Download Pre-Class Problems 4 for Wednesday, September 11 These are the

Pre-Class Problems 4 for Wednesday, September 11
These are the type of problems that you will be working on in class. These
problems are from Lesson 4.
Objective of the following problems: To find the exact value of the six
trigonometric functions for an angle, which is numerically greater than 2  or
360  , using the coterminal angle, which is numerically less than 2  or 360  .
1.
2.
Find the angle between 0 ( 0  ) and 2  ( 360  ) that is coterminal with the
following angles. Then use this coterminal angle to find the exact value of
the cosine, sine, and tangent of the original angle.
a.
960 
b.
11
3
c.
101
6
d.
89 
4
e.
115 
6
f.
183 
4
g.
57 
2
h.
79 
2
i.
206 
3
j.
42 
k.
25 
Find the angle between  2  (  360  ) and 0 ( 0  ) that is coterminal with
the following angles. Then use this coterminal angle to find the exact value
of the cosine, sine, and tangent of the original angle.
a.
 1305 
b.

43 
6
c.

67 
4
d.

106 
3
85 
6
f.

179 
6
g.

170 
3
h.

125 
6
j.
 17 
k.
 26 
e. 
i.

83 
2
Additional problems available in the textbook: Page 171 … 55 - 58, 65, 66, 68.
Page 154 … Example 5c.
Explanation of the calculations of the coterminal angles for Problems 1 and 2.
Solutions:
1a.
960 
Animation of the making of the 960  angle.
960   720   240 
NOTE: The angle 240  is the angle between 0  and 360  which is
coterminal with the angle 960  . Even though the coterminal angles are not
equal, the trigonometric functions of the coterminal angles are equal.
cos 960   cos 240    cos 60   
sin 960   sin 240    sin 60   
tan 960   tan 240   tan 60  
Answer: cos 960   
NOTE:
1
,
2
cos 960   
sin 960   
tan 960  
1
2
3
2
3
sin 960   
3
,
2
tan 960  
1
 sec 960    2
2
3
2
 csc 960   
2
3
3  cot 960  
1
3
3
NOTE: The terminal side of the angle 240  (and the angle 960  ) is in the
third quadrant.
NOTE: In the third quadrant, cosine is negative, sine is negative, and tangent
is positive.
NOTE: The reference angle of 240  is 60  .
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3
1
Trigonometry to find that cos 60   , sin 60  
, and tan 60   3 .
2
2
Back to Problem 1.
1b.
11
Animation of the making of the
angle.
3
11
3
11
6
5


3
3
3
NOTE: The angle
5
is the angle between 0 and 2  which is coterminal
3
11
. Even though the coterminal angles are not equal, the
3
trigonometric functions of the coterminal angles are equal.
with the angle
cos
11
5

1
 cos
 cos

3
3
3
2
sin
3
11
5

 sin
  sin
 
3
3
3
2
tan
11
5

 tan
  tan
 
3
3
3
3
Answer: cos
NOTE:
11
1
 ,
3
2
sin
3
11
 
,
3
2
tan
cos
11
1
11

 sec
 2
3
2
3
sin
3
11
11
2
 
 csc
 
3
2
3
3
tan
11
 
3
3  cot
11
 
3
3
11
1
 
3
3
NOTE: The terminal side of the angle
5
11
(and the angle
) is in the
3
3
fourth quadrant.
NOTE: In the fourth quadrant, cosine is positive, sine is negative, and tangent
is negative.
5

NOTE: The reference angle of
is .
3
3
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3


1

sin

cos

 3.
Trigonometry to find that
,
, and tan
3
2
3
2
3
Back to Problem 1.
1c.
101
6
Animation of the making of the
101
angle.
6
16
101
 6 101
6
6
41
36
5
101
5
101
5
5
 16 

 16  
 8 ( 2 ) 
6
6
6
6
6
5
is the angle between 0 and 2  which is coterminal
6
101
with the angle
. Even though the coterminal angles are not equal, the
6
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
cos
3
101
5

 cos
  cos
 
6
6
6
2
sin
101
5

1
 sin
 sin

6
6
6
2
tan
101
5

1
 tan
  tan
 
6
6
6
3
Answer: cos
NOTE:
3
101
 
,
6
2
cos
sin
101
1
 ,
6
2
tan
3
101
101
2
 
 sec
 
6
2
6
3
101
1
 
6
3
sin
101
1
101

 csc
 2
6
2
6
tan
101
1
101
 
 cot
 
6
6
3
NOTE: The terminal side of the angle
3
5
101
(and the angle
) is in the
6
6
second quadrant.
NOTE: In the second quadrant, cosine is negative, sine is positive, and
tangent is negative.
NOTE: The reference angle of
5

is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1


 , and tan

Trigonometry to find that cos
, sin
.
6
6
2
6
2
3
Back to Problem 1.
1d.
89 
4
Animation of the making of the
22
89
 4 89
4
8
9
8
1
89
1
89 


