LESSON 8 THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS
Topics in this lesson:
1.
SINE GRAPHS
2.
COSINE GRAPHS
3.
SINE AND COSINE GRAPHS WITH PHASE SHIFTS
4.
SECANT AND COSECANT GRAPHS
5.
TANGENT GRAPHS
6.
COTANGENT GRAPHS
1.
SINE GRAPHS
Example Use the Unit Circle to graph two cycles of the function y  sin x on the
interval [ 0 , 4  ] .
Example Use the Unit Circle to graph two cycles of the function y  sin x on the
interval [  4  , 0 ] .
Definition The amplitude of a trigonometric function is one-half of the difference
between the maximum value of the function and the minimum value of the function
if the function has both of these values.
NOTE: The maximum value of a function is the largest y-coordinate on the graph
of the function and the minimum value of a function is the smallest y-coordinate
on the graph of the function if the graph has both of these values.
The sine and cosine functions will have an amplitude. However, the tangent,
cotangent, secant, and cosecant functions do not have an amplitude because these
functions do not have a maximum value nor a minimum value.
Definition The period of a trigonometric function is the distance needed to
complete one cycle of the graph of the function.
All the trigonometric functions have a period.
For the function y  sin x , the amplitude of the function is 1 and the period is 2  .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Given the function y  a sin b x , the amplitude of this function is a and the
2
period is
.
b
Theorem The sine function is an odd function. That is, sin (   )   sin  for
all  in the domain of the function.
NOTE: The domain of the sine function is all real numbers.
Examples Sketch two cycles of the graph of the following functions. Label the
numbers on the x- and y-axes.
1.
y  5 sin 3 x
2
2
Period =
=
3
3
Amplitude = 5 = 5
1
1 2
1 

 period =


=
=
4
4
3
2 3
6
y
5
x
5

6

3

2
2
3
5
6

7
6
4
3
x
NOTE: The first cycle begins at 0. We do not need to label that number.
2
2
Since the period is
, the first cycle ends at
, which is obtained by
3
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
2
2
2

. That is, we add the period of
to the starting point of 0.
3
3
3
2
2
4
4


The second cycle ends at
, which is obtained by
. That
3
3
3
3
2
2
is, we add the period of
to the starting point of
.
3
3
0 
Now, the rest of the numbers on the x-axis were obtained in the following
manner:




0


The
was obtained by
. That is, we add , which is one6
6
6
6
fourth of the period, to the starting point of 0.
2







 . That is, we add , which is
was obtained by
6
6
6
3
3
6

one-fourth of the period, to the next starting point of .
6
The
2
3






 . That is, we add , which is
was obtained by
2
6
6
6
2
6
2
one-fourth of the period, to the next starting point of
.
6
The
2
3
4
2





We can check the
by
. That is, we add , which
6
6
6
3
6
3
3
is one-fourth of the period, to the next starting point of
.
6
5
4
5




was obtained by
. That is, we add , which is one6
6
6
6
6
4
fourth of the period, to the next starting point of
. Or, you can obtain the
6
The
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
5

2
by adding the period of
to the previous
in the first cycle. Thus,
6
6
3
2
4
5






.
6
3
6
6
6
5
6




  . That is, we add , which is
6
6
6
6
5
one-fourth of the period, to the next starting point of
. Or, you can obtain
6

2
the  by adding the period of
to the previous
in the first cycle.
3
3
2
3



 .
Thus,
3
3
3
The  was obtained by
6
7
7




was obtained by
. That is, we add , which is one6
6
6
6
6
6
fourth of the period, to the next starting point of
. Or, you can obtain the
6

7
2
by adding the period of
to the previous
in the first cycle. Thus,
2
6
3
2
3
4
7





.
2
3
6
6
6
The
7
8
4
4





We can check the
by
. That is, we add , which
6
6
6
3
3
6
7
is one-fourth, of the period to the next starting point of
.
6
The graph of two cycles of y  5 sin 3 x in blue compared with the graph of
two cycles of y  sin x in red.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
2.
y 
 x
2 sin  
6
Amplitude =
2
Period = 1 = 2   6 = 12 
6
2
12 
1
1
 period =
 12  =
= 3
4
4
4
y
2
3
x

6
9
12 
15 
18 
21
24 
x
2
NOTE: The first cycle begins at 0. We do not need to label that number.
Since the period is 12  , the first cycle ends at 12  , which is obtained by
0  12   12  . That is, we add the period of 12  to the starting point of
0. The second cycle ends at 24  , which is obtained by 12   12   24  .
That is, we add the period of 12  to the starting point of 12  .
Now, the rest of the numbers on the x-axis were obtained in the following
manner:
The 3  was obtained by 0  3  3 . That is, we add 3  , which is onefourth of the period, to the starting point of 0.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
The 6  was obtained by 3  3  6  . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 3  .
The 9  was obtained by 6   3  9  . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 6  .
We can check the 12  by 9   3  12  . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 9  .
The 15  was obtained by 12   3  15  . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 12  . Or, you can
obtain the 15  by adding the period of 12  to the previous 3  in the first
cycle. Thus, 3  12   15  .
The 18  was obtained by 15  3  18  . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 15  . Or, you can
obtain the 18  by adding the period of 12  to the previous 6  in the first
cycle. Thus, 6   12   18 .
The 21 was obtained by 18  3  21 . That is, we add 3  , which is
one-fourth of the period, to the next starting point of 18  . Or, you can obtain
the 21 by adding the period of 12  to the previous 9  in the first cycle.
Thus, 9   12   21 .
We can check the 24  by 21  3  24  . That is, we add 3  , which is
one-fourth, of the period to the next starting point of 21 .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
The graph of two cycles of y 
of two cycles of y  sin x in red.
3.
y  
 x
2 sin   in blue compared with the graph
6
4
sin 2  x
7
4
, then the
7
graph will be inverted. Thus, we will need to draw two inverted sine cycles.
2
4
4
Amplitude = 
=
Period =
= 1
2
7
7
1
1
1
 period =
1 =
4
4
4
NOTE: Since the sine function is being multiplied by a negative
y
4
7
x
1
4
x
1
2
3
4
1
5
4
3
2
7
4
2
- 74
NOTE: The first cycle begins at 0. We do not need to label that number.
Since the period is 1, the first cycle ends at 1, which is obtained by
0  1  1 . That is, we add the period of 1 to the starting point of 0. The
second cycle ends at 2, which is obtained by 1  1  2 . That is, we add the
period of 1 to the starting point of 1.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Now, the rest of the numbers on the x-axis were obtained in the following
manner:
1
1
1
1
 . That is, we add , which is one-fourth
was obtained by 0 
4
4
4
4
of the period, to the starting point of 0.
The
1
1
1
2
1
1


 . That is, we add , which is onewas obtained by
2
4
4
4
4
2
1
fourth of the period, to the next starting point of .
4
The
3
2
1
3
1

 . That is, we add , which is onewas obtained by
4
4
4
4
4
2
fourth of the period, to the next starting point of .
4
The
3
1
4
1


