251
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
a
127. sinA =- ==> a = c. sin A = 20 sin 28° ~ 9.39
129. a = ~/c2 - b2 = -~/12.542 - 6.22 ~ 10.90
c
b 6.2
sin B .... ==~ B ~ 29.63°
c 12.54
A = 90° - 29.63° = 60.37°
B = 90° -A° = 62°
b
cosA=- ==~ b=c.cosA~ 17.66
c
Verifying Trigonometric Identities
Section 5.2
[] You should know the difference between an expression, a conditional equation, and an identity.
[] You should be able to solve trigonometric identities, using the following techniques.
(a) Work with one side at a time. Do not "cross" the equal sign.
(b) Use algebraic techniques such as combining fractions, factoring expressions, rationalizing denominators,
and squaring binomials.
(c) Use the fundamental identities.
(d) Convert all the terms into sines and cosines.
Solutions to Odd-Numbered Exercises
csc2 x 1 sin x
cot X
sinE x cos x
=1
1. sin t csc t = sin
1
sin x ¯ cos x
CSCX " secx
7. tanEo+6=(secE0- 1)+6
5. COSEft- sinEft= (1 -sinEft)- sin2fl
= sec2 19 + 5
= 1 -- 2 sinEfl
sin x
9. cos x + sin x tan x = cos x + sin x ¯ ~
cos x
cos2 x + sin2 x
COS X
1
COS X
= sec X
11.
x
Yl
¸Y2
0.2
4.835
4.835
0.4
0.6
0.8
1.0
1.2
1.4
2.1785 1.2064 0.6767 0.3469 0.1409 0.0293
2.1785 1.2064 0.6767 0.3469 0.1409 0.0293
1
sec x tan x
cos x
= COS X * ~
sin x
COS2 x
sin x
1 - sin2 x
sin x
1
sin x
sin x
= csc x - sin x
13.
252
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
0.4
0.6
0.8
1.0
1.2
1
cscx - sinx = ~- sin x
sin x
1 - sin2 x
sin x
1.4
x
0.2
Yl
4.835
2.1785 1.2064 0.6767 0.3469 0.1409 0.0293
Y2
4.835
2.1785 1.2064 0.6767 0.3469 0.1409 0.0293
cos2 x
sin x
= COS X °
COS x
sin x
-" COSX " cotx
o
15.
o
iX
0.2
0.6
0.4
0.8
1.0
1.2
1.4
Yl
5,0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148
Y2
5.0335 2.5679 1.7710 1.3940 1.1884 1.0729 1.0148
’ COS X
sin x + cos x cot x = sin x + cos x .
sm x
2
2
_,_ sin x + cos x
sin x
1
sin x
= csc x
1.5
17.
x
0.2
0.4
0.6
0.8
1.0
1.2
1.4
Yl
5.1359 2.7880 2.1458
2.1995 2.9609 5.9704
Y2
5.1359 2.7880 2.i458
2.1995 2.9609 5.9704
o
~1 +~1 =c°tx+tanx
tanx cotx tanx.cotx
= cot x + tan x
o
19. The error is in line 1: cot(- x) #’cot x.
21. Missing step: (sec2x - 1)2 = (tan2 X)2 -" tarl4 x
23. sin~/2 x cos x - sin~/2 x cos x = sin1/2 x cos x(1 - sin2 x) = sint/2 x cos x ¯ cos2 x = coss x.,/~ffx
1
-- X SeCX = COtX" secx
27. see(-x.__._..~) = cos(-x) = sin(-x)
csc(-x)
1
cos(-x)
sin x
- ~ = - tan x
COS X
253
29.
cos(-0)
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
cos 0 1 + sin 0
1-sin0 1 +sin0
cos ~(1 + sin 0)
1 - sin2 0_
cos ~(1 + sin O)
cos2 ~
1 + sin O
1 + sin(-0)
~os 0
1
sin 0
cos 0 cos O
= sec 0 + tan 0
sin x cos y + cos x sin y
31.
cos x cos y - sin x sin y
33. tan x + cot y
--
tan x cot y
sin x cos y+ cos x sin y
cos x Cos y cos x cos y
cos x cos y
sin x sin y
cos x cos y cos x cos y
tan x + tan y
1 - tan x tan y
1
1
tan y + cot x
cot x tan y cot x ¯ tan y
= tan y + cot x
11o
1
cot x tan y cot x ¯ tan y
35. ~/~+sinO_ ~/~+sinO. l+sinO
-sinO
-sinO 1 +sinO
=~/(i
Note: Check your answer with a graphing utility. What
happens if you leave off the absolute value?