 22 

 22  
 11( 2  ) 
4
4
4
4
4
89 
angle.
4
NOTE: The angle

is the angle between 0 and 2  which is coterminal
4
89 
. Even though the coterminal angles are not equal, the
4
trigonometric functions of the coterminal angles are equal.
with the angle
cos
89 

 cos

4
4
2
2
sin
89 

 sin

4
4
2
2
tan
89 

 tan
1
4
4
Answer: cos
NOTE:
89 

4
2
,
2
sin
89 

4
2
,
2
cos
89 

4
2
89 
 sec

2
4
sin
89 

4
2
89 
 csc

2
4
tan
89 
89 
 1  cot
1
4
4
NOTE: The terminal side of the angle
tan
89 
1
4
2
2
89 

(and the angle
) is in the first
4
4
quadrant.
NOTE: In the first quadrant, cosine is positive, sine is positive, and tangent is
positive.
NOTE: The reference angle of


is .
4
4
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
2
2



tan
 1.
cos

sin

Trigonometry to find that
,
, and
4
2
4
4
2
Back to Problem 1.
1e.
115 
Animation of the making of the
angle.
6
115 
6
19
115
 6 115
6
6
55
54
1
115
1
7
115 
7
 19 
 18 

 18  
6
6
6
6
6
7
is the angle between 0 and 2  which is coterminal
6
115 
with the angle
. Even though the coterminal angles are not equal, the
6
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
cos
3
115 
7

 cos
  cos
 
6
6
6
2
sin
115 
7

1
 sin
  sin
 
6
6
6
2
tan
115 
7

 tan
 tan

6
6
6
Answer: cos
NOTE:
3
115 
 
,
6
2
1
3
sin
115 
1
  ,
6
2
tan
cos
3
115 
115 
2
 
 sec
 
6
2
6
3
sin
115 
1
115 
 
 csc
 2
6
2
6
tan
115
1
115

 cot

6
6
3
NOTE: The terminal side of the angle
115 

6
1
3
3
7
115 
(and the angle
) is in the
6
6
third quadrant.
NOTE: In the third quadrant, cosine is negative, sine is negative, and tangent
is positive.
7

NOTE: The reference angle of
is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1


 , and tan

Trigonometry to find that cos
, sin
.
6
6
2
6
2
3
Back to Problem 1.
1f.
183
4
Animation of the making of the
183
angle.
4
45
183
 4 183
4
16
23
20
3
183
3
7
183 
7
7
 45 
 44 

 44  
 22 ( 2  ) 
4
4
4
4
4
4
7
is the angle between 0 and 2  which is coterminal
4
183
with the angle
. Even though the coterminal angles are not equal, the
4
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
cos
183 
7

 cos
 cos

4
4
4
sin
2
183
7

 sin
  sin
 
4
4
4
2
tan
183 
7

 tan
  tan
 1
4
4
4
Answer: cos
183

4
2
,
2
sin
2
2
2
183 
 
,
4
2
tan
183 
 1
4
NOTE:
cos
183

4
2
183
 sec

2
4
sin
2
183 
183 
 
 csc
 
4
2
4
tan
183 
183
  1  cot
 1
4
4
NOTE: The terminal side of the angle
2
2
183
7
(and the angle
) is in the
4
4
fourth quadrant.
NOTE: In the fourth quadrant, cosine is positive, sine is negative, and tangent
is negative.
7

NOTE: The reference angle of
is .
4
4
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
2
2



 1.