 1 . That is, we add , which is one4
4
4
4
3
fourth of the period, to the next starting point of .
4
We can check the 1 by
5
4
1
5
1


The was obtained by
. That is, we add , which is one-fourth
4
4
4
4
4
4
5
of the period, to the next starting point of . Or, you can obtain the
by
4
4
1
adding the period of 1 to the previous
in the first cycle.
Thus,
4
1
1
4
5
 1 

 .
4
4
4
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
3
5
1
6
3
1


 . That is, we add , which is onewas obtained by
2
4
4
4
4
2
5
3
fourth of the period, to the next starting point of . Or, you can obtain the
4
2
1
by adding the period of 1 to the previous
in the first cycle. Thus,
2
1
1
2
3
 1 

 .
2
2
2
2
The
7
6
1
7
1

 . That is, we add , which is onewas obtained by
4
4
4
4
4
6
7
fourth of the period, to the next starting point of . Or, you can obtain the
4
4
3
by adding the period of 1 to the previous
in the first cycle. Thus,
4
3
3
4
7
1

 .
4
4
4
4
The
7
1
8
1


 2 . That is, we add , which is one4
4
4
4
7
fourth, of the period to the next starting point of .
4
We can check the 2 by
4
sin 2  x in blue compared with the
7
graph of two cycles of y   sin x in red.
The graph of two cycles of y  
4.
 7x
y  8 sin  

 3 
Copyrighted by James D. Anderson, The University of Toledo
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 7x
 =
Since the sine function is an odd function, then sin  
 3 
7 
 sin  x  . Thus, we have that
3 
NOTE:
 7x
y  8 sin  
 =  8 sin
3


7 
 x
3 
Since the sine function is being multiplied by a negative 8, then the graph
will be inverted. Thus, we will need to draw two inverted sine cycles.
2
6
3
Period = 7 = 2  
=
7
7
3
Amplitude = 8
3
1
1 6
1 3
 period =


=
=
14
4
4
7
2
7
y
8
x
8
3
14
3
7
9
14
6
7
15 
14
9
7
3
2
12 
7
x
NOTE: The first cycle begins at 0. We do not need to label that number.
6
6
Since the period is
, the first cycle ends at
, which is obtained by
7
7
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
6
6
6

. That is, we add the period of
to the starting point of 0.
7
7
7
12 
6
6
12 


The second cycle ends at
, which is obtained by
.
7
7
7
7
6
6
That is, we add the period of
to the starting point of
.
7
7
0 
Now, the rest of the numbers on the x-axis were obtained in the following
manner:
3
3
3
3

was obtained by 0 
. That is, we add
, which is one14
14
14
14
fourth of the period, to the starting point of 0.
The
3
3
6
3
3
3



The
was obtained by
. That is, we add
,
7
14
14
14
14
7
3
which is one-fourth of the period, to the next starting point of
.
14
9
6
3
9
3


was obtained by
. That is, we add
, which is
14
14
14
14
14
6
one-fourth of the period, to the next starting point of
.
14
The
9
3
12 
6
3
6



by
. That is, we add
,
14
14
14
7
14
7
9
which is one-fourth of the period, to the next starting point of
.
14
We can check the
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
15 
12 
3
15 
3


was obtained by
. That is, we add
, which is
14
14
14
14
14
12 
one-fourth of the period, to the next starting point of
. Or, you can
14
15 
3
6
obtain the
by adding the period of
to the previous
in the first
14
7
14
3
6
3
12 
15 




cycle. Thus,
.
14
7
14
14
14
The
9
15 
3
18 
9
3



was obtained by
. That is, we add
,
14
14
14
14
7
7
15 
which is one-fourth of the period, to the next starting point of
. Or, you
14
9
3
6
can obtain the
by adding the period of
to the previous
in the
7
7
7
3
6
9


first cycle. Thus,
.
7
7
7
3
3
18 
3
21
3



The
was obtained by
. That is, we add
,
2
14
14
14
2
14
18
which is one-fourth of the period, to the next starting point of
. Or, you
14
9
6
3
can obtain the
by adding the period of
to the previous
in the
2
7
14
9
6
9
12 
21
3





first cycle. Thus,
.
14
7
14
14
14
2
The
21
3
24 
12 
12 
3



We can check the
by
. That is, we add
,
7
14
14
14
14
7
21
which is one-fourth, of the period to the next starting point of
.
14
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
4
sin 2  x in blue compared with the
7
graph of two cycles of y   sin x in red.
The graph of two cycles of y  
Back to Topics List
2.
COSINE GRAPHS
Example Use the Unit Circle to graph two cycles of the function y  cos x on the
interval [ 0 , 4  ] .
Example Use the Unit Circle to graph two cycles of the function y  cos x on the
interval [  4  , 0 ] .
For the function y  cos x , the amplitude of the function is 1 and the period is
2 .
Given the function y  a cos b x , the amplitude of this function is a and the
2
period is
.
b
Theorem The cosine function is an even function. That is, cos (   )  cos  for
all  in the domain of the function.
NOTE: The domain of the cosine function is all real numbers.
Examples Sketch two cycles of the graph of the following functions.
1.
y 
3 cos 8 x
Amplitude =
3
=
3
2
2

Period =
=
=
8
8
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
1
1 

 period =

=
16
4
4 4
y
3

16
x


8
3
16

4
5
16
3
8
7
16

2
x
3
Since the period is


, the first cycle ends at
and the second cycle ends at
4
4
2

 . The other numbers on the x-axis were obtained by the following:
4
2
0 



16
16
2






16 16
16
8
2
3



16
16
16
3
4





Check:
16
16
16
4
4
5



16
16
16
OR
5
6
3




16
16
16
8
4
5







16
4
16
16
16
OR
2
3







8
4
8
8
8
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6
7



16
16
16
OR
3
3
4
7





16
4
16
16
16
7
8





Check:
16
16
16
2
The graph of two cycles of y  3 cos 8 x in blue compared with the graph
of two cycles of y  cos x in red.
2.
 x
y   4 cos  
5
Since the cosine function is being multiplied by a negative 4, then the graph
will be inverted. Thus, we will need to draw two inverted cosine cycles.
2
Period = 1 = 2   5 = 10 
5
Amplitude = 4
10 
5
1
1
 period =
 10  =
=
4
4
4
2
y
4
x
4
5
2
5
15 
2
10 
25 
15 
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
35 
2
x
20 
Since the period is 10  , the first cycle ends at 10  and the second cycle
ends at 20  . The other numbers on the x-axis were obtained by the
following:
0 
5
5

2
2
5
5
10 


 5
2
2
2
10 
5
15 


2
2
2
15
5
20 


 10 
Check:
2
2
2
20 
5
25 


2
2
2
OR
5
5
20 
25 
 10  


2
2
2
2
25
5
30 


 15
2
2
2
OR
30 
5
35 


2
2
2
15
15
20 
35
 10  


2
2
2
2
Check:
OR
5  10   15
35 
5
40 


 20 
2
2
2
 x
The graph of two cycles of y   4 cos   in blue compared with the graph
5
of two cycles of y   cos x in red.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
3.
y 
1
 7 x 
cos  

