Solutions Manual 21. cos 0- cos 0 tan 0 =0 (60) cos 0(1 - tan 0) =0 COS 0 = 0 o = 90°,270° I-tanO=O tan 0 =1 o = 45°,225° Since tan 90° and tan 270° are undefined, the expression does not make sense when 0 = 90° or o = 270°. o = 45°,225° 22. sin 40 + 1=0 (52) sin 48 = -1 40 = 270°,630°,990°,1350° o = 67.5°,157.5°,247.5°,337.5° 23. -90° < 0 < 90° ~~ Arctan [tan (-120°)] = Arctan (tan 60°) = 60° 24. csc 2 (- 37r) + tan347r _ sin2 (_ 37r) ~~ 2 2 = (csc ~r + (tan 0)3 - (sin ~r = (1)2 + (0)3 - (1)2 = 0 25. (f - g)(5100) = sec 2 (510°) - tan 2 (510°) (24,48) = sec 2 150° - tan 2 150° = (-sec 30°)2 - (-tan 30°)2 (-~r-(-~r 4 3 ~= 1 3 26. 21n (x - 1) = In (3x - 5) (59) In (x - 1)2 = In (3x - 5) (x - 1)2 = 3x - 5 x 2 - 2x + 1 = 3x - 5 x 2 - 5x + 6 = 0 (x - 2)(x - 3) = 0 x = 2,3 Advanced Mathematics, Second Edition Problem Set 69 27. 3 log2 5 - log2 x = log- x (59) . 3 log2 5 = log2 x + log2 x log2 125 = Iog 2 x 2 125 = x 2 x = 5-JS 28. e3 In5 _ 1OIog3+2 log 2 _ In e- 3 (59) e ln53 _ 1OIog3+log22 - (-3) eln 125 _ 1OIog(3)(4) + 3 125 - 12 + 3 = 116 29. (53,56) 5 in. H = 6 sin 50° = 4.5963 in. Area = !(5)(4.5963) = 11.491 in. 2 2 11.491 in. 2 x 1 ft 2 ~ = 0.080 ft 2 144 in. 30. (a) antilog, 3 = 53 = 125 (67) (b) antilog j d = 3 4 = 81 Problem Set 69 1. (a) {ReTe = Dc (38) (b) RBTB = DB Te = 2 + T B , Te + TB = 6 Te + TB = 6 (2 + T B ) + TB = 6 2TB = 4 TB = 2 To =2 + TB =2 + (2) = 4 (a) ReTe = Dc Re(4) = 160 Re = 40mph (b) RBTB = DB R B (2) = 160 RB = 80mph 217
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~r - s3.amazonaws.com fileSolutions Manual 21. cos 0 - cos 0 tan 0 = 0 (60) cos 0(1 - tan 0) = 0 COS 0 = 0 o = 90°,270° I-tanO=O tan 0 = 1 o = 45°,225° Since tan 90° and tan 270°
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Solutions Manual
21. cos 0 - cos 0 tan 0 = 0(60) cos 0(1 - tan 0) = 0
COS 0 = 0o = 90°,270°
I-tanO=O
tan 0 = 1o = 45°,225°
Since tan 90° and tan 270° are undefined, theexpression does not make sense when 0 = 90° oro = 270°.
In a normal distribution, 68% of the data would liebetween -1 and 1, or within a of the mean. Since thedata is evenly distributed, 34% lie between -1 and O.Likewise, 95% of the data would lie between -2 and2, or within 20' of the mean. So, 47.5% lie betweeno and 2. Therefore, the percentage between -1 and 2,is 34% + 47.5% = 81.5%.
