
= 2 + 2 sin x
2 sin x
1
cos x(1 + sin x) = (cos x + cos x) 2
2(1+ sin x) =
cos x(1 + sin x) = cos x
2
= 2 cos x
= 2 sec x
− 2π 2π
49.
sin 2 y
1 − cos y
− 2
1 − cos2 y =
1 − cos y
(1 + cos y)(1 − cos y) =
1 − cos y
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= 1 + cos y
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Section 2.1 U s i ng Fundam ental Ident i ties 217
52. y1 = sec x csc x − tan x
1 1 sin x
= −
53. y = tan x + 1
1 sec x + csc x
cos x sin x cos x sin x
+ 1
1 sin 2 x = − =
cos x 1 1
cos x sin x
1 − sin 2 x cos x sin x
cos x +
sin x
= cos x sin x
cos2 x =
cos x sin x
cos x =
sin x + cos x
= cos x sin x + cos x
sin x cos x
sin x + cos x
sin x cos x
sin x =
= cot x 6
= sin x
cos x sin x + cos x
2
− 2π 2π
− 2π 2π
− 6
54.
y1 =
1 1
− 2
− cos x = tan x
sin x cos x 5
1 1 1 cos x
− cos x = −
sin x cos x
sin x cos x sin x − 2π 2π
1 − cos2 x
sin 2 x
sin x
= = = = tan x − 5sin x cos x sin x cos x cos x
55. Let x = 3 cos θ . 57. Let x = 2 sec θ .
9 − x2 =
=
=
9 − (3 cos θ )2
9 − 9 cos2 θ
9(1 − cos2 θ )
x2 − 4 =
=
=
(2 sec θ )2
− 4
4(sec2 θ − 1)
4 tan 2 θ
= 9 sin 2 θ
= 3 sin θ = 2 tan θ
56. Let x = 7 sin θ . 58. Let 3x = 5 tan θ .
2 2
49 − x2 = 49 − (7 sin θ )2
9x + 25 = (3x) + 25
= 49 − 49 sin 2 θ
= 49(1 − sin 2 θ )
= 49 cos2 θ
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= (5 tan θ )2
+ 25
= 25 tan 2 θ + 25
= 25(tan 2 θ + 1)
= 7 cos θ
= 25 sec2 θ
= 5 sec θ
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= ± 1 − 1
218 Chapte r 2 A nalytic Trigo nomet ry
59. Let x = 2 sin θ .
4 − x2 = 2
4 − (2 sin θ )2
= 2
4 − 4 sin 2 θ = 2
4(1 − sin 2 θ ) = 2
4 cos2 θ = 2
2 cos θ = 2
cos θ = 2 2
2
sin θ = 1 − cos2 θ = 2 2
1 −
= ±
60. Let x = 2 cos θ .
2 2
61.
x = 6 sin θ
16 − 4x2 = 2 2 3 = 36 − x2
16 − 4(2 cos θ )2 = 2 2 = 36 − (6 sin θ )
2
16 − 16 cos2 θ
16(1 − cos2 θ )
= 2 2
= 2 2
=
=
36(1 − sin 2 θ )
36 cos2 θ
16 sin 2 θ = 2 2 = 6 cos θ
4 sin θ = ±2 2 cos θ = 3
= 1
sin θ = ± 2
2
sin θ
6 2
= ± 1 − cos2 θ
cos θ =
=
1 − sin 2 θ
1 − 1
2
2
2 = ±
3
= 1 4
2
= ± 3
2 2 =
2 62. x = 10 cos θ
5 3 =
5 3 =
5 3 =
5 3 =
100 − x2
100 − (10 cos θ )2
100(1 − cos2 θ )
100 sin 2 θ
5 3 = 10 sin θ
sin θ
= 5 3
= 3
cos θ = 10 2
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1 − sin 2 θ = 2
3 1
1 −
= 2 2
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Section 2.1 Usi ng Fundamenta l Ident i ties 219
63. sin θ = 1 − cos2 θ 67. μ W cos θ
μ
= W sin θ
W sin θ
=
= tan θ
Let y1 = sin x and y2 = 1 − cos2 x , 0 ≤ x ≤ 2π . W cos θ
y1 = y2 for 0 ≤ x ≤ π .
sec x tan x − sin x = 1
⋅ sin x
− sin x
So, sin θ = 1 − cos2 θ for 0 ≤ θ ≤ π . 68.
cos x
cos x
2
y2
0 2π
y1
sin x = − sin x
cos2 x
sin x − sin x cos2 x =
cos2 x
64.
− 2
cos θ
= − 1 − sin 2 θ
sin x(1 − cos2 x) =
cos2 x
2
sin x sin x =
2Let y1 = cos x and y2 = − 1 − sin 2 x , 0 ≤ x ≤ 2π . cos x
2
y1 = π y2 for
≤ x ≤ 3π
. = sin x tan x
So, cos θ
2 2
= − 1 − sin 2 θ
for π
3π
≤ θ ≤ .
69. True. sin u
2 2
2
0 2π
tan u = cos u
cot u = cos u sin u
sec u = 1 cos u
65.
− 2
sec θ =
1 + tan 2 θ
1
csc u = 1 sin u
Let y1 = and y2 = cos x
π
1 + tan 2 x , 0 ≤ x ≤ 2π .
3π
70. False. A cofunction identity can be used to transform a
tangent function so that it can be represented by a
cotangent function.
y1 = y2 for 0 ≤ x < and 2 2
< x ≤ 2π .
π −
71. As x → , tan x → ∞ and cot x → 0.
So, sec θ = 1 + tan2 θ for 0 ≤ θ < π 2
and 2
+ 1
3π < θ
2 < 2π . 4
72. As x → π , sin x → 0 and csc x = sin x
→ −∞.
y2
0 2π
y1
− 4
73. cos(−θ ) ≠ − cos θ
cos(−θ ) = cos θ
The correct identity is sin θ
= cos(−θ )
sin θ
cos θ
66. csc θ = 1 + cot 2 θ
1
= tan θ
Let y1 = and y2 = si
n x
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1 + cot 2 x , 0 ≤ x ≤ 2π .
74. Let u = a tan θ ,
then
y1 = y2 for 0 < x < π .
a 2 + u 2 =
a 2 + (a tan θ )2
So, csc θ = 1 + cot 2 θ for 0 < θ < π .
2 2 2
= a + a 2
tan θ
0 2π
= a 2 (1 + tan 2 θ )
= a 2 sec2 θ
= a sec θ . − 2
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= 1 + a
= 1 + b
2
220 Chapte r 2 A nalytic Trigo nomet ry
75. Because sin 2 θ + cos2 θ = 1, then cos2 θ = 1 − sin2 θ .
cos θ
= ± 1 − sin θ
tan θ = sin θ =
sin θ
cot θ
cos θ ±
= cos θ
= ±
sin θ
1 − sin 2 θ
1 − sin 2 θ
sin θ
sec θ = 1
= 1
csc θ
cos θ
1 =
sin θ
± 1 − sin 2 θ
76. To derive sin 2 θ + cos2 θ = 1, let sin θ = a
and cos θ = b
a2 + b2 a2 + b2 .
2 2
a b a b2
So, sin 2 θ + cos2 θ = + = +
a2 + b2 a2 + b2 a2 + b2
a2 + b2
= a2 + b2
a2 + b2
= 1.
To derive 1 + ta n 2 θ
= sec2 θ , let tan θ = a
and sec θ b
a 2 + b 2 = .
b
So, 1 + ta n 2 θ
2
b
a2
= 1 + = b2
b2 + a2
b2
2 2 a2 + b2 a2 + b2
= = b2 b
= sec2 θ .
To derive 1 + cot 2 θ
= csc2 θ , let cot θ = b
and csc θ a
a 2 + b 2 = .
a
So, 1 + cot 2 θ 2
a
= 1 + b2
a2
2
a2 + b2 a2 + b2
= = a2 a2
2
a2 + b2
= = csc2 θ .
a
Answers will vary.
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1 1 +
sin θ
Section 2.2 Verifying Tri gonom e t r ic Identities 221
77.
sec θ (1 + tan θ
) sec θ + csc θ
cos θ
cos θ =
1 +
1
cos θ sin θ
cos θ + sin θ
= cos2 θ sin θ + cos θ
sin θ cos θ
= sin θ + cos θ sin θ cos θ
cos2 θ
sin θ + cos θ
= sin θ cos θ
Section 2.2 Verifying Trigonometric Identities 1. identity
2. conditional equation
3. tan u
16.
cos (π 2) − x
sin (π 2) − x
π
sin x = = tan x
cos x
1
4. cot u 17. sin t csc
2 − t = sin t sec t = sin t
cos t
5. sin u
sin t =
cos t
= tan t
6. cot 2 u
7. −csc u
18.
sec2 y − cot 2 π
2
− y
= sec2 y − tan 2 y = 1
8. sec u 19.
1 +
1 =
cot x + tan x
9. tan t cot t =
sin t
cos t
⋅ cos t
= 1 sin t
tan x cot x tan x cot x
= cot x + tan x
1
10. tan x cot x
cos x =
1 = sec x
cos x
20.
= tan x + cot x
1 −
1 =
csc x − sin x
11. (1 + sin α )(1 − sin α ) = 1 − sin2 α = cos2 α sin x csc x sin x csc x
csc x − sin x
=
12. cos2 β − sin 2 β = cos2 β − (1 − cos2 β ) = 2 cos2 β − 1
1
= csc x − sin x
13.
14.
cos2 β − sin 2 β
sin 2 α − sin 4 α
= (1 − sin 2 β ) − sin 2 β
= 1 − 2 sin 2 β
= sin 2 α (1 − sin 2 α )
= (1 − cos2 α )(cos2 α )
= cos2 α − cos4 α
21.
1 + sin θ
cos θ
+ cos θ
= 1 + sin θ
=
=
(1 + sin θ )2
+ cos2 θ
cos θ (1 + sin θ )
1 + 2 sin θ + sin 2 θ + cos2 θ
cos θ (1 + sin θ )
2 + 2 sin θ
cos θ (1 + sin θ )
2(1+ sin θ )
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15. tan π 2 − θ
tan θ = cot θ tan θ =
cos θ (1 + sin θ )
1 = tan θ
tan θ
= 1
= 2
cos θ
= 2 sec θ
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( )
(
222 Chapte r 2 A nalytic Trigo nomet ry
cos θ cot θ cos θ cot θ − (1 − sin θ ) 1
22. 1 − sin θ
− 1 =
=
1 − sin θ
cos θ cos θ
sin θ −
1 + sin θ
⋅ 1 − sin θ
cos2 θ − sin θ + sin 2 θ
sin θ
sin θ
25.
26.
27.
sec y cos y = cos y = 1 cos y
cot 2 y(sec2 y − 1) = cot 2 y tan2 y = 1
tan =
( ) = sin θ tan θ
2 θ sin θ cos θ tan θ =
sin θ (1 − sin θ ) sec θ 1 cos θ
= 1 − sin θ sin θ (1 − sin θ )
= 1 sin θ
= csc θ
28.
cot 3 t
csc t
cot t cot 2 t =
csc t
cot t (csc2 t − 1) =
csc t
cos t csc2 t − 1
sin t23.
1 1 + =
cos x + 1 cos x − 1
=
cos x − 1 + cos x + 1
(cos x + 1)(cos x − 1)
2 cos x
cos2 x − 1
2 cos x
= 1
sin t
= cos t sin t
csc2
sin t
t − 1)
= −sin 2 x = cos t(csc2 t − 1)
= −2 ⋅ 1
⋅ cos x
1 1 + tan 2 β
sin x sin x 29. tan β
+ tan β = tan β
24.
= −2 csc x cot x
cos x − cos x
= cos x(1 − tan x) − cos x
sec2 β =
tan β
1 − tan x 1 − tan x
= −cos x tan x
1 − tan x
−cos x(sin x cos x)
cos x
= ⋅ 1 − (sin x cos x)
= −sin x cos x cos x − sin x
sin x cos x
cos x
= sin x − cos x
30. sec θ − 1
= sec θ − 1
⋅ sec θ sec θ (sec θ − 1)
=
= sec θ
1 − cos θ 1 − (1 sec θ ) sec θ sec θ − 1
cot 2 t cos2 t sin 2 t cos2 t 1 − sin 2 t 1
31. = = = 33. sec x − cos x = − cos x
csc t 1 sin t sin t sin t cos x
1 − cos2 x =
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32. sin x
cos x + sin x tan x = cos x + sin x cos x
cos2 x + sin 2 x
cos x
sin 2 x =
cos x
= cos x
= sin x ⋅
sin x
= 1 cos x
= sec x
cos x
= sin x tan x
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Section 2.2 Veri fying Tri gono met r ic Id enti ties 223
34.
cot x − tan x = cos x −
sin x
sin x cos x
cos2 x − sin 2 x =
sin x cos x 1 − sin 2 x − sin 2 x
= sin x cos x
1 − 2 sin 2 x =
sin x cos x
1 1 − 2 sin 2 x =
cos x sin x
1 1 2 sin 2 x =
cos x sin x
− sin x
= sec x(csc x − 2 sin x)
cot x cos x sin x cos2 x 1 − sin 2 x 1 sin 2 x
35. = = = = − = csc x − sin x
36.
sec x
csc(− x) sec(− x)
1 cos x
1 sin(− x) =
1 cos(− x) cos(− x)
sin x sin x sin x sin x
= sin(− x)
37.
= cos x −sin x
= −cot x
sin1 2 x cos x − sin5 2 x cos x = sin1 2 x cos x(1 − sin2 x) = sin1 2 x cos x ⋅ cos2 x = cos3 x
sin x
38.
sec6 x(sec x tan x) − sec4 x(sec x tan x) = sec4 x(sec x tan x)(sec2 x − 1) = sec4 x(sec x tan x) tan 2 x = sec5 x tan3 x
39. (1 + sin y)1 + sin(− y) = (1 + sin y)(1 − sin y)
= 1 − sin 2 y
41.
1 + sin θ =
1 − sin θ
1 + sin θ
1 − sin θ
1 + sin θ ⋅
1 + sin θ
40.
= cos2 y
1 +
1 tan x + tan y
= cot x cot y
⋅ cot x cot y
(1 + sin θ )2
= 1 − sin 2 θ
(1 + sin θ ) 2
=1 − tan x tan y 1 − 1
⋅ 1 cot x cot y cos2 θ
cot x cot y 1 + sin θ
== cot y + cot x
cot x cot y − 1 cos θ
42. cos x − cos y
+ sin x − sin y
= (cos x − cos y)(cos x + cos y) + (sin x − sin y)(sin x + sin y)
sin x + sin y cos x + cos y (sin x + sin y)(cos x + cos y) cos2 x − cos2 y + sin 2 x − sin 2 y
= (sin x + sin y)(cos x + cos y)
(cos2 x + sin 2 x) − (cos2 y + sin 2 y) =
(sin x + sin y)(cos x + cos y)
43. = 0
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cot(
− x)
≠
cot
x
44. The first line claims that sec(−θ )
= −sec θ
and
The correct substitution is cot(− x) = − cot x. sin(−θ ) = sin θ . The correct substitutions are
1
+ cot(− x) = cot x − cot x = 0 tan x
sec(−θ ) = sec θ and sin(−θ ) = −sin θ .
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( )( )
( 2 2 )
224 Chapte r 2 A nalytic Trigo nomet ry
45. (a)
3
− 2π 2π
− 1
Identity
(b)
Identity
(c) 1 + cot 2 x cos2 x = csc2 x cos2 x = 1
⋅ cos2 x = cot 2 x sin 2 x
3
46. (a) (b)
− 2π 2π
− 1
Identity
Identity
(c) csc x(csc x − sin x) + sin x − cos x
+ cot x = csc2 x − csc x sin x + 1 − cos x
+ cot x
sin x sin x
= csc2 x − 1 + 1 − cot x + cot x
= csc2 x
47. (a) 5
y2
y1
− 2π 2π
− 1
Not an identity
(b)
Not an identity
(c) 2 + cos2 x − 3 cos4 x = (1 − cos2 x)(2 + 3 cos2 x) = sin 2 x(2 + 3 cos2 x) ≠ sin2 x(3 + 2 cos2 x)
48. (a)
− π
5
y1 y2
π
(b)
− 5
Not an identity
sin 4 x
sin 2 x
Not an identity
(c) tan 4 x + tan 2 x − 3 = + − 3
cos4 x
1 cos2 x
sin 4 x
= cos2 x
cos2 x
+ sin 2 x − 3
1 sin 4 x + sin 2 x cos2 x =
cos2 x
cos2 x − 3
1 sin 2 x = sin x + cos x − 3
cos2 x cos2 x
1 sin 2 x =
cos2 x cos2 x
⋅ 1 − 3
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= sec2 x tan 2 x − 3
≠ sec2 x(4 tan 2 x − 3)
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.
cos2 x
( )
Section 2.2 Veri fying Tri gono met r ic Id entit ies 225
49. (a) 3 50. (a) 3
y1
− 2π 2π
− 2 2
y2
− 3 − 5
(b)
Identity
(b)
Not an identity
(c)
Identity
1 + cos x
(1 + cos x)(1 − cos x) =
Not an identity
(c) cot α
is the reciprocal of csc α + 1
sin x sin x(1 − cos x) csc α + 1 cot α
1 − cos2 x =
sin x(1 − cos x) sin 2 x
= sin x(1 − cos x)
sin x =
51.
They will only be equivalent at isolated points in
their respective domains. So, not an identity.
tan3 x sec2 x − tan3 x = tan3 x(sec2 x − 1)
= tan3 x tan 2 x
5
1 − cos x = tan x
2 4
52. (tan 2 x + tan 4 x) sec2 x = sin x
sin x 1 +
cos4 x cos2 x
1 = sin 2 x + sin 4 x
cos4 x cos2 x
1 sin 2 x cos2 x + sin 4 x =
cos4 x cos2 x
1 sin 2 x cos2 x + sin 2 x =
cos4 x cos2 x
1 sin 2 x =
⋅ 1 = sec4 x ⋅ tan 2 x
cos4 x cos2 x
53. (sin 2 x − sin 4 x) cos x = sin 2 x(1 − sin 2 x) cos x
= sin 2 x cos2 x cos x
= sin 2 x cos3 x
54. sin 4 x + cos4 x = sin 2 x sin 2 x + cos4 x
= (1 − cos2 x)(1 − cos2 x) + cos4 x
= 1 − 2 cos2 x + cos4 x + cos4 x
= 1 − 2 cos2 x + 2 cos4 x
55. sin 2 25° + sin 2 65° = sin 2 25° + cos2 (90° − 65°)
= sin 2 25° + cos2 25°
= 1
56. tan 2 63° + cot 2 16° − sec2 74° − csc2 27° = tan 2 63° + cot 2 16° − csc2 (90° − 74°) − sec2 (90° − 27°)
= tan 2 63° + cot 2 16° − csc2 16° − sec2 63°
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=
(t
a
n 2
6
3
°
−
s
e
c2
6
3
°
) +
(c
o
t 2
1
6
°
− c
s
c2
1
6
°
) =
−
1
+
(−
1
)
=
−2
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θ 15° 30° 45° 60° 75° 90°
s 18.66 8.66 5 2.89 1.34 0
1
226 Chapte r 2 A nalytic Trigo nomet ry
57. Let θ
= sin−1 x sin θ = x = x
.
60. Let θ = cos−1 x + 1 cos θ = x + 1
.
1 2 2
1 x
2 4 − (x + 1)2
θ
1 − x2
θ
x + 1
From the diagram,
From the diagram,
tan(sin −1 x) = tan θ = x
. −1
x + 1
4 − ( x + 1)2
1 − x2 tan cos 2
= tan θ = x 1
.
58. Let θ
= sin−1 x sin θ
= x = x
.
+
1 cos x
1 61. cos x − csc x cot x = cos x − sin x sin x
1 = cos x1 −
sin 2 x x
θ
1 − x2
= cos x(1 − csc2 x) = −cos x(csc2 x − 1)
= −cos x cot 2 x
From the diagram,
cos(sin −1 x) = cos θ
1 − x 2
= = 1
1 − x2 .
62. (a)
(b)
h sin(90° −θ ) sin θ
= h cos θ
= h cot θ sin θ
59. Let θ = sin −1 x − 1 sin θ = x − 1
.
4 4
4
x − 1
(c) Maximum: 15°
Minimum: 90°
(d) Noon
θ 63. False. tan x2 = tan(x ⋅ x) and
16 − (x − 1)2
From the diagram,
tansin−1 x − 1 = tan θ
x − 1
= .
tan 2 x = (tan x)(tan x), tan x2
64. True. Cosine is an even function,
≠ tan 2 x.
