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Exam 4 Review- Solutions to selected Practice Test Problems
Sec 5.3 to 6,3
rev1012 6th ed
d)Prove the identity: Put Statement on left |
Math 170
Reason on right
d) 1. 2 cot 2x = cot x – tan x
_____________________________________________________
1.
2.
3.
2
tan(2 x)
1. cot x =
2
2 tan( x)
1 − tan 2 ( x)
2. tan 2x=
1 − tan 2 ( x)
tan( x)
4. algebra
5. cot x - tan(x)
5. cot(x) =
cos 2 y=
d) 2.
2 tan( x)
1 − tan 2 ( x)
3. simplify
1
- tan(x)
tan( x)
4.
1
tan( x)
1
tan( x)
1 + cos(2 y )
2
_____________________________________________________
1.
1 + (2cos 2 y - 1)
2
1. cos 2y=2cos2 y -1
2.
2cos 2 y
2
2. Simplify
3.
cos 2y
3. Algebra
1
d) 3. Prove the identity: Put Statement on left | Reason on right
2 sin2 x + 2 sin2x cos2x = 1- cos 2x
_____________________________________________________
1. 2 sin2 x (sin2 x + cos2 x)
1. Factor
2
2. 2 sin x
2. sin 2 x+-cos2 x=1
3. 2(
1 − cos(2 x)
)
2
3. cos 2x=1 -2 sin2 x or
sin2 x=
4. 1- cos 2x
1 − cos(2 x)
2
4. simplify
d) 5. cos 3A = 4 cos3A - 3 cos A
_____________________________________________________
1. cos (A +2A)
1. Algebra
2. cosAcos2A - sinAsin2A
2. cos(A ± B)=cosAcosB m sinAsinB
2
3. cosA(2cos A – 1)
3. cos 2A=2cos 2A – 1
- sinA(2sinAcosA)
sin 2A=2sinAcosA
3
4. 2 cos A- cosA
4. Algebra
2
- 2sin A cos A
5. 2 cos3A- cosA
5. sin 2 A= 1-cos2 A
- 2(1-cos2 A) cos A
6. 2 cos3A- cosA
6. Algebra
- 2cosA+2cos3 A
7. 4 cos3A - 3 cos A
7. Algebra
2
g) 1. sin2
x
2
=
csc( x) − cot( x)
2 csc( x)
_____________________________________________________
1
cos x
−
)
sin
x
sin
x
1.
2
sin x
1 − cos x
2.
2
x
3. sin2
2
(
1. csc(x) =
1
sin( x)
cot x =
cos( x)
sin( x)
2. Simplify
x 1 − cos x
2
2
3. sin2 =
j) Solve the trigonometric equations: for 0 o ≤ θ < 360 o or radians
1. 2 cos θ + 3 = 0
cos θ = - 3 /2
cos < 0 in QII and QIII
)
ref ∠ = θ = 30 o
o
o
θ = 150 , 210
j) 2. 5 cos θ + 12 = cos θ
ans. θ = 150 o , 210 o
ans. θ =
5π 7π
,
6
6
5 cos θ - cos θ =- 12
4 cos θ = - 12
cos θ = - 12 /4 = - 3 /2
θ=
5π 7π
,
QII and QIII as above
6
6
3
j) 3. ( cos x – 1) (2 cos x – 1) = 0
. ( cos x – 1)= 0
( cos x – 1)= 0
cos x= 1
x=0
ans. x = 0,
π
3
,
5π
3
,
5π
3
(2 cos x – 1) = 0
Unit circle x=1 point(1,0)
(2 cos x – 1) = 0
cos x = 1/2
cos > 0 in QI and QIV
)
ref ∠ = θ = 60 o
o
o
θ = 60 , 300
θ=
π
3
,
5π
3
ans. x = 0,
π
3
k) Solve more complicated trigonometric equations: Use double or half angle
formulas for 0 o ≤ θ < 360 o or radians
2. cos 2 θ
-
cos θ - 2 = 0
Use cos 2 θ = 2 cos2 θ - 1
(2 cos2 θ - 1)- cos θ - 2 = 0
2 cos2 θ - cos θ - 3 = 0
(2 cos θ – 3) (cos θ + 1) = 0
( 2cos θ – 3)= 0
cos θ = 3/2 Never
|cos θ | ≤ 1
ans. θ = π
( cos θ + 1) = 0
cos θ = - 1 Unit circle x=- 1 point(-1,0)
ans. θ = π
4
k) Solve more complicated trigonometric equations: Use double or half angle
formulas for 0 o ≤ θ < 360 o or radians
3. sin x – cos x = 2
ans. x = 135 o
sin x = cos x + 2
Square both sides
sin2 x = cos2 x +2 2 cos x + 2
Use sin 2x = 1 - cos2 x
1 - cos2 x = cos2 x +2 2 cos x + 2
0 =2 cos2 x +2 2 cos x + 1
Use quadratic formula a = 2 b = 2 2 c = 1
COS X=( - 2 2 ± 8 − 8 )/4
cos x = - 2 /2 X=135 o, 225 o Only x=135 o checks
4 . cos
θ
2
- cos θ = 0
ans. θ = 0 o , 240 o
θ
cos = cos θ
2
θ
Use cos =
2
1 + cos ϑ
2
1 + cos ϑ
= cos θ
2
Square both sides
1 + cos ϑ
2
= cos θ
2
0=2 cos2 θ - cos θ - 1
(2 cos θ +1)( cos θ - 1)= 0
( 2cos θ + 1)= 0
cos θ = -1/2
o
o
θ =120 ,240
Only X=240 o checks
( cos θ - 1) = 0
cos θ = 1 Unit circle x= 1 point(1,0)
o
θ =0
ans. θ = 0 o , 240 o
5