+ sin 0)2
-- sin2 0
cos2 O
1 + sin 0
37. cos2x+cos
-x =cos2x+sin2x= 1
¯
-x =secx.cosx= 1
41. 2 sec2 x -- 2 secx x sin2 x - sin2 x -- COS2 x = 2 sec2 x(1 - sin2x) - (sin2 x + cos2 x)
= 2 sec2 x(cos2 x) -- 1
1
=2"~" cos2x1
2
cos x
=2-1
=1
43. 2 + cos2x - 3 cos’ix = (1 - cos2 x)(2 + 3 cos2 x)
= sin2 x(2 + 3 cos2 x)
254
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
45. csc4 x - 2 csc2 x + 1 = (csc2 x - 1)2
= (cot2 x)2 = cot4 x
47. see4 0 - tan4 0 = (see2 0 + tall20)(sec2 0 -- tan2 O)
-- (1 + tanz 0 + tan2 0)(1)
= 1 + 2tan20
49.
sin/3 1 + cos/3
1 -cos/3 1 +cos/3
--
sin f!(1 + cos f!)
1 - cos2/3
sin/3(1 + cos/3)
sin2/3
--
1 + cos/3
sin/3
51. tan3 a - 1 (tan
a - 1)(tan2 a + tan a + ~1) =tan2a+tana+ 1
tana- 1
tana- 1
53. It appears that Yl = 1. Analytically,
1
1
tanx+ 1 +cotx+ 1
+
cotx + 1 tan x + 1 (cot x + 1)(tanx + 1)
.--.
tanx + cotx + 2
cotxtanx + cotx + tanx + 1
tan x -t- cot x ÷ 2
tanx+cotx+2
=1.
55. It appears that Yl -" Sin X. Analytically,
1
COS2 X -- 1 -- COS2 X -- sin2x
- sinx.
sin x
sin x
sin x
sinx
2
57. Inlcot 01 - _ Icos 0
- tnlsin 01
- lnlcos 0l - Inlsin 01
-2
59. -ln(1 + cos 0) = In(1 + cos 0)-1
= lnl +cosO 1 -cos
=ln
1 - cos 0
1 - cos2 0
1 - cos 0
=In sin2 0
= In(1 - cos 0) - In sin2 0
= In(1 - cos o) - 2 Inlsin ol
61. sin2 25° + sin2 65° = sin2 25° + cos2 25° = 1
255
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
63. cos2 20° + cos2 52° + cos~ 38° + cos2 70° = c°s2 20° + c°s2 522 + sin2(90° - 38°) + sin2(90° - 70°)
= cos2 20° + cos~ 522 + sin252° + sinz 20°
= (cos2 20° + sin2 20°) + (cos~ 52° + sin~ 52°)
=1+1
=2
65. tanS x = tanax " tan2 x
= tan3 x(sec2 x - 1)
= tan3 x sec2 x -- tan3 x
67. (sinz x - sin~ x)cos x = sin~ x(1 - sinz x)cos x
69. /zW cos 0 = W sin 0
71. cos x - csc x. cot x = cos x
W sin 0 sin 0
tan 0, W 4:0
/z W cos 0 cos 0
-- sin2X ° COS2 X ° COS X
= COS3 X sin2 x
1 cos x
sin x sin x
sina
- cos x(1 - csc2 x)
cos
= cos x(-cot~ x)
73. True. f(x) = cos x and g(x) = sec x are even
75. False. For example, sin(l,z) 4: sinz (1)
79° ~/sin~ x + cos2x 4: sin x + cos x
The left side is 1 for any x, but the right side is not
necessarily 1. For example, the equation is not tree
for x = 7r/4.
+ 1)~! = sin[51-(12n~r + ~r)l
$1. sinI(12n 6
= sin(2n,rr + -~)
"tr 1
= sin 62
= ~1for all integers
Thus, sin[(12n
6 ÷ 1)qr]
¯
83. (x- i)(x + i)(x- 4i)(x + 40 = (x~ + 1)(x2 + 16)
=x*+ 17xz+ 16
256
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
87. f(x) = -2x-3
89, f(x) = 5-x - 2
y
y
246
91. s = rO
0-
s 26
r 11
95. Quadrant III
93. Quadrant III
~ 2.3636 radians
Section 5.3
Solving Trigonometric Equations
[] You should be able to identify and solve trigonometric equations.
[] A trigonometric equation is a conditional equation. It is true for a specific set of values.
[] To solve trigonometric equations, use algebraic techniques such as collecting like terms, taking square roots,
factoring, squaring, converting to quadratic form, using formulas, and using inverse functions. Study the
examples in this section.
I Use your graphing utility to calculate solutions and verify results.
Solutions to Odd-Numbered Exercises
1. 2cosx- 1 =0
3. 3tan22x- 1 =0
(a) 2cos~- 1 =2 - 1 =0
(a) 3 tan\-~-/j - 1 = 3~an2-~- 1
(b) 2cos’~’-- 1 = 2 - 1 =0
=3
-1
=0
~_~ -1 =3tan~ - 1
(b) [3(lO’n’]]
tank 212]J
=0