2
 17 
 7 x 
cos

 =
NOTE: Since the cosine function is an even function, then
 17 
 7 
cos 
x  . Thus, we have that
17


y 
1
1
 7 x 
 7 
cos  
cos 
x
 =
2
2
 17 
 17 
2
17
17
34
2


2

Period = 7  =
=
=
7
7
7
17
1
Amplitude =
2
1
1 34
1 17
17
 period =


=
=
2 7
4
4 7
14
y
1
2
x
17
14
x
17
7
51
14
34
7
85
14
- 12
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
51
7
119
14
68
7
Since the period is
34
34
, the first cycle ends at
and the second cycle ends at
7
7
68
. The other numbers on the x-axis were obtained by the following:
7
0 
17
17

14
14
17
17
34 17



14
14
14
7
34
17
51


14
14
14
51 17
68
34



Check:
14
14
14
7
68 17
85


OR
14
14
14
85 17
102
51



14
14
14
7
102
17
119


14
14
14
Check:
17
34
17
68
85




14
7
14
14
14
17
34
51


OR
7
7
7
OR
51
34
51
68
119




14
7
14
14
14
119 17 136
68



14
14
14
7
1
 7 x 
cos  
 in blue compared with the
2
 17 
graph of two cycles of y  cos x in red.
The graph of two cycles of y 
Back to Topics List
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
3.
SINE AND COSINE GRAPHS WITH PHASE SHIFTS
Definition A phase shift for a trigonometric function is a horizontal shift. That is,
it is a shift with respect to the x-axis. Thus, the shift is either right or left.
NOTE: In order to identify a horizontal shift, hence, a phase shift, the coefficient
of the x variable must be 1. If the coefficient is not 1, then you will need to factor
out the coefficient.
Given the function y  a sin ( b x  c ) , we may write this function as
 
c 
y  a sin b  x    by factoring out b.
b 
 
2
c
. The phase shift is
b
b
c
c
c
 0 or is
 0.
units to the right if
units to the left if
b
b
b
The amplitude of this function is a and the period is
Similarly, given the function y  a cos ( b x  c ) , we may write this function as
 
c 
y  a cos b  x    by factoring out b.
b 
 
2
c
. The phase shift is
b
b
c
c
c
 0 or is
 0.
units to the right if
units to the left if
b
b
b
The amplitude of this function is a and the period is
Examples Sketch one cycle of the graph of the following functions.
1.
y  3 sin ( 2 x   )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
NOTE: Since the coefficient of the x variable is not 1, the phase shift is not
 units to the right.
Since the coefficient of the x variable is 2, then we will need to factor the 2
out in order to identify the phase shift.
 
 
y  3 sin ( 2 x   )  3 sin  2  x   
2 
 
2
Period =
= 
2
Amplitude = 3
Phase Shift:

units to the right
2
y
3

2
X
3
4

5
4
3
2
x
3


units to the right, then the cycle starts at . Since
2
2
3
the period is  , then this cycle ends at
obtained by
2
2
3


  


.
2
2
2
2
Since the phase shift is
The other numbers on the x-axis were obtained by the following:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
1
1

 period =
 =
4
4
4
2
3
3








The
was obtained by
. That is, we add ,
4
2
4
4
4
4
4
which is one-fourth of the period, to the starting point of the cycle, which is

.
2
3
4




  . That is, we add , which is
4
4
4
4
3
one-fourth of the period, to the next starting point of
.
4
The  was obtained by
5
4
5




was obtained by
. That is, we add , which is one4
4
4
4
4
4
fourth of the period, to the next starting point of
.
4
The
Check:
2.
5
6
3




4
4
4
2
2 
 x
y  7 sin   

9 
 4
NOTE: Since the coefficient of the x variable is not 1, the phase shift is not
2
units to the left.
9
Since the coefficient of the x variable is 

1
, then we will need to factor the
4
1
out in order to identify the phase shift.
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
2 
 x
y  7 sin   
  7 sin
9 
 4
NOTE: The
 1
8

x


 4
9





2
2
8
8
1

 4 =
was obtained by
=
.
9
9
9
4
9
Since the sine function is an odd function, then
 1
1 
8  
8  
y  7 sin    x 
  =  7 sin   x 

9  
9 
4
 4
Since the sine function is being multiplied by a negative 7, then the graph
will be inverted. Thus, we will need to draw an inverted sine cycle.
2
Amplitude = 7
Period = 1 = 2   4 = 8 
4
Phase Shift:
8
units to the right
9
y
7
x
7
8
9
26 
9
44 
9
62 
9
80 
9
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
x
8
8
units to the right, then the cycle starts at
.
9
9
80 
Since the period is 8  , then this cycle ends at
obtained by
9
8
8
72 
80 
 8 


.
9
9
9
9
Since the phase shift is
The other numbers on the x-axis were obtained by the following:
8
1
1
 period =
 8 =
= 2
4
4
4
26 
8
8
18
26 

2




The
was obtained by
. That is, we add
9
9
9
9
9
2  , which is one-fourth of the period, to the starting point of the cycle,
8
which is
.
9
44 
26 
18
44 


was obtained by
. That is, we add
9
9
9
9
18 
2 
, which is one-fourth of the period, to the next starting point of
9
26 
.
9
The
62 
44 
18 
62 


The
was obtained by
. That is, we add
9
9
9
9
18 
2 
, which is one-fourth of the period, to the next starting point of
9
44 
.
9
62 
18 
80 


Check:
9
9
9
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
3.
 

6 sin  3 x  
4

y 
NOTE: Since the coefficient of the x variable is not 1, the phase shift is not

units to the left.
4
Since the coefficient of the x variable is 3, then we will need to factor the 3
out in order to identify the phase shift.
 

6 sin  3 x   
4

y 
Amplitude =
Phase Shift:
6
 
 
6 sin  3  x 

12  
 
2
Period =
3

units to the left
12
y
6

12
 6

x
7
12
3
4
11
12
13
12
5
4
x


units to the left, then the cycle starts at  .
12
12
2
7
Since the period is
, then this cycle ends at
obtained by
12
3
Since the phase shift is
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330

2
8
7



 


. This cycle starts to the left of the y-axis
12
3
12
12
12
and finishes to the right of the y-axis. We are sketching a graph that goes up,
down, and up again. If we allow our sketch to cross the y-axis, our picture
will probably contain misinformation about where the actual graph crosses
the y-axis. We do not want our picture to have misinformation in it. Our
sketches have not been drawn to scale, but all the numbers on the x- and yaxes have been correct.
The sketch, that we draw, does not have to cross the y-axis. We know that
where one cycle of the graph ends, another cycle begins.
7
. Let’s sketch the second cycle
12
7
5
that begins at
. This second cycle will end at
obtained by
4
12
7
8
15 
5
2
8




