LESSON 4 COTERMINAL ANGLES
Topics in this lesson:
1.
THE DEFINITION AND EXAMPLES OF COTERMINAL ANGLES
2.
FINDING COTERMINAL ANGLES
3.
TRIGONOMETRIC FUNCTIONS OF COTERMINAL ANGLES
1.
THE DEFINITION AND EXAMPLES OF COTERMINAL ANGLES
Definition Two angles are said to be coterminal if their terminal sides are the
same.
Examples Here are some examples of coterminal angles.
1.
The two angles of   140  and    220  are coterminal angles.


Animation of the making of these two coterminal angles.
2.
The two angles of  
7
31
and  
are coterminal angles.
6
6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330


Animation of the making of these two coterminal angles.
3.
The two angles of    1845  and   1395  are coterminal angles.


Animation of the making of these two coterminal angles.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
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2.
FINDING COTERMINAL ANGLES
Theorem The difference between two coterminal angles is a multiple (positive or
negative) of 2  or 360  .
Examples Find three positive and three negative angles that are coterminal with
the following angles.
1.
  660 
1  660   360   1020 
 2  1020   360   1380 
NOTE:  2  1380   1020   360   ( 660   360 )  360   660   2 ( 360 )
 3  660   360   300 
 4  300   360    60 
NOTE:  4   60   300   360   ( 660   360 )  360   660   2( 360 )
 5   60   360    420 
NOTE:  5   420    60   360   [660   2 (360 )]  360   660   3( 360 )
 6   420   360    780 
NOTE:  6   780    420   360   [660   3 (360 )]  360   660   4( 360 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
NOTE: There are other answers for this problem. If the difference between
the angle of   660  and another angle is a multiple of 360  , then this
second angle is an answer to problem.
2.

87 
5
1 
87 
87  10  77 
 2 


5
5
5
5
2 
77 
77  10  67 
 2 


5
5
5
5
NOTE:  2 
3 
67 
67  10  57 
 2 


5
5
5
5
NOTE:  3 
4 
67  77 
87 
 87 


 2  
 2   2 
 2 (2  )
5
5
5
5


57  67 
87 
 87 


 2  
 2 (2  )   2  
 3 (2  )
5
5
5
5


87 
87 
3
 10   87  90 
 9 (2  ) 
9



5
5
5
5
5
 5 
5  
3
3 10 
13
 2  


5
5
5
5
13
3
87 
 87 







2



9
(
2

)

2


 10 (2  )


NOTE: 5
5
5
5
 5

6  
13
13 10 
23
 2  


5
5
5
5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
NOTE:  6  
23
13
87 
 87 


 2  
 10 (2  )   2  
 11 (2  )
5
5
5
 5

NOTE: There are also other answers for this problem. If the difference
87 
between the angle of  
and another angle is a multiple of 2  , then
5
this second angle is an answer to problem.
87 
is made. We will need the following property of
5
arithmetic, which comes from the check for long division.
Let’s see how the angle  
b
q
a

a  q b  r
r
Dividing both sides of the equation a  q  b  r by b, we obtain the equation
a
r
 q  . For the work that we will do in order to find one particular coterminal
b
b
angle of a given angle in radians, we will want the number q above to be an even
number.
Now, consider the fraction of
87 
87
in the angle  
given above.
5
5
17
87
 5 87
5
5
37
35
2
87
2
 17  . The number 17 is an odd number. However, we may write 17
5
5
as 17  16  1 , where the number 16 is an even number. Thus, we have that
Thus,
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
87
2
2

 17   (16  1)   16   1 
5
5
5

Thus,
2
7
5 2
  16      16 
5
5
5 5
87
7
 16  . Now, multiply both sides of this equation by  to obtain that
5
5
87 

  16 
5

7
7
7
 8 ( 2 ) 
  16  
5
5
5
87 
. In order to make the angle
5
7
87 

, it will take eight complete revolutions and an additional rotation of
5
5
87 


going in the counterclockwise direction in order to make the angle
.
5
This equation tells us how to make the angle  
Thus, the two angles  
7
 2 .
5
7
87 
and
are coterminal angles.
5
5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Notice that
Animation of the making of these two coterminal angles.
Another animation of the making of these two coterminal angles.
Examples Find the angle between 0 and 2  or the angle between  2  and 0 that
is coterminal with the following angles.
1.
 
