Name:
Math 1060
Exam 4 Review
For exam 4 you should make sure you know all of the identities from §5.1,
and that you can compute trigonometric functions at special angles.
However, on this exam I will give you the following identities, which you had
to memorize on exam 3:
sin(a + b) = sin a cos b + cos a sin b
cos(a + b) = cos a cos b − sin a sin b
And the following, which you never had to memorize but should understand
how to use:
1
(cos(x − y) − cos(x + y))
2
1
cos(x) cos(y) = (cos(x − y) + cos(x + y))
2
1
sin(x) cos(y) = (sin(x − y) + sin(x + y))
2 x−y
x+y
cos
sin(x) + sin(y) = 2 sin
2
2
x+y
x−y
cos(x) + cos(y) = 2 cos
cos
2
2
sin(x) sin(y) =
The half-angle formulas require a little bit of thought to use:
r
θ
1 + cos θ
sin
=±
2
2
r
1 − cos θ
θ
=±
cos
2
2
You will be able to do all of these problems with just the identities from §5.1–
4, but some problems will be easier if you understand how to use the identities from §5.5.
1. Compute the values of the six trigonometric functions for each of the following angles:
(a)
π
3
I haven’t rationalized the denominators in these. We’re at the
Exam 4 Review
Math 1060
point in this class where I no longer care if you do that.
√
√
π
3
π
1
π
sin
=
cos
=
tan
= 3
3
2
3
2
3
π
π
2
π
1
csc
sec
=√
=2
cot
=√
3
3
3
3
3
(b)
3π
4
3π
1
=√
4
2
√
3π
csc
= 2
4
sin
−1
3π
=√
4
2
√
3π
sec
=− 2
4
cos
3π
= −1
4
3π
cot
= −1
4
tan
(c) − π3
−π
sin
=
3
−π
csc
=
3
√
− 3
2
−2
√
3
−π
1
=
3
2
−π
sec
=2
3
cos
√
−π
=− 3
3
−π
−1
cot
=√
3
3
tan
(d) − 7π
6
−7 π
1
sin
=
6
2
−7 π
csc
=2
6
(e)
−7 π
cos
=
6
−7 π
sec
=
6
√
− 3
2
−2
√
3
−7 π
−1
=√
6
3
√
−7 π
cot
=− 3
6
tan
17π
3
17 π
sin
=
3
17 π
csc
=
3
(f) 120◦
√
− 3
2
−2
√
3
17 π
1
=
3
2
17 π
sec
=2
3
cos
√
17 π
=− 3
3
17 π
−1
cot
=√
3
3
tan
Exam 4 Review
Math 1060
√
3
sin (120 ) =
2
2
csc (120◦ ) = √
3
◦
cos (120◦ ) =
−1
2
sec (120◦ ) = −2
√
tan (120◦ ) = − 3
−1
cot (120◦ ) = √
3
2. Compute
sin(15◦ ) sin(75◦ )
There are as many methods for solving this as stars in the sky. Here’s
the easy one:
sin(15◦ ) sin(75◦ ) =
=
=
=
=
1
[cos(15◦ − 75◦ ) − cos(15◦ + 75◦ )]
2
1
[cos(−60◦ ) − cos(90◦ )]
2
1
[cos(60◦ ) − 0]
2
11
22
1
4
3. Graph the following functions.
I’ll keep my answers to these brief, but please ask as many questions
as you like in class. I’ve also been careful to mark the x- and y-intercepts,
which you have to solve an equation for. I’ll add that work soon.
(a) y = 4 sin(2x + π) + 2
Amplitude: |a| = 4
Vertical Shift: d = 2
Period: 2π
b =π
Phase Shift: bc = − π2
So our sine wave starts at the point − π2 , 2 :
Exam 4 Review
Math 1060
6.0
5.0
4.0
3.0
2.0
1.0
−12
π
−13
π
−16
π
−1.0
1
6
π
1
3
π
1
2
π
2
3
π
5
6
π
π
7
6
π
4
3
π
3
2
π
−2.0
(b) y = 2 sec
x
2
− π2 − 1
Amplitude: |a| = 2
Vertical Shift: d = −1
Period: 2π
b = 4π
Phase Shift: bc = π
So our cosine wave starts at the point (π, 1). I’ve plotted the
cosine dashed, and the secant solid. The vertical lines are the asymptotes.
4.0
2.0
−π
π
2π
3π
4π
5π
6π
7π
−2.0
−4.0
−6.0
(c) y = tan x − π4 + 1
Amplitude doesn’t really make sense here, but the other properties do:
Vertical Shift: d = 1
Period: πb = π
Exam 4 Review
Math 1060
Phase Shift: bc = π4
So the "center" of this tangent graph is at π4 , 1 .