. That is, will add the period of
to the
3
12
12
12
12
4
7
starting point of the second cycle, which is
.
12
In this problem, our first cycle ends at
The other numbers on the x-axis were obtained by the following:
1
1 2
1 

 period =


=
=
2 3
4
4 3
6
3
7
7
2
9
3






The
was obtained by
. That is, we
4
12
6
12
12
12
4

add , which is one-fourth of the period, to the starting point of the second
6
7
cycle, which is
.
12
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
9
2
11
2
11




was obtained by
. That is, we add
,
12
12
12
12
6
12
9
which is one-fourth of the period, to the next starting point of
.
12
The
18 
11
2
13
13


was obtained by
. That is, we add 2  
,
12
12
12
12
9
11
which is one-fourth of the period, to the next starting point of
.
12
The
Check:
4.
y  
13
2
15 
5



12
12
12
4
12  
11
 6x
sin  


8
5
7


Since the coefficient of the x variable is 

6
, then we will need to factor the
5
6
out in order to identify the phase shift.
5
y  
11
11
 6 x 12  
sin  

   sin
8
7 
8
 5
 6
10   

x



 5
7  


12 
10 
6 12  5 2  5 10 

 =
 =
NOTE: The
was obtained by
=
.
7
7
7
5
7
6
7
1
Since the sine function is an odd function, then
y  
11
sin
8
 6
6 
10   
10   
11
  5  x  7   = 8 sin  5  x  7  



 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
2
5
5
5
Period = 6 = 2  
=  
=
6
3
3
5
11
Amplitude =
8
Phase Shift:
10 
units to the left
7
y
11
8

x
10 
7
- 118
5
21
55 
84
15
14
125
84
40 
21
x
10 
10 
units to the left, then the cycle starts at 
.
7
7
5
5
Since the period is
, then this cycle ends at
obtained by
3
21
10 
5
30 
35
5


 


. This cycle starts to the left of the y7
3
21
21
21
axis and finishes to the right of the y-axis. Again, since we are sketching the
graph of this function, we do not want our sketch to cross the y-axis because
our picture will probably contain misinformation about where the actual
graph crosses the y-axis.
Since the phase shift is
5
. So, let’s sketch the second cycle
21
40 
This second cycle will end at
obtained by
21
In this problem, our first cycle ends at
that begins at
5
.
21
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
5
35
40 


.
21
21
21
That is, will add the period of
starting point of the second cycle, which is
5
.
21
5
35 

to the
3
21
The other numbers on the x-axis were obtained by the following:
5
1
1 5
 period =

=
4 3
4
12
5
5
20 
35 
55 
55 




was obtained by
. That is, we
84
21
12
84
84
84
5
add
, which is one-fourth of the period, to the starting point of the second
12
5
cycle, which is
.
21
The
55
35
90 
45
15
15




was obtained by
. That is, we
84
84
84
42
14
14
5
35 

add
, which is one-fourth of the period, to the next starting point
12
84
55 
of
.
84
The
125
90 
35
125


The
was obtained by
. That is, we add
84
84
84
84
5
35 

, which is one-fourth of the period, to the next starting point of
12
84
90 
.
84
125 
35 
160 
40 



Check:
84
84
84
21
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
5.
 

y   2 cos  4 x  
6

Since the coefficient of the x variable is 4, then we will need to factor the 4
out in order to identify the phase shift.
 
 
 

y   2 cos  4 x     2 cos  4  x 

6
24




Since the cosine function is being multiplied by a negative 2, then the graph
will be inverted. Thus, we will need to draw an inverted cosine cycle.
Amplitude = 2
Phase Shift:
Period =
2

=
2
4

units to the right
24
y
2

24
x

6
7
24
5
12
13
24
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
x


units to the right, then the cycle starts at
.
24
24
13

Since the period is
, then this cycle ends at
obtained by
2
24
12 
13







.
24
2
24
24
24
Since the phase shift is
The other numbers on the x-axis were obtained by the following:
1
1 

 period =

=
4
4 2
8
3
4









 . That is, we add
The
was obtained by
24
8
24
24
24
6
6

, which is one-fourth of the period, to the starting point of the cycle, which
8

is
.
24
7
4
3
7
3




The
was obtained by
. That is, we add
,
8
24
24
24
24
24
4
which is one-fourth of the period, to the next starting point of
.
24
7
3
10 
5
5



The
was obtained by
. That is, we add
6
24
24
24
12
3
7


, which is one-fourth of the period, to the next starting point of
.
8
24
24
10 
3
13


Check:
24
24
24
6.
y 
5 
1
x
cos 


6
8 
 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Since the coefficient of the x variable is


, then we will need to factor the
3
3
out in order to identify the phase shift.
y 
NOTE: The
5 
 
1
1
15  
x
cos 

  cos   x 

6
8 
6
8  
 3
3 
5
5
15

3
5 3
15



was obtained by
=
=
=
.
8
8
3
8
8

8 1
2
3
3
2


2

Period =  =
=
= 6

1
3
1
Amplitude =
6
15
Phase Shift:
units to the left
8
y
1
6
x

x
15
8
33
8
45
8
57
8
69
8
81
8
- 16
15
15
units to the left, then the cycle starts at  .
8
8
33
Since the period is 6, then this cycle ends at
obtained by
8
15
15
48
33

 6  


. This cycle starts to the left of the y-axis and
8
8
8
8
Since the phase shift is
Copyrighted by James D. Anderson, The University of Toledo
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finishes to the right of the y-axis. Again, since we are sketching the graph of
this function, we do not want our sketch to cross the y-axis because our
picture will probably contain misinformation about where the actual graph
crosses the y-axis.
33
In this problem, our first cycle ends at
. So, let’s sketch the second cycle
8
33
81
that begins at
. This second cycle will end at
obtained by
8
8
33
48 81
48

 . That is, will add the period of 6 
to the starting point
8
8
8
8
33
of the second cycle, which is
.
8
The other numbers on the x-axis were obtained by the following:
1
1
6
3
 period =
6 =
=
4
2
4
4
5
45
33
3
33 12
45




was obtained by
. That is, we add
,
8
8
2
8
8
8
12
which is one-fourth of the period, to the starting point of the second cycle,
33
which is
.
8
The
57
45
12
57
3 12


The
was obtained by
. That is, we add 
, which is
8
2
8
8
8
8
45
one-fourth of the period, to the next starting point of
.
8
69
57
12
69
3 12



The
was obtained by
. That is, we add
, which is
8
8
8
8
2
8
57
one-fourth of the period, to the next starting point of
.
8
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Check:
7.
69
12
81


8
8
8
 2

y  cos 
 x
 5

Since the coefficient of the x variable is  1 , then we will need to factor the
 1 out in order to identify the phase shift.
 