85 
6
85 
85


Consider the fraction of
in the angle
.
6
6
14
85
 6 85
6
6
25
24
1
85
1

14

Thus,
. The 14 is an even number. So, multiply both sides of
6
6
this equation by  to obtain that
85 

  14 
6

1


 7 ( 2 ) 
  14  
6
6
6
85 
, it will take seven complete revolutions
6

and an additional rotation of
going in the counterclockwise direction.
6

Thus, the angle of
is the coterminal angle that we are looking for.
6
In order to make the angle  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Animation of the making of these two coterminal angles. Another animation
of the making of these two coterminal angles.
Answer:
2.
 

6
91
4
91
91



Consider the fraction of
in the angle
.
4
4
22
91
 4 91
4
8
11
8
3
91
3
 22  . The 22 is an even number. Now, multiply both sides of
4
4
this equation by   to obtain that
Thus,

91
3
3
3

    22     22  
 11(  2  ) 
4
4
4
4

91
, it will take eleven complete
4
3
revolutions and an additional rotation of
going in the clockwise
4
3
direction. Thus, the angle of 
is the coterminal angle that we are
4
looking for. Animation of the making of these two coterminal angles.
In order to make the angle   
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Answer: 
3.
 
3
4
112 
3
112 
112



Consider the fraction of
in the angle
.
3
3
37
112
 3 112
3
9
22
21
1
112
1

37

Thus,
. The 37 is an odd number. So, we may write 37 as
3
3
37  36  1 , where the number 36 is an even number. Thus, we have that
112
1
1

 37   ( 36  1)   36   1 
3
3
3

1
4
 3 1
  36      36 
3
3
 3 3
112
4
 36  . Now, multiply both sides of this equation by   to
3
3
obtain that
Thus,

112 
4
4
4

    36     36  
 18 (  2  ) 
3
3
3
3

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
112 
, it will take eighteen complete
3
4
revolutions and an additional rotation of
going in the clockwise
3
4

direction. Thus, the angle of
is the coterminal angle that we are
3
looking for.
In order to make the angle   
Answer: 
4.

4
3
59 
2
59 
59


Consider the fraction of
in the angle
.
2
2
29
59
 2 59
2
4
19
18
1
59
1
 29  . The 29 is an odd number. So, we may write 29 as
2
2
29  28  1 , where the number 28 is an even number. Thus, we have that
Thus,
59
1
1

 29   ( 28  1 )   28   1 
2
2
2

Thus,
1
3
2 1
  28      28 
2
2
 2 2
59
3
 28  . Now, multiply both sides of this equation by  to obtain
2
2
that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
59 
3
3
3

   28    28  
 14 ( 2  ) 
2
2
2
2

59 
, it will take fourteen complete
2
3
revolutions and an additional rotation of
going in the counterclockwise
2
3
direction. Thus, the angle of
is the coterminal angle that we are looking
2
for. Animation of the making of these two coterminal angles.
In order to make the angle  
Answer:
5.
 
3
2
17 
2
17 
17



Consider the fraction of
in the angle
.
2
2
8
17
 2 17
2
16
1
17
1
 8  . The 8 is an even number. Now, multiply both sides of
2
2
this equation by   to obtain that
Thus,

17 
1



    8     8 
 4 (  2 ) 
2
2
2
2

Copyrighted by James D. Anderson, The University of Toledo
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In order to make the angle   
and an additional rotation of
angle of 
17 
, it will take four complete revolutions
2

going in the clockwise direction. Thus, the
2

is the coterminal angle that we are looking for. Animation of
2
the making of these two coterminal angles.
Answer: 
6.

