Name:
Math 1060
Exam 3 Review
1. Graph the following functions:
(a) f (x) = 2 sin(πx + π) − 1
It’s easiest to start by computing the period, amplitude, &c. Here they are:
amplitude = 2
period = 2
phase shift = −1
vertical shift = −1
That gives the following lovely graph:
1
-1
-2
-1
-2
-3
1
2
(b) f (x) = tan(3x) − 1
x=
Here we want to find the vertical asymptotes. We know that tan(x) has vertical asymptotes at
+ πn, so to find the asymptotes for this beast:
π
2
π
+ πn
2
π π
x= + n
6 3
3x =
Now just graph one period and recopy the graph to get a nice graph of f (x). You can plug in to get
for instance:
π
π
f −
= −2
f (0) = −1
f
=0
12
12
And graph:
2.0
1.0
−13
π
−16
π
−1.0
1
6
π
1
3
π
−2.0
−3.0
(c) f (x) = 2 sec4 (5x) − 8
Note that the original problem had a typo. This is the one I intended to ask, with −8 rather
than −4.
The process is similar to above, but much more subtle. I’m going to answer it very, very thor-
Exam 3 Review
Math 1060
oughly. First let’s find the x-intercepts:
2 sec4 (5x) − 8 = 0
2 sec4 (5x) = 8
sec4 (5x) = 4
Break up into two equations:
√
sec(5x) = 2
1
cos(5x) = √
2
√
sec(5x) = − 2
1
cos(5x) = − √
2
Now we have to break each of these up again:
7π
+ 2πn
4
7π 2π
x=
+
n
20
5
π
+ 2πn
4
π 2π
+
n
x=
20
5
5x =
3π
+ 2πn
4
3π 2π
x=
+
n
20
5
5x =
5π
+ 2πn
4
5π 2π
x=
+
n
20
5
5x =
5x =
By being immensely clever we can notice that we can rewrite this as:
x=
π
π
+ n
20 10
Now. We know that cos(x) = 0 when x =
vertical asymptotes when:
π
2
+ πn. That means that 2 sec4 (5x) − 8 = 2 cos41(5x) − 8 has
cos(5x) = 0
π
5x = + πn
2
π π
x=
+ n
10 5
We just need to plug in some points now to figure out what the graph actually looks like. I’m going to pick a point halfway between each point of vertical asymptotes. So for instance 0, π5 , 2π
5 ,
and so forth:
π
π
f
n = 2 sec4 5 − 8 = 2 sec4 (π) − 8 = 2(−1)4 − 8 = −6
5
5
All of that information taken together gives us:
1.0
−103
π
−15
π
−101
1
10
π
−1.0
−2.0
−3.0
−4.0
−5.0
−6.0
−7.0
π
1
5
π
3
10
π
Exam 3 Review
Math 1060
2. Suppose sin u = 35 , sin v = 12 , cos u > 0, cos v < 0. Compute the following:
The following two graphs will help quite a bit with all of these problems:
2
5
1
3
v
u
√
3
√
52 − 32 = 4
Note that, although you can figure out what the angle v is, you don’t need it.