2 
2  
 2


y  cos 
 x   cos   x 
  cos    x 

5
5
5





 
Since the cosine function is an even function, then
 
2  
2 

y  cos    x 
   cos  x 

5 
5 

 
Amplitude = 1
Period = 2 
2
Phase Shift:
units to the right
5
y
1
2
5
x
9
10
7
5
19 
10
12 
5
1
Copyrighted by James D. Anderson, The University of Toledo
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x
2
2
units to the right, then the cycle starts at
.
5
5
12 
Since the period is 2  , then this cycle ends at
obtained by
5
2
2
10 
12 
 2 


.
5
5
5
5
Since the phase shift is
The other numbers on the x-axis were obtained by the following:
2
1

 period =
=
4
2
4
9
2
4
5
9






The
was obtained by
. That is, we add ,
5
2
10
10
10
10
2
which is one-fourth of the period, to the starting point of the cycle, which is
2
.
5
7
9
5
14 
7



was obtained by
. That is, we add
10
10
10
5
5
5
9


, which is one-fourth of the period, to the next starting point of
.
2
10
10
The
19 
5
14 
5
19 




was obtained by
. That is, we add
,
2
10
10
10
10
10
14 
which is one-fourth of the period, to the next starting point of
.
10
The
Check:
19 
5
24 
12 



10
10
10
5
Back to Topics List
4.
SECANT AND COSECANT GRAPHS
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www.math.utoledo.edu/~janders/1330
Given the function y  a csc ( b x  c ) , we may write this function as
 
c 
y  a csc b  x    by factoring out b. The cosecant function does not have
b 
 
2
c
c
 0
an amplitude, the period is
, and the phase shift is units to the right if
b
b
b
c
c
 0 . In order to obtain a sketch of the graph of the
units to the left if
b
b
cosecant function, you will make use of the sketch of the graph of sine function.
First sketch the graph of y  a sin ( b x  c ) and then locate the x-intercepts of the
sketch. These are the locations of the vertical asymptotes of the cosecant function.
Draw these vertical asymptotes and then use the sketch of the graph of the sine
function to sketch the graph of the cosecant function.
or is
Similarly, given the function y  a sec ( b x  c ) , we may write this function as
 
c 
y  a sec b  x    by factoring out b. The secant function does not have an
b 
 
2
c
c
amplitude, the period is
, and the phase shift is units to the right if  0 or
b
b
b
c
c
 0 . In order to obtain a sketch of the graph of the
units to the left if
b
b
secant function, you will make use of the sketch of the graph of cosine function.
First sketch the graph of y  a cos ( b x  c ) and then locate the x-intercepts of the
sketch. These are the locations of the vertical asymptotes of the secant function.
Draw these vertical asymptotes and then use the sketch of the graph of the cosine
function to sketch the graph of the secant function.
is
Examples Sketch two cycles of the graph of the following functions. Label the
numbers on the y-axis. On the x-axis, only label where the cycles begin and end.
1.
y  2 csc 6 x
First, sketch the graph of y  2 sin 6 x . For this sine function, we have the
following:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Amplitude = 2
Period =
2

=
3
6
Phase Shift: None

, then the first cycle starts at
3
2


0 and end at . The second cycle starts at
and ends at
.
3
3
3
Since there is no phase shift and the period is
y
2

3
x
2
2.
y 
2
3
x
8
 x
csc  

5
4 

8
 x
sin  
 . Since the sine function is an
5
4 

  x
 
sin



sin


 x  . Thus, we have that
odd function, then
 4 
 4 
First, sketch the graph of y 
y 
8
8
 x
 
sin  
 =  sin  x 
5
5
4 
 4 

Copyrighted by James D. Anderson, The University of Toledo
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8
, then the graph
5
will be inverted. Thus, we will need to draw two inverted sine cycles.
Since the sine function is being multiplied by a negative
For this sine function, we have the following:
2
4
4
Period =  = 2  
= 2 
= 8

1
4
8
Amplitude =
5
Phase Shift: None
Since there is no phase shift and the period is 8, then the first cycle starts at 0
and end at 8. The second cycle starts at 8 and ends at 16.
y
8
5
x
8
-
3.
y 
8
5
5 

5 csc  x 

6 

Copyrighted by James D. Anderson, The University of Toledo
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16
x
5 

5 sin  x 
 . Notice the coefficient of
6 

the x variable is 1. For this sine function, we have the following:
First, sketch the graph of y 
Phase Shift:
Period = 2 
5
Amplitude =
5
units to the right
6
5
5
units to the right, then the first cycle starts at
.
6
6
17 
Since the period is 2  , then this first cycle ends at
, obtained by
6
5
5
12 
17 
29 
 2 


, and the second cycle ends at
,
6
6
6
6
6
17 
12 
29 


obtained by
.
6
6
6
Since the phase shift is
y
5
5
6
x

17 
6
5
Copyrighted by James D. Anderson, The University of Toledo
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29 
6
x
4.
 3x 
y  12 sec 

 8 
3 
y

12
cos
 x  . For this cosine function, we
First, sketch the graph of
8 
have the following:
2
16 
8
Amplitude = 12
Period = 3 = 2  
=
3
3
8
Phase Shift: None
16 
, then the first cycle starts
3
16 
16 
32 
at 0 and end at
. The second cycle starts at
and ends at
.
3
3
3
Since there is no phase shift and the period is
y
12
x
x
5.
 12
16 
3
y   sec  x
First, sketch the graph of y   cos  x .
Copyrighted by James D. Anderson, The University of Toledo
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32 
3
x
Since the cosine function is being multiplied by a negative 1, then the graph
will be inverted. Thus, we will need to draw two inverted cosine cycles.
For this cosine function, we have the following:
2
Period =
= 2

Amplitude = 1
Phase Shift: None
Since there is no phase shift and the period is 2, then the first cycle starts at 0
and end at 2. The second cycle starts at 2 and ends at 4.
y
1
x
x
6.
2
4
x
1
3 

y  5 sec  7 x 

4 

3 

 . Since the coefficient of
First, sketch the graph of y  5 cos  7 x 
4 

the x variable is 7, then we will need to factor the 7 out in order to identify
the phase shift. Thus, we have that
Copyrighted by James D. Anderson, The University of Toledo
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3 

y  5 cos  7 x 
  5 cos
4 

 
3  
7
x




28



For this cosine function, we have the following:
Amplitude = 5
Period =
2
7
3
Phase Shift:
units to the left
28
3
units to the left, then the first cycle starts at
28
3
2
5

. Since the period is
, then this first cycle ends at
, obtained by
28
28
7
3
2
3
8
5


 


. This cycle starts to the left of the y-axis
28
7
28
28
28
and finishes to the right of the y-axis. Since we are sketching the graph of
this function, we do not want our sketch to cross the y-axis because our
picture will probably contain misinformation about where the actual graph
crosses the y-axis.
Since the phase shift is
5
. So, let’s sketch the second cycle
28
5
13
that begins at
as our first cycle. This cycle will end at
obtained by
28
28
5
8
13
3


.
The next cycle will end at
, obtained by
4
28
28
28
13
8
21
3



.
28
28
28
4
In this problem, our first cycle ends at
Copyrighted by James D. Anderson, The University of Toledo
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y
5
x