2
  82 
The number 82 is an even number. Thus, 82   41( 2  ) . In order to make
the angle   82  , it will take forty-one complete revolutions. Thus, the
angle of 0 is the coterminal angle that we are looking for.
Answer: 0
7.
   21 
The 21 is an odd number. So, we may write 21 as 21  20  1 , where the
number 20 is an even number. Now, multiply both sides of this equation by
  to obtain that
 21     ( 20  1)   20     10 (  2  )  
In order to make the angle    21  , it will take ten complete revolutions
and an additional rotation of  going in the clockwise direction. Thus, the
angle of   is the coterminal angle that we are looking for.
Answer:  
8.
  4110 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Use your calculator to find the whole number of times that 360 will divide
into 4110 . Since 4110  360  11.41666 .... and 360 times 11 equals
3960, then we have that
11
360 4110
3960
150
Thus, 4110  11 ( 360 )  150 . When working in degrees, the whole number of
11, in this case, does not have to be even. Thus,
4110  11 ( 360 )  150 
4110   11( 360  )  150 
In order to make the angle   4110  , it will take eleven complete
revolutions and an additional rotation of 150  going in the counterclockwise
direction. Thus, the angle of 150  is the coterminal angle that we are
looking for.
Answer: 150 
9.
   8865 
Use your calculator to find the whole number of times that 360 will divide
into 8865 . Since 8865  360  24.625 and 360 times 24 equals 8640, then
we have that
24
360 8865
8640
225
Copyrighted by James D. Anderson, The University of Toledo
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Thus, 8865  24 ( 360 )  225 . Multiplying both sides of this equation by  1 ,
we obtain that  8865  24 (  360 )  225 . Thus,
 8865  24 (  360 )  225 
 8865   24 (  360  )  225 
In order to make the angle    8865  , it will take twenty-four complete
revolutions and an additional rotation of 225  going in the clockwise
direction. Thus, the angle of  225  is the coterminal angle that we are
looking for.
Answer:  225 
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3.
TRIGONOMETRIC FUNCTIONS OF COTERMINAL ANGLES
Theorem Let  and  be two coterminal angles. Then
1.
cos   cos 
4.
sec   sec 
2.
sin   sin 
5.
csc   csc 
3.
tan   tan 
6.
cot   cot 
We can use this theorem to find any one of the six trigonometric functions of an
angle that is numerically bigger than 2  or 360  . When given an angle that is
numerically bigger than 2  or 360  , we will want to find the angle that is
coterminal to it and is numerically smaller than 2  or 360  .
Examples Use a coterminal angle to find the exact value of the six trigonometric
functions of the following angles.
1.
 
11 
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
Doing one subtraction of 2  , we obtain that
11 
 2
3
=
11 
6
5


3
3
3
11 
5
5
and
are coterminal. The angle
is
3
3
3

in the IV quadrant and has a reference angle of . Thus, by the theorem
3
above, we have that
Thus, the two angles of  
2.

1
 11 
 5 
cos 

  cos    cos
3
2
 3 
 3 
 11 
sec 
  2
3


3

 11 
 5 
sin 
 
  sin     sin
3
2
 3 
 3 
2
 11 
csc 
  
3
 3 

 11 
 5 
tan 
 
  tan     tan
3
 3 
 3 
1
 11 
cot 
  
3
 3 
 
3
19 
4
Doing one addition of 2  , we obtain that

19 
 2
4
= 
19 
8
11 

 
4
4
4
Doing a second addition of 2  , we obtain that

11 
 2
4
= 
11 
8
3

 
4
4
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
19 
3
and 
are coterminal. The angle
4
4
3


is in the III quadrant and has a reference angle of . Thus, by the
4
4
theorem above, we have that
Thus, the two angles of   
3.
2

