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Numbers

In: Business and Management

Submitted By lolabbunny
Words 557
Pages 3
7.1
2. Fill in the blank If cosθ=-.65, cos-θ= -.65
8. Find sin. θ in quadrant IV cotθ=-13 1+cot2θ=csc2θ
1+-132=csc2θ
1+19=csc2θ
103=cscθ
sinθ=1cscθ=1103=-310=-31010
26. Find the five remaining trig functions. cosθ=15, θ in quadrant I sin2θ=1-cos2θ=1-152=1-125=2425=265 secθ=1cosθ=115=5 tanθ= sinθcosθ=26515=26 cotθ=cosθsinθ=15265=126=612 cscθ=1sinθ=12612=526=5612
34.Choose the expression that completes the identity. tanx=sinxcosx , (D.)
38. –tanxcosx=-sinxcosxcosx=-sinx=sin-x, (C.)
56. Write expression in terms of sin and cos, simplify so that no quotients appear in the final expression. cot2θ(1+tan2θ) cos2θsin2θ1cos2θ
1sin2θ=csc2θ

7.2
2. Secxcscx+cscxsecx
1cosx1sinx+1sinx1cosx
sinxcosx+cosxsinx sin2x+cos2xcosxsinx 1cosxsinx=secx

18. 4tan2β+tanβ-3=(tanβ+1)(4tanβ-3)
26. cott+tant=1tant(tant)=1
36. tanαsecα=sinα
* working w/ left side tanαsecα=sinαcosα1cosα=sinα 46. cosθsinθcotθ=1
*working w/ left side cosθsinθcotθ=cosθsinθ1cotθ=1tanθtanθ=1 68. sinθ+cosθ=sinθ1-cosθsinθ+cosθ1-sinθcosθ
84. 1+cot2x=sec2xsec2x

7.3
6. cos-15° cos30°-45° cos30°cos45°+sin30°sin45°
3222+1222
5+24
72. tanπ4+x=tanπ4+tanx1-tanπ4tanx=1+tanx1-tanx
106. sins+tcosscost=tans+tant
* working w/left side sinscost+cosssintcosscost sinscostcosscost+cosssintcosscost sinscoss+sintcost=tans+tant 7.4
20. 1-2sin2221°2=cos[1(221°2)]=cos45°=22
30. cos3x cos2x+x cos2xcosx-sin2xsinx

56. 8sin7xsin9x
8×12[cos7x-9x-cos(7x+9)]
4cos-2x-cos16x
4cos2x-cos16x
58. cos5x+cos8x
2cos(5x+8x2)cos(5x-8x2)
2cos13x2cos(-3x2)
2cos13x2cos(3x2)
72. tan195°=tan390°2=1-3212=2-3212=2-3
112. sinθ2=1m;m=32,sin2θ2=1-cosθ2 sinθ2=132=23 232=1-cosθ2
49=1-cosθ2
89=1-cosθ
-19=-cosθ
cos-1-19≈96°

7.5
20. y=arcsin(-32) siny=-32 y=-π3
28. y=sec-1(-2) secy=-2 y=3π4
38. θ=arcsin(-22) sinθ=-22 θ=-45°
62. y=arccos .44624593 cos-1 .44624593=y y=1.108230308 26
80. sec(sin-1(-15)) ; quadrant IV, *let A=sin-1(-15), so sinA=-15 sec(sin-1(-15)) =secA=526=5612

A
-1
5

94. sin(cos-1.25)=.9682458366

7.6
2. sinx+2=3 solve by linear method
20. 2cos2x-3cosx=0 cosx(2cos-3)=0 cosx=0 cosx=32 solution set:π2,3π2,π6,11π6
28. cos2θ=sin2θ+1 sin2θ-cos2θ+1=0 sin2θ-(-sin2θ)=1
2sin2θ=0
sinθ=0 solution set:90°, 270°
36. 4cos2θ+4cosθ=1
4cos2θ+4cosθ-1=0
cosθ=-4±42-44-124 cosθ=-4±328 cosθ=-4±428 solution set:78°, 282°
44. 2cos2θ+cosθ=1
2cos2θ+cosθ-1=0
cosθ=-1±12-42-122 cosθ=-1±94 cosθ=-1±34 solution set:π3+nπ,5π3+nπ, π+nπ,3π2+nπ

64. 23sin2x=3
2sin2x=0
sin2x=0 solution set:{π2, 0, π,3π2}
76. -2cos2θ=3 cos2θ=-32 solution set:{75°, 105°, 225°, 285°}

7.7
6. 4y=sinx x=sin-14y 28. 4π+4tan-1y=π
4tan-1y=-3π
tan-1y=-3π4 tan-3π4=y y=1
A
12
13
5
32. arctanx=arccos513 *let arccos513=A arctanx=A cosA=513 tanA=x x=125

38. arccosx+2arcsin32=π3 arccosx=π3-2arcsin32 arccosx=π3-2π3 arccosx=π3-2π3 arccosx=-π3 x=cos(-π3) x=12

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