12. y = logs x(65) 3"
y
321
-1I I f I I • x
13. Function = cos x(66)
Centerline = 5
Amplitude = 4
Period = n
Phase angle =3n8
Coefficient = 2n = 2n
y = 5 + 4cos2(X + 3:)Advanced Mathematics, Second Edition
Problem Set 69
14. Function = sin ()(66)
Centerline = -3
Amplitude = 10
Period = 120°
Phase angle = 20°
360°Coefficient = -- = 3
120°y = -3 + 10 sin 3«() - 20°)
15. (6 cis 215°)(2 cis 205°) = 12 cis 420°
(64) = 12 cis 60° = 12(cos 60° + i sin 60°)
=12(±+i~)=6+6F3i
16. (a) 6 - 2i(64)
y
Ie::: ,~ I I • X
2
R = ~62 + (_2)2 =
26
() = 18.43°
2M = 6.32
tan ()
The polar angle is 360° - 18.43° = 341.57°.
6.32 cis 341.57°
(b) 5 cis (_ 1~n ) = 5 cis ( _ 5: )
5( n ., n)= - cos "4 + I Sill "4
= 5[(- '7) + {'7)] =5-5 5-5.
--+-12 2
17.(62)
(a) {ax + by = c(b) px + qy = d
-pea) -pax - pby = -pc
a(b) pax + qay = ady(aq - pb) = ad - pc
ad - pcy =
aq - pb
219
Problem Set 69
18. {y ~ (x - 3)2 + 1(56) y < X
(parabola)
(line)
Solutions Manual
The area is beneath the line and on or above theparabola.
y y=x/
.;
21.(60)
(tan e - .J3)(tan e + .J3) = 0
tane-.J3=O tane+.J3=O
tan e = .J3 tan e = -.J3
54321
e ,. 4,.3'3
. e = 2,. 5,.3 '3
~IIII.; 1 2 • x)I 3 4
19. x - y + 3 = 0(58)
y = x + 3Equation of the perpendicular line:
y = -x + b(8) = -(-3) + b
b = 5
y = -x + 5
Point of intersection:
-x + 5 = x + 3
2x = 2x = 1
y = x + 3 = (1) + 3 = 4
(1,4) and (-3,8)
D = ~[1 - (_3)]2 + (4 - 8)2 f32 4.fi
e ,. 2,. 4,. 5,.3'3'3'3
20.(56)
22. sin x + 2 sin2 x = 0(60)
sinx(1 + 2 sin x) = 0
sin x = 0x = 0,,.
+ 2 sin x = 0
2sinx =-1
H = 60 sin 70° = 56.3816 em
Asegment = Asector - Atriangle
= (1l00)(,.)(60)2 _ (!)(60)(56.3816)360° 2
= 3455.76 - 1691.45 = 1764.31 cm2
220
sinx =12
7,. 11,.6'6x =
x = 0," 7,. 11,., ,-6 6
23.(60)
2 1tanx--=O3
(tan x - ~ )(tan x + ~) = 0
1tan x - .J3
1tan x + .J3o o
tan x1
.J3
,. 7,.6'6 x
tan x1
-.J3
5,. 11,.6'6x =
,. 5,. 7,. 11,.x------
- 6' 6 ' 6' 6
24. 0° ~ e ~ 180°(32)
Arccos (cos 210°) = Arccos (- ~) = 150°
25.(48)
cot2 (-510°) - csc3 (-510°) - tan2 (-510°)
= cot2 (210°) - csc ' (210°) - tan2 (210°)
= (cot 30°)2 - (- csc 30°)3 - (tan 30°)2
= (.J3)2 - (_2)3 - (~ r1 2 32
= 3 - (-8) - - = 10- = -3 3 3
Advanced Mathematics, Second Edition
Solutions Manual Problem Set 70
26. [x - (-J2 + i)][ x - (-J2 - i)] = 0(46)
(a) 20 = ml000 + b-(b) -30 = -m3000 - b
-10 = -m2000
m = 0.005
(x - -J2 - i)(x - -J2 + i) = 0
x2 - -J2x + ix - -J2x + 2 - -J2 i. r;;2. ·2 0- IX + -YL.I - I =