Name:
Math 1060
Final Exam Review
1. Complete the following table:
θ (radians)
θ (degrees)
sin(θ)
cos(θ)
√
−1
2
3
2
2π
3
120
7π
3
420
3
2
π
6
30
1
2
−5 π
4
−225
3π
4
tan(θ)
√
− 3
√
√
1
2
√
sec(θ)
csc(θ)
cot(θ)
−2
√2
3
−1
√
3
2
√2
3
√1
3
2
3
√
3
2
√1
3
√2
3
√1
2
−1
√
2
−1
√
− 2
√
135
√1
2
−1
√
2
−1
√
− 2
√
5π
2
450◦
1
0
undef
undef
1
0
π
180◦
0
−1
0
−1
undef
undef
3π
2
270◦
−1
0
undef
undef
−1
0
3π
4
135
√1
2
−1
√
2
−1
√
− 2
√
3
2
−1
2
−1
2
−1
2. Place all of the angles from problem 1 on a unit circle.
−54
π
3
4
π
2
3
π
5
2
π
1
7
3
1
6
0.5
π
-1
π
-0.5
0.5
π
1
-0.5
-1
3
2
π
3. For each of the following angles, find the complementary angle and the supplementary angle assuming they exist.
Final Exam Review
Math 1060
(a) 60◦
Complementary: 90◦ − 60◦ = 30◦
Supplementary: 180◦ − 60◦ = 120◦
(b)
2π
3
Complementary: None; 2π
3 >
π
Supplementary: π − 2π
=
3
3
(c)
π
2
3π
8
π
Complementary: π2 − 3π
8 = 8
3π
5π
Supplementary: π − 8 = 8
4. Suppose sin(θ) =
5
13
and θ is in the second quadrant. Find cos(θ).
There are a number of ways to do this. Here’s one:
cos2 θ + sin2 θ = 1
2
5
cos2 θ +
=1
13
25
=1
cos2 θ +
169
144
cos2 θ =
169
r
cos θ = ±
144
12
=±
169
13
But since θ is in the second quadrant:
cos θ = −
12
13
5. My giant novelty watch has a 2 inch long hour hand. How far does the tip of it move between 6:00pm and 10:00pm?
Between 6:00pm and 10:00pm the hand sweeps out an angle of
2π
3 .
The radius is 2 inches. So:
s = θr
2π
=
2
3
4π
=
3
So the hand sweeps out a distance of
4π
3
inches.
6. You are standing 500 feet from Paul Bunyan. The angle of elevation to the top of his head is 30◦ . How tall is Paul?
–2–
Final Exam Review
Math 1060
h
30◦
500 ft
We can see that
h
500
1
h
√ =
500
3
500
√ =h
3
tan 30◦ =
500
√
3
So Paul is
feet tall.
7. For each of the following functions, find the amplitude, period, vertical shift, and phase shift, and graph at least
two periods of the function.
(a) y = sin(x − π)
Amplitude: 1
Period: 2π
Phase Shift: π
Vertical Shift: 0
1.0
0.5
−0.5
1
2
π
π
3
2
2π
π
−1.0
(b) y = cos(π − x)
First, rearrange things a bit:
y = cos(π − x) = cos(−(x − π)) = cos(x − π)
Then:
Amplitude: 1
Period: 2π
Phase Shift: π
Vertical Shift: 0
–3–
5
2
π
3π
7
2
π
4π
Final Exam Review
Math 1060
1.0
0.5
1
2
−0.5
π
π
3
2
2π
π
5
2
3π
π
7
2
4π
π
−1.0
(c) y = 4 cos x +
π
4
+4
Amplitude: 4
Period: 2π
Phase Shift: − π4
Vertical Shift: 4
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
1
2
(d) y = sin
x
2
π
π
−
π
2
3
2
π 2π
5
2
π 3π
7
2
π 4π
9
2
π 5π
11
2
π 6π
13
2
π 7π
15
2
π 8π
+1
Amplitude: 1
Period: 4π
Phase Shift: π
Vertical Shift: 1
2.0
1.5
1.0
0.5
π
2π
3π
4π
5π
6π
7π
8. Graph the following functions: Note that to get full credit, your x-intercepts do have to be accurate.
(a) y = 6 sin π4 x + π2 − 3
–4–
8π
Final Exam Review
Math 1060
2.0
−2.0
2π
4π
6π
8π
10 π
12 π
14 π
16 π
−4.0
−6.0
−8.0
(b) y = − csc(4x − π)
3.0
2.0
1.0
1
4
1
2
π
−1.0
−2.0
−3.0
(c) y = 2 tan(πx) + 2
–5–
π
3
4
π
π
Final Exam Review
Math 1060
6
4
2
1
2
4
3
-2
(d) y = 2 sec(x + π)
6.0
4.0
2.0
1
2
π
π
3
2
2π
π
5
2
π
3π
7
2
π
4π
−2.0
−4.0
−6.0
9. You are riding a ferris wheel. It takes 150 seconds for the ferris wheel to go all the way around once. You start on
the ground; at the top of the ferris wheel you are 100 feet off the ground. Consider the function h(t), your height
off the ground after t seconds.
(a) Find the amplitude, period, phase shift, and vertical shift for the function h(t).
Amplitude: 50 feet (half the total vertical distance you travel)
Period: 150 seconds
Vertical Shift: 50 feet, the middle of your range of heights.
–6–
Final Exam Review
Math 1060
Phase Shift: This one’s tricky. It depends on if you model the problem using a sine or a cosine. If we’re
clever, we can use a cosine and fix the phase shift to be 0.
(b) Write down a rule for the function h(t) and graph it.
Since you start at the bottom, and the phase shift is 0, you can see that if we use a cosine we need to set
a to be negative. So using the above information:
π h(t) = −50 cos
t + 50
75
And here are two periods of the graph:
feet
100
80
60
40
20
30
60
90
120
150
180
210
240
270
300
seconds
10. Wonder Woman is flying her invisible jet over you. Her altitude is 20 miles. You are looking at her through your
death ray scope. If the angle of elevation to Wonder Woman from your telescope is θ, graph the horizontal distance d(θ) from you to Wonder Woman over the interval 0 < θ < π2 .
20 miles
θ
d miles
From the picture we can see that
d
20
20 cot(θ) = d
cot(θ) =
d = 20 cot(θ)
We don’t know how to graph cot(θ), though, so let’s do this:
π
−θ
2 π d = 20 tan − θ −
2
π
d = −20 tan θ −
2
d = 20 tan
–7–
Final Exam Review
Math 1060
And that’s easy to graph (relatively speaking):
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
1
8
1
4
π
π
3
8
π
1
2
π
11. Compute:
(a) arcsin
(b) arctan
√ 2
2
√
=
3=
(c) arccos cos
π
4
π
3
3π
2
= arccos(0) =
π
2
12. Wonder Woman is flying her invisible jet over you. Her altitude is 20 miles. You are looking at her through your
death ray scope. If the horizontal distance from you to Wonder Woman is x, graph the angle of elevation a(x)
from you to Wonder Woman over the interval 0 < x < 20.