4.0
3.0
2.0
1.0
−32
π
−π
−12
1
2
π
π
π
3
2
π
−1.0
−2.0
4. Prove the following identities:
(a) sec2 (θ) tan2 (θ) = sec4 (θ) − sec2 (θ)
sec2 (θ) tan2 (θ) = sec4 (θ) − sec2 (θ)
There are a lot of ways to do this. I’m going to do it the slow way:
sec2 (θ) tan2 (θ) = sec2 (θ)(sec2 (θ) − 1)
sec2 (θ) tan2 (θ) = sec2 (θ) tan2 (θ)
(b) cos(π − 2x) = 2 sin2 (x) − 1
This one is harder. Here’s my favorite way to do it:
π
cos(π − 2x) = cos 2
−x
2
π
−x −1
= 2 cos2
2
2
= 2 sin (x) − 1
5. If u
~ = h2, 1i, v~ = h−1, −3i, compute:
(a) k~
uk
Exam 4 Review
Math 1060
√
√
k~
uk = 22 + 12 = 5
(b) k~
vk
q
k~
vk =
√
(−1)2 + (−3)2 = 10
(c) k3~
vk
√
k3~
v k = |3|k~
vk = 3 10
(d) 2~
u + 3~
v
2~
u + 3~
v = h2(2) + 3(−1), 2(1) + 3(−3)i = h1, −7i
(e) u
~ − 2~
v
u
~ − 2~
v = h2 − 2(−1), 1 − 2(−3)i = h4, 7i
(f) A unit vector in the direction of u
~
There’s a simple procedure here. We just need to scale u
~ down by
its own length. So:
*
+
1
1
2 1
u
~ = √ h2, 1i = √ , √
k~
uk
5
5 5
6. If u
~ is the vector with magnitude 5 and direction θ =
ponent form?
Here’s a diagram to help out:
3π
4 ,
what is it in com-
Exam 4 Review
Math 1060
~u
θ = 34π
|~v|sin(θ) =5sin34π
|~v|cos(θ) =5cos34π
D
E √ √ 5 2 5 2
3π
So the component form is 5 cos 3π
,
5
sin
4
4 = − 2 , 2 .
7. Hilbert walked 7 miles north, then turned east and walked 5 miles. Then
he turned 30◦ to his left and walked 3 miles.
(a) Write down the component form of a vector describing his displacement at the end of his walk.
The diagonal vector is
π
π
v~ = 3 sin
, 3 cos
6
6
√ +
*
3 3 3
= ,
2 2
And the final answer is:
√ +
3 3 3
h0, 7i + h5, 0i + ,
2 2
√ +
*
3
3 3
= 5 + ,7 +
2
2
√ +
*
13 14 + 3 3
=
,
2
2
*
(b) How far is he from his starting point?
Exam 4 Review
Math 1060
The distance is simply the magnitude of his displacement vector:
s
*
√ +
√ !2
2
13 14 + 3 3 13
14 + 3 3
,
=
+
2
2
2
2
r
√ 2
1
(13)2 + 14 + 3 3
=
2
q
√
1
=
169 + 196 + 84 3 + 27
2
q
√
1
=
392 + 84 3
2
q
√
1
= 2 98 + 21 3
2
q
√
= 98 + 21 3
8. If my friend Ug and I are trying to lift up a 100 pound mammoth steak,
and he is pulling up and to the left 30◦ from vertical with 60 pounds of
force, and I’m pulling up and to the right 30◦ from vertical with 60 pounds
of force, will we pulling with enough force to lift up the mammoth steak?
If u
~ is Ug’s lifting force, and m
~
is mine, then:
Ug+Me
Ug
Me
π π
, 60 sin
m
~ = 60 cos
3
3
√ +
*
1
3
= 60 , 60
2
2
D
√ E
= 30, 30 3
2π 2π u
~ = 60 cos
, 60 sin
3
3
√ +
*
1
3
= −60 , 60
2
2
D
√ E
= −30, 30 3
Exam 4 Review
Math 1060
And the total lifting force is:
D
√ E D
√ E
m
~ +u
~ = 30, 30 3 + −30, 30 3
D
√ E
= 0, 60 3
≈ h0, 103.923i
So we’re pulling up with about 104 pounds of force. The steak only
weighs 100 pounds, so we can lift it.
9. I’m driving north in my convertible at 75 miles per hour with the top
down, so I can feel the wind in my hair. But there’s also a wind blowing
from east to west at 20 miles per hour. How fast does the wind feel to me?
Real Wind
Total Wind
Wind (from driving)
The total wind is
h−20, 0i + h0, −75i = h−20, −75i
where the first vector is the actual wind, and the second is the wind from
driving. So the wind speed I feel is just the magnitude of this vector:
q
√
√
√
kh−20, −75ik = (−20)2 + (−75)2 = 5 42 + 152 = 5 16 + 225 = 5 241
√
So the total windspeed I feel is 5 241 ≈ 77.62 miles per hour.