Trigonometry to find that cos
, sin
, and tan
4
2
4
4
2
Back to Problem 1.
1g.
57 
2
Animation of the making of the
28
57
 2 57
2
4
17
16
1
57 
angle.
2
57
1
57 


 28 

 28  
 14 ( 2  ) 
2
2
2
2
2
NOTE: The angle

is the angle between 0 and 2  which is coterminal
2
57 
. Even though the coterminal angles are not equal, the
2
trigonometric functions of the coterminal angles are equal.
with the angle
cos
57 

 cos
 0
2
2
sin
57 

 sin
1
2
2
tan
57 

1
 tan

 undefined
2
2
0
Answer: cos
NOTE:
57 
 0,
2
sin
57 
 1,
2
tan
57 
 undefined
2
cos
57 
57 
 0  sec
 undefined
2
2
sin
57 
57 
 1  csc
1
2
2
cot
57 
0

 0
2
1
NOTE: The terminal side of the angle
positive y-axis. The angle
57 

(and the angle
) is on the
2
2

does not have a reference angle. You will have
2
to use
Unit Circle Trigonometry to find that cos
tan

 undefined.
2


 0 , sin
 1 , and
2
2
Back to Problem 1.
1h.
79 
2
Animation of the making of the
79 
angle.
2
39
79
 2 79
2
6
19
18
1
79
1
3
79 
3
3
 39 
 38 

 38  
 19 ( 2  ) 
2
2
2
2
2
2
3
is the angle between 0 and 2  which is coterminal
2
79 
with the angle
. Even though the coterminal angles are not equal, the
2
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
cos
79 
3
 cos
 0
2
2
sin
79 
3
 sin
 1
2
2
tan
79 

1
 tan

 undefined
2
2
0
Answer: cos
NOTE:
79 
 0,
2
sin
79 
  1,
2
tan
79 
 undefined
2
cos
79 
79 
 0  sec
 undefined
2
2
sin
79 
79 
 1  csc
1
2
2
cot
79 
0

 0
2
1
NOTE: The terminal side of the angle
3
79 
(and the angle
) is on the
2
2
3
does not have a reference angle. You will
2
3
 0,
have to use Unit Circle Trigonometry to find that cos
2
3
3
sin
  1 , and tan
 undefined.
2
2
negative y-axis. The angle
Back to Problem 1.
1i.
206 
3
Animation of the making of the
206 
angle.
3
68
206
 3 206
3
18
26
24
2
206
2
206 
2
2
 68 

 68  
 34 ( 2  ) 
3
3
3
3
3
2
is the angle between 0 and 2  which is coterminal
3
206 
with the angle
. Even though the coterminal angles are not equal, the
3
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
cos
206 
2

1
 cos
  cos  
3
3
3
2
sin
206 
2

 sin
 sin

3
3
3
tan
206 
2

 tan
  tan
 
3
3
3
Answer: cos
NOTE:
206 
1
  ,
3
2
3
2
sin
3
206 

3
3
,
2
cos
206 
1
206 
 
 sec
 2
3
2
3
sin
206 

3
3
206 
 csc

2
3
2
3
tan
206 
 
3
3
tan
206 
 
3
3  cot
206 
1
 
3
3
206 
2
NOTE: The terminal side of the angle
(and the angle
) is in the
3
3
second quadrant.
NOTE: In the second quadrant, cosine is negative, sine is positive, and
tangent is negative.
NOTE: The reference angle of
2