 19  
 3 
cos  
 
  cos  
   cos
4 
4
2

 4 
 19  
sec  
  
4 

2
2

 19  
 3 
sin  
 
  sin  
   sin
4 
4
2

 4 
 19  
csc  
  
4 

2

 19  
 3 
tan  
1
  tan  
  tan
4
4
4




 19  
cot  
 1
4


 
149 
6
Consider the fraction of
149 
149
in the angle  
.
6
6
24
149
 6 149
6
12
29
24
5
Thus,
149
5
 24  . This implies that
6
6
149 
5
5
 24  
 12 ( 2  ) 
6
6
6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
149 
, it will take twelve complete
6
5
revolutions and an additional rotation of
going in the counterclockwise
6
5
direction. Thus, the angle of
is the coterminal angle that we are looking
6
5

for. The angle
is in the II quadrant and has a reference angle of .
6
6
Thus, by the theorem above, we have that
In order to make the angle  
4.
cos
3
149 
5

 cos
  cos
 
6
6
6
2
sec
149 
2
 
6
3
sin
149 
5

1
 sin
 sin

6
6
6
2
csc
149 
 2
6
tan
149 
5

1
 tan
  tan
 
6
6
6
3
cot
149 
 
6
 
58 
3
58 
58


Consider the fraction of
in the angle
.
3
3
19
58
 3 58
3
3
28
27
1
Copyrighted by James D. Anderson, The University of Toledo
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3
Thus,
58
1
4
 19   18  . This implies that
3
3
3
58 
4
4
 18  
 9 ( 2 ) 
3
3
3
58 
, it will take nine complete revolutions
3
4
and an additional rotation of
going in the counterclockwise direction.
3
4
Thus, the angle of
is the coterminal angle that we are looking for. The
3
4

angle
is in the III quadrant and has a reference angle of . Thus, by the
3
3
theorem above, we have that
In order to make the angle  
5.
cos
58 
4

1
 cos
  cos
 
3
3
3
2
sec
58 
 2
3
sin
3
58 
4

 sin
  sin
 
3
3
3
2
csc
58 
2
 
3
3
tan
58 
4

 tan
 tan

3
3
3
cot
58 
1

3
3

3
169 
4
169 
169



Consider the fraction of
in the angle
.
4
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
42
169
 4 169
4
16
9
8
1
Thus,
169
1
 42  . This implies that
4
4

169 


  42  
 21 (  2  ) 
4
4
4
169 
, it will take twenty-one complete
4

revolutions and an additional rotation of
going in the clockwise direction.
4


Thus, the angle of
is the coterminal angle that we are looking for. The
4


angle  is in the IV quadrant and has a reference angle of . Thus, by the
4
4
theorem above, we have that
In order to make the angle   

 169  
 
cos  

  cos     cos
4
4
4




2
2
 169  
sec  
 
4


2
2

 169  
 
sin  
 
  sin      sin
4 
4
2
 4

 169  
csc  
  
4



 169  
 
tan  
 1
  tan      tan
4 
4
 4

 169  
cot  
  1
4 

Copyrighted by James D. Anderson, The University of Toledo
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2
6.
 
527 
6
Consider the fraction of
527 
527
in the angle   
.
6
6
87
527
 6 527
6
48
47
42
5
Thus,
527
5
11
 87   86 
. This implies that
6
6
6

527 
11 
11 
  86  
 43 (  2  ) 
6
6
6
527 
, it will take forty-three complete
6
11 
revolutions and an additional rotation of
going in the clockwise
6
11 
direction. Thus, the angle of 
is the coterminal angle that we are
6
11 
looking for. The angle 
is in the I quadrant and has a reference angle
6

of . Thus, by the theorem above, we have that
6
In order to make the angle   
3

 527  
 11  
cos  

  cos  
  cos
6 
6 
6
2


Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
2
 527  
sec  
 
6 
3

7.

