Download 6.5 Assess Your Understanding

490
CHAPTER 6 Analytic Trigonometry
Then
1 — cos a
- =
sin a
ia
a
2 sin —
sm —
2
2
a
=
= tan —
a
a
a
2
2 sm — cos —cos —
Since it also can be shown that
sin a
1 - cos a
sin a
1 + cos a
we have the following two Half-angle Formulas:
a
Half-angle Formulas for tan —
a
1 - cos a
tan — = sm a
sin a
1 + cos a
(11)
With this formula, the solution to Example 6(c) can be obtained as follows
3
37T
cos a = -- 7 T < a < . —
sin a = - V1 -
4
116
S 2 a = - , / l - — = -J~= -'?
25
Then, by equation (11),
a
1 — cos a
tan - = sm a
\
= -2
4
6.5 Assess Your Understanding
Concepts and Vocabulary
1. cos(20) = cos2 8 -
3. tan - =
4. True or False: cos(20) has three equivalent forms:
cos2 0 - sin2 6, 1 - 2 sin2 9, 2 cos2 9 - 1
5. True or False: sin(2#) has two equivalent forms:
2 sin 8 cos 8 and sin2 8 - cos2 0
6. True or False: tan(20) + tan(20) = tan(46»)
-1 = 1-
1 - cos 6
Skill Building
In Problems 7-18, use the information given about the angle 8, 0 s 0 < 2-n-, to //nrf ?Ae ejcacf value of
(a) sin(20)
\.
(b) cos(20)
sin 0 = -, 0 < 6 < —
10. tan 0 = -,
2
TT < 0 < —
2
(c) sin-
(d) cos-
8. cos 8 = -, 0 < 8 < ~
V6 TT
11. cos 9 = --»-| -r < 9 < TT
4
37T
9. tan 6> = -, TT < 0 < —
12. sin 0 = - ——, ~ < 6 <2-!T
SECTION 6.5
491
15. cot 0 = -2, sec 6 < 0
18. cot 0 = 3, cos 0 < 0
14. csc0 = -V5, cos0 < 0
17. tan 6 = -3, sin 0 < 0
13. sec 0 = 3, sin 6 > 0
16. sec 9 = 2, esc 6 < 0
Double-angle and Half-angle Formulas
I InProblems 19-28, use the Half-angle Formulas to find the exact value of each trigonometric function.
119. sin 22.5°
20. cos 22.5°
24. sin 195°
25. sec
15ir
21. tan —-
22. tan-r
26. csc—-
27. sin( -
23. cos 165C
28. cosf-^
o
1
3 1
y. Show that sin4 0 = - - - cos(20) + - cos(40).
o 2
o
30. Show that sin(40) = (cos0)(4sin0 - 8sin 3 0).
31. Develop a formula for cos (30) as a third-degree polynomial
in the variable cos 0.
32. Develop a formula for cos (40) as a fourth-degree polynomial in the variable cos 0.
33. Find an expression for sin(50) as a fifth-degree polynomial
in the variable sin 0.
34. Find an expression for cos(50) as a fifth-degree polynomial
in the variable cos 0.
In Problems 35-56, establish each identity.
35. cos4 6 - sin 4 0 = cos(20)
38. cot(20) = -(cot 0 - tan 0)
. cot 0 - tan e
36. - = cos(20)
cot 0 + tan 0
sec 2 0
39. sec(20) =
2 - sec2 0
37. cot(20) =
cot2 0 - 1
2cot0
40. csc(20) = - sec 0 csc 0
cos(20)
1 + sin(20)
cot 0 - 1
cot 0 + 1
41. cos2(20) - sin2(20) = cos(40)
42. (4 sin 0 cos 0)(1 - 2 sin2 0) = sin(40)
43.
