Answer Key for Trig Test
Honors Math 4, Ms. Tracy
November 2016
Problem 1
A)
sin−1 ( 12 ) =
π
6
B) cos−1 (− √12 ) =
C)
cot−1 (−
D)
π
6
E)
π
6
F)
5π
6
√
3
3 )
=
3π
4
2π
3
Problem 2
θ = sin−1 ( 27 ) − 2π and θ = − sin−1 ( 27 ) − π
Problem 3
cot−1 (x) = y
cot(y) = x
1
=x
tan(y)
tan(y) = x1
tan−1 ( x1 ) = y
∴ cot−1 (x) = tan−1 ( x1 ) by the transitive property.
This principle holds true ALGEBRAICALLY for the other functions:
sec−1 (x) = cos−1 ( x1 )
csc−1 (x) = sin−1 ( x1 )
1
Problem 4
For this problem, n ∈ R
4 csc(3πx + 6) + 7 = −1
A)
csc(3πx + 6) = −2
3πx + 6 =
x=
7
18
−
2
π
7π
6
+
+ 2πn and 3πx + 6 =
2n
3
and x =
11
18
−
2
π
+
11π
6
2n
3
+ 2πn
B) tan2 (5x) = 3
√
tan(5x) = ± 3
π
2π
3 + πn and 5x = 3 + πn
π
πn
2π
πn
15 + 5 and x = 15 + 5
5x =
x=
C)
2 sin(θ) cos θ − cos θ + 2 sin θ − 1 = 0
cos θ(2 sin θ − 1) + 1(2 sin θ − 1) = 0
(2 sin θ − 1)(cos θ + 1) = 0
(2 sin θ − 1) = 0 and (cos θ + 1) = 0
sin θ =
θ=
D)
π
6
1
2
and cos θ = −1
+ 2πn and θ =
5π
6
+ 2πn and θ = π + 2πn
sin x = −5/6
Because sin x is negative in quadrants III and IV, we use the relationship that sin x = sin(π − x)
x = sin−1 (− 65 ) + 2πn and x = π − sin−1 (− 56 ) + 2πn
E) sec2 (3x) = −6 sec(3x)
sec2 (3x) + 6 sec(3x) = 0
sec 3x(sec(3x) + 6) = 0
(note: sec(3x) could never be 0.)
sec(3x) + 6 = 0
cos(3x) = − 16
3x = 2πn ± cos−1 ( 16 )
x=
2πn
3
±
cos−1 ( 16 )
3
F)
tan(4x) = 6
x=
tan−1 (6)
4
+
πn
4
Problem 5
For this problem, see ”Supplemental Graphs” or please check your work in Desmos, or with a calculator, or
with any other graphing implement.
2
Problem 6
A)
Period =
π
157 ,
Maximum = 10, Minimum = 4
B) Period = 10, Maximum = 96, Minimum = −102
Problem 7
A)
Period = 6π, Asymptotes:
B) Period =
π
6,
Asymptotes:
9π
4
πn
6
+ 3πn
−
π
2
Problem 8
y
F
B
π−θ
E
θ
O
θ
A
x
A)
For this problem, we are on the unit circle. Therefore, OF = OB = 1. The coordinates of Point B are
(cos(θ), sin(θ)). The coordinates of Point F are (cos(π − θ), sin(π − θ)). ∠F OE ∼
= ∠BOA since they are both
∼
= θ. These triangles are both right triangles, therefore ∠OEF = ∠OAB. Triangle OAB and triangle OEF
3
are congruent by AAS. Because of the congruence, we could also describe the coordinates of point F as the
reflection of point B over the y axis, or (− cos(θ), sin(θ)). This works because both triangles are congruent,
and F must have the same y-coordinate as B and the negative x-coordinate of B. Since both order pairs
represent the same point, they must agree. Therefore, cos(π − θ) = (− cos(θ)).
B) Confirm cos(π − 0) = (− cos(0)). cos(π) = −1. − cos(0) = −1. ∴ cos(π − θ) = (− cos(θ)) if θ = 0
Confirm cos(π − π/2) = (− cos(π/2). cos(π − π/2) = 0. − cos(π/2) = 0. ∴ cos(π − θ) = (− cos(θ)) if θ = π/2
Problem 9
Answers will vary, samples provided:
y = 3 cos( 12 (x + π)) − 4
y = 3 sin( 12 (x + 2π)) − 4
y = −3 sin( 12 (x)) − 4
y = −3 cos( 12 (x − π)) − 4
Problem 10
A local min occurs at x = 8, because it’s half-way between the x-intercepts. Therefore the period is
local max occurs at x = −16. Therefore, the phase shift for cosine is left 16.
Use (−4, 7) to see that k = 7, and (0, 0) to see that A = 14.
π
f (x) = 14 cos( 24
(x + 16)) + 7
Problem 11
1
√
1 − x2
cos−1 (x)
x
A)
tan(cos
−1
√
(x)) =
1−x2
x
1
√
1−x
√
cos−1 ( x)
√
x
B)
√
√
sin(cos−1 x) = 1 − x
4
π
24
The
Problem 12
• Right
π
2.
Formula y = cos(x − π2 ) = sin x
• Take reciprocal. Formula y =
1
sin x
= csc x.
• Right 2. Formula y = csc(x − 2).
• Horizontal shrink by 13 . Formula y = csc(3x − 2).
Problem 13
See supplementary graphs.
A)
Inverse, then involute.
B) Reciprocal, then inverse.
C)
sin−1 ( x1 ) = csc−1 (x). See question 3.
Problem 14
A)
π
See supplemental graphs. y = 4000 cos( 45
(t − 10))
B) When t = 25, y = 2000. When t = 41, y = −2237. When t = 163, y = −1236.
C)
When y = −1600, the first time this happens is t = 38.39. Then t = 71.61, and t = 128.39.
D)
When t = 0, y = 3064. So Cape Canaveral is 3064 km from the equator.
E)
On your own :)
5