π π 4 16 − (x − 1)
2 cosθ − = cos− − θ
2
= cos π
2
− θ
2
= sin θ .
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65. False. For the equation to be an identity, it must be true
for all values of θ in the domain.
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2
b 2
b
2 2
2
Section 2.3 Solving Trigo nomet ric Equ a tion s 227
66. If sin θ = a , sec θ =
c , and 68. tan θ = sec2 θ − 1
c b True identity: tan θ
= ± sec2 θ − 1
a2 + b2 = c2 a2
c
= c2 − b2 , then tan θ =
sec2 θ − 1 is not true for π 2 < θ < π
sec2 θ − 1 − 1
=
or 3π 2 < θ
< 2π . So, the equation is not true for
sec2 θ c
c2
θ
69.
= 3π 4.
1 − cos θ
= sin θ
b2 − 1
= c2
(1 − cos θ )
= (sin θ )
2 2
b2 1 − 2 cos θ + cos θ = sin θ
c2 − b2
= b2
c2
1 − 2 cos θ + cos2 θ
2 cos2 θ − 2 cos θ
= 1 − cos2 θ
= 0
b2
c2 − b2 b2
= ⋅ b2 c2
c2 − b2
2 cos θ (cos θ − 1) = 0
The equation is not an identity because it is only true
when cos θ = 0 or cos θ = 1. So, one angle for which
the equation is not true is − π
.
= c2
a2 70. =
c2
1 + tan θ
(1 + tan θ )2
2
= sec θ
= (sec θ )2
a 2 2=
1 + 2 tan θ + tan θ = sec θ
c
= sin 2 θ . 1 + 2 tan θ + tan 2 θ
2 tan θ
= 1 + tan 2 θ
= 0
67. Because sin 2 θ
= 1 − cos2 θ , then tan θ = 0
sin θ = ± 1 − cos2 θ ; sin θ ≠ 1 − cos2 θ
if θ This equation is not an identity because it is only true
lies in Quadrant III or IV. when tan θ = 0. So, one angle for which the equation
One such angle is θ = 7π
. 4
is not true is π
. 6
Section 2.3 Solving Trigonometric Equations 1. isolate 6. sec x − 2 = 0
2. general
3. quadratic
(a) x =
π 3
sec π
− 2 = 1
− 2
4. extraneous
5.
ta
n
x −
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3 = 0
3
cos(π
3)
=
1
−
2
=
2
−
2
=
0 1 2
(a) x =
π 3
tan π
− 3 =
3 − 3 = 0
(b) x =
5π 3
sec 5π
− 2 = 1
− 2
(b)
3
x = 4π 3
3 cos(5π 3)
= 1
− 2 = 2 − 2 = 0 1 2
tan 4π
− 3 = 3
3 − 3 = 0
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=
2 cos2 2
2
2 cos2 2
2 = −
228 Chapte r 2 A nalytic Trigo nomet ry
7. 3 tan 2
2 x − 1 = 0 10. csc4
x − 4 csc2
x = 0
(a) x =
π 12
2
3 tan 2 π
− 1 = 3 tan 2 π − 1
(a) x =
π 6
csc4 π
− 4 csc2 π
1 4
= −
12
6 4 2
2
1 3 − 1
6 6 sin (π 6) 1 4
= −
sin (π 6)
3 (1 2)4
(1 2)2
(b)
= 0
x = 5π 12
2
(b)
x = 5π 6
= 16 − 16 = 0
3 tan 2 5π
− 1 = 3 tan 2 5π − 1
csc4 5π
4 csc 5π
1 4
12
6
− = 4
− 2
2
1 = 3 − − 1
6 6 sin (5π 6)
1 4 = −
sin (5π 6)
3 (1 2)4
(1 2)2
= 0
8. 2 cos2 4 x − 1 = 0
11. 3 csc x − 2 = 0
= 16 − 16 = 0
(a) x = π 16
π π 4 − 1 = 2 cos − 1
3 csc x = 2
csc x = 2
3
16 4
2
x = π
+ 2nπ
3
= 2 2 − 1
2
(b)
x = 3π
1 = 2 − 1 = 1 − 1 = 0
12. tan x +
or x = 2π
3
3 = 0
+ 2nπ
16
3π 3π 4 − 1 = 2 cos − 1
tan x = − 3
x = 2π
+ nπ
16 4
2
2 = 2 − − 1
3
13. cos x + 1 = − cos x
2
2 cos x + 1 = 0
= 1
− 1 = 0 2
cos x 1 2
x = 2π + 2nπ or x =
4π + 2nπ
9. 2 sin 2 x − sin x − 1 = 0 3 3
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2
(a)
x = π 2
2 sin 2 π
− sin π
− 1 = 2(1)2
− 1 − 1
14. 3 sin x + 1 = sin x
2 sin x + 1 = 0
1
2 2 sin x = −
= 0 7π
(b) x = 7π 6
2 sin 2 7π
− sin 7π
1
2
− 1 = 2 −
1
− − − 1
x = + 2nπ or 6
x = 11π
+ 2nπ 6
6 6 2 2
= 1
+ 1
− 1 2 2
= 0
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Section 2.3 Solving Trigo nomet ric Equ a t ion s 229
15. 3 sec2 x − 4 = 0 20. (2 sin 2 x − 1)(tan 2 x − 3) = 0
sec2 x = 4 3
2 sin 2 x − 1 = 0 or tan 2 x = 3
sec x = ± 2 3
sin 2 x = 1 2
tan x = ± 3
x = π 6
+ nπ sin x = ±
1 2
x = π 3
+ nπ
or x = 5π
+ nπ sin x = ±
2 x = 2π + nπ
16.
6
3 cot 2 x − 1 = 0
cot 2 x = 1
x = π 4
x = 3π
2 3
+ 2nπ
+ 2nπ
3
cot x = ± 1 3
4
x = 5π 4
x = 7π
+ 2nπ
+ 2nπ
x = π 3
+ nπ 4
or x = 2π 3
+ nπ
21.
cos3 x − cos x = 0
cos x(cos2 x − 1) = 0
17. 4 cos2 x − 1 = 0 cos x = 0
or cos2 x − 1 = 0
cos2 x = 1 4
cos x 1
x = π
+ nπ 2
cos x = ±1
x = nπ
= ± 2 Both of these answers can be represented as x = nπ
.
x = π + nπ or x =
2π 2
+ nπ
18.
3 3
2 − 4 sin 2 x = 0
sin 2 x = 1
2
22.
sec2 x − 1 = 0
sec2 x = 1
sec x = ±1
x = nπ
sin x = ± 1 2
= ± 2
2
23.
3 tan3 x = tan x
3
x = π + 2nπ 3 tan x − tan x = 0
4
x = 3π
+ 2nπ
tan x(3 tan 2 x − 1) = 0
4
x = 5π 4
x = 7π
+ 2nπ
+ 2nπ
tan x = 0
x = nπ
or 3 tan 2 x − 1 = 0
tan x = ±
x = π
3
3
+ nπ , 5π
+ nπ
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4
These answers can be represented as x = π
+ nπ
.
24.
6 6 sec x csc x = 2 csc x
19.
sin x(sin x + 1) = 0
4 2 sec x csc x − 2 csc x = 0
csc x(sec x − 2) = 0
sin x = 0 or sin x = −1
csc x = 0 or sec x − 2 = 0
x = nπ x = 3π 2
+ 2nπ No solution sec x = 2
x = π
+ 2nπ , 5π
+ 2nπ
3 3
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= −
230 Chapte r 2 A nalytic Trigo nomet ry
25. 2 cos2 x + cos x − 1 = 0
(2 cos x − 1)(cos x + 1) = 0
2 cos x − 1 = 0
or cos x + 1 = 0
cos x = 1 cos x = −1
2
x = π
+ 2nπ , 5π
+ 2nπ
x = π + 2nπ
3 3
26. 2 sin 2 x + 3 sin x + 1 = 0
(2 sin x + 1)(sin x + 1) = 0
2 sin x + 1 = 0
or sin x + 1 = 0
sin x 1 2
sin x = −1
x = 3π
+ 2nπ
x = 7π + 2nπ ,
11π + 2nπ 2
6 6
27. sec2 x − sec x = 2
sec2 x − sec x − 2 = 0
(sec x − 2)(sec x + 1) = 0
sec x − 2 = 0
sec x = 2
or sec x + 1 = 0
sec x = −1
x = π + 2nπ ,
5π
+ 2nπ x = π + 2nπ
3 3
28. csc2 x + csc x = 2
csc2 x + csc x − 2 = 0
(csc x + 2)(csc x − 1) = 0
csc x + 2 = 0
csc x = − 2
or csc x − 1 = 0
csc x = 1
x = 7π + 2nπ ,
11π
+ 2nπ x = π
+ 2nπ
6 6 2
29. sin x − 2 = cos x − 2
sin x = cos x
30. cos x + sin x tan x = 2
cos x + sin x sin x
= 2
sin x = 1
cos x
cos x
cos2 x + sin 2 x
tan x = 1
x = tan −1 1
x = π
, 5π
4 4
cos x = 2
1
= 2 cos x
1
cos x = 2
x = π
, 5π
3 3
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Section 2.3 Solving Trigo nomet ric Equ a tion s 231
31. 2 sin 2 x = 2 + cos x
2 − 2 cos2 x = 2 + cos x
2 cos2 x + cos x = 0
36. 3 sec x − 4 cos x = 0
3 − 4 cos x = 0
cos x
cos x(2 cos x + 1) = 0 3 − 4 cos 2 x
= 0 cos x
cos x = 0 or 2 cos x + 1 = 0 3 − 4 cos2 x = 0
x = π
, 3π 2 cos x = −1
3
2 2 1
cos x = − 2
cos2 x = 4
3
32.
tan 2 x = sec x − 1
sec2 x − 1 = sec x − 1
x = 2π
, 4π
3 3
37.
cos x = ± 2
x = π
, 5π
, 7π
, 11π
6 6 6 6
csc x + cot x = 1
sec2 x − sec x = 0 (csc x + cot x)2
= 12
33.
sec x(sec x − 1) = 0
sec x = 0 or sec x − 1 = 0
No Solutions sec x = 1
x = 0
sin 2 x = 3 cos2 x
csc2 x + 2 csc x cot x + cot 2 x = 1
cot 2 x + 1 + 2 csc x cot x + cot 2 x = 1
2 cot 2 x + 2 csc x cot x = 0
2 cot x(cot x + csc x) = 0
2 cot x = 0 or cot x + csc x = 0
x π
, 3π cos x 1
sin 2 x − 3 cos2 x = 0 = = − 2 2 sin x
sin x
sin 2 x − 3(1 − sin 2 x) = 0
4 sin 2 x = 3
sin x = ± 3
3π 2
is extraneous.
cos x = −1
x = π
(π is extraneous.)
2
x = π
, 2π
, 4π
, 5π
3 3 3 3
34. 2 sec2 x + tan 2 x − 3 = 0
38.
x = π 2 is the only solution.
sec x + tan x = 1
1 sin x + = 1
cos x cos x
2(tan 2 x + 1) + tan 2 x − 3 = 0 1 + sin x =
cos x
3 tan 2 x − 1 = 0 (1 + sin x)2
= cos2 x
tan x = ± 3
3 1 + 2 sin x + sin 2 x = cos2 x
1 + 2 sin x + sin 2 x = 1 − sin 2 x
x = π
, 5π
, 7π
, 11π
6 6 6 6
2 sin 2
x + 2 sin x = 0
35.
2 sin x + csc x = 0
2 sin x + 1
= 0
2 sin x(sin x + 1) = 0
sin x = 0 or sin x + 1 = 0
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.
sin x
2 sin 2 x + 1 = 0
sin 2 x
1
No solution
x = 0, π
(π is extraneous.)
sin x = −1
x = 3π 2
= − 2
x = 0 is the only solution.
3π 2
is extraneous
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2
2
232 Chapte r 2 A nalytic Trigo nomet ry
39. 2 cos 2x − 1 = 0 45. 3 tan
x −
3 = 0
cos 2x = 1 2
2
tan x
= 3
2 x = π
+ 2nπ
or 2 x = 5π 2 3
+ 2nπ
3 3 x = π + nπ x =
π + 2nπ
x = π
+ nπ x = 5π
+ nπ 2 6 3
6 6 x
46. tan +
3 = 0
40. 2 sin 2 x + 3 = 0 2
x
sin 2 x = − 3
2
tan 2
x
= − 3
= 2π
+ nπ
x = 4π
+ 2nπ
2 x = 4π + 2nπ or 2x =
5π + 2nπ 2 3 3
3 3
x = 2π
+ nπ x =
5π
+ nπ
47. y = sin π x
+ 1
3 6 2
π x + =41. tan 3x − 1 = 0
tan 3x = 1
sin 1 0
π x
3x = π
+ nπ sin = −1
4
x = π
+ nπ
π x
2 = 3π 2
+ 2nπ
12 3
42. sec 4 x − 2 = 0
sec 4x = 2
cos 4x = 1 2
48.
x = 3 + 4n
For −2 < x < 4, the intercepts are −1 and 3.
y = sin π x + cos π x
sin π x + cos π x = 0
sin π x = −cos π x
4 x = π + 2nπ or 4 x =
5π + 2nπ π
3 3 π x = − + nπ 4
x = π
+ nπ x =
5π +
nπ 1
12 2 12 2 x = − + n 4
43. 2 cos x
=
2 = 0
For 1 x
3, the intercepts are − 1 ,
3 ,
7 , 11
2
cos x
= 2
− < < . 4 4 4 4
2 2 49. 5 sin x + 2 = 0 8
x = π + 2nπ or
x =
7π + 2nπ
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2 4 2 4
x = π
+ 4nπ x = 7π
+ 4nπ 0 2π
2 2
44. 2 sin x
= 2
3 = 0
− 5
x ≈ 3.553 and x ≈ 5.872
sin x
= − 3 50. 2 tan x + 7 = 0
2 2 15
x =
4π
+ 2nπ or x 5π
=
+ 2nπ
2 3 2 3
x = 8π
+ 4nπ x = 10π
+ 4nπ 0 2π
3 3 − 5
x ≈ 1.849 and x ≈ 4.991
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Section 2.3 Solving Trigo nomet ric Equ a t ion s 233
51. sin x − 3 cos x = 0
5
55. sec2 x − 3 = 0
5
0 2π
0 2π
− 5
x ≈ 1.249 and x ≈ 4.391
52. sin x + 4 cos x = 0
5
56.
− 4
x ≈ 0.955, x ≈ 2.186, x ≈ 4.097 and x ≈ 5.328
csc2 x − 5 = 0
5
0 2π 0 2π
− 5
x ≈ 1.816 and x ≈ 4.957
− 5
x ≈ 0.464, x ≈ 2.678, x = 3.605 and x ≈ 5.820
53. cos x = x
4
57.
2 tan 2 x = 15 6
0 2π 0 2π
− 8
x ≈ 0.739
54. tan x = csc x
− 18
x ≈ 1.221, x ≈ 1.921, x ≈ 4.362 and x ≈ 5.062
10
58.
6 sin 2
x = 5
6
0 2π
0 2π
− 10
x ≈ 0.905 and x ≈ 5.379
− 18
x ≈ 1.150, x ≈ 1.991, x ≈ 4.292 and x ≈ 5.133
59. tan 2 x + tan x − 12 = 0
(tan x + 4)(tan x − 3) = 0
tan x + 4 = 0 or tan x − 3 = 0
tan x = −4 tan x = 3
x = arctan(−4) + nπ x = arctan 3 + nπ
60. tan 2 x − tan x − 2 = 0
(tan x + 1)(tan x − 2) = 0
tan x + 1 = 0 or tan x − 2 = 0
tan x = −1 tan x = 2
x = 3π 4
+ nπ
x = arctan 2 + nπ
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3 3
234 Chapte r 2 A nalytic Trigo nomet ry
61. sec2 x − 6 tan x = − 4
1 + tan 2 x − 6 tan x + 4 = 0
tan 2 x − 6 tan x + 5 = 0
(tan x − 1)(tan x − 5) = 0
tan x − 1 = 0 tan x − 5 = 0
tan x = 1 tan x = 5
x = π 4
+ nπ
x = arctan 5 + nπ
62. sec2 x + tan x − 3 = 0
1 + tan 2 x + tan x − 3 = 0
tan 2 x + tan x − 2 = 0
(tan x + 2)(tan x − 1) = 0
tan x + 2 = 0 tan x − 1 = 0
tan x = −2 tan x = 1
x = arctan(−2) + nπ x = arctan(1) + nπ
63.
≈ −1.1071 + nπ
2 sin 2 x + 5 cos x = 4
2(1 − cos2 x) + 5 cos x − 4 = 0
− 2 cos2 x + 5 cos x − 2 = 0
− (2 cos x − 1)(cos x − 2) = 0
= π
+ nπ 4
2 cos x − 1 = 0 or cos x − 2 = 0
cos x = 1 2
x = π
+ 2nπ , 5π
+ 2nπ
cos x = 2
No solution
3 3
64. 2 cos2 x + 7 sin x = 5
2(1 − sin 2 x) + 7 sin x − 5 = 0
− 2 sin 2 x + 7 sin x − 3 = 0
− (2 sin x − 1)(sin x − 3) = 0
2 sin x − 1 = 0 or sin x − 3 = 0
sin x = 1 2
sin x = 3
x = π + 2nπ ,
5π
+ 2nπ
No solution
6 6
65. cot 2 x − 9 = 0
cot 2 x = 9
1 = tan 2 x
9
± 1
= tan x 3
x = arctan 1 + nπ , arctan(− 1 ) + nπ
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5 5
=
Section 2.3 Solving Trigo nomet ric Equ a t ion s 235
66. cot 2 x − 6 cot x + 5 = 0
(cot x − 5)(cot x − 1) = 0
cot x − 5 = 0 or cot x − 1 = 0
cot x = 5 cot x = 1
1 = tan x
5
1 = tan x
x = arctan 1
+ nπ x = π
+ nπ
5 4
67. sec2 x − 4 sec x = 0
sec x(sec x − 4) = 0
sec x = 0 sec x − 4 = 0
No solution sec x = 4
1 = cos x
4
x = arccos 1
+ 2nπ , − arccos 1
+ 2nπ 4 4
68. sec2 x + 2 sec x − 8 = 0
(sec x + 4)(sec x − 2) = 0
sec x + 4 = 0 or sec x − 2 = 0
sec x = −4 sec x = 2
− 1
= cos x 1 = cos x
4 2
x = arccos −
1 + 2nπ , − arccos
−
1 + 2nπ
x =
π + 2nπ ,
5π
+ 2nπ
4
4
3 3
69. csc2 x + 3 csc x − 4 = 0
(csc x + 4)(csc x − 1) = 0
csc x + 4 = 0 or csc x − 1 = 0
csc x = −4 csc x = 1
− 1
= sin x 4
1 = sin x
x = arcsin 1
+ 2nπ , arcsin −
1 + 2nπ
x =
π
+ 2nπ
4
4
2
70. csc2 x − 5 csc x = 0
csc x(csc x − 5) = 0
csc x = 0 or csc x − 5 = 0
No solution csc x = 5
1 sin x 5
x = arcsin(1 ) + 2nπ , arcsin(− 1 ) + 2nπ
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3
236 Chapte r 2 A nalytic Trigo nomet ry
71. 12 sin 2 x − 13 sin x + 3 = 0
−(−13) ±
(−13)2
− 4(12)(3)
30
13 ± 5
sin x = = 2(12) 24
0 2π
sin x = 1 or sin x =
3 − 10
3 4
x ≈ 0.3398, 2.8018 x ≈ 0.8481, 2.2935 The x-intercepts occur at x ≈ 0.3398,
x ≈ 0.8481, x ≈ 2.2935, and x ≈ 2.8018.
72. 3 tan 2 x + 4 tan x − 4 = 0
−4 ±
42 − 4(3)(−4)
−4 ±
64 2
tan x = = = −2,
tan x = −2 tan x =
2(3) 6 3
2
3 50
x = arctan(−2) + nπ
≈ −1.1071 + nπ
x = arctan 2
+ nπ
≈ 0.5880 + nπ
0 2
The values of x in [0, 2π ) are 0.5880, 3.7296, 2.0344, 5.1760.
−10
73. tan 2 x + 3 tan x + 1 = 0
−3 ±
32 − 4(1)(1)
−3 ± 5
tan x = = 10
2(1) 2
tan x =
−3 − 5 or tan x =
−3 + 5
0 2π
2 2
x ≈ 1.9357, 5.0773 x ≈ 2.7767, 5.9183
− 5
The x-intercepts occur at x ≈ 1.9357, x ≈ 2.7767,
x ≈ 5.0773, and x ≈ 5.9183.