3
28
5
13
28
5
28
3
4
x
Back to Topics List
5.
TANGENT GRAPHS
Example Find the x-intercepts of the graph of y  tan x in the interval
[  2 , 2 ] .
NOTE: The interval [  2  , 2  ] of angles are the angles going one time around
the Unit Circle clockwise for the subinterval [  2  , 0 ] and the angles going one
time around the Unit Circle counterclockwise for the subinterval [ 0 , 2  ] .
To find the x-intercepts, set y equal to 0: tan x  0 .
sin x
sin x
tan
x

tan
x

0

 0.
Since
cos x , then
cos x
Since a fraction can only equal zero when the numerator of the fraction equals zero,
sin x
 0  sin x  0 . By Unit Circle Trigonometry, we are looking for
then
cos x
angles in the interval [  2  , 2  ] that intersect the Unit Circle so that the yCopyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
coordinate of the point of intersection is 0. Going around the Unit Circle
clockwise, these angles are
x  0 ,   ,  2  . Going around the Unit Circle counterclockwise, these angles
are x  0 ,  , 2  .
Thus, the x-intercepts of the graph of y  tan x in the interval [  2  , 2  ] are the
points (  2  , 0 ) , (   , 0 ) , ( 0 , 0 ) , (  , 0 ) , and ( 2  , 0 ) .
Example Find the vertical asymptotes of the graph of y  tan x in the interval
[  2 , 2 ] .
NOTE: The interval [  2  , 2  ] of angles are the angles going one time around
the Unit Circle clockwise for the subinterval [  2  , 0 ] and the angles going one
time around the Unit Circle counterclockwise for the subinterval [ 0 , 2  ] .
sin x
tan
x

Since
cos x , then the vertical asymptotes will occur where the
denominator of this fraction is equal to zero. Thus, we want to solve the equation
cos x  0 in the interval [  2  , 2  ] .
By Unit Circle Trigonometry, we are looking for angles in the interval [  2  , 2  ]
that intersect the Unit Circle so that the x-coordinate of the point of intersection is
3

x


,

0. Going around the Unit Circle clockwise, these angles are
.
2
2
 3
x

,
Going around the Unit Circle counterclockwise, these angles are
.
2
2
Thus, the vertical asymptotes of the graph of y  tan x in the interval [  2  , 2  ]
3
3


x


,
x


,
x

,
x

are
and
.
2
2
2
2
Example Sketch two cycles of the graph of y  tan x using the x-intercepts and
vertical asymptotes of the function.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
y


2

2

3
2
x
The function y  tan x does not have an amplitude and the period of the function
is  .
function y  a tan ( b x  c ) , we may write this function as

  by factoring out b. The tangent function does not have an


c
c
 0 or
amplitude, the period is
, and the phase shift is units to the right if
b
b
b
c
c
 0.
is
units to the left if
b
b
In general, given the
 
c
y  a tan b  x 
b
 
Theorem The tangent function is an odd function. That is, tan (   )   tan 
for all  in the domain of the function.
A sketch of the graph of the tangent function can be obtained from the x-intercepts
and vertical asymptotes of the function. Two consecutive vertical asymptotes are a
distance of the period from each other. The x-coordinates of the x-intercepts are
Copyrighted by James D. Anderson, The University of Toledo
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midway between two consecutive vertical asymptotes. That is, the x-coordinates
of the x-intercepts are the midpoint of two consecutive vertical asymptotes.
So to sketch the graph of a tangent function, you only need to know where the first
vertical asymptote is located. Then use the period to find the next consecutive
vertical asymptote. For an unshifted tangent function, the first two consecutive
vertical asymptotes are symmetric about the y-axis. For a shifted tangent function,
first shift one vertical asymptote for the unshifted graph. Once you have the
vertical asymptotes, you find the x-coordinates of the x-intercepts by finding the
midpoint of two consecutive vertical asymptotes.
Recall the midpoint of the numbers a and b is the number
a  b
.
2
Examples Sketch two cycles of the graph of the following functions.
1.
y  8 tan 5 x

Period =
5
Amplitude: None
Phase Shift: None
1

period =
10
2
y


10

10

5
Copyrighted by James D. Anderson, The University of Toledo
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3
10
x
3
2
3








was obtained by
. That is, we add ,
10
5
10
5
10
10
10

which is the period, to the starting point of
, which is where the first
10
vertical asymptote to the right of the y-axis crosses the x-axis.
The
3 
1
1  4 
1  2  









  . That
The
was obtained by
2  10
10 
2  10 
2 5 
5
5
3

is, we found the midpoint of
and
. Of course, you could also use the
10
10
fact that two consecutive x-intercepts are a distance of the period from each
other. Since the first x-intercept is 0 and the period is
intercept is
2.
y  

.
5

, then the second x5
 9x 
21 tan 

8


Since the tangent function is being multiplied by a negative 21 , then the
graph will be inverted. Thus, we will need to draw two inverted tangent
cycles.

8
8
Period = 9 =  
=
9
9
8
Amplitude: None
Phase Shift: None
4
1
1 8

period =
=
2
2
9
9
Copyrighted by James D. Anderson, The University of Toledo
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y

4
9
4
9
8
9
12  4 

9
3
x
8
4
4
8
12 
4



The
was obtained by
. That is, we add
,
9
9
9
3
3
9
4
which is the period, to the starting point of
, which is where the first
9
vertical asymptote to the right of the y-axis crosses the x-axis.
12  
8
1  4
1  16  
8

  
 
was obtained by 
. That is, we
2 9
9 
2 9 
9
9
4
12 
found the midpoint of
and
. Of course, you could also use the fact
9
9
that two consecutive x-intercepts are a distance of the period from each other.
8
Since the first x-intercept is 0 and the period is
, then the second x9
8
intercept is
.
9
The
3.
y 
4
tan (   x )
7
Copyrighted by James D. Anderson, The University of Toledo
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NOTE: Since the tangent function is an odd function, then tan (   x ) =
 tan  x . Thus, we have that
y 
4
4
tan (   x )   tan  x
7
7
4
, then the
7
graph will be inverted. Thus, we will need to draw two inverted tangent
cycles.
Since the tangent function is being multiplied by a negative
Amplitude: None
Period =

= 1

Phase Shift: None
1
1
period =
2
2
y
x

The
1
2
1
2
1
3
2
3
1
1
2
3

 . That is, we add 1, which is
was obtained by  1 
2
2
2
2
2
Copyrighted by James D. Anderson, The University of Toledo
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1
, which is where the first vertical
2
asymptote to the right of the y-axis crosses the x-axis.
the period, to the starting point of
11
3
14
1
       ( 2 )  1 . That is, we
22
2
22
2
1
3
found the midpoint of
and . Of course, you could also use the fact that
2
2
two consecutive x-intercepts are a distance of the period from each other.
Since the first x-intercept is 0 and the period is 1, then the second x-intercept
is 1.
The 1 was obtained by
4.
4 

y  12 tan  3 x 

7 

Since the coefficient of the x variable is 3, then we will need to factor the 3
out in order to identify the phase shift. Thus, we have that
 
4 
4  

y  12 tan  3 x 
  12 tan  3  x 

7 
21





Period =
3
Amplitude: None
Phase Shift:
4
units to the right
21
1
1 


period =
=
2 3
2
6
For the unshifted tangent graph the first two consecutive vertical asymptotes
are symmetric about the y-axis separated by a distance of the period, which is

for this function. Thus, the first vertical asymptote to the right of the
3
Copyrighted by James D. Anderson, The University of Toledo
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y-axis for the unshifted tangent graph is x 