is .
3
3
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3


1

sin

cos

tan
 3.
Trigonometry to find that
,
, and
3
2
3
2
3
Back to Problem 1.
1j.
42 
Animation of the making of the 42  angle.
42   21 ( 2  )
NOTE: The angle 0 is the angle between 0 and 2  which is coterminal with
the angle 42  . Even though the coterminal angles are not equal, the
trigonometric functions of the coterminal angles are equal.
cos 42   cos 0  1
sin 42   sin 0  0
tan 42   tan 0 
0
 0
1
Answer: cos 42   1 ,
NOTE:
sin 42   0 ,
tan 42   0
cos 42   1  sec 42   1
sin 42   0  csc 42   undefined
tan 42   0  cot 42   undefined
NOTE: The terminal side of the angle 0 (and the angle 42  ) is on the
positive x-axis. The angle 0 does not have a reference angle. You will have
to use Unit Circle Trigonometry to find that cos 0  1 , sin 0  0 , and
tan 0  0 .
Back to Problem 1.
1k.
25 
Animation of the making of the 25  angle.
25   24     12 ( 2  )  
NOTE: The angle  is the angle between 0 and 2  which is coterminal
with the angle 25  . Even though the coterminal angles are not equal, the
trigonometric functions of the coterminal angles are equal.
cos 25   cos    1
sin 25   sin   0
tan 25   tan  
0
 0
1
Answer: cos 25    1 ,
NOTE:
sin 25   0 ,
tan 25   0
cos 25    1  sec 25    1
sin 25   0  csc 25   undefined
tan 25   0  cot 25   undefined
NOTE: The terminal side of the angle  (and the angle 25  ) is on the
negative x-axis. The angle  does not have a reference angle. You will have
to use Unit Circle Trigonometry to find that cos    1 , sin   0 , and
tan   0 .
Back to Problem 1.
2a.
 1305 
Animation of the making of the  1305  angle.
 1305    1080   225   3 (  360  )  225 
NOTE: The angle  225  is the angle between  360  and 0  which is
coterminal with the angle  1305  . Even though the coterminal angles are
not equal, the trigonometric functions of the coterminal angles are equal.
cos (  1305  )  cos (  225  )   cos 45   
sin (  1305  )  sin (  225  )  sin 45  
2
2
tan (  1305  )  tan (  225  )   tan 45    1
2
2
2
,
2
Answer: cos (  1305  )  
tan (  1305  )   1
NOTE:
cos (  1305  )  
sin (  1305  ) 
sin (  1305  ) 
2
,
2
2
 sec (  1305  )  
2
2
 csc (  1305  ) 
2
2
2
tan (  1305  )   1  cot (  1305  )   1
NOTE: The terminal side of the angle  225  (and the angle  1305  ) is in
the second quadrant.
NOTE: In the second quadrant, cosine is negative, sine is positive, and
tangent is negative.
NOTE: The reference angle of  225  is 45  .
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
2
2
Trigonometry to find that cos 45  
, sin 45  
, and tan 45   1 .
2
2
Back to Problem 2.
2b.

43 
6

43 
36 
7
7
 

 3(  2 ) 
6
6
6
6
Animation of the making of the 
43 
angle.
6
7
is the angle between  2  and 0 which is
6
43 
coterminal with the angle 
. Even though the coterminal angles are not
6
equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 
3

 43 
 7 
cos  
  cos  
   cos  
6 
6
2

 6 

1
 43  
 7 
sin  

  sin  
  sin
6 
6 
6
2



1
 43 
 7 
tan  
 
  tan  
   tan
6 
6 
6
3



Answer: cos  


tan  

NOTE:
43 
6
43
6
3

  
,
2

1

  
3

1
 43  
sin  
  ,
6 
2

3
2
 43 
 43 
cos  
 sec  
  
  
6 
2
6 
3


1
 43  
 43  
sin  
 csc  
 
  2
6
2
6




1
 43 
 43 
tan  
 cot  
  
  
6
6
3




NOTE: The terminal side of the angle 
second quadrant.
3
43 
7

(and the angle
) is in the
6
6
NOTE: In the second quadrant, cosine is negative, sine is positive, and
tangent is negative.
NOTE: The reference angle of 
7

is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1

tan

sin

cos

Trigonometry to find that
,
, and
.
6
6
2
6
2
3
Back to Problem 2.
2c.

67 
4
Animation of the making of the 
67 
angle.
4
16
67
 4 67
4
4
27
24
3
67
3
67 
3
3
 16 
 
  16  
 8 (  2 ) 
4
4
4
4
4
3
is the angle between  2  and 0 which is
4
67 
coterminal with the angle 
. Even though the coterminal angles are not
4
equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 
2

 67  
 3 
cos  
 
  cos  
   cos
4 
4
2

 4 
2

 67  
 3 
sin  
 
  sin  
   sin
4 
4
2

 4 

 67  
 3 
tan  
1
  tan  
  tan
4 
4 
4



Answer: cos  


tan  

NOTE:
67 
4
67 
4
2

  
,
2


 1

2
 67  
sin  
  
,
4 
2

2
 67  
 67  
cos  
 sec  
  
  
4 
2
4 


2
2
 67  
 67  
sin  
 csc  
  
  
4
2
4




2
 67  
 67  
tan  
  1  cot  
 1
4
4




NOTE: The terminal side of the angle 
3
67 
(and the angle 
) is in the
4
4
third quadrant.
NOTE: In the third quadrant, cosine is negative, sine is negative, and tangent
is positive.
NOTE: The reference angle of 
3

is .
4
4
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
2
2



 1.