74. 4 cos2 x − 4 cos x − 1 = 0
4 ±
(−4)2
− 4(4)(−1)
4 ± 32 1 ± 2
cos x = = =
cos x = 1 − 2
2(4) 8 2
cos x = 1 + 2
7
2 2
− 1 2 x = arccos
No solution 2
0 2π
≈ 1.7794
1 + 2 − 3
> 1
2
− 1 − 2
Solutions in [0, 2π ) are arccos and 2π − arccos
1 2
: 1.7794, 4.5038.
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2 2
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−
−
−
Section 2.3 Solving Trigo nomet ric Equ a t ion s 237
75. 3 tan 2 x + 5 tan x − 4 = 0,
3
π , π
2 2
77. 4 cos2 x − 2 sin x + 1 = 0,
6
π , π
2 2
−π π
2 2
− p p 2 2
76.
− 7
x ≈ −1.154, 0.534
cos2 x − 2 cos x − 1 = 0, [0, π ] 3
78.
− 2
x ≈ 1.110
2 sec2 x + tan x − 6 = 0,
π
, π
2 2
4
0 π
− 3
x ≈ 1.998
−π π 2 2
− 6
x ≈ −1.035, 0.870
79. (a) f (x) = sin2 x + cos x
2
(b) 2 sin x cos x − sin x = 0
sin x(2 cos x − 1) = 0
0 2π
sin x = 0 or 2 cos x − 1 = 0
x = 0, π
− 2 ≈ 0, 3.1416 cos x =
1 2
Maximum: (1.0472, 1.25)
Maximum: (5.2360, 1.25)
Minimum: (0, 1)
Minimum: (3.1416, −1)
x = π
, 5π
3 3
≈ 1.0472, 5.2360
80. (a) f (x) = cos2 x − sin x 2
(b) −2 sin x cos x − cos x = 0
−cos x(2 sin x + 1) = 0
−cos x = 0 2 sin x + 1 = 00 2π
− 2
cos x = 0 sin x = − 1 2
x = π
, 3π
x = 7π
, 11π
Maximum: (3.6652, 1.25) 2 2 6 6
81. (a)
Maximum: (5.7596, 1.25)
Minimum: (1.5708, −1)
Minimum: (4.7124, 1)
f (x) = sin x + cos x
3
≈ 1.5708, 4.7124
(b) cos x − sin x = 0
cos x = sin x
≈ 3.6652, 5.7596
0 2π
1 = sin x cos x
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− 3
Maximum: (0.7854, 1.4142)
Minimum: (3.9270, −1.4142)
tan x = 1
x = π
, 5π
4 4
≈ 0.7854, 3.9270
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2
+ =
238 Chapte r 2 A nalytic Trigo nomet ry
82. (a) f (x) = 2 sin x + cos 2 x
3
(b) 2 cos x − 4 sin x cos x = 0
2 cos x(1 − 2 sin x) = 0
2 cos x = 0 1 − 2 sin x = 00 2π
x = π
, 3π sin x =
1
2 2 2− 3
Maximum: (0.5236, 1.5)
Maximum: (2.6180, 1.5)
Minimum: (1.5708, 1.0)
Minimum: (4.7124, − 3.0)
≈ 1.5708, 4.7124 x = π
, 5π
6 6 ≈ 0.5236, 2.6180
83. (a) f (x) = sin x cos x
2
84. (a) f (x) = sec x + tan x − x 6
0 2π 0 2π
− 2
Maximum: (0.7854, 0.5)
Maximum: (3.9270, 0.5)
Minimum: (2.3562, − 0.5)
(b)
− 8
Maximum: (3.1416, − 4.1416)
Minimum: (0, 1)
sec x tan x + sec2 x − 1 = 0
Minimum: (5.4978, − 0.5) 1 ⋅
sin x + 1
− 1 = 0
(b) −sin 2 x + cos2 x = 0 cos x cos x cos x
−sin 2 x + 1 − sin 2 x = 0
−2 sin 2 x + 1 = 0
sin x + 1 − 1 = 0
cos2 x
sin x + 1 cos2 x − = 0
sin 2 x = 1 cos2 x cos2 x
2 sin x + 1 − cos2 x
2 = 0
sin x = ± 1
= ± 2 cos x
2 2 sin x + sin 2 x = 0
x = π
, 3π
, 5π
, 7π
4 4 4 4
≈ 0.7854, 2.3562, 3.9270, 5.4978
cos2 x
sin x + sin 2 x = 0
sin x(1 + sin x) = 0
sin x = 0 or 1 + sin x = 0
x = 0, π
≈ 0, 3.1416
sin x = −1
x = 3π 2
3π is undefined in original function. So, it is not
2
a solution.
85. The graphs of y1 = 2 sin x and y2
equation 2 sin x = 3x + 1.
= 3x + 1 appear to have one point of intersection. This implies there is one solution to the
86. The graphs of y1 = 2 sin x and y2 1 x 1 appear to have three points of intersection. This implies there are three solutions 2
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+ to the equation 2 sin x = 1 x 1. 2
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x
32
32
Month
ly s
ales
(in t
housa
nds
of
do
llar
s)
Section 2.3 Solving Trigo nomet ric Equ a tion s 239
87. f (x) = sin x 90. Graph the following equations. 4
x y1 = 1.56t −1 2 cos 1.9t 0 10
(a) Domain: all real numbers except x = 0. y2 = 1
(b) The graph has y-axis symmetry.
(c) As x → 0, f (x) → 1.
y3 = −1 − 4
(d) sin x
x
= 0 has four solutions in the interval [−8, 8]. The rightmost point of intersection is at approximately
(1.91, −1).
The displacement does not exceed one foot from
sin x 1
= 0
equilibrium after t ≈ 1.91 seconds.
π t
sin x = 0
x = −2π , −π , π , 2π
91. Graph y1 = 58.3 + 32 cos 6
y2 = 75.
88. f (x) = cos
1
x
(a) Domain: all real numbers x except x = 0.
(b) The graph has y-axis symmetry and a horizontal
asymptote at y = 1.
(c) As x → 0, f (x) oscillates between −1 and 1.
Left point of intersection: (1.95, 75)
Right point of intersection: (10.05, 75)
So, sales exceed 7500 in January, November,
and December. S
(d) There are infinitely many solutions in the interval
2
100
75
[−1, 1]. They occur at x = (2n + 1)π
any integer.
where n is 50
25
x
89.
(e) The greatest solution appears to occur at
x ≈ 0.6366.
y = 1
(cos 8t − 3 sin 8t )
92.
2 4 6 8 10 12
Month (1 ↔ January)
Range = 300 feet
12 v0
= 100 feet per second
1 (cos 8t − 3 sin 8t ) = 0
12
r = 1 v02 sin 2θ
cos 8t = 3 sin 8t
1 = tan 8t
3
8t ≈ 0.32175 + nπ
t ≈ 0.04 + nπ 8
1 (100)2
sin 2θ
sin 2θ
2θ
θ
or
= 300
= 0.96
= arcsin(0.96) ≈ 73.74°
≈ 36.9°
In the interval 0 ≤ t ≤ 1, t ≈ 0.04, 0.43, and 0.83. 2θ = 180° − arcsin(0.96) ≈ 106.26°
θ ≈ 53.1°
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0
16 2
16 2
16 2
6
1 1
6 6
240 Chapte r 2 A nalytic Trigo nomet ry
93. (a) and (c) 100
94. h(t ) = 53 + 50 sin π
t − π
(a) h(t ) = 53 when 50 sin
π t −
π = 0.
1 12
0 π
t − π
= 0 or π
t − π
= π
The model fits the data well. 16 2 16 2
π π π 3π
(b) C = a cos(bt − c) + d t = t =
16 2 16 2
a = 1[high − low] =
1[84.1 − 31.0] = 26.55 t = 8 t = 24
2 2
p = 2[high time − low time] = 2[7 − 1] = 12 A person on the Ferris wheel will be 53 feet
above ground at 8 seconds and at 24 seconds
b = 2π p
= 2π
= π
12 6
(b) The person will be at the top of the Ferris wheel
when
The maximum occurs at 7, so the left end point is sin π
t − π
= 1
c = 7 c = 7
π =
7π b 6
π t −
π = π
16 2 2
d = [high + low] = [93.6 + 62.3] = 57.55 2 2
π
16
t = π
C = 26.55 cos π
t − 7π
+ 57.55
(d) The constant term, d, gives the average maximum
t = 16.
The first time this occurs is after 16 seconds.
2π
temperature. The period of this function is π 16
= 32.
95.
The average maximum temperature in Chicago is
57.55°F.
(e) The average maximum temperature is above 72°F
from June through September. The average
maximum temperature is below 70°F from October
through May.
A = 2 x cos x, 0 < x < π 2
During 160 seconds, 5 cycles will take place and
the person will be at the top of the ride 5 times,
spaced 32 seconds apart. The times are: 16 seconds,
48 seconds, 80 seconds, 112 seconds, and
144 seconds.
(a) 2
π 2
− 2
(b)
The maximum area of A ≈ 1.12 occurs when x ≈ 0.86.
A ≥ 1 for 0.6 < x < 1.1
96. f (x) = 3 sin(0.6 x − 2)
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(a) Zero : sin(0.6x − 2) = 0 (b) g (x) =
0.6 x − 2 = 0
0.6x = 2
x = 2
= 10
4
0
−0.45x2 + 5.52 x − 13.70
6
0.6 3 f
g
− 4
For 3.5 ≤ x ≤ 6 the approximation appears to be good.
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=
= −
=
Section 2.3 Solving Trigo nomet ric Equ a tion s 241
(c) −0.45x2 + 5.52x − 13.70 = 0
−5.52 ±
(5.52)2
− 4(−0.45)(−13.70)
x =
x ≈ 3.46, 8.81
2(−0.45)
The zero of g on [0, 6] is 3.46. The zero is close to the zero 10
3
≈ 3.33 of f.
97. f (x) = tan π x 4
Because tan π 4 = 1, x = 1 is the smallest nonnegative
fixed point.
100. False.
sin x = 3.4 has no solution because 3.4 is outside the
range of sine.
98. Graph y = cos x and y = x on the same set of axes.
101. cot x cos2 x = 2 cot x
cos2 x = 2
Their point of intersection gives the value of c such that
f (c) = c cos c = c.
2 (0.739, 0.739)
− 3 3
− 2
c ≈ 0.739
cos x = ± 2
No solution
Because you solved this problem by first dividing by
cot x, you do not get the same solution as Example 3.
When solving equations, you do not want to divide each
side by a variable expression that will cancel out because
you may accidentally remove one of the solutions.
102. The equation 2 cos x − 1 = 0 is equivalent to
99. True. The period of 2 sin 4t − 1 is π
and the period of
cos x = 1 . So, the points of intersection of y cos x 2
2 and y 1 2
represent the solutions of the equation
2 sin t − 1 is 2π . 2 cos x − 1 = 0. In the interval (− 2π , 2π ) the solutions
In the interval [0, 2π ) the first equation has four cycles
whereas the second equation has only one cycle, so the
first equation has four times the x-intercepts (solutions)
as the second equation.
of the equation are x 5π
, − π
, π
, and 3 3 3
5π .
3
103. (a)
3
0 2π
− 2
The graphs intersect when x = π 2
and x = π .
(b)
3
0 2π
− 2
The x-intercepts are π
, 0
and (π , 0).
2
(c) Both methods produce the same x-values. Answers will vary on which method is preferred.
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4 6 4 6
242 Chapte r 2 A nalytic Trigo nomet ry
Section 2.4 Sum and Difference Formulas 1. sin u cos v − cos u sin v π π π π π π
7. (a) cos + = cos cos − sin sin
2. cos u cos v − sin u sin v
3. tan u + tan v 1 − tan u tan v
4 3 4 3 4 3
2 1 2 3 = ⋅ − ⋅
2 2 2 2
2 − 6 =
4
4. sin u cos v + cos u sin v (b)
cos
π + cos
π
= 2
+ 1
= 2 + 1
5. cos u cos v + sin u sin v 4 3 2 2 2
7π π
5π π 1
6. tan u − tan v 8. (a) sin
6 −
3 = sin
6 = sin =
6 21 + tan u tan v
(b) sin 7π − sin
π = − 1
− 3 =
−1 − 3
9. (a)
sin(135° − 30°) = sin 135° cos 30° − cos 135° sin 30°
6 3 2 2 2
2 3 2 1 6 + 2
=
− − =
2 2 2 2 4
(b)
sin 135° − cos 30° = 2
− 3
= 2 − 3
10. (a)
2 2 2
cos(120° + 45°) = cos 120° cos 45° − sin 120° sin 45°
1 2 3 2 − 2 − 6
= −
−
=
2 2 2 2 4
(b)
cos 120° + cos 45° = − 1
+ 2
= −1 + 2
11.
sin 11π
= sin 3π
+ π
2 2 2
tan 11π
= tan 3π
+ π
12
4
= sin 3π
cos π + cos
3π sin
π tan 3π + tan
π
4 6 4 6 = 4 6
2 3 2 1 1 − tan 3π
tan π
= ⋅ + −
4 62 2 2 2
−1 + 3
co
s
11π
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(
4 6
(
2 =
4
= cos 3π
3 − 1)
+ π
=
3
1
−
(−
1
)
3
3 12 =
−3 + 3 ⋅
3 − 3
= cos 3π
cos π − sin
3π sin
π 3 + 3 3 − 3
4 6 4 6 =
−12 + 6 3 = −2 + 3
2 3 2 1 2 = − ⋅ − ⋅ = − 3 + 1)
6
2 2 2 2 4
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4 6
3 4
(
(
3 4
(
4 6
3 4
)
)
)
Section 2.4 Sum a nd Dif f erence Formu l as 243
12. 7π
= π
+ π
13.
sin 17π
= sin 9π 5π
−
12 3 4 12
sin 7π
= sin π
+ π
= sin 9π
cos 5π
− cos 9π
sin 5π
12
= sin π
cos π
+ sin π
cos π
4 6 4 6
2 3 2 1 =
−
−
3 4 4 3
3 2 2 1
2 2 2 2
= ⋅ + ⋅ 2 2 2 2 = −
2 3 + 1
4
= 2
3 + 1 4
cos
17π
12 = cos
9π
4
5π −
6
cos 7π
= cos π
+ π
= cos 9π
cos 5π
+ sin 9π
sin 5π
12
4 6 4 6
= cos π
cos π
− sin π
sin π 2 3 2 1
3 4 3 4 = 2
− 2
+ 2
2
= 1
⋅ 2
− 3
⋅ 2
2
2 2 2 2 = 4
(1 − 3)
2 =
4 1 − 3
tan
17π
= tan 9π
− 5π
tan 7π
= tan π
+ π
12
12 tan(9π 4) − tan(5π 6)
= 1 + tan(9π 4) tan(5π 6)
tan π
+ tan π
= 3 4 1 − (− 3 3)
1 − tan π
tan π
3 4
= 1 + (− 3 3)
3 + 1 = =
3 + 3 ⋅
3 + 3
1 − 3 3 − 3 3 + 3
= −2 − 3 =
12 + 6 3 6
= 2 + 3
14. − π
= π
− π
12 6 4
sin − π
= sin π
− π
cos − π
= cos π
− π
tan − π
= tan π
− π
12
6 4
12
6 4
12
6 4
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( ( ) )
= sin π
cos π − sin
π cos
π = cos π
cos π + sin
π sin
π tan π − tan
π
6 4 4 6
1 2 2 3 = ⋅ − ⋅
2 2 2 2
2 = 1 − 3
4
6 4 6 4
3 2 1 2 = ⋅ + ⋅
2 2 2 2
= 2
3 + 1 4
= 6 4
1 + tan π
tan π
6 4
3 − 1
= 3
1 + 3
3
= −2 + 3
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(
(
(
(
(
(
(
)
)
)
)
)
244 Chapte r 2 A nalytic Trigo nomet ry
15. sin 105° = sin(60° + 45°)
= sin 60° cos 45° + cos 60° sin 45°
3 2 1 2 = ⋅ + ⋅
2 2 2 2
17. sin(−195°) = sin(30° − 225°) = sin 30° cos 225° − cos 30° sin 225°
= sin 30°(− cos 45°) − cos 30°(− sin 45°)
= 1
− 2
− 3
− 2
2 2 2 2
= 2
3 + 1)
4
cos 105° = cos(60° + 45°) = −
2 1 − 3
4
= cos 60° cos 45° − sin 60° sin 45°
1 2 3 2
= 2
4 3 − 1)
= ⋅ − ⋅ 2 2 2 2
2 = 1 − 3
4
cos(−195°) = cos(30° − 225°)
= cos 30° cos 225° + sin 30° sin 225°
= cos 30°(− cos 45°) + sin 30°(− 45°)
tan 105° = tan(60° + 45°) 3 2 1 2 =
−
+
−
tan 60° + tan 45° =
2 2 2 2
1 − tan 60° tan 45° = −
2 3 + 1
4
3 + 1 3 + 1 1 + 3 = = ⋅ tan(−195°) = tan(30° − 225°)
1 − 3 1 − 3 1 + 3
= tan 30° −tan 225°
= 4 + 2 3
−2 = −2 − 3 1 + tan 30° tan 225°
= tan 30° −tan 45°
16. 165° = 135° + 30°
sin 165° = sin (135° + 30°)
= sin 135° cos 30° + sin 30° cos 135°
1 + tan 30° tan 45°
3
3 − 1
3 − 3 3 − 3 = = ⋅
= sin 45° cos 30° − sin 30° cos 45° 3
1 + 3 + 3 3 − 3
3 2 3 1 2
= ⋅ − ⋅ 2 2 2 2
= −12 + 6 3
= − 2 + 3 6
= 2
3 − 1 4
18. 225° = 300° − 45°
cos 165° = cos (135° + 30°)
= cos 135° cos 30° − sin 135° sin 30°
= −cos 45° cos 30° − sin 45° cos 30°
sin 255° = sin(300° − 45°)
= sin 300° cos 45° − sin 45° cos 300°
= − sin 60° cos 45° − sin 45° cos 60°
2 3 2 1 = − ⋅ − ⋅ = −
3 ⋅
2 − 2
⋅ 1 = −
2 ( 3 + 1)
2 2 2 2 2 2 2 2 4
= − 2
3 + 1 4
tan 165° = tan (135° + 30°)
cos 255° = cos(300° − 45°)
= cos 300° cos 45° + sin 300° sin 45°
= cos 60° cos 45° − sin 60° sin 45°
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( ) tan 135° + tan 30° =
1 2 3 2 = ⋅ − ⋅
2 = 1 − 3
1 − tan 135° tan 30°
= − tan 45° + tan 30° 1 + tan 45° tan 30°
−1 + 3
= 3
1 + 3
2 2 2 2 4
tan 255° = tan(300° − 45°)
tan 300° −tan 45° =
1 + tan 300° tan 45°
= − tan 60° − tan 45° 1 − tan 60° tan 45°
3 − 3 − 1
= −2 + 3 =
1 − 3 = 2 + 3
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4 3 4 3
4 3
(
3 4
(
3 4
(
3 4
)
)
)
Section 2.4 Sum a nd Dif f erence Formu l as 245
19. 13π =
3π + π
12 4 3
sin 13π
= sin 3π
+ π
tan 13π
= tan 3π
+ π
12
= sin 3π
cos π
+ cos 3π
sin π
12
tan 3π
+ tan π
4 3 4 3 =
2 1 2 3 1 − tan 3π
tan π
= ⋅ + −
4
3 2 2
= 2 (1 −
2 2
3)
= −1 + 3
( )4 1 − −1 ( 3)
cos 13π
= cos 3π
+ π 1 − 3 1 − 3
12 4 3
= − ⋅
1 + 3 1 − 3
= cos 3π
cos π
− sin 3π
sin π 4 − 2 3
4 3 4 3 = −
−2
2 1 2 3 = − ⋅ − ⋅
2 = − 1 + 3 = 2 − 3
20. 19π
2 2 2 2 4
π 5π = +
12 3 4
sin 19π
= sin π 5π
+ 12
= sin π
cos 5π
+ sin 5π
cos π
3 4 4 3
3 =
−
2
+
− 2 1
⋅
2 2 2 2
= − 2
3 + 1 4
cos 19π
= cos π 5π
+ 12
= cos π
cos 5π
− sin π
sin 5π
3 4 3 4
1 2 =
−
−
3 2
−
2 2 2 2
2 =
4 −1 + 3
tan 19π
= tan π
+ 5π
12
tan π + tan
5π
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= 3 4
1 − tan π
tan 5π
3 4
tan π + tan
π
= 3 4
1 − tan π
tan π
3 4
3 + 1 1 + 3 = ⋅
1 − 3 1 + 3
= 4 + 2 3
− 2
= − 2 − 3
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(
(
− =
(
= + − = (1 − )
246 Chapte r 2 A nalytic Trigo nomet ry
21. − 5π = −
π − π
12 4 6
sin − π
− π
= sin − π
cos π
− cos − π
sin π
4 6
4
6
4
6
2 3 2 1 2 =
−
−
= −
3 + 1) 2 2 2 2 4
cos − π
− π
= cos − π
cos π
+ sin − π
sin π
4 6
4
6
4
6
2 3 2 1 2 =
+
− =
3 − 1) 2 2 2 2 4
tan − π tan
− π
− tan π
π 4
6
4 6 1 + tan −
π tan
π
4 6
−1 − 3
= 3 =
− 3 − 3
−
1 + (−1) 3
3 3
3
= − 3 − 3
⋅ 3 + 3
3 − 3 3 + 3
22. − 7π
= − π
= −12 − 6 3
6
− π
= − 2 − 3
12 3 4
sin −
7π = sin
− π
− π
= sin − π
cos π
− cos − π
sin π
12
3 4
3
4
3
4
3 2 1 2 2
= − − = − 3 + 1)
2 2 2 2 4
cos −
7π = cos
− π
− π
= cos − π
cos π
+ sin − π
sin π
12
3 4
3
4
3
4
1 2 3 2 2
3 2 2 2 2 4
tan
−
7π = tan
− π
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π 3 4 tan
− π
− tan π
− =
= − 3 − 1
= 2 + 3 12 3 4
1 + tan − π
tan π 1 + (− 3)(1)
3 4
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− = − (
− − = (
(
(
(
)
Section 2.4 Sum a nd Dif f erence Formu l as 247
23. 285° = 225° + 60°
sin 285° = sin(225° + 60°) = sin 225° cos 60° + cos 225° sin 60°
2 1 2 3 2 = −
3 + 1)
2 2 2 2 4
cos 285° = cos(225° + 60°) = cos 225° cos 60° − sin 225° sin 60°
2 1 2 3 2 = −
3 − 1)
2 2 2 2 4
tan 285° = tan(225° + 60°) = tan 225° + tan 60° 1 − tan 225° tan 60°
1 + 3 1 + 3 = ⋅ =
4 + 2 3 = −2 − 3 = −(2 + 3)
1 − 3 1 + 3 −2
24. 15° = 45° − 30°
sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°
2 3 2 1 2 ( 3 − 1) 2 =
−
= =
3 − 1) 2 2 2 2 4 4
cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30°
2 3 2 1 2 ( 3 + 1) 2 =
+
= =
3 + 1) 2 2 2 2 4 4
tan 15° = tan(45° − 30°) = tan 45° −tan 30° 1 + tan 45° tan 30°
1 − 3 3 − 3
= 3 = 3 = 3 − 3
⋅ 3 − 3
12 − 6 3
=
= 2 − 3
3 3 + 3 3 + 3 3 − 3 6
1 + (1) 3 3
25.