. Let’s shift this vertical
6
4
units to the right. Thus, the first vertical asymptote for the
21
5
4
7
8
15 
5

x






shifted graph is
, obtained by
.
14
6
21
42
42
42
14
4
That is, we add
, which is the amount of the shift, to the starting point of
21

, which is the first unshifted vertical asymptote to the right of the y-axis
6
crosses the x-axis.
asymptote
y
5
14
11
21
29 
42
6
7
43 
42
x
5
15 
14 
29 
29 





was obtained by
. That is, we
42
14
3
42
42
42
5

add , which is the period, to the starting point of
, which is where the
3
14
first shifted vertical asymptote to the right of the y-axis crosses the x-axis.
The
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
43
29 
14 
43
14 




was obtained by
. That is, we add
,
42
42
42
42
3
42
29 
which is the period, to the starting point of
, which is where the second
42
vertical asymptote crosses the x-axis.
The
29  
1  5
1
11




The
was obtained by
2  14
42 
2
21
29  
 15 


 
42 
 42
1  44  
1  22   11
5

  
 
. That is, we found the midpoint of
and
2  42 
2  21 
21
14
29 
.
42
The
43 
1  29 
1  72  
1  36  
6

  
  
 
was obtained by 
2  42
42 
2  42 
2  21 
7
6
1  12  
43 
29 

 
. That is, we found the midpoint of
and
. Of
2 7 
7
42
42
course, you could also use the fact that two consecutive x-intercepts are a
distance of the period from each other. Since the first x-intercept, that we
11
7


found using the midpoint formula, is
and the period is
, then
3
21
21
11
7
18 
6



the second x-intercept is
.
21
21
21
7
5.
11 
 6x
y  tan 


21 
 7
6
6
Since the coefficient of the x variable is , then we will need to factor the
7
7
out in order to identify the phase shift. Thus, we have that
Copyrighted by James D. Anderson, The University of Toledo
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6 
11 
11
 6x
y  tan 

  tan   x 
21 
18
 7
7 
NOTE: The
11
.
18
11
11 7
11 1
11
6



was obtained by
=
=
=
3
6
18
21
7
21
6

7
7
Period = 6 =  
=
6
6
7
Amplitude: None
Phase Shift:



11
units to the left
18
7
1
1 7

period =
=
2
6
2
12
For the unshifted tangent graph the first two consecutive vertical asymptotes
are symmetric about the y-axis separated by a distance of the period, which is
7
for this function. Thus, the first vertical asymptote to the left of the
6
7
x


y-axis for the unshifted tangent graph is
. Let’s shift this vertical
12
11
asymptote
units to the left. Thus, the first vertical asymptote for the
18
43
shifted graph is x  
, obtained by
36
7  11
21
22 
43
11


 

 
. That is, we subtract
, which
12
18
36
36
36
18
7
is the amount of the shift, to the starting point of 
, which is the first
12
unshifted vertical asymptote to the right of the y-axis crosses the x-axis.
Note that we subtract the amount of the shift because we are moving to the
left.
Copyrighted by James D. Anderson, The University of Toledo
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y
127 

36
85 
36
x
53

18
85
16 


36
9
43

36
43
7
43
42 
85

 

 
.
36
6
36
36
36
7
43
That is, we subtract
, which is the period, to the starting point of 
,
36
6
which is where the first shifted vertical asymptote to the left of the y-axis
crosses the x-axis.
The 
was obtained by 
127 
85 
42 
127 

 
was obtained by 
. That is, we
36
36
36
36
85 
7
42 

subtract
, which is the period, to the starting point of 
,
6
36
36
which is where the second vertical asymptote crosses the x-axis.
The 
The 
1  85
1  128  
16 
 43  







 
was obtained by 

2  36
2
36 
9
 36  
16 
1  32  

  
.
2
9 
9
43

.
36
That is, we found the midpoint of 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
85 
and
36
The 
1  127 
1  212  
53
 85  
 
   
 
was obtained by  
2
36
36
2
36
18




53
1  53 
127 


  
. That is, we found the midpoint of
and
2
9 
18
36
85 

. Of course, you could also use the fact that two consecutive x36
intercepts are a distance of the period from each other. Since the first x16 
intercept, that we found using the midpoint formula, is 
and the period
9
7
16 
7
32 
21

 


is
, then the second x-intercept is 
9
6
18
18
6
53

.
18
Back to Topics List
6.
COTANGENT GRAPHS
Example Find the x-intercepts of the graph of y  cot x in the interval
[  2 , 2 ] .
NOTE: The interval [  2  , 2  ] of angles are the angles going one time around
the Unit Circle clockwise for the subinterval [  2  , 0 ] and the angles going one
time around the Unit Circle counterclockwise for the subinterval [ 0 , 2  ] .
To find the x-intercepts, set y equal to 0: cot x  0 .
Since cot x 
cos x
cos x
 0.
, then cot x  0 
sin x
sin x
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Since a fraction can only equal zero when the numerator of the fraction equals zero,
cos x
 0  cos x  0 . By Unit Circle Trigonometry, we are looking for
then
sin x
angles in the interval [  2  , 2  ] that intersect the Unit Circle so that the xcoordinate of the point of intersection is 0. Going around the Unit Circle
3

x


,

clockwise, these angles are
. Going around the Unit Circle
2
2
 3
,
counterclockwise, these angles are x 
.
2
2
Thus, the x-intercepts of the graph of y  cot
 3
  
 

, 0  ,   , 0  ,  , 0  , and
points  
2

  2   2 
x in the interval [  2  , 2  ] are the
 3

, 0 .

 2

Example Find the vertical asymptotes of the graph of y  cot x in the interval
[  2 , 2 ] .
NOTE: The interval [  2  , 2  ] of angles are the angles going one time around
the Unit Circle clockwise for the subinterval [  2  , 0 ] and the angles going one
time around the Unit Circle counterclockwise for the subinterval [ 0 , 2  ] .
cos x
sin x , then the vertical asymptotes will occur where the
denominator of this fraction is equal to zero. Thus, we want to solve the equation
sin x  0 in the interval [  2  , 2  ] .
Since cot x 
By Unit Circle Trigonometry, we are looking for angles in the interval [  2  , 2  ]
that intersect the Unit Circle so that the y-coordinate of the point of intersection is
0. Going around the Unit Circle clockwise, these angles are x  0 ,   ,  2  .
Going around the Unit Circle counterclockwise, these angles are x  0 ,  , 2  .
Thus, the vertical asymptotes of the graph of y  cot x in the interval [  2  , 2  ]
are x   2  , x    , x  0 , x   , and x  2  .
Copyrighted by James D. Anderson, The University of Toledo
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NOTE: The vertical line given by the equation x  0 is the y-axis.
Example Sketch two cycles of the graph of y  cot x using the x-intercepts and
vertical asymptotes of the function.
y