20 miles
θ
d miles
From the picture we can see that
d
cot(θ) =
20
π
d
tan
−θ =
2
20
π
d
−1
− θ = tan
2
20
d
π
−θ = tan−1
−
20
2
π
d
θ = − tan−1
+
20
2
And the graph:
–8–
Final Exam Review
1
2
π
3
8
π
1
4
π
1
8
π
Math 1060
5
10
13. Prove the following identities:
(a)
tan x+cot y
tan x cot y
= tan y + cot x
tan x + cot y
= tan y + cot x
tan x cot y
tan x
cot y
+
= tan y + cot x
tan x cot y tan x cot y
1
1
+
= tan y + cot x
cot y tan x
tan y + cot x = tan y + cot x
(b) (1 + sin(θ))(1 + sin(−θ)) = cos2 (θ)
(1 + sin(θ))(1 + sin(−θ)) = cos2 (θ)
(1 + sin(θ))(1 − sin(θ)) = cos2 (θ)
1 − sin2 (θ) = cos2 (θ)
cos2 (θ) = cos2 (θ)
–9–
15
20
Final Exam Review
Math 1060
5
13
14. Suppose sin u =
and cos v = − 53 , and cos u and sin v are both positive. Compute:
(a) sin(u + v)
s
2
r
r
25
144
12
= 1−
1−
=
=
169
169
13
s
r
r
2
p
3
9
16
4
sin v = 1 − cos2 v = 1 − −
= 1−
=
=
5
25
25
5
p
cos u = 1 − sin2 u =
5
13
And now:
sin(u + v) = sin u cos v + cos u sin v
3
12 4
5
−
+
=
13
5
13 5
15 48
=− +
65 65
33
f=
65
(b) cos(u − v)
cos(u − v) = cos u cos v + sin u sin v
12
3
5 4
=
−
+
13
5
13 5
36 20
=− +
65 65
16
=−
65
15. Compute sin
sin
π
12
11π 12
sin
11π
12
.
π 1
11π
π
11π
π
sin
=
cos
−
− cos
+
12
2
12
12
12
12
1
5π
=
cos
− cos (π)
2
6
" √
#
1
3
=
−
− (−1)
2
2
" √
#
1
3 2
=
−
+
2
2
2
–10–
Final Exam Review
Math 1060
√
2− 3
=
4
16. Solve the following equations:
(a) 4 cos2 x − 1 = 0
4 cos2 x − 1 = 0
4 cos2 x = 1
1
cos2 x =
4
cos x = ±
1
2
The solutions between 0 and 2π are π3 ,
x=
π
+ 2πn
3
2π 4π 5π
3 , 3 , 3 :
x=
2π
+ 2πn
3
x=
4π
+ 2πn
3
x=
5π
+ 2πn
3
(b) 12 sin2 x − 13 sin x + 3 = 0
12 sin2 x − 13 sin x + 3 = 0
(3 sin x − 1)(4 sin x − 3) = 0
1
sin x =
3
sin x =
3
4
This one is good practice for your inverse functions. There are four sets of answers:
1
x = sin
+ 2πn
3
3
x = sin−1
+ 2πn
4
−1
1
x = π − sin
+ 2πn
3
3
x = π − sin−1
+ 2πn
4
−1
And that’s it. You should understand how to give answers with inverse trig functions in them.
(c) cos(2x)(2 cos x + 1) = 0
cos(2x)(2 cos x + 1) = 0
–11–
Final Exam Review
Math 1060
This splits up into two equations:
cos(2x) = 0
π
2x = + 2πn
2
π
+ πn
x=
4
3π
+ 2πn
2
3π
x=
+ πn
4
2x =
2 cos x + 1 = 0
1
2
2π
x=
+ 2πn
3
cos x = −
x=
17. Suppose ~u = h1, −1i and ~v = h0, 2i. Compute and draw:
(a) 2~u + ~v
2~u + ~v = h2 · 1 + 0, 2 · −1 + 2i = h2, 0i
2~u + ~v
(b) ~v − ~u
~v − ~u = h0 − 1, 2 − (−1)i = h−1, 3i
~v − ~u
(c) 3~v + 6~u
–12–
4π
+ 2πn
3
Final Exam Review
Math 1060
3~v + 6~u = h3 · 0 + 6 · 1, 3 · 2 + 6 · −1i = h6, 0i
Notice that you can do this faster by using the answer to part (a).
3~v + 6~u
18. Find a unit vector in the direction of w
~ = 3ı̂ − 4̂.
Simply compute the magnitude of w:
~
p
√
√
kwk
~ = 32 + (−4)2 = 9 + 16 = 25 = 5
Then scale w
~ down by its own magnitude:
1
1
3
4
w
~ = (3ı̂ − 4̂) = ı̂ − ̂
kwk
~
5
5
5
19. My friends and I are playing three-way free-for-all tug-of-war. We are pulling on ropes with forces of 70 pounds,
50 pounds, and 100 pounds, with angles of 0◦ , 135◦ , and 270◦ , respectively. Find the direction and magnitude of
the total force.