−165° = −(120° + 45°)
sin(−165°) = sin −(120° + 45°) = − sin(120° + 45°) = −[sin 120° cos 45° + cos 120° sin 45°]
3 2 1 2 2 (
3 1)= −
⋅ − ⋅
2 2 2 2 = − − 4
cos(−165°) = cos−(120° + 45°) = cos(120° + 45°) = cos 120° cos 45° − sin 120° sin 45°
1 2 3 2 = − ⋅ − ⋅
2 = − 1 + 3
2 2 2 2 4
tan(−165°) = tan −(120° + 45°) = −tan(120° + tan 45°) = − tan 120° + tan 45°
1 − tan 120° tan 45°
= − − 3 + 1
= − 1 − 3
⋅ 1 − 3
= − 4 − 2 3
= 2 − 3
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1 − (− 3)(1) 1 + 3 1 − 3 −2
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= − − = − (1 +
(
7 5
8 8
)
)
248 Chapte r 2 A nalytic Trigo nomet ry
26. −105 = 30° − 135°
sin(30° − 135°) = sin 30° cos 135° − cos 30° sin 135° = sin 30°(−cos 45°) − cos 30° sin 45°
1 2 3 2 2
3 2 2 2 2 4
cos(30° − 135°) = cos 30° cos 135° + sin 30° sin 135° = cos 30°(−cos 45°) + sin 30° sin 45°
3 2 1 2 2
= − + = 1 − 3
2 2 2 2 4
tan(30° − 135°) = tan 30° −tan 135°
= tan 30° −(− tan 45°)
1 + tan 30° tan 135°
3 − (−1)
= 3 = 2 + 3
1 + (−1)
1 + tan 30°(− tan 45°) 3
3
27.
sin 3 cos 1.2 − cos 3 sin 1.2 = sin(3 − 1.2) = sin 1.8
36.
cos π
cos 3π
− sin π
sin 3π
= cos π
+ 3π
16 16 16 16 16 16
28. cos π
cos π
− sin π
sin π
= cos π
+ π π 2
7 5 7 5
= cos12π
= cos = 4 2
29.
30.
35
sin 60° cos 15° + cos 60° sin 15° = sin(60° + 15°)
= sin 75°
cos 130° cos 40° − sin 130° sin 40° = cos(130° + 40°)
= cos 170°
37.
38.
cos 130° cos 10° + sin 130° sin 10° = cos(130° − 10°)
= cos 120°
1 = −
2
sin 100° cos 40° − cos 100° sin 40° = sin (100° − 40°)
= sin 60°
3
31. tan (π 15) + tan(2π 5) =
= tan(π 15 + 2π 5) 2 1 − tan (π 15) tan (2π 5)
= tan(7π 15) 39.
tan(9π 8) − tan(π 8) 9π
= tan −
π
32.
tan 1.1 − tan 4.6
= tan(1.1 − 4.6) = tan(− 3.5) 1 + tan 1.1 tan 4.6
1 + tan(9π 8) tan(π 8)
= tan π
= 0
33.
34.
cos 3x cos 2 y + sin 3x sin 2 y = cos(3x − 2 y)
sin x cos 2x + cos x sin 2 x = sin(x + 2x) = sin(3x)
40.
tan 25° + tan 110° = tan(25° + 110°)
1 − tan 25° tan 110°
= tan 135°
= −1
35. sin π
cos π
+ cos π
sin π
= sin π
+ π
12 4 12 4 12 4
= sin π 3
3
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= 2
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(15, 8
17
v
17 17 15
− −
− +
5 85
4
15
−
25 25 24
Section 2.4 Sum a n d Diff erence Formu l as 249
For Exercises 41 – 46, you have:
sin u = – 3 , u in Quadrant IV cos u = 4 , tan u = – 4
5 5 3
cos v = 15 , v in Quadrant I sin v = 8 , tan v = 8
y y
)
u
x x
5
(4, − 3)
41.
Figures for Exercises 41–46
sin(u + v) = sin u cos v + cos u sin v
= −
3 15 +
4 8
44.
csc(u − v) = 1
= 1
( )
5 17
5 17
sin u − v
1 sin u cos v − cos u sin v
13 = 3 15 4 8
= − 85
5 17 5 17 1 85
42.
cos(u − v) = cos u cos v + sin u sin v
= 4 15
+ −
3 8
= = −
− 77 77 85
5 17
5 17 1 1
45. sec(v − u) = cos(v − u)
= cos v cos u + sin v sin u
60 − 24 36 = + = 1 1
85 85 85 = 15 4 8
= 3 60
24
3 8
+ − 17 5 17 1 85
+ −
85
43. tan(u + v) = tan u + tan v
= 4 15 =
36 =
36
1 − tan u tan v 1 −
−
3 8 85
4 15
− 13
60
13 5
13
tan u + tan v
−
3 +
8
= = −
= − 46. tan(u + v) = =
3 8
1 + 32
60
60 7 84 1 − tan u tan v 1 −
−
4 15
13
= 60 7
5
= − 13 84
1 1 84
For Exercises 47– 52, you have:
sin u = – 7 , u in Quadrant III cos u = – 24 , tan u = 7
cot(u + v) = = tan(u + v)
= −
− 13 13 84
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u
25
(− 24, − 7)
cos v = – 4 , v in Quadrant III sin v = – 3 , tan v = 3 5 5 4
y y
v
x x
5
(− 4, − 3)
Figures for Exercises 47–52
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25 5 25 5
25 5 25 5
25 5 25 5
24
4
− 5 25 5
4
24
(
=
=
250 Chapte r 2 A nalytic Trigo nomet ry
47. cos(u + v) = cos u cos v − sin u sin v
51. csc(u − v) = 1
= 1
= (− 24 )(− 4 ) − (− 7 )(− 3 ) sin(u − v) sin u cos v − cos u sin v
1
3 = 5 7
− −
4 − −
24 3 −
48. sin(u + v) = sin u cos v + cos u sin v
= (− 7 )(− 4 ) + (− 24 )(− 3 ) 1
= 44
− 125
= 28 + 72 = 100 = 4
49.
125 125 125 5
tan(u − v) = tan u − tan v 1 + tan u tan v
7 3 11
52.
125 = −
44
sec(v − u) = 1 cos(v − u)
− − 24 4 24 44 1
= = = −
1 + 7 3 39
32
117 = cos v cos u + sin v sin u
1 =
50.
tan(v − u) = tan v − tan u
=
3 7 4 24
4 − −
24 + −
3 −
7
25
1 + tan v tan u
11
1 + 3 7
1 =
117
125
125
= 24 44 =
39 117
32
117
1 1 117cot(v − u) =
tan(v − u) =
44 =
44
117
53. sin(arcsin x + arccos x) = sin(arcsin x) cos(arccos x) + sin(arccos x) cos(arcsin x)
= x ⋅ x +
1 − x2 ⋅
1 − x2
= x2 + 1 − x2
= 1
1
x 1
1 − x2
θ θ
1 − x2 x
54.
sin(arctan 2x − arccos x) = sin(arctan 2x − arccos x)
θ = arcsin x
θ = arccos x
= sin(arctan 2x) cos(arccos x) − cos(arctan 2x) sin(arccos x)
= 2 x
(x) − 1
4x2 + 1 4x2 + 1
1 − x2 )
2 x2 − =
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1 − x2
4 x2 + 1
4x2 + 1 2x
1 1 − x2
θ
1
θ = arctan 2x
θ
x
θ = arccos x
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(
Section 2.4 Sum a nd Dif f erence Formu l as 251
55. cos(arccos x + arcsin x) = cos(arccos x) cos(arcsin x) − sin(arccos x) sin(arcsin x)
= x ⋅
= 0
(Use the triangles in Exercise 53.)
1 − x2 −
1 − x2 ⋅ x
56. cos(arccos x − arctan x) = cos(arccos x − arctan x)
= cos(arccos x) cos(arctan x) + sin(arccos x) sin(arctan x)
= (x)
1 + ( 1 − x2 ) x
= x + x
1 + x2
1 − x2
1 + x2 1
1 − x2
1 + x2
x
1 + x2
θ
x
θ = arccos x
θ
1
θ = arctan x
57.
sin π
− x
= sin π
cos x − cos π
sin x
59.
sin π
+ x
= sin π
cos x + cos π
sin x
2
2 2
6
6 6
= (1)(cos x) − (0)(sin x)
= cos x
=
5π
1
cos x + 2
5π
3 sin x)
5π
58.
sin π
+ x
= sin π
cos x + sin x cos π 60. cos
4 − x = cos cos x + sin
4 sin x
4
2
2 2
θ + π
= (1)(cos x) + (sin x)(0)
= cos x
tan θ + tan π tan θ + 0
= = =
tan θ
= θ
= − 2 (cos x + sin x)
2
61.
tan( ) 1 − tan θ tan π 1 − (tan θ )(0) tan 1
π tan
π − tan θ
4
1 − tan θ
62. tan 4
− θ = 1 + tan
π tan θ
4
= 1 + tan θ
63.
cos(π − θ ) + sin π
+ θ
= cos π cos θ + sin π sin θ + sin π
cos θ + cos π
sin θ
2
2 2
= (−1)(cos θ ) + (0)(sin θ ) + (1)(cos θ ) + (sin θ )(0)
= −cos θ + cos θ
= 0
64. cos(x + y) cos(x − y) = (cos x cos y − sin x sin y)(cos x cos y + sin x sin y)
= cos2 x cos2 y − sin 2 x sin 2 y
= cos2 x(1 − sin 2 y) − sin 2 x sin 2 y
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=
c
o
s2
x
− c
o
s2
x
s
i
n 2
y
−
s
i
n 2
x
s
i
n 2
y
=
c
o
s2
x
−
s
i
n 2
y
(c
o
s2
x
+ s
i
n 2
x
)
=
c
o
s2
x
−
s
i
n
2 y
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252 Chapte r 2 A nalytic Trigo nomet ry
65.
cos
3π − θ
= cos
3π cos θ + sin
3π sin θ
68.
tan(π + θ ) = tan π + tan θ
2
2 2
1 − tan π tan θ
= (0)(cos θ ) + (−1)(sin θ )
= − sin θ
2
0 + tan θ =
1 − (0) tan θ
= tan θ
1 1
− 2π 2π cot(π + θ ) = =
tan(π + θ )
tan θ = cot θ
5
− 2
The graphs appear to coincide, so
− 2π 2π
cos 3π 2
− θ
= − sin θ .
66.
sin (π + θ ) = sin π cos θ + cos π sin θ
= (0) cos θ + (−1) sin θ
= − sin θ
69.
− 5
The graphs appear to coincide, so cot (π + θ ) = cot θ
sin(x + π ) − sin x + 1 = 0
sin x cos π + cos x sin π − sin x + 1 = 0
2
− 2π 2π
(sin x)(−1) + (cos x)(0) − sin x + 1 = 0
−2 sin x + 1 = 0
sin x = 1 2
− 2
The graphs appear to coincide, so
sin (π + θ ) = − sin(θ ).
70.
x = π
, 5π
6 6
cos(x + π ) − cos x − 1 = 0
67.
sin 3π
+ θ
= sin 3π
cos θ + cos 3π
sin θ
cos x cos π − sin x sin π − cos x − 1 = 0
2
2 2
= (−1)(cos θ ) + (0)(sin θ )
= − cos θ
(cos x)(−1) − (sin x)(0) − cos x − 1 = 0
−2 cos x − 1 = 0
cos x 1
csc 3π = −
+ θ
= 1
= 1
= − sec θ 2
2
sin(3π
+ θ )
− cos θ
2π 4π 2
5
x = , 3 3
− 2π 2π
− 5
The graphs appear to coincide, so
csc 3π 2 + θ
= − sec θ .
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Section 2.4 Sum a nd Dif f erence Formu l as 253
71. cos x +
π − cos
x −
π = 1
4
4
cos x cos π
− sin x sin π
− cos x cos
π + sin x sin
π = 1
4 4
4 4
2 −2 sin x
= 1
2
− 2 sin x = 1
sin x = − 1 2
sin x = − 2
2
x = 5π
, 7π
4 4
72.
sin x +
π − sin
x −
7π =
3
6
6
2
sin x cos π
+ cos x sin π
− sin x cos 7π
− cos x sin 7π
= 3
6 6
6 6
2
(sin x) 3
+ (cos x) 1
− (sin x) − 3
+ (cos x) − 1 =
3 2
2 2
2 2
73.
tan(x + π ) + 2 sin(x + π ) = 0
tan x + tan π + 2(sin x cos π + cos x sin π ) = 0
1 − tan x tan π
3 sin x = 3 2
sin x = 1 2
x = π
, 5π
6 6
tan x + 0 +
− + =
1 − tan x(0) 2 sin x( 1) cos x(0) 0
tan x − 2 sin x = 0
1
sin x
cos x
= 2 sin x
sin x = 2 sin x cos x
sin x(1 − 2 cos x) = 0
sin x = 0 or cos x = 1 2
x = 0, π x = π
, 5π
3 3
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1 4 4
1
3 4
3
254 Chapte r 2 A nalytic Trigo nomet ry
74.
sin x +
π − cos2 x = 0
76.
tan(x + π ) − cos x +
π = 0
2
2
sin x cos π
+ cos x sin π
− cos2 x = 0
x = 0, π
2 2
(sin x)(0) + (cos x)(1) − cos2 x = 0
cos x − cos2 x = 0
cos x(1 − cos x) = 0
cos x = 0 or 1 − cos x = 0
4
0 2π
− 4
x = π
, 3π
2 2
cos x = 1 77.
sin x +
π + cos2 x = 0
x = 0
2
75. cos x +
π + cos
x −
π = 1
4
4
Graph y = cos x +
π + cos x −
π and y2
= 1. 0 2π
x = π
, 7π
4 4 2
− 1
x = π
, π , 3π
2 2
0 2π 78.
cos
x −
π − sin 2 x = 0
2
− 2
1
0 2
79.
y = 1
sin 2t + 1
cos 2t 3 4
− 3
x = 0, π
, π 2
(a) a =
1, b =
1 , B = 2
3 4
C = arctan b
a = arctan
3 ≈ 0.6435
4
y ≈ 1
2 2
+
sin(2t + 0.6435) = 5
sin(2t + 0.6435) 12
(b) Amplitude: 5 12
feet
(c) Frequency: 1
= B
= 2
= 1
cycle per secondperiod 2π 2π π
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λ
λ
λ T λ
λ λ T T
Section 2.4 Sum a nd Dif f erence Formu l as 255
t x 80. y1 = A cos 2π −
T
t x y2 = A cos 2π +
T
t x t x y1 + y2 = A cos 2π
T − + A cos 2π +
t x t x t x t x t x
y1 + y2 = Acos 2π
cos 2π λ
+ sin 2π T
sin 2π + Acos 2π
cos 2π λ
− sin 2π T
sin 2π = 2 A cos 2π T
cos 2π λ
81. True.
sin(u + v) = sin u cos v + cos u sin v
sin(u − v) = sin u cos v − cos u sin v
So, sin(u ± v) = sin u cos v ± cos u sin v.
82. False.
cos(u + v) = cos u cos v − sin u sin v
cos(u − v) = cos u cos v + sin u sin v
So, cos(u ± v) = cos u cos v sin u sin v.
83. sin (α + β ) = sin α cos β + sin β cos α = 0
sin α cos β + sin β cos α
sin α cos β
= 0
= − sin β cos α
False. When α and β are supplementary, sin α cos β = − cos α sin β .
84. cos( A + B) = cos(180° − C )
= cos(180°) cos(C ) + sin(180°) sin(C )
= (−1) cos(C ) + (0) sin(C )
= − cos(C )
True. cos( A + B) = − cos C. When A, B and C form Δ ABC , A + B + C
= 180°, so A + B = 180° − C.
85. The denominator should be 1 + tan x tan (π 4).
tan x −
π =
tan x − tan(π 4)
87. cos(nπ + θ ) = cos nπ cos θ − sin nπ sin θ
= (−1)n (cos θ ) − (0)(sin θ )
4
1 + tan x tan(π 4)
= tan x − 1 1 + tan x
88.
sin(nπ
= (−1)n (cos θ ), where n is an integer.
+ θ ) = sin nπ cos θ + sin θ cos nπ
n
86. (a) Using the graph, sin(u + v) ≈ 0 and
sin u + sin v ≈ 0.7 + 0.7 = 1.4. Because
= (0)(cos θ ) + (sin θ )(−1)
= (−1)n (sin θ ), where n is an integer.
0 ≠ 1.4, sin(u + v) ≠ sin u + sin v.
(b) Using the graph, sin(u − v) ≈ −1 and
sin u − sin v ≈ 0.7 − 0.7 = 0. Because
89.
