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MATH 116 EXAM 4
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to degree.
Problem 1(5 point each) a) Convert 5
4
solution)
5  180  900  225 degree.

4
4
b) Convert 4 radian to degree.
solution)
720
4  180
   degree.
Problem 2(7 point each)
a) Let   3 . Find the reference angle and sin , cos , tan .
solution) We see that the angle is in the Quad 1. The reference angle is 3 or 60
degree.
So, we have a triangle with hypotenuse 2, opposite 3 and adjacent 1. Hence,
sin  
opp
 1,
2
hypo
cos  
adj
 1,
2
hypo
tan  
opp

adj
3

1
3,
b) Let   135 0 . Find the reference angle and csc , sec , cot .
solution) Move 135 degree from the positive x-axis clockwise. So, we have a
triangle on the Quad 3. And we see that the reference angle is 45 degree. The
hypotenuse is 2 , opposite is 1 and the adjacent is 1. Thus,
csc  
hypo
2

  2.
1
opp
sec  
hypo
2

  2.
1
adj
adj
 1  1.
1
opp
cot  
1
3
Problem 3(7 point each) Let sin x  
and cos x  0.
a) Find cos x and tan x by drawing method.
solution) Since cos is positive and sin is negative, we know that the angle x is in the
Quad 4. We also know that hypotenuse is 3 and the opposite is 1, since
sin x   1 . Now, the adjacent is
3
 31   2.
But x is in the Quad 4. So, the adjacent is 2 . Thus,
2
3
cos x 

6
. ,
3
tan x  1 .
2
b) Find cos x and tan x by using identities.
solution)
sin 2 x  cos 2 x  1.
So,
cos 2 x  1  sin 2 x.
2
cos x  1 
 1
3
cos x  
2
 1 1  2.
3
3
2 .
3
Since cos is positive, we get
cos x 
2
3

2

3
To find tan x, let’s use
sin x .
tan x  cos
x
6
3
.
tan x 

1
3

2
3
1 3
 1 .
2
3  2
Problem 4
a) (6 points) Sketch y  sin x over two periods.
solution)
Note: You should indicate x  0,  2 , , 3
, 2 on the x-axis.
2
y
1
0.5
0
-5
-2.5
0
2.5
5
x
-0.5
-1
b) (6 points) Find sin and sin
solution)
3
2
sin  0,
.
sin 3   1.
2
c) (6 points) Sketch the graph of csc x and find the period.
solution) The period is 2.
Vertical asymptoes are where sin x  0. So, x  0, , 2, . . . .
Here, the vertical lines are the vertical asymptotes x  , 0, .
y
5
2.5
0
-5
-2.5
0
2.5
5
x
-2.5
-5
d) (8 points) Let y  5 sin 13 x. Sketch the graph over two periods. What are the
amplitude and the period?
solution) The amplitude is 5 and the period is 2
 2  3  6.
1
3
y
5
2.5
0
-15
-10
-5
0
5
10
15
x
-2.5
-5
e) (8 points) Sketch y  sinx  . Also find an identity for sinx  . (You may
use one of the following identities: sinA  B  sin A cos B  sin B cos A,
cosA  B  cos A cos B  sin A sin B.)
solution) By translating the graph of sin x to the right by , we have the following
graph.
y
1
0.5
0
-5
-2.5
0
2.5
5
x
-0.5
-1
You may find an identity sinx     cos x by examining the graph. Or, by
making use of the identity given, we have
sinx    sin x cos   sin  cos x   sin x.
f) (4 points) Is sinx  sin x? Explain your answer.
solution) No, since the graph is symmetric with respect to the origin (that means
odd function, so sinx   sin x).
Problem 5 (6 points) Sketch the graph of y  tan x and find the period. Also find
the period of y  tan 4x.
solution) The period of tan x is  and the period of tan 4x is  .
4
.
.
.
.
Vertical lines are the vertical asymptotes x   2 , 2 , 3
2
y
50
25
0
-1.25
0
1.25
2.5
3.75
5
x
-25
-50
Problem 6(6 point each)
a) Show that sin 3 x  sin 5 x cos x  sin 3 x cos 3 x.
solution) Here the useful identity is sin 2 x  cos 2 x  1.
sin 3 x  sin 5 x cos x  sin 3 x1  sin 2 x cos x  sin 3 x cos 2 x cos x  sin 3 x cos 3 x.
b) Substitute x  3 sec  and simplify x 2  9
solution) The useful identity in this problem is 1  tan 2   sec 2 .
x2  9 
9 sec 2   9 
9sec 2   1  3 sec 2   1  2 tan 2   2 tan .
4
4
c) Simplify cos x  sin x as much as possible (hint: factor the numerator).
cos x  sin x
solution) Note that
cos 4 x  sin 4 x  cos 2 x  sin 2 xcos 2 x  sin 2 x  cos 2 x  sin 2 xcos x  sin xcos x  sin x.
Since cos 2 x  sin 2 x  1, we have
cos 4 x  sin 4 x  cos x  sin xcos x  sin x.
Thus,
cos 4 x  sin 4 x  cos x  sin xcos x  sin x  cos x  sin x.
cos x  sin x
cos x  sin x