A diagram may help:
F~1 = 70 cos 0◦ ı̂ + 70 sin 0◦ ̂
F~2
= 70ı̂
F~2 = 50 cos 135◦ ı̂ + 50 sin 135◦ ̂
√
√
= −25 2ı̂ + 25 2̂
F~3 = 100 cos 270◦ ı̂ + 100 sin 270◦ ̂
F~1
= −100̂
F~3
So apparently the total force is
√
√
F~t = F~1 + F~2 + F~3 = (70 − 25 2)ı̂ + (25 2 − 100)̂
There is no more thinking now. You can see, either from the computation above or from the diagram, that F~t is
–13–
Final Exam Review
Math 1060
in the fourth quadrant. So first, the magnitude:
q
kF~t k =
√
√
(70 − 25 2)2 + (25 2 − 100)2 =
q
q
√
√
17400 − 8500 2 = 10 174 − 85 2
Obviously that’s awful (and the numbers on the test will be nicer) but once you’ve got the vector worked out,
there’s no thinking left. It’s a similar situation for the direction: since the vector is in the fourth quadrant, and
that’s in the range of tan−1 (x), we can just take an arctangent:
−1
θ = tan
!
√
25 2 − 100
√
= tan−1
70 − 25 2
!
√
5 2 − 20
√
14 − 5 2
√ 2
That’s the answer I’m looking for. If you like, you can simplify that further to tan−1 − 115+15
, but it’s not
73
necessary. The angle is about −0.3434π radians.
20. I am trying to take two dogs for a walk. We are walking east. Suddenly, each dog sees a different cat and runs in a
different direction. If one bolts 60◦ north of east, pulling on his leash with 100 pounds of force, and one bolts 60◦
south of east, pulling on her leash with 100 pounds of force, how hard do I have to pull on the leash to keep them
in place? (Hint: draw a diagram!)
You can see from the diagram that the vertical
components cancel out. We just need to add the horizontal components to get the total force:
F~1
100 cos(60◦ ) + 100 cos(−60◦ ) = 50 + 50 = 100
60◦
So the dogs are pulling with a total of 100 pounds of
force; I need to pull back with a total of 100 pounds of
force.
−60◦
F~2
21. Suppose ~u = h1, −1i and ~v = h0, 2i. Compute:
(a) k~v k
k~v k =
p
√
02 + 2 2 = 4 = 2
(b) ~u · ~v
~u · ~v = 1 · 0 + −1 · 2 = −2
(c) A unit vector in the direction of ~u
–14–
Final Exam Review
Math 1060
1
1
1
~u = p
h1, −1i = √ h1, −1i =
2
2
k~uk
2
1 + (−1)
1
1
√ , −√
2
2
*√
=
√ +
2
2
,−
2
2
(d) A unit vector in the direction of ~v
1
1
~v = h0, 2i = h0, 1i
k~v k
2
(e) The vector component of ~v in the direction of ~u
We
need
D √ first √
E to compute a unit vector in the direction of ~u. But luckily we already did! I’ll call it
2
2
û = 2 , − 2 . Now, the length of the component of ~v in the direction of û is:
√
2
+2·
~v · û = 0 ·
2
√ !
√
2
−
=− 2
2
But I asked for the vector component, which we get by multiplying that length by the unit vector û:
*√
√ + √
2
2
2 2
− 2
,−
=
,−
= h1, −1i
2
2
2 2
22. The other day I parked my awesome pickup truck on a 60◦ incline. If my truck weighs 6500 pounds, how much
force did the brakes need to apply in order to prevent the truck from rolling downhill?
I’ve diagrammed the forces at the left: F~g is the
downward force of gravity, F~r is the force that the
ramp has to take care of, and F~b is the force that the
brakes have to take care of. Since kF~g k = 6500, it’s
easy to see that the brakes have to take care of
√
√
3
◦
6500 sin 60 = 6500
= 3250 3pounds
2
F~r
60◦
F~g
F~b
–15–