−1 ≠ 0,
C = arctan b
sin(u − v) ≠ sin u − sin v.
sin C = b
, cos C = a
a a2 + b2 a2 + b2
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a2 + b2 sin(Bθ + C ) = a b
a2 + b2 sin Bθ ⋅ + ⋅ cos Bθ = a sin Bθ + b cos Bθ
a2 + b2 a2 + b2
90. C = arctan a sin C =
a , cos C =
b
b a2 + b2 a2 + b2
a2 + b2 cos(Bθ − C ) = b a
a2 + b2 cos Bθ ⋅ + sin Bθ ⋅
= b cos Bθ + a sin Bθ
a2 + b2
= a sin Bθ + b cos Bθ
a2 + b2
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θ
256 Chapte r 2 A nalytic Trigo nomet ry
91. sin θ + cos θ
a = 1, b = 1, B = 1
94. sin 2θ + cos 2θ
a = 1, b = 1, B = 2
(a)
C = arctan b
a = arctan 1 =
π 4
(a)
C = arctan b
a = arctan(1) =
π 4
sin θ + cos θ = a2 + b2 sin(Bθ + C )
sin 2θ + cos 2θ = a2 + b2 sin(Bθ + C )
π = 2 sinθ +
= 2 sin 2θ +
π
(b)
C = arctan
a
b
4
= arctan 1 = π 4
(b)
C = arctan a
4
= arctan(1) = π
sin θ + cos θ =
=
a2 + b2 cos(Bθ − C ) b 4
π ( ) 2 cosθ −
4
sin 2θ + cos 2θ =
=
a2 + b2 cos Bθ − C
2 cos 2θ −
π
4
92. 3 sin 2θ + 4 cos 2θ
a = 3, b = 4, B = 2 b π
(a) C = arctan b = arctan
4
≈ 0.9273 95. C = arctan
a = a = b, a > 0, b > 0
4
a
3 sin 2θ + 4 cos 2θ =
3
a2 + b2 sin(Bθ + C ) a2 + b2
B = 1
= 2 a = b = 2
≈ 5 sin(2θ + 0.9273) 2 sin
+ π
=
2 sin θ +
2 cos θ
(b) C = arctan a
b = arctan
3 4
≈ 0.6435 4
3 sin 2θ + 4 cos 2θ = a2 + b2 cos(Bθ − C )
96. C = arctan b
= π
a 4 a = b, a > 0, b > 0
93. 12 sin 3θ + 5 cos 3θ
a = 12, b = 5, B = 3
≈ 5 cos(2θ − 0.6435) a2 + b2
B = 1
= 5 a = b = 5 2
2
(a) C = arctan
b = arctan 5
≈ 0.3948 π 5 2 5 2 5 cosθ − = sin θ + cos θ
a
12 sin 3θ + 5 cos 3θ
12
= a2 + b2 sin(Bθ + C )
97.
4 2 2
y
(b)
C = arctan a
b
≈ 13 sin(3θ + 0.3948)
= arctan 12
≈ 1.1760 5
≈ 13 cos(3θ − 1.1760)
y1 = m1 x + b1
θ
δ
α β x
y2 = m2 x + b2
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m1 = tan α and m2 = tan β
β + δ = 90° δ = 90° − β
α + θ + δ = 90° α + θ + (90° − β )
= 90° θ = β − α
So, θ = arctan m2 − arctan m1. For y = x and
y = 3x you have m1 = 1 and m2 = 3.
θ = arctan 3 − arctan 1 = 60° − 45° = 15°
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2 2
B
Section 2.4 Sum a nd Dif f erence Formu l as 257
98. For m2 > m1 > 0, the angle θ between the lines is: 99. y1 = cos(x + 2), y2
2
= cos x + cos 2
m2 − m1
θ = arctan 1 + m1m2
m2 = 1
1
y2
0 2π y1
m1 =
3
1 −
1
− 2
No, y1 ≠ y2 because their graphs are different.
θ = arctan 3 = arctan(2 −
1 + 1
3) = 15° 100. y1 = sin(x + 4), y2
2
= sin x + sin 4
3
y1
0 2π
y2
− 2
No, y1 ≠
y2 because their graphs are different.
101. (a) To prove the identity for sin(u + v) you first need to prove the identity for cos(u − v).
y
C 1 u − v
Assume 0 < v < u < 2π and locate u, v, and u − v on the unit circle. B
The coordinates of the points on the circle are: D
A = (1, 0), B = (cos v, sin v), C = (cos(u − v), sin(u − v)), and D = (cos u, sin u). Because ∠DOB = ∠COA, chords AC and BD are equal. By the Distance Formula:
− 1
u
v A x
O 1
2 2 2 2
cos(u − v) − 1 + sin(u − v) − 0 = (cos u − cos v) + (sin u − sin v) − 1
cos2 (u − v) − 2 cos(u − v) + 1 + sin 2 (u − v) = cos2 u − 2 cos u cos v + cos2 v + sin 2 u − 2 sin u sin v + sin 2 v
cos2 (u − v) + sin 2 (u − v) + 1 − 2 cos(u − v) = (cos2 u + sin 2 u) + (cos2 v + sin 2 v) − 2 cos u cos v − 2 sin u sin v
2 − 2 cos(u − v) = 2 − 2 cos u cos v − 2 sin u sin v
−2 cos(u − v) = −2(cos u cos v + sin u sin v) cos(u − v) = cos u cos v + sin u sin v
Now, to prove the identity for sin(u + v), use cofunction identities.
sin(u + v) = cosπ
− (u + v)
= cos π
− u − v
2
2
= cos π
2 − u
cos v + sin
π
2 − u
sin v
= sin u cos v + cos u sin v
(b) First, prove cos(u − v) = cos u cos v + sin u sin v using the figure containing points
A(1, 0)
B(cos(u − v), sin(u − v)) C(cos v, sin v)
y
1 D u − v
C
u
v A x
D(cos u, sin u)
on the unit circle.
− 1 u − v 1
Because chords AB and CD are each subtended by angle u − v, their lengths are equal. Equating − 1
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2 2
2
d ( A, B) = d (C, D) you have (cos(u − v) − 1) + sin 2 (u − v) = (cos u − cos v) + (sin u − sin v) .
Simplifying and solving for cos(u − v), you have cos(u − v) = cos u cos v + sin u sin v.
Using sin θ = cos π
2
− θ ,
sin(u − v) = cos π
2 − (u − v) = cos
π 2
− u − (−v) = cos π
2 − u cos(−v) + sin
π
2 − u sin(−v)
= sin u cos v − cos u sin v
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h 0.5 0.2 0.1 0.05 0.02 0.01
f (h) 0.267 0.410 0.456 0.478 0.491 0.496
g (h) 0.267 0.410 0.456 0.478 0.491 0.496
= −
= −
258 Chapte r 2 A nalytic Trigo nomet ry
102. (a) The domains of f and g are the same, all real numbers h, except h = 0.
(b) (c) 2 (d) As h → 0*,
f → 0.5 and− 3 3
− 2
Section 2.5 Multiple-Angle and Product-to-Sum Formulas
g → 0.5.
1. 2 sin u cos u
2. cos2 u − sin 2 u = 2 cos2 u − 1 = 1 − 2 sin 2 u
8. sin 2x sin x = cos x
2 sin x cos x sin x − cos x = 0
cos x(2 sin 2 x − 1) = 0
1 3.
2 sin(u + v) + sin(u − v)
cos x = 0 or 2 sin 2
x = π
+ 2nπ
x − 1 = 0
sin 2 x = 1
4. tan 2 u 2 2
sin x = ± 2
25. ±
1 − cos u
2
x = π
+ nπ
6. −2 sin u + v
sin u − v
4 2
9. cos 2 x − cos x = 0
2
2
7. sin 2 x − sin x = 0
2 sin x cos x − sin x = 0
sin x(2 cos x − 1) = 0
sin x = 0 or 2 cos x − 1 = 0
cos 2 x = cos x
cos2 x − sin 2 x = cos x
cos2 x − (1 − cos2 x) − cos x = 0
2 cos2 x − cos x − 1 = 0
(2 cos x + 1)(cos x − 1) = 0
x = nπ cos x = 1 2
2 cos x + 1 = 0 or cos x − 1 = 0
x = π
+ 2nπ , 5π
+ 2nπ cos x 1 cos x = 1
10.
3 3
cos 2x + sin x = 0
1 − 2 sin 2 x + sin x = 0
2 sin 2 x − sin x − 1 = 0
(2 sin x + 1)(sin x − 1) = 0
2
x = 2nπ 3
x = 0
2 sin x + 1 = 0 or sin x − 1 = 0
sin x 1 2
sin x = 1
x = 7π + 2nπ ,
11π
+ 2nπ x = π
+ 2nπ
6 6 2
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( )
Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 259
11. sin 4 x = −2 sin 2x
sin 4 x + 2 sin 2x = 0
2 sin 2x cos 2 x + 2 sin 2 x = 0
2 sin 2x(cos 2x + 1) = 0
2 sin 2x = 0 or cos 2 x + 1 = 0
sin 2 x = 0 cos 2x = −1
2x = nπ 2x = π + 2nπ
x = nπ x =
π
+ nπ
2 2
12. (sin 2x + cos 2x)2
= 1
sin 2 2 x + 2 sin 2 x cos 2x + cos2 2 x = 1
2 sin 2 x cos 2 x = 0
sin 4 x = 0
4 x = nπ
x = nπ 4
13. tan 2x − cot x = 0
2tan x = cot x
1 − tan 2 x
2 tan x = cot x(1 − tan 2 x)
2 tan x = cot x − cot x tan 2 x
2 tan x = cot x − tan x
3 tan x = cot x
3 tan x − cot x = 0
3 tan x − 1
= 0 tan x
3 tan 2 x − 1= 0
tan x
1 3 tan 2 x − 1 = 0
tan x
cot x(3 tan 2 x − 1) = 0
cot x = 0 or 3 tan 2 x − 1 = 0
x = π
+ nπ tan 2 x = 1
2 3
tan x = ±
x = π
3
3
+ nπ , 5π
+ nπ
6 6
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tan 2x − 2 cos x 0
2tan x
1 − tan 2 x 2 cos x
2 tan x
2 tan x
2 cos x(1 − tan 2 x)
2 cos x − 2 cos x tan 2 x
cos x
2
2
=
260 Chapte r 2 A nalytic Trigo nomet ry
14. =
=
=
=
2 tan x = 2 cos x − 2 cos x
sin 2 x
cos2 x
2 tan x = 2 cos x − 2
sin 2 x
cos x
sin x
sin 2 x
tan x = cos x −
sin x
= cos x − cos x
sin 2 x
cos x
sin 2 x
cos x
cos x + − cos x = 0
cos x
sin x + sin 2 x − cos2 x = 0
cos x
1 sin x + sin 2 x − (1 − sin 2 x) = 0
sec x 2 sin 2 x + sin x − 1 = 0
sec x(2 sin x − 1)(sin x + 1) = 0
sec x = 0 or 2 sin x − 1 = 0 or sin x + 1 = 0
No solution
sin x = 1 sin x = −1
2 3π x =
x = π
, 5π 2
6 6
Also, values for which cos x = 0 need to be checked.
π ,
3π are solutions.
2 2
x = π + 2nπ ,
π + nπ , 5π
+ 2nπ
15.
6 2 6
6 sin x cos x = 3(2 sin x cos x)
= 3 sin 2x
19.
4 − 8 sin 2 x = 4(1 − 2 sin 2 x) = 4 cos 2x
16. sin x cos x =
=
1 (2 sin x cos x)
1 sin 2 x
20.
10 sin 2 x − 5 = 5(2 sin 2 x − 1) 2
17.
2
6 cos2 x − 3 = 3(2 cos2 x − 1) = 3 cos 2x
= −5(1 − 2 sin x) = −5 cos 2 x
18. cos2 x − 1 = 1 2(cos2 x − 1 )2 2 2
= 1 (2 cos2 x − 1) 1 cos 2 x
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5 3
u
− 4
u
4
5 −
34
3
5
, ,
2 2
3
tan 2u = =
2
2
= −
Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 261
21.
sin u 3 3π
= − 5 2
y
< u < 2π
22. cos u 4 π 5 2
y
< u < π
x x
3
sin 2u = 2 sin u cos u = −
3 4 = −
24 sin 2u = 2 sin u cos u = 3
− 4 24 = −
5 5 25 5 5 25
cos 2u = cos2 u − sin 2 u = 16
− 9
= 7
25 25 25 3
2 −
cos 2u = cos2 u − sin 2 u = 16
− 9
= 7
25 25 25
2 −
tan 2u = 2tan u
= 4 3 16 24 2tan u
4 3 16 24= − = − tan 2u = =
9 = − = −
23.
1 − tan 2 u
tan u = 3
, 0 < u < π
5 2
1 − 9 16
2 7 7 1 − tan 2 u 1 −
16
2 7 7
y 3 5 15 sin
2u = 2 sin u cos u = 2 = 34 34 17
cos 2u = cos2 u − sin 2 u = 25
− 9
= 8
34 34 17
3 2u
x 2tan u 5 =
6 25 =
15
24.
sec u = − 2, π
< u < 3π 2
1 − tan 2 u 1 − 9 25
5 16
8
y
sin 2u = 2 sin u cos u = −
3 −
1 =
3 2 2 2
cos 2u = cos2 u − sin 2 u = −
1
2
3
− −
= 1
− 3
= − 1
u
2 2 4 4 2
− 1 x 3
2
2 tan u 1 2 3
− 3 2 tan 2u = 1 − tan 2 u
= 2
= = − 3 3 − 2
1 − 1
25.
cos 4x = cos(2x + 2x) = cos 2x cos 2x − sin 2x sin 2x
= cos2 2x − sin 2 2x
= cos2 2x − (1 − cos2 2x) = 2 cos2 2x − 1
=
2
(c
o
s
2
x)2
− 1
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1 − tan 2 x
=
2
=
2
6.
tan 3x = tan(2x + x)
=
tan
2x +
tan x 1 − tan 2x tan x
2 tan x
1 − tan 2 x + tan x
1 − 2 tan x
(tan x)
2 tan x + tan x − tan3 x= 2(2 cos2
x − 1) − 1 1 − tan 2 x 2 2
= 2(4 cos4 x − 4 cos x + 1) − 1 1 − tan x − 2 tan x
1 − tan 2 x
= 8 cos4 x − 8 cos x + 1 3 tan x
tan3 x
= −
1 − 3 tan 2 x
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2
= 1 cos 2 x 1 cos 2 x
3
2
262 Chapte r 2 A nalytic Trigo nomet ry
27.
cos4 x = (cos2 x)(cos2 x) =
1 + cos 2 x 1 + cos 2 x =
1 + 2 cos 2 x + cos 2 x
2
2
4
1 + 2 cos 2x + 1 + cos 4 x
= 2 4
= 2 + 4 cos 2 x + 1 + cos 4 x
8
= 3 + 4 cos 2 x + cos 4 x
8
= 1(3 + 4 cos 2x + cos 4x)
8
8 4 4 2 2
2 2
28. sin x = (sin x)(sin x) = (sin x) (sin x)
2 2
− − 2 2
1 − 2 cos 2 x + cos2 2 x 1 − 2 cos 2x + cos2 2x =
4 4
1 − 2 cos 2 x + cos2 2 x − 2 cos 2 x + 4 cos2 2 x − 2 cos3 2 x + cos2 2x − 2 cos3 2 x + cos4 2 x =
16
1 − 4 cos 2 x + 6 cos2 2x − 4 cos3 2 x + cos4 2x =
16 2 3 2
2
1 − 4 cos 2 x + 6 cos 2 x − 4 cos 2 x + (cos 2 x) =
16
1 + cos 4 x 1 + cos 4 x 1 − 4 cos 2 x + 6 − 4 cos 2x +
= 2 2 16
1 + 2 cos 4x + cos2 4x 1 − 4 cos 2 x + 3 + 3 cos 4x − 4 cos3 2 x +
= 4
16
4 − 16 cos 2x + 12 + 12 cos 4x − 16 cos3 2 x + 1 + 2 cos 4x + cos2 4x =
64
17 − 16 cos 2 x + 14 cos 4x − 16 cos3 2x + 1 + cos 8x
= 2 64
34 − 32 cos 2x + 28 cos 4x − 32 cos3 2x + 1 + cos 8x =
128
35 − 32 cos 2x + 28 cos 4x − 32 cos3 2x + cos 8x =
128
35 − 32 cos 2x + 28 cos 4 x − 32 cos2 2x cos 2x + cos 8x =
128
1 + cos 4 x 35 − 32 cos 2x + 28 cos 4x − 32
2 cos 2x + cos 8x
= 128
= 35 − 32 cos 2 x + 28 cos 4 x − 16 cos 2 x − 16 cos 4 x cos 2 x + cos 8x
128
= 35 − 48 cos 2 x + 28 cos 4 x − 16 cos 4 x cos 2 x + cos 8x
128
= 1
(35 − 48 cos 2 x + 28 cos 4 x + cos 8x − 16 cos 2x cos 4x) 128
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= 1 cos 4 x
( 2 )
= 1 cos 4 x
( 2 )
( 2 )
2 2
2
Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 263
29. sin 4 2 x = (sin 2 2 x)
31. tan 4 2 x = (tan 2 2x)
2
−
2
1 − cos 4 x =
2 1 + cos 4 x
1 = 1 − 2 cos 4x + cos 4x
4
= 1 1 − 2 cos 4 x +
1 + cos 8x
1 − 2 cos 4 x + cos2 4 x =
1 + 2 cos 4x + cos2 4x
1 + cos 8x
4 2
1 1 1 1 = − cos 4 x + + cos 8x
4 2 8 8
3 1 1 = − cos 4x + cos 8x
8 2 8
1 = (3 − 4 cos 4x + cos 8x)
8
1 − 2 cos 4 x +
= 2
1 + 2 cos 4x + 1 + cos 8x
2
1 (2 − 4 cos 4x + 1 + cos 8x)
= 2 1 (2 + 4 cos 4x + 1 + cos 8x)
2
= 3 − 4 cos 4 x + cos 8 x
30. cos4 2x = (cos2 2 x) 3 + 4 cos 4x + cos 8x
2
+
32.
tan 2 2 x cos4 2 x = 1 − cos 4x
(cos2 2x)2
2
= 1 (1 + 2 cos 4 x + cos2 4x)
1 + cos 4 x
1 − cos 4 x 1 + cos 4 x 2
4 = 1 + cos 4 x
2
1 1 + cos 8x
= 1 + 2 cos 4 x +
(1 − cos 4 x)(1 + cos 4 x)(1 + cos 4 x)
4 2
1 1 1 1 = + cos 4x + + cos 8x
4 2 8 8
3 1 1 = + cos 4 x + cos 8x
8 2 8
1 = (3 + 4 cos 4 x + cos 8x)
8
= 4(1 + cos 4 x)
(1 − cos 4 x)(1 + cos 4 x) =
4
1 = 1 − cos 4 x
4
= 1
− 1 + cos 8x
14 2
33.
sin 2 2 x cos2 2 x = 1 − cos 4 x 1 + cos 4 x
1 1 1 = − − cos 8x
4 8 8
1 1 = − cos 8x
8 8
1 = (1 − cos 8x)
8
2
2
1 = 1 − cos 4 x
4
1 1 + cos 8 x
= 1 −
4 2
= 1
− 1
− 1
cos 8x 4 8 8
= 1
− 1
cos 8x
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8 8
= 1(1 − cos 8x)
8
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( 2 )
( 2 3 )
[ ]
[ ]
2 +
3
264 Chapte r 2 A nalytic Trigo nomet ry
34. sin 4 x cos2 x = sin 2 x sin 2 x cos2 x
= 1 − cos 2 x 1 − cos 2 x 1 + cos 2 x
2
2
2
1 = (1 − cos 2 x) 1 − cos 2x
8
1 = 1 − cos 2x − cos 2 x + cos 2 x
8
+
+ 1 1 cos 4 x 1 cos 4 x = 1 − cos 2x − + cos 2x
8 2 2
1 = 2 − 2 cos 2x − 1 − cos 4 x + cos 2 x + cos 2 x cos 4x
16
1 = 1 − cos 2x − cos 4 x + cos 2 x cos 4 x
16
1
1 − cos 150° 1 + ( 3 2)
35. sin 75° = sin ⋅ 150° = =
2 2 2
= 1
2 + 3 2
1
1 + cos 150° 1 − ( 3 2)
cos 75° = cos ⋅ 150° = =
2 2 2
= 1
2 − 3 2
tan 75° = tan 1
⋅ 150°
= sin 150°
= 1 2
2
1 + cos 150°
1 ( 3 2) −
1 2 + 3 = ⋅
= 2 + 3
= 2 + 3
2 − 3 2 +
1
3 4 − 3
1 − cos 330°
1 − ( 3 2)36. sin 165° = sin ⋅ 330° = =
2 2 2
= 1
2 − 3 2
1
1 + cos 330° 1 + ( 3 2)
cos 165° = cos ⋅ 330° = − = −
2 2 2
1 = −
2
tan 165° = tan 1
⋅ 330°
= sin 330°
= −1 2
2
1 + cos 330°
1 ( 3 2) +
−1 2 − 3
− 2 + 3
= ⋅ = = − 2 + 32 + 3 2 − 3 4 − 3
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1
Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 265
1
1 − cos 225° 1 − (− 2 2) 1
37. sin 112° 30′ = sin ⋅ 225° = = = 2 + 2
2 2 2 2
1
1 + cos 225° 1 + (− 2 2) 1
cos 112° 30′ = cos ⋅ 225° = − = − = − 2 − 2
2 2 2 2
tan 112° 30′ = tan 1
⋅ 225°
= sin 225°
= − 2 2
= − 2
⋅ 2 + 2
= − 2 2 − 2
= −1 − 2
2
1 + cos 225° 1 (
2 2)
2 − 2 2 + 2 2
+ −
1
1 − cos 135° 1 + ( 2 2) 1
38. sin 67° 30′ = sin ⋅ 135° = = = 2 + 2
2 2 2 2
1
1 + cos 135° 1 − ( 2 2) 1
cos 67° 30′ = cos ⋅ 135° = = = 2 − 2
2 2 2 2
tan 67° 30′ = tan 1
⋅ 135°
= sin 135°
= 2 2
= 1 + 2 2
1 + cos 135°
1 ( 2 2) −
π 1 π 1 − cos π 4 1
39. sin = sin = = 2 − 2
8 2 4 2 2
ππ 1 π 1 + cos
4
cos = cos = = 2 + 2
8 2 4 2 2
sin π 2
tan π
= tan 1 π
= 4 = 2 =
2 − 1
8 2 4 1 + cos
π 1 +
2
4 2
40.