44. sin2 0 cos2 0 = -[1 - cos(40)]
2
45. secz- =
2 1 + cos 0
2
46. csc - = ,
2 1 - cos I
48. tan - = csc 0 - cot 0
49. cos 0 =
,0
2
1-tan2^
sec 0 + 1
sec 0 - 1
47. cot2 - = -
1 + tan 2 sm(30)
cos(30)
51. —— - -—• = 2
sin 0
cos 0
V
3 tan 0 - tan3 0
V 53. tan (36) = —:—
1 - 3 tan2 0
1 .
sin3 0 + cos3 0
50. 1 - - sm(20) = sin 6 + cos 0
„ cos 0 + sin 0 cos 0 - sin 0
52. - = 2 tan(20)
cos 0 - sin 0 cos 0 + sin 0
54. tan0 + tan(0 + 120°) + tan(0 + 240°) = 3 tan(36)
55. In|sin0| = -(ln|l - cos(20)| - In 2)
56. In|cos0| = |(ln|l + cos(20)| - In 2)
In Problems 57-68, find the exact value of each expression.
A/3
57. sin! 2 sin ' 58. sin 2 sin )
2 J
61. tan 2 cos"1 (
65. sin2! — cos"1
-1)1
!)
62. tan! 2 tan"1
!)
I)
66. cos2! — sin"1
59. cosl 2 sin
l
63. sin! 2 cos"1
1
4'
5,
67. seel 2 tan"1 |
60. cos( 2 cos ' 64. cos 2 tan"1 ( -~
I
V ~
68. csc 2 sin"11 --
Applications and Extensions
69. If x = 2 tan 0, express sin(20) as a function of x.
70. If x — 2 tan 0, express cos (20) as a function of x.
492
CHAPTER 6 Analytic Trigonometry
71. Find the value of the number C:
72. Find the value of the number C:
— sin2 x + C = - — cos(2.r)
— cos2 x + C = — cos(2:c)
73. If z = tan —, show that sin a = — —r.
2
1+ z
74. If z = tan —, show that cos a =
75. Area of an Isosceles Triangle Show that the area A of an
isosceles triangle whose equal sides are of length s and 6 is
the angle between them is
a
n
82. If tan 9 = a tan — , express tan — in terms of a.
-s 2 sin0
[Hint: See the illustration. The height h bisects the angle 8
and is the perpendicular bisector of the base.]
83. Projectile Motion An object is propelled upward at an
angle 9, 45° < 6 < 90°, to the horizontal with an initial velocity of Vo feet per second from the base of a plane thai
makes an angle of 45° with the horizontal. See the illustration. If air resistance is ignored, the distance R that it travels
up the inclined plane is given by
R=
76. Geometry A rectangle is inscribed in a semicircle of radius 1. See the illustration.
(a) Show that
(a) Express the area A of the rectangle as a function of the
angle 6 shown in the illustration.
(b) Show that A = sin (20).
(c) Find the angle d that results in the largest area A.
(d) Find the dimensions of this largest rectangle.
1 - cos(2x)
77. Graph f ( x ) = sin2 x = for 0 < x s 277 by
using transformations.
78. Repeat Problem 77 for g ( x ) = cos2 x.
R =
[sin(20) - cos(20) - 1]
(b) Graph R = R(0). (Use v0 = 32 feet per second.)
(c) What value of 0 makes R the largest? (Use vn = 32 feel
per second.)
84. Sawtooth Curve An oscilloscope often displays a saw
tooth curve. This curve can be approximated by sinusoida
curves of varying periods and amplitudes. A first approximation to the sawtooth curve is given by
y = -sin(277x) + -sin(477z)
79. Use the fact that
cos ^
77
= !(v^ +
Show that y = sin(277.x) cos2(77x).
77
to find sin — and cos —.
80. Show that
and use it to find sin — and cos —.
16
16
81. Show that
sin 3 6» + sin3(0 + 120°) + sin3(6» + 240°) = --sin(30)
Discussion and Writing
85. Go to the library and research Chebyshev polynomials
Write a report on your findings.