2

3
2
x
2
The function y  cot x does not have an amplitude and the period of the function
is  .
In general, given the function y  a cot ( b x  c ) , we may write this function as
 
c 
y  a cot b  x    by factoring out b. The cotangent function does not have
b 
 

c
c
 0
an amplitude, the period is
, and the phase shift is units to the right if
b
b
b
c
c
 0.
or is
units to the left if
b
b
Theorem The cotangent function is an odd function. That is, cot (   )   cot 
for all  in the domain of the function.
Copyrighted by James D. Anderson, The University of Toledo
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Like the tangent function, a sketch of the graph of the cotangent function can be
obtained from the x-intercepts and vertical asymptotes of the function. Two
consecutive vertical asymptotes are a distance of the period from each other. The
x-coordinates of the x-intercepts are midway between two consecutive vertical
asymptotes. That is, the x-coordinates of the x-intercepts are the midpoint of two
consecutive vertical asymptotes.
So to sketch the graph of a cotangent function, you only need to know where the
first vertical asymptote is located. Then use the period to find the next consecutive
vertical asymptote. For an unshifted cotangent function, the first vertical asymptote
is the y-axis. For a shifted cotangent function, first shift one vertical asymptote for
the unshifted graph. Once you have the vertical asymptotes, you find the xcoordinates of the x-intercepts by finding the midpoint of two consecutive vertical
asymptotes.
Examples Sketch two cycles of the graph of the following functions.
1.
y 
7 cot 8 x

Period =
8
Amplitude: None
Phase Shift: None
y

16

8
3
16
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2 

8
4
x
1
  1

0


 
The
was obtained by
2
8
2
16

the midpoint of 0 and .
8

 
  
. That is, we found
 8  16
2 
1
1  3  3
3





 
The
was obtained by
. That is, we
2 8
8 
2 8 
16
16
2



found the midpoint of
and
. Of course, you could also use the
8
4
8
fact that two consecutive x-intercepts are a distance of the period from each
other. Since the first x-intercept, that we found using the midpoint formula,
2



is
and the period is
, then the second x-intercept is
16
8
16
2
3



.
16
16
16
2.
 x
y  15 cot   
 2
 x
cot
  =
NOTE: Since the cotangent function is an odd function, then
 2
1 
 cot  x  . Thus, we have that
2 
 x
1 
y  15 cot      15 cot  x 
 2
2 
Since the cotangent function is being multiplied by a negative 15, then the
graph will be inverted. Thus, we will need to draw two inverted cotangent
cycles.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330

Period = 1 = 2 
2
Amplitude: None
Phase Shift: None
y

The  was obtained by
midpoint of 0 and  .
2
3
4
x
1
1
( 0  2  )  ( 2  )   . That is, we found the
2
2
1
1
( 2   4  )  ( 6  )  3 . That is, we found
2
2
the midpoint of 2  and 4  . Of course, you could also use the fact that two
consecutive x-intercepts are a distance of the period from each other. Since
the first x-intercept, that we found using the midpoint formula, is  and the
period is 2  , then the second x-intercept is   2   3 .
The 3  was obtained by
3.
y 
3
 9

cot 
 6x 
5
 2

Copyrighted by James D. Anderson, The University of Toledo
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Since the coefficient of the x variable is  6 , then we will need to factor the
 6 out in order to identify the phase shift. Thus, we have that

9  3
3  
3
 9
 3


cot 
 6 x   cot   6 x 
  cot   6  x 

5
2  5
4 
 2
 5



3
9
9 1
3 1
3
 6 =


NOTE: The
was obtained by
=
=
.
4
2
2
6
2
2
4
y 
NOTE:
Since the cotangent function is an odd function, then

 
3 
3 

cot   6  x 
 =  cot  6  x 
 . Thus, we have that
4 
4 


 
y 

 
3 
3 
3
3

cot   6  x 
   cot  6  x 

5
4
5
4




 
3
, then the
5
graph will be inverted. Thus, we will need to draw two inverted cotangent
cycles.
Since the cotangent function is being multiplied by a negative
Amplitude: None
Phase Shift:
Period =

6
3
units to the right
4
For the unshifted cotangent graph the first vertical asymptote is the y-axis.
3
Let’s shift this vertical asymptote
units to the right. Thus, the first
4
3
vertical asymptote for the shifted graph is x 
, obtained by
4
3
3
3
0

. That is, we add
, which is the amount of the shift, to the
4
4
4
starting point of 0.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
y
3
4
5
6
11
12

13 
12
x
11
3

9
2
11




The
was obtained by
. That is, we add
4
6
12
12
12
12
3

, which is the period, to the starting point of
, which is where the first
4
6
shifted vertical asymptote to the right of the y-axis crosses the x-axis.
13
11
2
13
2




was obtained by
. That is, we add
,
6
12
12
12
12
12
11
which is the period, to the starting point of
, which is where the second
12
vertical asymptote crosses the x-axis.
The
11 
11 
1  3
1  9
5






 
The
was obtained by
2 4
12 
2  12
12 
6
5
1  20  
1  5 
3

  
 
. That is, we found the midpoint of
and
2  12 
2 3 
6
4
11
.
12
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
13 
1  11
1  24  
1


  
  ( 2  )   . That
2  12
12 
2  12 
2
11
13
is, we found the midpoint of
and
. Of course, you could also use
12
12
the fact that two consecutive x-intercepts are a distance of the period from
each other. Since the first x-intercept, that we found using the midpoint
5

formula, is
and the period is
, then the second x-intercept is
6
6
5

6


.
6
6
6
The  was obtained by
4.
x

y  2 cot   4  
7

1
1
, then we will need to factor the
7
7
out in order to identify the phase shift. Thus, we have that
Since the coefficient of the x variable is
 x

y  2 cot   4    2 cot
7

Amplitude: None
1

 7 ( x  28  )

Period = 1 = 7 
7
Phase Shift: 28 units to the left
For the unshifted cotangent graph the first vertical asymptote is the y-axis.
Let’s shift this vertical asymptote 28 units to the right. Thus, the first
vertical asymptote for the shifted graph is x   28 , obtained by
0  28   28 . That is, we subtract 28 , which is the amount of the
shift, to the starting point of 0. Note that we subtract the amount of the shift
because we are moving to the left.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
y
 42 

77 
2
 35 

63 
2
x
 28 
The  35  was obtained by  28  7    35 . That is, we subtract
7  , which is the period, to the starting point of  28  , which is where the
first shifted vertical asymptote to the left of the y-axis crosses the x-axis.
The  42  was obtained by  35  7    42  . That is, we subtract
7  , which is the period, to the starting point of  35  , which is where the
second shifted vertical asymptote to the left of the y-axis crosses the x-axis.
The 
63
63
1
1
was obtained by [  35   (  28  ) ]  (  63 )  
.
2
2
2
2
That is, we found the midpoint of  35  and  28  .
The 
77 
77 
1
1
was obtained by [  42   (  35 ) ]  (  77  )  
.
2
2
2
2
That is, we found the midpoint of  42  and  35  . Of course, you could
also use the fact that two consecutive x-intercepts are a distance of the period
from each other. Since the first x-intercept, that we found using the midpoint
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
63
14 
and the period is 7  
, then the second x-intercept
2
2
63
14 
77 

 
is 
.
2
2
2
formula, is 
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Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330