1
 527  
 11  
sin  

  sin  
  sin
6 
6 
6
2


 527  
csc  
  2
6 


1
 527  
 11  
tan  

  tan  
  tan
6 
6 
6
3


 527  
cot  
 
6 

 
3
17 
2
Consider the fraction of
17 
17
17
1
 8 ,
in the angle   
. Since
2
2
2
2
then

17 


  8   4 (  2  ) 
2
2
2
In order to make the angle   
17 
, it will take four complete revolutions
2

going in the clockwise direction. Thus, the
2


angle of  is the coterminal angle that we are looking for. The angle 
2
2
and an additional rotation of
lies on the negative y-axis. Thus, by the theorem above, we have that
 17  
 
cos  
  cos     0
2 
 2

 17  
sec  
 = undefined
2


 17  
 
sin  
  sin      1
2 
 2

 17  
csc  
  1
2


1
 17  
 
tan  
  tan    
= undefined
2 
0
 2

0
 17  
cot  
0
 
2 
1

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
8.
  65 
Since 65   64     32 ( 2  )   , then in order to make the angle
  65  , it will take thirty-two complete revolutions and an additional
rotation of  going in the counterclockwise direction. Thus, the angle of 
is the coterminal angle that we are looking for. The angle  lies on the
negative x-axis. Thus, by the theorem above, we have that
cos 65   cos    1
sec 65    1
sin 65   sin   0
csc 65  = undefined
tan 65   tan  
9.
0
 0
1
cot 65  = undefined
   30 
Since  30   15 (  2  ) , then in order to make the angle    30  , it
will take fifteen complete revolutions going in the clockwise direction. Thus,
the angle of 0 is the coterminal angle that we are looking for. The angle 0
lies on the positive x-axis. Thus, by the theorem above, we have that
cos (  30  )  cos 0  1
sec (  30  )  1
sin (  30  )  sin 0  0
csc (  30  ) = undefined
tan (  30  )  tan 0 
10.
0
0
1
cot (  30  ) = undefined
  9840 
Use your calculator to find the whole number of times that 360 will divide
into 9840 . Since 9840  360  27.3333 .... and 360 times 27 equals 9720,
then we have that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1330
27
360 9840
9720
120
Thus, 9840  27 ( 360 )  120 . Thus,
9840  27 ( 360 )  120 
9840   27 ( 360  )  120 
In order to make the angle   9840  , it will take twenty-seven complete
revolutions and an additional rotation of 120  going in the counterclockwise
direction. Thus, the angle of 120  is the coterminal angle that we are
looking for. The angle 120  is in the II quadrant and has a reference angle
of 60  . Thus, by the theorem above, we have that
cos 9840   cos 120    cos 60   
sin 9840   sin 120   sin 60  
sec 9840    2
3
csc 9840  
2
tan 9840   tan 120    tan 60   
11.
1
2
cot 9840   
3
2
3
1
3
  900 
Subtracting two times 360  , we obtain that 900   720  = 180  .
Thus, the two angles of   900  and 180  are coterminal. The angle 180 
is on the negative x-axis. Thus, using Unit Circle Trigonometry, we have that
cos 900   cos 180    1
sec 900    1
Copyrighted by James D. Anderson, The University of Toledo
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12.
sin 900   sin 180   0
csc 900   undefined
tan 900   tan 180   0
cot 900   undefined
   1290 
Adding three times 360  , we obtain that  1290   1080  =  210  .
Thus, the two angles of    1290  and  210  are coterminal. The angle
 210  is in the II quadrant and has a reference angle of 30  . Thus, by the
theorem above, we have that
cos (  1290  )  cos (  210  )   cos 30   
3
2
sec (  1290  )  
sin (  1290  )  sin (  210  )  sin 30  
1
2
tan (  1290  )  tan (  210  )   tan 30   
2
3
csc (  1290  )  2
1
3
cot (  1290  )  
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3