sin 7π
= sin 1 7π
= 1 − cos
7π 6 =
1 + 3
2 = 1
2 + 3
12 2
6
2 2 2
1 + cos 7π
1 − 3
cos 7π
= cos1 7π
= − 6 = −
2 = − 1
2 − 3
12 2
6
2 2 2
sin 7π
− 1
tan 7π
= tan 1 7π
= 6 = 2 = −2 − 3
12 2 6 1 + cos 7π
1 − 3
6 2
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2
2
,
2
2
= −
266 Chapte r 2 A nalytic Trigo nomet ry
41. cos u = 7
, 0 < u < π
42. sin u = 5
, π
< u < π cos u 12
25 2
(a) Because u is in Quadrant I, u
is also in Quadrant I. 2
1 − 7
13 2 13
(a) Because u is in Quadrant II, u
is in Quadrant I. 2
1 + 12
(b)
sin u
= 1 − cos u
= 25 =
9 3
= (b)
sin
u = 1 − cos u
= 13 5 26
=
2 2 2 25 5
1 + 7
2 2 26
1 − 12
cos u
= 1 + cos u
= 25 =
16 4 =
cos u
= 1 + cos u
= 13 26
=
2 2 2 25 5
1 − 7
2 2 26
5
tan u
= 1 − cos u
= 25 = 3
tan u
= sin u
= 13 = 5
2 sin u
24 4
2
1 + cos u
1 12
43.
tan u
5 3π = −
12 2
25
< u < 2π
− 13
(a) Because u is in Quadrant IV, u
is in Quadrant II. 2
1 − 12
(b) sin u
= 1 − cos u = 13 = 1 26
=
2 2 2 26 26
1 + 12
cos u 1 + cos u
= − = − 13 25 5 26 = − = −
2 2 2 26 26
1 − 12
tan u =
1 − cos u = 13 = − 1
2 sin u −
5 5
13
44.
cot u = 3, π
< u < 3π 2
(a) Because u is in Quadrant III, u
is in Quadrant II. 2
1 + 3
(b)
sin u
= 1 − cos u
= 10
= 10 + 3 10
= 1 10 + 3 10
2 2 20 2 5
1 − 3
cos u
= − 1 + cos u
= − 10 10 − 3 10
= −
1 10 − 3 10
= −
2 2 20 2 5
1 + 3
tan u
= 1 − cos u
= 10
= −
10 − 3
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2
sin u
1 −
10
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Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 267
45. sin x
+ cos x = 0 2
47. cos x
− sin x = 0 2
± 1 − cos x
2
1 − cos x
2
= −cos x
= cos2 x
± 1 + cos x
2
1 + cos x
2
= sin x
= sin 2 x
cos x = 1 2
0 = 2 cos2 x + cos x − 1
= (2 cos x − 1)(cos x + 1)
or cos x = −1
1 + cos x = 2 sin 2 x
1 + cos x = 2 − 2 cos2 x
2 cos2 x + cos x − 1 = 0
(2 cos x − 1)(cos x + 1) = 0
x = π
, 5π
3 3 x = π
2 cos x − 1 = 0 or cos x + 1 = 0
2 1 cos x =
2
cos x = −1
0 2π
x = π
, 5π
3 3
x = π
− 2
By checking these values in the original equation,
x = π 3 and x = 5π 3 are extraneous, and x = π
is the only solution.
x = π
, π , 5π
3 3
π 3, π , and 5π 3 are all solutions to the equation.
2
46. h(x) = sin x
+ cos x − 1 2
sin x
+ cos x − 1 = 0 2
0 2π
− 2
± 1 − cos x
2
1 − cos x
= 1 − cos x
= 1 − 2 cos x + cos2 x
48.
g (x) = tan x
− sin x 2
x
2
1 − cos x = 2 − 4 cos x + 2 cos2 x
tan − sin x = 0 2
1 − cos x
2 cos2 x − 3 cos x + 1 = 0
sin x = sin x
(2 cos x − 1)(cos x − 1) = 0
2 cos x − 1 = 0 or cos x − 1 = 0
1 − cos x = sin 2 x
1 − cos x = 1 − cos2 x
2
cos x = 1 cos x = 1 cos x − cos x = 0
2
x = π
, 5π
3 3
x = 0
cos x(cos x − 1) = 0
cos x = 0 or cos x − 1 = 0
0, π
, and 5π
are all solutions to the equation. x =
π ,
3π 2 2 cos x = 1
3 3 x = 0
1
0, π
, and 3π
are all solutions to the equation. 0 2π 2 2
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3
− 2
0 2π
− 3
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2
π π
π
268 Chapte r 2 A nalytic Trigo nomet ry
49. sin 5θ sin 3θ = 1 cos(5θ − 3θ ) − cos(5θ + 3θ ) = 1 (cos 2θ − cos 8θ )2 2
50. 7 cos(−5β ) sin 3β = 7 ⋅ 1 sin(−5β + 3β ) − sin(−5β − 3β ) = 7 (sin(−2β ) − sin(−8β ))2 2
51. cos 2θ cos 4θ = 1 cos(2θ − 4θ ) + cos(2θ + 4θ ) = 1 cos(−2θ ) + cos 6θ
2 2
52. sin(x + y) cos(x − y) = 1 (sin 2 x + sin 2 y)
54.
sin 3θ + sin θ = 2 sin 3θ + θ
cos 3θ − θ
2 2
53.
sin 5θ − sin 3θ = 2 cos
5θ + 3θ sin
5θ − 3θ = 2 sin 2θ cos θ
2
2
= 2 cos 4θ sin θ 55.
cos 6 x + cos 2 x = 2 cos 6 x + 2 x
cos 6 x − 2 x
2
2
56.
cos x + cos 4 x = 2 cos x + 4 x
cos x − 4 x
= 2 cos 4x cos 2x
2
2
= 2 cos 5x
cos − 3x
2
2
° + °
° − °
57.
75 15 75 15 2 3 6 sin 75° + sin 15° = 2 sin cos = 2 sin 45° cos 30° = 2 =
2 2 2 2 2
° + °
° − ° 58.
120 60 120 60 3 cos 120° + cos 60° = 2 cos cos = 2 cos 90° cos 30° = 2(0) = 0
2 2 2
3π
+ π 3π
− π
59. cos 3π − cos
π = −2 sin 4 4
sin 4 4
= −2 sin sin
4 4 2
2
2 4
cos 3π − cos
π 2 2 = − − = − 2
4 4 2 2
5π +
3π 5π −
3π
60. sin 5π − sin
3π = 2 cos 4 4
sin 4 4
= 2 cos π sin
4 4 2
2
4
sin 5π − sin
3π 2 2 = − − = − 2
4 4 2 2
61. sin 6x + sin 2x = 0
2 sin 6 x + 2 x
cos 6 x − 2 x
= 0 2
2
2(sin 4 x) cos 2x = 0
sin 4x = 0 or cos 2x = 0
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− 2
2
4x = nπ 2x = π 2
+ nπ 0 2π
x = nπ x =
π + nπ
4 4 2
In the interval [0, 2π )
x = 0, π
, π
, 3π
, π , 5π
, 3π
, 7π
. 4 2 4 4 2 4
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Section 2.5 Mult iple Angl e and Product - to-S um Fo rmulas 269
62. h(x) = cos 2x − cos 6x
cos 2x − cos 6 x = 0
−2 sin 4x sin(−2x) = 0
2 sin 4 x sin 2 x = 0
sin 4x = 0 or sin 2x = 0 2
4x = nπ 2x = nπ 0 2π
x = nπ x =
nπ
4 2 − 2
x = 0, π
, π
, 3π
, π , 5π
, 3π
, 7π x = 0,
π , π ,
3π
4 2 4 4 2 4 2 2
63. cos 2 x
− 1 = 0 sin 3x − sin x
cos 2 x = 1
sin 3x − sin x 2
cos 2 x = 1
65.
csc 2θ 1
= sin 2θ
1 =
2 sin θ cos θ
= 1
⋅ 1
2 cos 2 x sin x 0 2π
2 sin x = 1
sin θ
csc θ =
2 cos θ
sin x = 1
− 2
2
2 cos θ
x = π
, 5π 66. ( )( )
cos4 x − sin 4 x =
6 6 = (
cos2 x − sin 2 x
cos 2x)(1)
cos2 x + sin 2 x
64.
f (x) = sin 2 3x − sin 2 x
sin 2 3x − sin 2 x = 0
(sin 3x + sin x)(sin 3x − sin x) = 0
(2 sin 2x cos x)(2 cos 2x sin x) = 0
67. (sin x + cos x)2
= cos 2x
= sin 2 x + 2 sin x cos x + cos2 x
= (sin 2 x + cos2 x) + 2 sin x cos x
= 1 + sin 2x
sin 2x = 0 x = 0, π
, π , 3π
or 2 2
68.
tan u
1 − cos u
=
cos x = 0 x = π
, 3π
or 2 2
2 sin u
1 cos u = −
cos 2x = 0 x = π
, 3π
, 5π
, 7π
or sin u sin u
4 4 4 4
sin x = 0 x = 0, π
1
= csc u − cot u
2 sin x ± y
cos x y
0 2π
69. sin x ± sin y
cos x + cos y = 2 2
2 cos x + y
cos x − y
2
2
− 1 = tan x ± y
2
π + x +
π
− x
π + x −
π
− x
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3
70. π
cosπ
+ x + cos − x = 2 cos 3 3
cos 3 3
3 3 2
2
= 2 cos π
cos(x)
1 = 2 cos x = cos x
2
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2
= 1
2
270 Chapte r 2 A nalytic Trigo nomet ry
71. (a) sin θ
= ±
1 − cos θ
2
1 =
M
(c) When M = 4.5, cos θ (4.5)
2 − 2
= (4.5)
2
2
1 − cos θ
±
cos θ ≈ 0.901235.
2 M So, θ ≈ 0.4482 radian.
1 − cos θ
2
1 =
M 2 (d) When M = 2, speed of object
= M speed of sound
M 2 (1 − cos θ ) = 2 speed of object = 2
1 − cos θ
− cos θ
2 =
M 2
2 = − 1
760 mph
speed of object = 1520 mph.
speed of object
cos θ
M 2
= 1 − 2 M 2
M 2 − 2
When M = 4.5,
speed of sound
speed of object
760 mph
= M
= 4.5
cos θ =
M 2
22 − 2 1 π
speed of object = 3420 mph.
(b) When M = 2, cos θ = = 22 2
. So, θ = . 3
72.
1 (75)
2 sin 2θ
32
sin 2θ
= 130
130(32) =
752
75. True. Using the double angle formula and that sine is an
odd function and cosine is an even function,
sin(− 2 x) = sin 2(− x)
= 2 sin(− x) cos(− x)
θ = 1
sin −1130(32)
= 2(− sin x) cos x
2
752
θ ≈ 23.85° = − 2 sin x cos x.
76. False. If 90° < u < 180°,
73. x
= 2r sin 2 θ = 2r
1 − cos θ
2 2
2
u is in the first quadrant and
= r(1 − cos θ ) 2
So, x = 2r(1 − cos θ ). sin u
= 1 − cos u .
74. (a) Using the graph, sin 2u ≈ 1 and
2 2
77. Because φ and θ are complementary angles,
2 sin u cos u ≈ 2(0.7)(0.7) ≈ 1.
Because 1 = 1, sin 2u = 2 sin u cos u. sin φ = cos θ and cos φ = sin θ .
(b) Using the graph, cos 2u ≈ 0 and
cos2 u − sin 2 u ≈ (0.7)2
− (0.7)2
= 0.
(a) sin(φ − θ ) = sin φ cos θ − sin θ cos φ
= (cos θ )(cos θ ) − (sin θ )(sin θ )
= cos2 θ − sin 2 θ
Because 0 = 0, cos 2u = cos2 u − sin 2 u.
Rev
iew Exercises for Chapter 2
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(b)
= c
o
s
2
θ
c
o
s
(φ
−
θ
)
=
c
o
s
φ
c
o
s
θ
+
s
i
n
φ
s
i
n
θ
=
(s
i
n
θ
)
(c
o
s
θ ) + (cos θ )(sin θ )
= 2 sin θ cos θ
= sin 2θ
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1. cot x 3. cos x
2. sec x 4. cot 2 x + 1 = csc2 x =
csc x
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= −
= −
tan 2 x
2
Review Exer cis e s for Chapter 2 271
5. cos θ 2
, tan θ 5
> 0, θ is in Quadrant III.
sec θ
= 1
= − 5
cos θ 2
sin θ
= − 1 − cos2 θ = −
1 − 4
= −
21 = −
21
csc θ
25 25 5
= 1
= − 5
= − 5 21
sin θ 21 21
− 21
tan θ = sin
= 5 = 21
cos θ −
2 2 5
cot θ = 1
= 2
= 2 21
tan θ 21 21
6. cot x 2
, cos x < 0, x is in Quadrant II. 3
tan x = 1
= − 3
cot x 2
csc x =
1 + cot 2 x =
1 + 4
=
13 13 =
9 9 3
sin x = 1
= 3
= 3 13
csc x 13 13
cos x = − 1 − sin 2 x = −
1 − 9
4 2 2 13 = − = − = −
sec x = 1
= − 13
13 13 13 13
cos x 2
1 1
7. = = sin 2 x cot 2 x + 1 csc2 x
13.
cos2 x + cos2 x cot 2 x = cos2 x(1 + cot 2 x)
= cos2 x(csc2 x)sin θ
1
8. tan θ
= cos θ
= 1
= cos2 x sin 2 x
1 − cos2 θ sin 2 θ sin θ cos θ
cos2 x
= csc θ sec θ = sin 2 x
2
9. tan 2 x(csc2 x − 1) = tan 2 x(cot 2 x) = cot x
= tan 2 x 1
14. (tan x + 1)2
cos x = (tan 2 x + 2 tan x + 1) cos x
= 1 = (sec
x + 2 tan x) cos x
10.
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( ) cos x
=
2
cot 2
x sin 2 x =
sin 2 x =
cos2
x sin 2
x
sin x = sec2 x cos x + 2 cos x
cos x
= sec x + 2 sin x
cot π
− u
11. 2
cos u
tan u
cos u
= tan u sec u
12.
sec (−θ )
sec θ
1 cos θ
sin θ 2
2 2 2 2
= = = = tan θcsc2 θ csc2 θ 1 sin 2 θ cos2 θ
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2
272 Chapte r 2 A nalytic Trigo nomet ry
15.
1 1 (csc θ − 1) − (csc θ + 1) − =
16.
tan x
sec x − 1
2 2
=csc θ + 1 csc θ − 1 (csc θ + 1)(csc θ − 1)
= −2 csc2 θ − 1
= −2 cot 2 θ
= −2 tan 2 θ
1 + sec x 1 + sec x
(sec x + 1)(sec x − 1) =
sec x + 1
= sec x − 1
17. Let x = 5 sin θ , then
25 − x2 =
25 − (5 sin θ )2
=
25 − 25 sin 2 θ =
25(1 − sin 2 θ ) =
25 cos2 θ
= 5 cos θ .
18. Let x = 4 sec θ , then
x2 − 16 =
(4 sec θ )2
− 16 =
16 sec2 θ − 16 =
16(sec2 θ − 1) =
16 tan 2 θ
= 4 tan θ .
19. cos x(tan 2 x + 1) = cos x sec2 x 25. sin5 x cos2 x = sin 4 x cos2 x sin x
1 = sec2 x
sec x = (1 − cos2
x) cos2
x sin x
= sec x = (1 − 2 cos
x + cos
x) cos
x sin x
20.
sec2 x cot x − cot x = cot x(sec2 x − 1)
= cot x tan 2 x
26.
2 4 2
= (cos2 x − 2 cos4 x + cos6 x) sin x
cos3 x sin 2 x = cos x cos2 x sin 2 x
1 = tan 2 x = tan x
tan x
= cos x(1 − sin 2 x) sin 2 x
4
= cos x(sin 2
x − sin x)
21. sin π 2
− θ tan θ
= cos θ tan θ
= (sin 2 x − sin 4 x) cos x
sin θ = cos θ
cos θ
= sin θ
27. sin x =
sin x =
3 − sin x
3
2
x = π + 2π n,
2π + 2π n
22. cot π 2 − θ
csc θ
3 3 = tan θ csc θ
sin θ 1 =
cos θ sin θ
28. 4 cos θ
2 cos θ
= 1 + 2 cos θ
= 1
1
1 =
cos θ
cos θ = 2
π 5π
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= sec θ θ = + 2nπ or
3 + 2nπ
3
23. 1
= 1
= cos θ 29. 3 3 tan u = 3
tan θ csc θ sin θ
cos θ ⋅
1 sin θ
tan u = 1
3
π24.
1 1 1 = =
u = + nπ 6
tan x csc x sin x (tan x)
1 (sin x)
sin x tan x
= cot x
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Review Exer cis e s for Chapter 2 273
30. 1
sec x − 1 = 0 2
1 sec x = 1
2
35. cos2 x + sin x = 1
1 − sin 2 x + sin x − 1 = 0
−sin x(sin x − 1) = 0
sec x = 2 sin x = 0 sin x − 1 = 0
cos x = 1 2
x = 0, π sin x = 1
x = π
x = π + 2nπ or
5π 2 + 2nπ
31.
3 3
3 csc2 x = 4
36. sin 2 x + 2 cos x = 2
1 − cos2 x + 2 cos x = 2
2
csc2 x = 4 0 = cos x − 2 cos x + 1
3
sin x = ± 3
2
0 = (cos x − 1)2
cos x − 1 = 0
cos x = 1
x = π + 2π n,
2π + 2π n, 4π + 2π n,
5π + 2π n x = 0
3 3 3 3
These can be combined as:
37. 2 sin 2x −
2 = 0
x = π
+ nπ or x = 2π
+ nπ
sin 2x = 2
3 3
2 x = π
+ 2π n, 3π
2
+ 2π n
32. 4 tan 2 u − 1 = tan 2 u 4 4
3 tan 2 u − 1 = 0 x =
π + π n,
3π
+ π n
tan 2 u = 1 3
tan u = ± 1 3
= ± 3 3
8 8
x = π
, 3π
, 9π
, 11π
8 8 8 8
x
u = π
+ nπ or 5π
+ nπ 38. 2 cos + 1 = 0
2
6 6 x 1 cos = −
33. sin3 x = sin x 2 2
sin3 x − sin x = 0
sin x(sin 2 x − 1) = 0
sin x = 0 x = 0, π
sin 2 x = 1
x =
2π 2 3
x = 4π 3
x
sin x = ±1 x = π
, 3π
2 2
39. 3 tan 2 − 1 = 0 3
tan 2 x =
1
34.
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3 2
c
o
s2
x
+
3
c
o
s
x
=
0
c
o
s
x
(2
c
o
s
x
+
3
)
=
0
cos
x =
0
or
2
cos
x +
3 =
0
3
tan x
=
± 1
3 3
tan x
=
± 3
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= −
3 3
x = π
, 3π
2 2 2 cos x = −3 x
= π
, 3 6
5π ,
7π 6 6
cos x 3 2 x =
π ,
2
5π ,
7π 2 2
No solution 5π and
7π 2 2
are greater than 2π , so they are not
solutions. The solution is x = π
. 2
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2
(
(
= −
274 Chapte r 2 A nalytic Trigo nomet ry
40. 3 tan 3x = 0
tan 3x = 0
3x = 0, π , 2π , 3π , 4π , 5π
43. tan 2 x − 2 tan x = 0
tan x(tan x − 2) = 0
tan x = 0 or tan x − 2 = 0
x = 0, π
, 2π
, π , 4π
, 5π
3 3 3 3 x = nπ tan x = 2
x = arctan 2 + nπ
41. cos 4x(cos x − 1) = 0
44.