Trigonometry to find that cos
, sin
, and tan
4
2
4
2
4
Back to Problem 2.
2d.

106 
3
Animation of the making of the 
106 
angle.
3
35
106
 3 106
3
9
16
15
1
106
1
4
106 
4
4
 35 
 34 
 
  34  
 17 (  2  ) 
3
3
3
3
3
3
4
is the angle between  2  and 0 which is
3
106 
coterminal with the angle 
. Even though the coterminal angles are
3
not equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 

1
 106  
 4 
cos  
 
  cos  
   cos
3 
3 
3
2



 106  
 4 
sin  

  sin  
  sin
3 
3

 3 
3
2

 106  
 4 
tan  
 
  tan  
   tan
3
3
3




 106 
Answer: cos  
3

 106 
tan  
3

NOTE:
1

   ,
2


   3

3
 106  
sin  
 
3


3
,
2
1
 106  
 106  
cos  
 sec  
  
  2
3
2
3




 106  
sin  
 
3


3
 106  
 csc  
 
2
3


 106  
tan  
  
3 

2
3
1
 106  
3  cot  
  
3 
3

NOTE: The terminal side of the angle 
106 
4
(and the angle 
) is in the
3
3
second quadrant.
NOTE: In the second quadrant, cosine is negative, sine is positive, and
tangent is negative.
NOTE: The reference angle of 
4

is .
3
3
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1



 , sin
 3.
Trigonometry to find that cos
, and tan
3
2
3
2
3
Back to Problem 2.
2e.

85 
6
Animation of the making of the 
85 
angle.
6
14
85
 6 85
6
6
25
24
1
85
1
85 


 14 
 
  14  
 7 (  2 ) 
6
6
6
6
6

is the angle between  2  and 0 which is coterminal
6
85 

with the angle
. Even though the coterminal angles are not equal, the
6
NOTE: The angle 
trigonometric functions of the coterminal angles are equal.

 85  
  
cos  

  cos     cos
6
6
6




3
2

1
 85  
  
sin  
 
  sin      sin
6 
6
2

 6

1
 85  
  
tan  
 
  tan      tan
6 
6
3

 6

Answer: cos  


tan  

85 
6
85 
6
3

 
,
2

1

  
3

1
 85  
sin  
   ,
6 
2

NOTE:
 85  
cos  
 
6


3
 85  
 sec  
 
2
6


2
3
1
 85  
 85  
sin  
 csc  
  
  2
6 
2
6 


1
 85  
 85  
tan  
 cot  
  
  
6
6
3




NOTE: The terminal side of the angle 
3
85 

(and the angle 
) is in the
6
6
fourth quadrant.
NOTE: In the fourth quadrant, cosine is positive, sine is negative, and tangent
is negative.
NOTE: The reference angle of 


is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1


 , and tan

Trigonometry to find that cos
, sin
.
6
6
2
6
2
3
Back to Problem 2.
2f.

179 
6
Animation of the making of the 
179 
angle.
6
29
179
 6 179
6
12
59
54
5
179
5
11
179 
11
11
 29 
 28 
 
  28  
 14 (  2  ) 
6
6
6
6
6
6
11
is the angle between  2  and 0 which is
6
179 
coterminal with the angle 
. Even though the coterminal angles are
6
not equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 

 179  
 11 
cos  

  cos  
  cos
6 
6 
6


3
2

1
 179  
 11 
sin  

  sin  
  sin
6 
6 
6
2



 179  
 11 
tan  

  tan  
  tan
6
6
6




 179 
Answer: cos  
6

 179 
tan  
6


 


 

3
,
2
1
3
1
3
1
 179  
sin  
  ,
6 
2

NOTE:
 179  
cos  
 
6 

3
 179  
 sec  
 
2
6 

2
3
 179   1
 179  
sin  
 csc  
 
  2
6 
2
6 


 179  
tan  
 
6 

1
 179  
 cot  
 
6 
3

NOTE: The terminal side of the angle 
3
11
179 
(and the angle 
) is in
6
6
the first quadrant.
NOTE: In the first quadrant, cosine is positive, sine is positive, and tangent is
positive.
NOTE: The reference angle of 
11

is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1


 , and tan

Trigonometry to find that cos
, sin
.
6
6
2
6
2
3
Back to Problem 2.
2g.