(a) cos u =
4
5
(b) tan u =
3
4
(c) sec u =
1
cos u
=
5
4
(d) csc u =
1
sin u
=
5
3
1
tan u
√
= − 23
=
4
3
(e) cot u =
(f) cos v
√
(g) tan v =
(h) sec v =
(i) csc v =
3
3
√
1
√2 = − 2 3
=
−
cos v
3
3
sin v
cos v
= − √1 = −
1
sin v
=
3
2
1
=2
√
(j) cot v =
1
tan v
=−
3
1
√
=− 3
(k) sin(u + v)
sin(u + v) = sin u cos v + cos u sin v
√ 3
4 1
3
=
−
+
5
2
5 2
√
4−3 3
=
10
(l) tan(u − v)
Exam 3 Review
Math 1060
tan(u − v) =
tan u − tan v
1 + tan
u tan
v
√
− − 33
√ =
1 + 34 − 33
√ 3
−
− 33
4
12
√ ·
=
12
3
3
1+ 4 − 3
√
9+4 3
=
√
12 − 3 3
3
4
You can end there, but:
=
=
=
=
√
√
9 + 4 3 12 + 3 3
√ ·
√
12 − 3 3 12 + 3 3
√
144 + 75 3
144 − 27
√
144 + 75 3
117
√
48 + 25 3
39
(m) cot(u − v)
1
tan(u − v)
39
=
√
48 + 25 3
cot(u − v) =
You can end there, but:
√
48 − 25 3
√
48 + 25 3 48 − 25 3
√
1872 − 975 3
=
429
√
624 − 325 3
=
143
=
(n) sin(u + π)
39
√
Exam 3 Review
Math 1060
sin(u + π) = sin u cos π + cos u sin π
3
4
= (−1) + (0)
5
5
3
=−
5
(o) cos(u − π)
cos(u − π) = cos u cos π + sin u sin π
3
4
= (−1) + (0)
5
5
4
=−
5
(p) cot u + π2
If we try to compute this using the sum and difference formulas we’ll get stuck, since tan
So instead compute sin u + π2 and cos u + π2 :
π
π
π
sin u +
= sin u cos + cos u sin
2
2
2
3
4
= (0) + (1)
5
5
4
=
5
π
π
π
= cos u cos − sin u sin
cos u +
2
2
2
4
3
= (0) − (1)
5
5
3
=−
5
Now, combine:
4
π
cot
= 53
2
−5
4
=−
3
(q) sec u + π3
π
2 .
Exam 3 Review
Math 1060
Let’s compute this instead:
π
π
π
cos u +
= cos u cos − sin u sin
3
3
3
√
41 3 3
=
−
52 5 2
√
4 3 3
=
−
10
10
√
4−3 3
=
10
π
1
sec u +
=
3
cos u + π
3
=
=
=
=
10
√
4−3 3
√
10 4 + 3 3
√
√
4−3 3 4+3 3
√
10(4 + 3 3)
16 − 27
√
10(4 + 3 3)
9
3. Simplify:
(a)
1−sin2 x
csc2 x−1
1 − sin2 x cos2 x
=
csc2 x − 1 cot2 x
cos2 x
=
2
cos x
sin2 x
2
= sin x
(b) sin y sec y + cos y csc y
sin y sec y + cos y csc y =
=
sin y cos y
+
cos y sin y
sin2 y
cos2 y
+
cos y sin y cos y sin y
sin2 y + cos2 y
cos y sin y
1
=
cos y sin y
=
Exam 3 Review
(c) cot π2 − φ cos φ
Math 1060
cot
π
− φ cos φ = tan φ cos φ
2
sin φ
=
cos φ
cos φ
= sin φ
4. Verify:
(a) sin2 θ − cos2 θ = 2 sin2 θ − 1
sin2 θ − cos2 θ = 2 sin2 θ − 1
sin2 θ − (1 − sin2 θ) = 2 sin2 θ − 1
2 sin2 θ − 1 = 2 sin2 θ − 1
(b) csc α − sin α = cos α cot α
csc α − sin α = cos α cot α
1
cos α
− sin α = cos α
sin α
sin α
1 − sin2 α cos2 α
=
sin α
sin α
cos2 α cos2 α
=
sin α
sin α
(c) sin4 x − sin2 x = cos4 x − cos2 x
sin4 x − sin2 x = cos4 x − cos2 x
− sin2 x 1 − sin2 x = − cos2 x 1 − cos2 x
− sin2 x cos2 x = − cos2 x sin2 x
5. Solve:
(a) 2 sin ξ + 1 = 0
Exam 3 Review
Math 1060
2 sin ξ + 1 = 0
2 sin ξ = −1
1
sin ξ = −
2
π
ξ = − + 2πn
6
ξ =−
5π
+ 2πn
6
(b) 2 sin2 x + sin x − 1 = 0
2 sin2 x + sin x − 1 = 0
(2 sin x − 1)(sin x + 1) = 0
2 sin x − 1 = 0
1
sin x =
2
π
x = + 2πn
6
sin x + 1 = 0
sin x = −1
x=
5π
+ 2πn
6
x=
3π
+ 2πn
2
(c) sin x − cos x = 1
sin x − cos x = 1
sin x = 1 + cos x
sin2 x = 1 + 2 cos x + cos2 x
1 − cos2 x = 1 + 2 cos x + cos2 x
0 = 2 cos x + 2 cos2 x
cos2 x + cos x = 0
cos x(cos x + 1) = 0
cos x = 0
π
x = + 2πn
2
Now let’s check these:
π
π
sin
− cos
= 1 − 0 = 1X
2
2
3π
3π
sin
− cos
= −1 − 0 = −1×
2
2
sin π − cos π = 1 − −1 = 1X
cos x = −1
3π
x=
+ 2πn
2
x = π + 2πn
Exam 3 Review
Math 1060
So the actual solutions are:
x=
π
+ 2πn
2
x = π + 2πn
6. Using fundamental trigonometric identities (those from §5.1) and sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
show that cos(a − b) = cos(a) cos(b) + sin(a) sin(b).