2 tan 2 x − 3 tan x = −1
cos 4x = 0 cos x − 1 = 0 2 tan 2 x − 3 tan x + 1 = 0
4x = π + 2π n,
3π + 2π n cos x = 1 (2 tan x − 1)(tan x − 1) = 0
2 2 2 tan x − 1 = 0 or tan x − 1 = 0
x = π +
π n,
3π + π
n x = 0 2 tan x = 1 tan x = 1
8 2 8 2
x = 0, π
, 3π
, 5π
, 7π
, 9π
, 11π
, 13π
, 15π
8 8 8 8 8 8 8 8
tan x = 1 2
x = π
+ nπ 4
42.
3 csc2 5x = −4
x = arctan 1
+ nπ
csc2 5x 4 3
csc 5x = ± − 4 3
No real solution
45. tan 2 θ + tan θ − 6 = 0
(tan θ + 3)(tan θ − 2) = 0
tan θ + 3 = 0 or tan θ − 2 = 0
tan θ
θ
= −3 tan θ
= arctan(−3) + nπ θ
= 2
= arctan 2 + nπ
46.
sec2 x + 6 tan x + 4 = 0
1 + tan 2 x + 6 tan x + 4 = 0
tan 2 x + 6 tan x + 5 = 0
47. sin 75° = sin(120° − 45°) = sin 120° cos 45° − cos 120° sin 45°
3 2 1 2
= − −
(tan x + 5)(tan x + 1) = 0 2 2 2 2
tan x + 5 = 0 or tan x + 1 = 0 = 2
4 3 + 1)
tan x = −5 tan x = −1 cos 75° = cos (120° − 45°)
x = arctan(−5) + nπ x = 3π + nπ = cos 120° cos 45° + sin 120° sin 45°
4
= − 1 2 3 2
+
2 2 2 2
= 2
4
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3 − 1)
tan 75° = tan(120° − 45°) = tan 120° −tan 45°
1 + tan 120° tan 45°
= − 3 − 1
= − 3 − 1
1 + (− 3)(1) 1 − 3
= − 3 − 1
⋅ 1 + 3
1 − 3 1 + 3
= − 4 − 2 3
− 2
= 2 + 3
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(
(
= − + = (
(
)
)
Review Exer cis e s for Chapter 2 275
48. sin(375°) = sin(135° + 240°)
= sin 135° cos 240° + cos 135° sin 240°
2 1 2 3
=
− + −
−
2 2 2 2
= 2
3 − 1 4
cos(375°) = cos(135° + 240°)
= cos 135° cos 240° − sin 135° sin 240°
2 1 2 3
= − − −
−
2 2 2 2
2 = 1 + 3
4
tan(375°) = tan(135° + 240°)
= tan 135° + tan 240° 1 − tan 135° tan 240°
= −1 + 3
1 − (−1)( 3)
= −1 + 3
⋅ 1 − 3
= − 4 + 2 3
= 2 − 31 + 3 1 − 3 1 − 3
49.
sin 25π
= sin 11π
+ π
= sin 11π
cos π
+ cos 11π
sin π
12
6 4
6 4 6 4
1 2 3 2 2
3 − 1) 2 2 2 2 4
cos 25π
= cos11π
+ π
= cos 11π
cos π
− sin 11π
sin π
12
6 4
6 4 6 4
3 2 1 2 2
= − − =
3 + 1)
2 2 2 2 4
tan 11π
+ tan π
tan 25π
= tan11π
+ π
= 6 4
12
6 4
1 tan 11π
tan π
− 6 4
3 − + 1
3
= = 2 − 3 3
1 − − (1) 3
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5 3
4
− 4
v
− 3 5
(
=
4 4
276 Chapte r 2 A nalytic Trigo nomet ry
50.
sin19π
= sin11π
− π
cos19π
= cos11π
− π
12
6 4
12
6 4
= sin 11π
cos π
− cos 11π
sin π
= cos 11π
cos π
+ sin 11π
sin π
6 4 6 4
1 2 3 2
6 4 6 4
3 2 1 2
= − ⋅ − ⋅ = ⋅ + −
2 2 2 2 2 2 2 2
= − 2
(1 + 3) = − 2
( 3 + 1) = 2
3 − 1)4 4 4
tan19π
= tan11π
− π
12
6 4
tan 11π − tan
π
= 6 4
1 + tan 11π
tan π
6 4
− 3
− 1
= 3 =
− 3 − 3
⋅ 3 + 3
3 1 + − (1)
3 − 3 3 + 3
3
−(12 + 6 3) = = −2 − 3
6
51. sin 60° cos 45° − cos 60° sin 45° = sin(60° − 45°)
= sin 15°
52. tan 68° −tan 115°
= tan(68° − 115°) 1 + tan 68° tan 115°
= tan(−47°)
y y
u x x
53. ( )
Figures for Exercises 53–56
3 ( 4 )
4 ( 3 ) 24
sin u + v = sin u cos v + cos u sin v = 5
− 5
3
+ 3 3
+ 5
− 5
= − 25
54. tan(u + v) = tan u + tan v
= 4 4 = 2 = 3 16 24
1 − tan u tan v 1 −
3 3
7
16
2 7 7
55. ( )
4 ( 4 )
3 ( 3 )cos u − v = cos u cos v + sin u sin v =
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5 −
5 +
5 −
5 = −1
56. ( )
3 ( 4 )
4 ( 3 )sin u − v = sin u cos v − cos u sin v = 5
− 5
− 5
− 5
= 0
57.
cos x +
π = cos x cos
π − sin x sin
π = cos x(0) − sin x(1) = −sin x
2
2 2
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, π
2
4
1 − 4
= −
= 2
= −
Review Exer cis es for Chapter 2 277
58. tan x −
π = −tan
π − x
= −cot x
61. sin
x +
π − sin
x −
π = 1
2
2
4
4
59.
tan(π − x) = tan π − tan x
= −tan x 1 − tan π tan x
2 cos x sin π
= 1 4
60.
sin(x − π ) = sin x cos π − cos x sin π
= sin x(−1) − cos x(0)
= − sin x
cos x = 2
2
x = π
, 7π
4 4
62. cos x +
π − cos
x −
π = 1
6
6
cos x cos
π − sin x sin
π −
cos x cos
π + sin x sin
π = 1
6 6
6 6
−2 sin x sin π
= 1 6
63.
sin u 4
< u < 3π
−2 sin x 1
= 1
sin x = −1
x = 3π 2
65.
sin 4x = 2 sin 2x cos 2 x
5 2 = 22 sin x cos x(cos2 x − sin 2 x)cos u = − 1 − sin 2 u =
−3 5 = 4 sin x cos x(2 cos2
x − 1)
tan u = sin u
= 4
cos u 3
= 8 cos3 x sin x − 4 cos x sin x
2
4 3 24 sin
2u = 2 sin u cos u = 2 − − = 5 5 25 − 2π 2π
cos 2u = cos2 u − sin 2 u = −
3 2 2
− −
7
= −
5
5
25 − 2
4 2
1 − cos 2 x
1 − (1 − 2 sin 2 x)
tan 2u = 2 tan u
1 − tan 2 u
3 = − 24
7
3
66. 1 + cos 2 x
= 1 + (2 cos x2 − 1) 2 sin 2 x
= 2 cos2 x
64. cos u = − 2
, π
5 2 < u < π sin u =
1 and 5
= tan 2 x
tan u 1 4
2
1 2 4 sin 2u = 2 sin u cos u = 2 − = −
5 5 5 − 2π 2π
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2
2 4
cos 2u = cos2 u − sin 2 u = −
2 2 2
− 1
= 3 − 1
5
5
5
2 sin 2 3x 1 − cos 6x
2
1 − cos 6 x
1 2 − 67. tan 3x =
2 = =
tan 2u = 2tan u
= 2
= −1
= − 4 cos 3x 1 + cos 6 x 1 + cos 6 x
1 − tan 2 u 1 3 3 2
1 − −
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2
12
2
3
3
=
278 Chapte r 2 A nalytic Trigo nomet ry
68.
sin 2 x cos2 x = 1 − cos 2 x 1 + cos 2 x
2
2
1 − cos2 2 x =
4
1 − 1 + cos 4 x
= 4
= 1 − cos 4 x
8
1 − cos 150°
3 1 − − 2
2 + 3 1
69. sin(−75°) = − = − = − = − 2 + 3 2 2 2 2
3 1 + −
1 + cos 150° 2 2 − 3 1
cos(−75°) = − = = = 2 − 3 2 2 2 2
3
1 − −
− °
tan(−75°) 1 cos 150
2 = −(2 + 3) = −2 − 3
= −
sin 150°
= − 1
2
5π 1 − −
−
5π 1 cos 6 2 2 + 3 1
70. sin = = = = 2 + 3
12 2 2 2 2
5π 1 + − +
5π 1 cos 6 2 2 − 3 1
cos = = = = 2 − 3
12 2 2 2 2
tan 5π
=
1 − cos 5π
6
sin 5π
3 1
− −
= 1
= 2 + 3
71.
tan u =
4
, π
6 2
< u < 3π
(b)
sin u
=
1 − cos u =
1 − −
3
5 4
3 2 2 2 2 5 y
2 5 =
5
1 + −
3
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u cos u 1 + cos u
= − = − 5
= − 1
− 3 x 2 2 2 5
− 4 5
5 = −
5
1 − −
3
(a) Because u is in Quadrant III, u
is in Quadrant II. tan u 1 − cos u
= = 5 = −2
2 2 sin u −
4
5
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5 3
4
1 −
1 +
1 −
,
= −
=
Review Exer cis e s for Chapter 2 279
72. sin u = 3
, 0 < u < π
5 2 y
u x
(a) Because u is in Quadrant I, u
is in Quadrant I. 2
4
(b) sin u
= 1 − cos u = 5 = 1
= 10
2 2 2 10 10
4
cos u
= 1 + cos u = 5 = 9
= 3 10
2 2 2 10 10
4
tan u =
1 − cos u = 5 = 1
2 sin u 3 3
5
73. cos u 2 π 7 2
y
< u < π
7
3 5 u
x − 2
(a) Because u is in Quadrant II, u
is in Quadrant I. 2
1 − −
2
(b) sin u
= 1 − cos u = 7 = 9
2 2 2 14
3 14=
cos u
=
14
1 + cos u =
1 + −
2
7 5
2 2 2 14
70 =
14
1 − −
2
tan u 1 − cos u
= = 7 3 5
=
2 sin u 3 5 5
7
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1 −
1 +
1 −
= −
2
2
2
cos
θ
280 Chapte r 2 A nalytic Trigo nomet ry
74.
tan u = − 21
, 3π
2 2 y
< u < 2π
u
2 x
5 − 21
(a) Because u is in Quadrant IV, u
is in Quadrant II. 2
2
(b) sin u
= 1 − cos u = 5 = 3
= 30
2 2 2 10 10
2
cos u = −
1 + cos u = − 5 = −
7 = −
70
2 2 2 10 10
2
tan u =
1 − cos u = 5 3 3 21 = − = −
21
2 sin u 21
− 21 21 7
5
75. cos 4θ sin 6θ = 1 sin(4θ + 6θ ) − sin(4θ − 6θ ) = 1 sin 10θ − sin(−2θ )
76.
77.
2 sin 7θ cos 3θ
cos 6θ + cos 5θ
= 2 ⋅ 1 sin(7θ + 3θ ) + sin(7θ − 3θ ) = sin 10θ + sin 4θ
= 2 cos 6θ + 5θ
cos 6θ − 5θ
= 2 cos 11θ
cos θ
2
2
2 2
78. sin 3x − sin x = 2 3x + x
sin 3x − x
79. r = 1
v02
sin 2θ
2 2 32
= 2 cos 2 x sin x range = 100 feet
v0 = 80 feet per second
r = 1
(80)2 sin 2θ
32
= 100
sin 2θ
2θ
θ
= 0.5
= 30°
= 15° or π 12
80. Volume V of the trough will be the area A of the isosceles triangle times the length l of the trough.
V =
(a)
A ⋅ l
A = 1 bh
2
cos θ =
h h = 0.5 cos
θ 4 m
2 0.5 2 b b
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sin θ = 2
b = 0.5 sin θ h
2 0.5 2 2 0.5 m 0.5 m
A = 0.5 sin θ
0.5 cos θ = (0.5)
2 sin
θ cos
θ = 0.25 sin θ
cos θ
square meters
2 2 2 2 2 2
V = (0.25)(4) sin θ
cos θ
cubic meters = sin θ
cos θ
cubic meters 2 2 2 2
Not drawn to scale
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=
Proble m Solv i ng fo r Cha p t e r 2 281
(b) V
= sin θ
cos θ
= 1 2 sin
θ cos
θ = 1
sin θ cubic meters
2 2 2
2 2
2
Volume is maximum when θ = π
. 2
81. False. If π < θ < π , then
π < θ
< π
, and θ
is in 84. True. It can be verified using a product-to-sum formula.
2
Quadrant I. cos θ
> 0
4 2 2 2 4 sin 45° cos 15° = 4 ⋅
1[sin 60° + sin 30°]
2
2 3 1
82. True. cot x sin 2 x = cos x
sin 2 x = cos x sin x.
sin x
= 2
+ = 2 2
3 + 1
85. Yes. Sample Answer. When the domain is all real
83. True. 4 sin(− x)cos(− x) = 4(−sin x) cos x
= −4 sin x cos x
numbers, the solutions of sin x =
5π
1 are x =
π 2 6
+ 2nπ
= −2(2 sin x cos x)
= −2 sin 2x
Problem Solving for Chapter 2
and x = + 2nπ , so there are infinitely many 6
solutions.
1. sin θ = ± 1 − cos2 θ You also have the following relationships:
tan θ
= sin θ
= ± cos θ
1 − cos2 θ
cos θ sin θ = cos
π 2
− θ
csc θ
= 1
= ± sin θ
1
1 − cos2 θ
tan θ
csc θ
cos (π 2) − θ
cos θ
1 =
sec θ
cot θ
= 1 cos θ
= 1
= ± cos θ
sec θ
cos(π 2) − θ
1 =
cos θ
tan θ 1 − cos2 θ
cot θ cos θ
= cos(π 2) − θ
(2n + 1)π 2nπ + π
2. cos = cos (12n + 1)π 1
3. sin = sin (12nπ + π )
2
2
6
6
= cos nπ +
π
= sin 2nπ +
π
2
6
= cos nπ cos π − sin nπ sin
π = sin π
= 1
2 2
= (±1)(0) − (0)(1) 6 2
(12n + 1)π 1
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= 0 So, sin 6
= for all integers n. 2
(2n + 1)π
So, cos = 0 for all integers n.
2
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4π 1 2 3 5 6
1
2
282 Chapte r 2 A nalytic Trigo nomet ry
4. p(t) = 1
p (t ) + 30 p (t) + p (t ) + p (t) + 30 p (t) 1.4
1.4
p1(t) p2(t)
(a) p1(t ) = sin(524π t ) 1
− 0.006 0.006
− 0.006 0.006
p2 (t) =
p3 (t ) =
p5 (t) =
sin(1048π t) 2
1 sin(1572π t )
3
1 sin(2620π t )
5
1.4
p3(t)
− 1.4
1.4
p5(t)
− 1.4
1.4
p
6(t)
p6 (t) = 1
sin(3144π t) 6
− 0.006 0.006 − 0.006 0.006 − 0.006 0.006
The graph of − 1.4
− 1.4
− 1.4
p(t) = 1
sin(524π t ) + 15 sin(1048π t) + 1
sin(1572π t ) + 1
sin(2620π t ) + 5 sin(3144π t )4π 3 5
yields the graph shown in the text below.
y
1.4
y = p(t)
t
0.006
(b)
− 1.4
Function Period
p (t ) 2π
524π
p (t) 2π
1
= ≈ 0.0038 262
= 1
≈ 0.0019
(c)
1.4 Max
0 0.00382
p3 (t )
1048π
2π
524
= 1
≈ 0.0013
− 1.4 Min
1
1572π
2π 786
1
Over one cycle, 0 ≤ t < , you have five t-intercepts: 262
p5 (t)
p6 (t )
2620π
2π
3144π
= ≈ 0.0008 1310
1 = ≈ 0.0006
1572
t = 0, t ≈ 0.00096, t ≈ 0.00191, t ≈ 0.00285,
t ≈ 0.00382
(d) The absolute maximum value of p over one cycle is
p ≈ 1.1952, and the absolute minimum valueThe graph of p appears to be periodic with a
period of 1
≈ 0.0038. 262
of p over one cycle is p ≈ −1.1952.
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2
2
= −
b
Proble m Solv i ng fo r Cha p t e r 2 283
5. From the figure, it appears that u + v = w. Assume that u, v, and w are all in Quadrant I.
From the figure:
tan u = s
= 1
3s 3
tan v = s
= 1
2s 2
tan w = s
= 1 s
tan u + v tan u + tan v 1 3 + 1 2 5 6
= = = = 1 = tan w.
( ) 1 − tan u tan v
1 − (1 3)(1 2) 1 − (1 6)
So, tan(u + v) = tan w. Because u, v, and w are all in Quadrant I, you have
arctan tan(u + v) = arctan[tan w]u + v = w.
6. y = − 16 x2 + (tan θ )x + h0
Let h0
v0 2 cos2 θ
= 0 and take half of the horizontal distance:
1 1 v
sin 2θ
= 1
v
2 sin θ cos θ
1
= v
sin θ cos θ
2
2 32
0 0 ( ) 0
2
64 32
Substitute this expression for x in the model.
2
y = − 16 1
v 2 sin θ cos θ sin θ 1
+ v sin θ cos θ
v 2 cos2 θ 32
0
cos θ
32
0
0
1 v 2 sin 2 θ +
1 v 2 sin 2 θ
64 0
1 = v0
2 sin 2 θ 64
32 0
7. (a) 10 θ 10
h
sin θ
b
1 b
= 2
1 2
and cos θ
= h
2 10 2 10
b = 20 sin θ h = 10 cos
θ
2 2
A = 1 bh
2
1 θ θ
= 20 sin 10 cos
2
2
2
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2
(b)
= 100 sin θ
cos θ
2 2
θ θ A = 50 2 sin cos
2 2
θ
= 50 sin 2
= 50 sin θ
Because sin π
= 1 is a maximum, θ = π
. So, the area is a maximum at A = 50 sin π
= 50 square meters.
2 2 2
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2
284 Chapte r 2 A nalytic Trigo nomet ry
8. The hypotenuse of the larger right triangle is:
θ
2 1 2(1 + cos θ )
1
sin 2 θ + (1 + cos θ )2
=
=
sin 2 θ + 1 + 2 cos θ + cos2 θ
2 + 2 cos θ
cos θ θ = 2(1 + cos θ )
sin θ
sin θ
= sin θ
= sin θ
⋅ 1 − cos θ
= sin θ 1 − cos θ
= sin θ 1 − cos θ
=
1 − cos θ
2
θ
2(1 + cos θ )
1 + cos θ
2(1 + cos θ )
(1 + cos θ )2
1 − cos θ
1 + cos θ
2(1 − cos2 θ ) 2 sin θ 2
cos = = = 2 2(1 + cos θ ) 2(1 + cos θ ) 2
tan θ
= sin θ
1 + cos θ
9. F
0.6W sin(θ + 90°) =
sin 12°
(a) F =
0.6W (sin θ cos 90° +cos θ sin 90°) sin 12°
= 0.6W (sin θ )(0) + (cos θ )(1)
sin 12°
(b) Let y1 =
550
0.6(185) cos x.
sin 12°
= 0.6W cos θ
sin 12°
0 90
0
(c) The force is maximum (533.88 pounds) when θ = 0°.
The force is minimum (0 pounds) when θ = 90°.
π (t + 0.2)10. Seward: D = 12.2 − 6.4 cos
11.
d = 35 − 28 cos π
t when t = 0 corresponds to
182.6 6.2
π (t + 0.2) New Orleans: D = 12.2 − 1.9 cos
12:00 A.M.
(a) The high tides occur when cos π
t = −1. Solving
182.6 6.2
(a)
20
0 365 0
yields t = 6.2 or t = 18.6.
These t-values correspond to 6:12 A.M. and 6:36 P.M.
The low tide occurs when cos π
t = 1. Solving 6.2
yields t = 0 and t = 12.4 which corresponds to
12:00 A.M. and 12:24 P.M.
(b) The graphs intersect when t ≈ 91 and when
t ≈ 274. These values correspond to April 1 and
(b) The water depth is never 3.5 feet. At low tide, the
depth is d = 35 − 28 = 7 feet.
October 1, the spring equinox and the fall equinox.
(c) Seward has the greater variation in the number of
daylight hours. This is determined by the
a
m
p
litudes, 6.4 and
1.9.
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(d) Period: 2π
= 365.2 days π 182.6
(c)
70
0 24 0
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4π
2
2
2
2
2
2
2
2
2
2
.