170 
3
Animation of the making of the 
170 
angle.
3
56
170
 3 170
3
15
20
18
2
170
2
170 
2
2
 56 
 
  56  
 28 (  2  ) 
3
3
3
3
3
2
is the angle between  2  and 0 which is
3
170 

coterminal with the angle
. Even though the coterminal angles are
3
not equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 

1
 170  
 2 
cos  
 
  cos  
   cos
3 
3 
3
2


3

 170  
 2 
sin  
 
  sin  
   sin
3 
3 
3
2



 170  
 2 
tan  

  tan  
  tan
3
3
3




 170 
cos

Answer:
3

 170 
tan  
3

1

   ,
2


  3

3
3
 170  
sin  
  
,
3 
2

NOTE:
1
 170  
 170  
cos  
 sec  
  
  2
3
2
3




3
2
 170  
 170  
sin  
 csc  
  
  
3 
2
3 
3


 170  
tan  
 
3 

 170  
3  cot  
 
3 

NOTE: The terminal side of the angle 
1
3
170 
2
(and the angle 
) is in the
3
3
third quadrant.
NOTE: In the third quadrant, cosine is negative, sine is negative, and tangent
is positive.
NOTE: The reference angle of 
2

is .
3
3
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1


 , sin

 3.
Trigonometry to find that cos
, and tan
3
2
3
2
3
Back to Problem 2.
2h.

125 
6
Animation of the making of the 
20
125
 6 125
6
12
5
125 
angle.
6
125
5
125 
5
5
 20 
 
  20  
 10 (  2  ) 
6
6
6
6
6
5
is the angle between  2  and 0 which is
6
125 
coterminal with the angle 
. Even though the coterminal angles are
6
not equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 
3

 125  
 5 
cos  
 
  cos  
   cos
6 
6
2

 6 

1
 125  
 5 
sin  
 
  sin  
   sin
6 
6 
6
2



 125  
 5 
tan  

  tan  
  tan
6
6
6




 125 
cos

Answer:
6

 125 
tan  
6

NOTE:
3

  
,
2

1

 
3

1
3
1
 125  
sin  
   ,
6 
2

3
2
 125  
 125  
cos  
 sec  
  
  
6 
2
6 
3


1
 125  
 125  
sin  
 csc  
  
  2
6 
2
6 


 125  
tan  
 
6 

1
 125  
 cot  
 
6 
3

3
NOTE: The terminal side of the angle 
5
125 
(and the angle 
) is in the
6
6
third quadrant.
NOTE: In the third quadrant, cosine is negative, sine is negative, and tangent
is positive.
NOTE: The reference angle of 
5

is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle
3

1

1

tan

sin

cos

Trigonometry to find that
,
, and
.
6
6
2
6
2
3
Back to Problem 2.
2i.

83 
2
Animation of the making of the 
83 
angle.
2
41
83
 2 83
2
8
3
2
1
83
1
3
83 
3
3
 41 
 40 
 
  40  
 20 (  2  ) 
2
2
2
2
2
2
3
is the angle between  2  and 0 which is
2
83 
coterminal with the angle 
. Even though the coterminal angles are not
2
equal, the trigonometric functions of the coterminal angles are equal.
NOTE: The angle 
 83  
 3 
cos  
  cos  
  0
2 
2 


 83 
 3 
sin  
  sin  
 1
2 

 2 
1
 83  
 3 
tan  
 undefined
  tan  
 
2 
2 
0


 83  
 83 
  1,
  0 , sin  
Answer: cos  
2
2




 83 
tan  
  undefined
2


NOTE:
 83  
 83  
cos  
  0  sec  
  undefined
2
2




 83  
 83  
sin  
  1  csc  
 1
2
2




0
 83  
cot  
 0
 
2 
1

NOTE: The terminal side of the angle 
3
83 
(and the angle 
) is on the
2
2
3
does not have a reference angle. You will
2
 3 
  0,
have to use Unit Circle Trigonometry to find that cos  
2


