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Exam 2
Math 1151, Precalculus 2
Name:
Summer 2012
Instructions.
1. Show all the steps of your work clearly.
2. No graphing calculators, iPods, phones or computers
Question Points Your Score
Q1
30
Q2
25
Q3
15
Q4
10
Q5
20
Q6
20
Q7
20
Q8
20
Q9
20
TOTAL
180
Q1]. . . [30 points]
Let f (x) = −2 sin(2πx + π) − 1.
(a) What is the amplitude? 2
(b) What is the period T ? 1
(c) What is the phase shift
φ
ω
? −1/2
(d) Graph one cycle of f (x) on the interval [ ωφ , ωφ + T ] and label the five key points.
We are going to graph one cycle on the interval [−1/2, 1/2] To get the five key points we need to divide
[−1/2, 1/2] into four equal subintervals: [−1/2, −1/4]∪[−1/4, 0]∪[0, 1/4]∪[1/4, 1/2]. The endpoints give us
the x-coordinates of the key points. So the key points are: (−1/2, −1), (−1/4, −3), (0, −1), (1/4, 1), (1/2, −1).
Q2]. . . [25 points]
Let f (x) = −4 sec(4x).
(a) What is the period of the function? T = π/2
(b) What are the asymptotes in the interval (−π, π)?
The asymptotes occur when cos(4x) = 0 and this happens when 4x =
x=
πk
2
π
2
+ πk where k ∈ Z. Solving for
+ π8 . So the asymptotes in the interval (−π, π) are: ± π8 , ± 3π
, ± 5π
, ± 7π
.
8
8
8
Q3]. . . [15 points]
Find the exact value of cot(sec−1 (−3))? (DO NOT write the answer as a decimal)
Let θ = sec−1 (−3) so that sec θ = −3 where 0 ≤ θ ≤ π but θ cannot be π/2. Then cos θ = − 31 , and thus
√
√
√
p
θ
θ ∈ ( π2 , π]. A trigonometric identity gives us sin θ = 1 − 1/9 = 38 . Then cot θ = cos
= − 88 = − 42 .
sin θ
Q4]. . . [10 points]
What is the sum formula for cos(α + β)? cos(α + β) = cos α cos β − sin α sin β
Q5]. . . [20 points]
Given that tan α =
5
12
where π < α <
3π
2
and sin β = − 12 where π < β <
3π
,
2
find cos(α + β) using the
formula in Q4?
p
We need to find cos α, cos β and sin α. Using a trigonometric identity we find that cos β = − 1 − 1/4 =
√
−
3
.
2
Now to find the other two trig functions we let y = −5 and x = −12 from tan α and the location of
5
the angle α. Then r = 13 and cos α = − 12
and sin α = − 13
. So cos(α + β) =
13
√
12 3−5
.
26
Q6]. . . [20 points]
Write the trigonometric expression cot(csc−1 u) as an algebraic expression in u.
√
Let θ = csc−1 u so that csc θ = u where −π
≤ θ ≤ π2 but θ cannot be 0. Then cot θ = ± csc2 θ − 1 =
2
√
± u2 − 1. Now we do not know whether to choose positive or the negative because cot is positive in the
first quadrant but negative in the fourth quadrant.
Q7]. . . [20 points]
Prove the following identity by making the left side of the equation look like the right side.
1−
1−
sin2 (θ)
1+cos(θ)
=
1+cos θ−(1−cos2 θ)
1+cos θ
=
sin2 (θ)
= cos(θ).
1 + cos(θ)
cos θ(1+cos θ)
1+cos θ
= cos θ.
Q8]. . . [20 points]
Find the inverse of the function f (x) = cos(x + 2) + 1. State the domain of f −1 .
f −1 (x) = arccos(x − 1) − 2. Recall that the domain of the inverse cosine function is [−1, 1] and so
−1 ≤ x − 1 ≤ 1 and so the domain of f −1 is 0 ≤ x ≤ 2
Q9]. . . [20 points]
Solve the trigonometric equation cos2 θ − sin2 θ + sin θ = 0 on the interval [0, 2π].
Using the trigonometric identity cos2 θ = 1−sin2 θ we find that cos2 θ −sin2 θ +sin θ = −2 sin θ +sin θ +1 =
−(2 sin θ + 1)(sin θ − 1) = 0. Therefore it must be that sin θ = 1 or sin θ = − 21 . The former has solution
and the latter has solutions
11π 7π
, 6.
6
π
2