First we use the cofunction identities:
π
− (a − b)
2
π
= sin
−a+b
2
π
= sin
−a +b
2
cos(a − b) = sin
Now the sum formula for cos(a + b):
= sin
π
π
− a cos b + cos
− a sin b
2
2
Now the cofunction identities again:
cos(a − b) = cos a cos b + sin a sin b
7. Your brother drags your sled up a hill with a 15◦ incline. If he dragged you 100 feet, how much elevation did you gain?
A good diagram is always the first step:
100’
h
15◦
Ok. We know that 15◦ =
π
12 ,
so:
h
π
= sin
100
12
h = 100 sin
π
12
Exam 3 Review
Math 1060
π
Now let’s go ahead and compute sin 12
:
sin
π π
π
= sin
−
12
3 4
π
π
π
π
= sin cos − cos sin
3
4
3
4
√ √
√
3 2 1 2
=
−
2 2
2 2
√ √
√
3 2
2
=
−
4
4
√ √
6− 2
=
4
Putting everything together:
π
h = 100 sin
12
√ √
6− 2
= 100
√ 4√ = 25 6 − 2 ft
8. You are standing 50 feet from your brother’s model rocket launchpad. The rocket launches vertically.
If θ is the angle of elevation from you to the rocket, and s is the distance from you to the rocket, write
θ as a function of s.
Here’s a diagram:
Rocket
s
θ
50 ft
So then we have that:
cos θ =
50
s
θ = cos−1
50
s
We do need to think a bit now. Is this actually giving us the correct angle? Well, cos−1 (x) ∈ (0, π), so
this should give us angles between 0 and π2 . That’s exactly what we’d expect.
9. You tie your brother to the ceiling of the living room using a bungee cord. He starts out 6 feet from
Exam 3 Review
Math 1060
the ground, falls down to 2 feet from the ground, then bounces up and down indefinitely in simple
harmonic motion (meaning it can be described in terms of sin or cos). If it takes him 34 seconds each
bounce to fall from 6 feet to 2 feet:
(a) Find the amplitude and period of your brother’s motion.
The period is twice
that’s 2 feet.
3
4
seconds, or
3
2
seconds. The amplitude is half the distance he travels, so
(b) Write down his height h as a function of seconds passed t.
He starts at the top, so we want to use cos. Now, we know that the formula for period is
2π
b
3 2π
=
2
b
2π
b= 3
p=
2
4π
b=
3
4π
h = 2 cos
t +4
3
(c) When is he 3 feet off the ground?
Now we just solve an equation:
4π
h = 2 cos
t +4
3
4π
3 = 2 cos
t +4
3
4π
−1 = 2 cos
t
3
1
4π
− = cos
t
2
3
4π
4π
t=
+ 2πn
3
3
4π
3
t=
+ 2πn
3
4π
3
t = 1+ n
2
So the answer is 1 + 32 n seconds and
1
2
4π
2π
t=
+ 2πn
3
3
2π
3
t=
+ 2πn
3
4π
1 3
t= + n
2 2
+ 32 n seconds. He hits 3 feet many times!
(d) Graph at least one full period of your brother’s height function.
Exam 3 Review
Math 1060
6
5
4
3
2
1
0.25
0.5
0.75
1
1.25
1.5