Proble m Solv i ng fo r Cha p t e r 2 285
12. h1 = 3.75 sin 733t + 7.5 θ α
sin +
h = 3.75 sin 733 t + + 7.5
13. (a) n =
2 2
2 3
θ
(a) 15
sin 2
sin θ
cos α
+ cos θ
sin α
=
sin θ
0 1
0
(b) The period for h1 and h2 is
2π
733
≈ 0.0086.
= cos α
+ cot θ
sin α
θ
12 For α = 60°, n = cos 30° + cot sin 30° 2
n = 3
+ 1
cot θ
0 2π
733 3
2 2
(b) For glass, n = 1.50.
3 1
θ
The graphs intersect twice per cycle. 1.50 = + 2 2
cot 2
There are 1
≈ 116.66 cycles in the interval
3 θ2π 733
21.50 −
= cot
[0, 1], so the graphs intersect approximately 2 2
233.3 times. 1 θ
= tan
3 − 3
θ
= 2 tan −1 1
3 − 3
θ ≈ 76.5°
14. (a)
(b)
sin(u + v + w) = sin (u + v) + w
= sin(u + v) cos w + cos(u + v) sin w
= [sin u cos v + cos u sin v] cos w + [cos u cos v − sin u sin v] sin w
= sin u cos v cos w + cos u sin v cos w + cos u cos v sin w − sin u sin v sin w
tan(u + v + w) = tan (u + v) + w
tan(u + v) + tan w =
1 − tan(u + v) tan w
tan u + tan v + tan w
= 1 − tan u tan v ⋅
(1 − tan u tan v) tan u + tan v
1 − tan w 1 − tan u tan v
(1 − tan u tan v)
tan u + tan v + (1 − tan u tan v) tan w =
(1 − tan u tan v) − (tan u + tan v) tan w
= tan u + tan v + tan w − tan u tan v tan w 1 − tan u tan v − tan u tan w − tan v tan w
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6 6 3 3
= 1 cos 2x
+ 1 cos 2 x
4
2
4
.
2
2
2
286 Chapte r 2 A nalytic Trigo nomet ry
15. (a) Let y1 = sin x and y2 = 0.5. (b) Let y1 = cos x and y2 = −0.5.
2
0 2π
2
0 2π
− 2
sin x ≥ 0.5 on the interval π
, 5π
− 2
cos x ≤ −0.5 on the interval
2π
, 4π
.
(c) Let y1 = tan x and y2
= sin x.
(d) Let y1 = cos x and y2
= sin x.
2
0 2π
2
0 2π
− 2 − 2
π
5π
tan x < sin x on the intervals π
, π and
3π , 2π .
cos x ≥ sin x on the intervals 0,
and
, 2π .
2
2
4 4
16. (a) f (x) = sin 4 x + cos4 x
= (sin 2 x) + (cos2 x)
2
−
2
+
2 2
= 1 (1 − 2 cos 2x + cos2 2 x) + (1 + 2 cos 2 x + cos2 2x)
= 1 (2 + 2 cos2 2 x)
= 1 (1 + cos2 2x)
1 cos 4x
= 1 +
2 2
= 1 (3 + cos 4x)
4
(b) f (x) = sin 4 x + cos4 x
= (sin 2 x) + cos4 x
2 2
4
= (1 − cos x) + cos x
(c)
= 1 − 2 cos2 2x + cos4 x + cos4 x
= 2 cos4 x − 2 cos2 x + 1
f (x) = sin 4 x + cos4 x
= sin 4 x + 2 sin 2 x cos2 x + cos4 x − 2 sin 2 x cos2 x
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2
= (sin 2 x + cos2 x)
− 2 sin 2 x cos2 x
= 1 − 2 sin 2 x cos2 x
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2
( 2 )
Proble m Solv i ng fo r Cha p t e r 2 287
(d) f (x) = 1 − 2 sin 2 x cos2 x
1 − cos 2x 1 + cos 2x
= 1 − 2
2
2
= 1 − 1 (1 − cos2 2x)
1 1 = + cos2 2x
2 2
1 1 = + 1 − sin 2x
2 2
= 1 − 1
sin 2 2x 2
(e) No; there is often more than one way to rewrite a trigonometric expression, so your result and your friend’s result
could both be correct.
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= <
288 Chapte r 2 A nalytic Trigo nomet ry
Practice Test for Chapter 2
1. Find the value of the other five trigonometric functions, given tan x 4 , sec x 0. 11
2. Simplify sec2 x + csc2 x
. csc2 x(1 + tan 2 x)
3. Rewrite as a single logarithm and simplify ln tan θ − ln cot θ .
4. True or false:
cos π
− x
= 1
2
csc x
5. Factor and simplify: sin 4 x + (sin 2 x) cos2 x
6. Multiply and simplify: (csc x + 1)(csc x − 1)
7. Rationalize the denominator and simplify:
cos2 x
1 − sin x
8. Verify:
1 + cos θ
sin θ
sin θ
+ = 2 csc θ 1 + cos θ
9. Verify:
tan 4 x + 2 tan 2 x + 1 = sec4 x
10. Use the sum or difference formulas to determine:
(a) sin 105°
(b) tan 15°
11. Simplify: (sin 42°) cos 38° − (cos 42°) sin 38°
12. Verify tan
+ π
= 1 + tan θ
. θ
4 1 − tan θ
13. Write sin(arcsin x − arccos x) as an algebraic expression in x.
14. Use the double-angle formulas to determine:
(a) cos 120°
(b) tan 300°
15. Use the half-angle formulas to determine:
(a) sin 22.5°
(b) tan π 12
16. Given sin θ = 4 5, θ lies in Quadrant II, find cos(θ 2).
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1
Practice Te st for C hapte r 2 289
17. Use the power-reducing identities to write (sin 2 x) cos2 x in terms of the first power of cosine.
18. Rewrite as a sum: 6(sin 5θ ) cos 2θ .
19. Rewrite as a product: sin(x + π ) + sin(x − π ).
20. Verify sin 9 x + sin 5x cos 9 x − cos 5x
= −cot 2 x.
21. Verify:
(cos u) sin v =
2 sin(u + v) − sin(u − v).
22. Find all solutions in the interval [0, 2π ):
4 sin 2 x = 1
23. Find all solutions in the interval [0, 2π ):
tan 2 θ + ( 3 − 1) tan θ −
3 = 0
24. Find all solutions in the interval [0, 2π ):
sin 2 x = cos x
25. Use the quadratic formula to find all solutions in the interval [0, 2π ):
tan 2 x − 6 tan x + 4 = 0
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Analytic Trigonometry
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2.1 Using Fundamental Identities
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Objectives
Recognize and write the fundamental
trigonometric identities.
Use the fundamental trigonometric identities to
evaluate trigonometric functions, simplify
trigonometric expressions, and rewrite
trigonometric expressions.
3
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Introduction
4
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Introduction
You will learn how to use the fundamental identities to do
the following.
1. Evaluate trigonometric functions.
2. Simplify trigonometric expressions.
3. Develop additional trigonometric identities.
4. Solve trigonometric equations.
5
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Introduction
6
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Introduction cont’d
7
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Introduction
Pythagorean identities are sometimes used in radical form
such as
or
where the sign depends on the choice of u.
8
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Using the Fundamental Identities
9
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10
Using the Fundamental Identities
One common application of trigonometric identities is to
use given values of trigonometric functions to evaluate
other trigonometric functions.
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11
Example 1 – Using Identities to Evaluate a Function
Use the values and tan u > 0 to find the values of all six trigonometric functions.
Solution:
Using a reciprocal identity, you have
.
Using a Pythagorean identity, you have
Pythagorean identity
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12
Example 1 – Solution
cont’d
Substitute for cos u.
. Simplify.
Because sec u < 0 and tan u > 0, it follows that u lies in
Quadrant III.
Moreover, because sin u is negative when u is in
Quadrant III, choose the negative root and obtain
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13
Example 1 – Solution
Knowing the values of the sine and cosine enables you to
find the values of all six trigonometric functions.
cont’d
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Example 1 – Solution cont’d
14
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15
Example 2 – Simplifying a Trigonometric Expression
Simplify
sin x cos2 x – sin x.
Solution:
First factor out a common monomial factor and then use a fundamental identity.
sin x cos2 x – sin x = sin x(cos2 x – 1)
= –sin x(1 – cos2 x)
= –sin x(sin2 x)
= –sin3 x
Factor out common
monomial factor.
Factor out –1.
Pythagorean identity
Multiply.
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16
Using the Fundamental Identities
When factoring trigonometric expressions, it is helpful to
find a special polynomial factoring form that fits the
expression.
On occasion, factoring or simplifying can best be done by
first rewriting the expression in terms of just one
trigonometric function or in terms of sine and cosine only.
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17
Example 7 – Rewriting a Trigonometric Expression
Rewrite so that it is not in fractional form.
Solution:
From the Pythagorean identity
cos2 x = 1 – sin2 x = (1 – sin x)(1 + sin x)
multiplying both the numerator and the denominator by
(1 – sin x) will produce a monomial denominator.
Multiply numerator and
denominator by (1 – sin x).
Multiply.
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18
Example 7 – Solution
cont’d
Pythagorean identity
Write as separate fractions.
Product of fractions
Reciprocal and quotient
identities
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Analytic Trigonometry
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2.2 Verifying Trigonometric
Identities
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Objective
Verify trigonometric identities.
3
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Introduction
4
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5
Introduction
In this section, you will study techniques for verifying
trigonometric identities.
Remember that a conditional equation is an equation that is
true for only some of the values in its domain.
For example, the conditional equation
sin x = 0
is true only for
x = nπ
Conditional equation
where n is an integer. When you find these values, you are
solving the equation.
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6
Introduction
On the other hand, an equation that is true for all real values in the domain of the variable is an identity. For example, the familiar equation
sin2x = 1 – cos2x
Identity
is true for all real numbers x. So, it is an identity.
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Verifying Trigonometric Identities
7
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8
Verifying Trigonometric Identities
Although there are similarities, verifying that a trigonometric
equation is an identity is quite different from solving an
equation. There is no well-defined set of rules to follow in
verifying trigonometric identities, and it is best to learn the
process by practicing.
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9
Verifying Trigonometric Identities
Verifying trigonometric identities is a useful process when
you need to convert a trigonometric expression into a form
that is more useful algebraically.
When you verify an identity, you cannot assume that the
two sides of the equation are equal because you are trying
to verify that they are equal.
As a result, when verifying identities, you cannot use
operations such as adding the same quantity to each side
of the equation or cross multiplication.
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Example 1 – Verifying a Trigonometric Identity
Verify the identity
Solution:
Start with the left side because it is more complicated.
Pythagorean identity
Simplify.
Reciprocal identity
Quotient identity
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11
Example 1 – Solution
cont’d
Simplify.
Notice that you verify the identity by starting with the left
side of the equation (the more complicated side) and using
the fundamental trigonometric identities to simplify it until
you obtain the right side.
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Analytic Trigonometry
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2.3 Solving Trigonometric
Equations
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Objectives
Use standard algebraic techniques to solve
trigonometric equations.
Solve trigonometric equations of quadratic type.
Solve trigonometric equations involving multiple
angles.
Use inverse trigonometric functions to solve
trigonometric equations.
3
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Introduction
4
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5
Introduction
To solve a trigonometric equation, use standard algebraic
techniques (when possible) such as collecting like terms
and factoring.
Your preliminary goal in solving a trigonometric equation is
to isolate the trigonometric function on one side of the
equation.
For example, to solve the equation 2 sin x = 1, divide each
side by 2 to obtain
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Introduction
To solve for x, note in the figure below that the equation
has solutions x = π /6 and x = 5π /6 in the interval
[0, 2π).
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Introduction
Moreover, because sin x has a period of 2π, there are
infinitely many other solutions, which can be written as
and General solution
where n is an integer, as shown above.
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8
Introduction
The figure below illustrates another way to show that the equation has infinitely many solutions.
Any angles that are coterminal with π /6 or 5π /6 will also be
solutions of the equation.
When solving trigonometric
equations, you should write
your answer(s) using exact
values rather than decimal
approximations.
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9
Example 1 – Collecting Like Terms
Solve
Solution:
Begin by isolating sin x on one side of the equation.
Write original equation.
Add sin x to each side.
Subtract from each side.
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10
Example 1 – Solution
cont’d
Combine like terms.
Divide each side by 2.
Because sin x has a period of 2π, first find all solutions in the interval [0, 2π).
These solutions are x = 5π /4 and x = 7π /4. Finally, add multiples of 2π to each of these solutions to obtain the general form
and
General solution
where n is an integer.
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Equations of Quadratic Type
11
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12
Equations of Quadratic Type
Many trigonometric equations are of quadratic type
ax2 + bx + c = 0, as shown below.
Quadratic in sin x Quadratic in sec x
2 sin2 x – sin x – 1 = 0 sec2 x – 3 sec x – 2 = 0
2(sin x)2 – sin x – 1 = 0 (sec x)2 – 3(sec x) – 2 = 0
To solve equations of this type, factor the quadratic or,
when this is not possible, use the Quadratic Formula.
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13
Example 4 – Factoring an Equation of Quadratic Type
Find all solutions of 2 sin2 x – sin x – 1 = 0 in the interval
[0, 2π).
Solution:
Treat the equation as a quadratic in sin x and factor.
2 sin2 x – sin x – 1 = 0
(2 sin x + 1)(sin x – 1) = 0
Write original equation.
Factor.
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14
Example 4 – Solution
Setting each factor equal to zero, you obtain the following
solutions in the interval [0, 2π).
2 sin x + 1 = 0 and sin x – 1 = 0
cont’d
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Functions Involving Multiple Angles
15
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16
Functions Involving Multiple Angles
The next example involve trigonometric functions of
multiple angles of the forms cos ku and tan ku.
To solve equations of these forms, first solve the equation
for ku, and then divide your result by k.
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17
Example 7 – Solving a Multiple-Angle Equation
Solve 2 cos 3t – 1 = 0.
Solution:
2 cos 3t – 1 = 0
2 cos 3t = 1
Write original equation.
Add 1 to each side.
Divide each side by 2.
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18
Example 7 – Solution
cont’d
In the interval [0, 2π), you know that 3t = π /3 and 3t = 5π /3
are the only solutions, so, in general, you have
and
Dividing these results by 3, you obtain the general solution
and General solution
where n is an integer.
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Using Inverse Functions
19
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20
Example 9 – Using Inverse Functions
sec2 x – 2 tan x = 4
1 + tan2 x – 2 tan x – 4 = 0
tan2 x – 2 tan x – 3 = 0
(tan x – 3)(tan x + 1) = 0
Original equation
Pythagorean identity
Combine like terms.
Factor.
Setting each factor equal to zero, you obtain two solutions
in the interval (–π /2, π /2). [We know that the range of the
inverse tangent function is (–π /2, π /2).]
x = arctan 3 and
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Example 9 – Using Inverse Functionscont’d
Finally, because tan x has a period of π, you add multiples
of π to obtain
x = arctan 3 + nπ and
where n is an integer.
General solution
You can use a calculator to approximate the value of
arctan 3.
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Analytic Trigonometry
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2.4 Sum and Difference Formulas
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3
Objective
Use sum and difference formulas to evaluate
trigonometric functions, verify identities, and
solve trigonometric equations.
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4
Using Sum and Difference
Formulas
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5
Using Sum and Difference Formulas
In this and the following section, you will study the uses of
several trigonometric identities and formulas.
Example 1 shows how sum and difference formulas can
enable you to find exact values of trigonometric functions
involving sums or differences of special angles.
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6
Example 1 – Evaluating a Trigonometric Function
Find the exact value of .
Solution:
To find the exact value of sin π /12, use the fact that
Consequently, the formula for sin(u – v) yields
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7
Example 1 – Solution
cont’d
Try checking this result on your calculator. You will find that
sin π /12 ≈ 0.259.
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8
Example 5 – Proving a Cofunction identity
Use a difference formula to prove the cofunction identity
= sin x.
Solution:
Using the formula for cos(u – v), you have
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Using Sum and Difference Formulas
Sum and difference formulas can be used to rewrite
expressions such as
and , where n is an integer
as expressions involving only sin θ or cos θ.
The resulting formulas are called reduction formulas.
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10
Example 7 – Solving a Trigonometric Equation
Find all solutions of
in the interval .
Solution:
Algebraic Solution
Using sum and difference formulas, rewrite the equation as
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11
Example 7 – Solution
So, the only solutions in the interval [0, 2π) are x = 5π /4
and x = 7π /4.
cont’d
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Analytic Trigonometry
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2.5 Multiple-Angle and
Product-to-Sum Formulas
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3
Objectives
Use multiple-angle formulas to rewrite and
evaluate trigonometric functions.
Use power-reducing formulas to rewrite and
evaluate trigonometric functions.
Use half-angle formulas to rewrite and evaluate
trigonometric functions.
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4
Objectives
Use product-to-sum and sum-to-product
formulas to rewrite and evaluate trigonometric
functions.
Use trigonometric formulas to rewrite real-life
models.
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Multiple-Angle Formulas
5
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6
Multiple-Angle Formulas
In this section, you will study four other categories of
trigonometric identities.
1. The first category involves functions of multiple angles
such as sin ku and cos ku.
2. The second category involves squares of trigonometric
functions such as sin2 u.
3. The third category involves functions of half-angles such
as sin(u / 2).
4. The fourth category involves products of trigonometric
functions such as sin u cos v.
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7
Multiple-Angle Formulas
You should learn the double-angle formulas because
they are used often in trigonometry and calculus.
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8
Example 1 – Solving a Multiple-Angle Equation
Solve 2 cos x + sin 2x = 0.
Solution:
Begin by rewriting the equation so that it involves functions
of x (rather than 2x). Then factor and solve.
2 cos x + sin 2x = 0
2 cos x + 2 sin x cos x = 0
2 cos x(1 + sin x) = 0
2 cos x = 0 and 1 + sin x = 0
Write original equation.
Double-angle formula.
Factor.
Set factors equal to zero.
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Example 1 – Solution
cont’d
So, the general solution is
and
Solutions in [0, 2π)
where n is an integer. Try verifying these solutions
graphically.
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Power-Reducing Formulas
10
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11
Power-Reducing Formulas
The double-angle formulas can be used to obtain the
following power-reducing formulas.
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12
Example 4 – Reducing a Power
Rewrite sin4 x in terms of first powers of the cosines of
multiple angles.
Solution:
Note the repeated use of power-reducing formulas.
Property of exponents
Power-reducing formula
Expand.
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13
Example 4 – Solution
cont’d
Power-reducing formula
Distributive Property
Simplify.
Factor out common factor.
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14
Example 4 – Solution
You can use a graphing utility to check this result, as
shown below. Notice that the graphs coincide.
cont’d
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Half-Angle Formulas
15
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16
Half-Angle Formulas
You can derive some useful alternative forms of the
power-reducing formulas by replacing u with u / 2. The
results are called half-angle formulas.
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17
Example 5 – Using a Half-Angle Formula
Find the exact value of sin 105°.
Solution:
Begin by noting that 105° is half of 210°. Then, using the
half-angle formula for sin(u / 2) and the fact that 105° lies in
Quadrant II, you have
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18
Example 5 – Solution
.
cont’d
The positive square root is chosen because sin θ is positive
in Quadrant II.
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Product-to-Sum Formulas
19
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20
Product-to-Sum Formulas
Each of the following product-to-sum formulas can be
verified using the sum and difference formulas.
Product-to-sum formulas are used in calculus to solve
problems involving the products of sines and cosines of two
different angles.
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21
Example 7 – Writing Products as Sums
Rewrite the product cos 5x sin 4x as a sum or difference.
Solution:
Using the appropriate product-to-sum formula, you obtain
cos 5x sin 4x = [sin(5x + 4x) – sin(5x – 4x)]
= sin 9x – sin x.
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22
Product-to-Sum Formulas
Occasionally, it is useful to reverse the procedure and write
a sum of trigonometric functions as a product. This can be
accomplished with the following sum-to-product
formulas.
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Application
23
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24
Example 10 – Projectile Motion
Ignoring air resistance, the range of a projectile fired at an
angle θ with the horizontal and with an initial velocity of
v0 feet per second is given by
where r is the horizontal distance (in feet) that the projectile
travels.
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25
Example 10 – Projectile Motion
A football player can kick a football
from ground level with an initial
velocity of 80 feet per second
a. Write the projectile motion model in a simpler form.
b. At what angle must the player kick the football so that
the football travels 200 feet?
cont’d
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26
2
Example 10 – Solution
a. You can use a double-angle formula to rewrite the projectile motion model as
r = v02 (2 sin θ cos θ)
= v02 sin 2θ.
Rewrite original projectile motion model.
Rewrite model using a double-angle formula.
b. r = v0 sin 2θ
Write projectile motion model.
200 = (80)2 sin 2θ
Substitute 200 for r and 80 for v0.
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