 3 
 3 
sin  
  undefined.
  1 , and tan  
2
2




positive y-axis. The angle 
Back to Problem 2.
2j.
 17 
Animation of the making of the  17  angle.
 17    16     8 (  2  )  
NOTE: The angle   is the angle between  2  and 0 which is coterminal
with the angle  17  . Even though the coterminal angles are not equal, the
trigonometric functions of the coterminal angles are equal.
cos (  17  )  cos (   )   1
sin (  17  )  sin (   )  0
tan (  17  )  tan (   ) 
0
 0
1
Answer: cos (  17  )   1 ,
NOTE:
sin (  17  )  0 ,
tan (  17  )  0
cos (  17  )   1  sec (  17  )   1
sin (  17  )  0  csc (  17  )  undefined
tan (  17  )  0  cot (  17  )  undefined
NOTE: The terminal side of the angle   (and the angle  17  ) is on the
negative x-axis. The angle   does not have a reference angle. You will
have to use Unit Circle Trigonometry to find that cos (   )   1 ,
sin (   )  0 , and tan (   )  0 .
Back to Problem 2.
2k.
 26 
Animation of the making of the  26  angle.
 26   13 (  2  )
NOTE: The angle 0 is the angle between  2  and 0 which is coterminal
with the angle  26  . Even though the coterminal angles are not equal, the
trigonometric functions of the coterminal angles are equal.
cos (  26  )  cos 0  1
sin (  26  )  sin 0  0
tan (  26  )  tan 0 
0
 0
1
Answer: cos (  26  )  1 ,
NOTE:
sin (  26  )  0 ,
tan (  26  )  0
cos (  26  )  1  sec (  26  )  1
sin (  26  )  0  csc (  26  )  undefined
tan (  26  )  0  cot (  26  )  undefined
NOTE: The terminal side of the angle 0 (and the angle  26  ) is on the
positive x-axis. The angle 0 does not have a reference angle. You will have
to use Unit Circle Trigonometry to find that cos 0  1 , sin 0  0 , and
tan 0  0 .
Back to Problem 2.
Solution to Problems on the Pre-Exam:
2.
Find the angle between 0 and 2  that is coterminal with the angle
then find the exact value of cot
167 
.
6
Animation of the making of the
167 
angle.
6
167 
and
6
27
167
 6 167
6
12
47
42
5
167
5
11
167 
11
11
 27 
 26 

 26  
 13 ( 2  ) 
6
6
6
6
6
6
11
is the angle between 0 and 2  which is coterminal
6
167 
with the angle
. Even though the coterminal angles are not equal, the
6
trigonometric functions of the coterminal angles are equal.
NOTE: The angle
tan
167 
11

1
 tan
  tan
 
6
6
6
3
tan
167 
1
167 
 
 cot
 
6
6
3
11
Coterminal Angle:
6
3
cot
167 
 
6
3
NOTE: The terminal side of the angle
11
167 
(and the angle
) is in the
6
6
fourth quadrant.
NOTE: In the fourth quadrant, tangent is negative.
NOTE: The reference angle of
11

is .
6
6
NOTE: You can use either Unit Circle Trigonometry or Right Triangle

1

Trigonometry to find that tan
.
6
3
3.
Find the angle between  2  and 0 that is coterminal with the angle 
 55  
.
and then find the exact value of cos  
3 

55 
Animation of the making of the 
angle.
3
18
55
 3 55
3
3
25
24
1
55
1
55 


 18 
 
  18  
 9 (  2 ) 
3
3
3
3
3
55 
3

is the angle between  2  and 0 which is coterminal
3
55 
with the angle 
. Even though the coterminal angles are not equal, the
3
NOTE: The angle 
trigonometric functions of the coterminal angles are equal.

1
 55  
  
cos  

  cos     cos
3 
3
2

 3
Coterminal Angle: 
1
 55  
cos  
 
3 
2


3
NOTE: The terminal side of the angle 
55 

(and the angle 
) is in the
3
3
fourth quadrant.
NOTE: In the fourth quadrant, cosine is positive.
NOTE: The reference angle of 


is .
3
3
NOTE: You can use either Unit Circle Trigonometry or Right Triangle

1
cos

Trigonometry to find that
.
3
2
4.
Find the angle between 0  and 360  that is coterminal with the angle 810 
and then find the exact value of sec 810  .
Animation of the making of the 810  angle.
810   720   90 
NOTE: The angle 90  is the angle between 0  and 360  which is coterminal
with the angle 810  . Even though the coterminal angles are not equal, the
trigonometric functions of the coterminal angles are equal.
cos 810   cos 90   0
cos 810   0  sec 810   undefined
Coterminal Angle: 90 
sec 810   undefined
NOTE: The terminal side of the angle 90  (and the angle 810  ) is on the
positive y-axis. The angle 90  does not have a reference angle. You will
have to use Unit Circle Trigonometry to find that cos 90   0 .