Finding the Exact Value of
Trigonometric Functions
Review: Special Right Triangles
Find the missing side Lengths:
π60°
/3
1
1
2
π45°
/4
1
π30°
/6
3
2
π /°4
45
2
2
2
2
Important Points on Unit Circle
1  0,1
π90°
/2
2π / 3
120°
3π / 4
135°
π60°
/3
π45°
/4
π 30°
/6
5π / 6
150°
 1, 0 180°
π
-1
0
0°
11π
/6
330°
7π / 6
210°
5π / 4
225°
4π
/3
240°
7π315°
/4
5π
/3
300°
3π
/2
270°
-1  0, 1
1,0
1
Important Points on Unit Circle
 1 3
  ,

 2 2 

2 2
,
 

2
2 

π/2
2π / 3
3π / 4

3 1
, 
 
 2 2
 1, 0
-1

3 1
,  
 
2
2

1  0,1
5π / 6
π
1 3
 ,

2 2 
π/3
π/4
1
1
 2 2
,


 2 2 

1
45°
60°
30°
1/2 2 3
0
2 2
11π / 6
7π / 6

2
2
,
 

2 
 2
5π / 4
4π / 3
 1
3
  , 

2 
 2
7π / 4


5π / 3

3π / 2 

-1  0, 1

3 1
3π / 6
, 

2
 2 2
2
2 1/2
1
3
 , 

2 
2
1,0
1
 3 1
,  

2
 2
2
2
,

2
2 
Important Points on Unit Circle
 1 3
  ,

 2 2 

2 2
,
 

2
2 
120°


3 1
, 
 
 2 2
1  0,1
90°
1 3
 ,

2 2 
 2 2
,


 2 2 
60°
135°
45°
150°
 1, 0 180°
-1

3 1
,  
 
2
2

0°
330°
210°

2
2
,
 

2 
 2
 3 1
, 

 2 2
30°
315°
225°
240°
 1
3
  , 

2 
 2
270°
1,0
1
 3 1
,  

2
 2
 2
2
,



300°
2 
 2
1
3
 , 

2 
2
-1  0, 1
Important Points on Unit Circle
 1 3
  ,

 2 2 

2 2
,
 

2
2 

π/2
2π / 3
3π / 4

3 1
, 
 
 2 2
 1, 0
-1

3 1
,  
 
2
2

1  0,1
1 3
 ,

2 2 
π/3
π/4
 2 2
,


 2 2 
 3 1
, 

 2 2
π/6
5π / 6
π
0
11π / 6
7π / 6

2
2
,
 

2 
 2
5π / 4
4π / 3
 1
3
  , 

2 
 2
7π / 4


5π / 3

3π / 2 

-1  0, 1
1
3
 , 

2 
2
1,0
1
 3 1
,  

2
 2
2
2
,

2
2 
Reference Angle
On the left are 3 reference angles that we know exact trig values
for. To find the reference angle for angles not in the 1st quadrant
(the angles at right), ignore the integer in the numerator.

6

4

3
: 30
: 45
: 60
5 7 11
,
,
6 6 6
3 5 7
, ,
4 4 4
2 4 5
,
,
3 3 3
Then
multiply the
number in
the
numerator
by the
degree to
find the
angle’s
quadrant.
Stewart’s Table: Finding Exact Values of
Trig Functions
R.A.
Sin
Cos
0
0
0
2
1 1

2
2
2
2
1
3
2
1
2

6

4

3

2
3
2
2
2
4 2
 1
2
2
Each time the square root
number goes up by 1
0
Reverse the order of the
values from sine
Tan
1. Find the
value of the
Reference
Angle.
2. Find the
angles
quadrant to
figure out the
sign (+/-).
Example 1
Find the exact value of the following:
cos  34 

Reference Angle:
4
Cosine of Reference Angle: cos  4  
2
2
Quadrant of Reference Angle: 3  45  135  Second Quadrant
Sign of Cosine in Second Quadrant:
Therefore:
cos  34   
2
2
 ,  
 ,  
 , 
 , 
 Negative
Example 2
Solve:
2sin  x   1  0
2sin  x   1
1
sin  x    2
Reference Angle that Makes sin  x  =  12 = 12 True:
Solutions in the interval 0  x  2 :
π/6
All solutions:
7
6
11
6

6
  6  76
π/6
2 n
where n is an integer
2 n
2  6  116
Slope on the Unit Circle
1
(cosӨ,sinӨ)
Ө
-1
cosӨ
sinӨ
What is the slope
of the terminal side
of an angle on the
unit circle?
Opposite
1
Adjacent
sin  
opposite
 tan  
Slope = cos   -1
adjacent
A Definition of Tangent
The tangent function is defined as:
tan   
sin  
cos  
There are values for which the tangent function are undefined:
Any Θ that makes cos(Θ)=0.
  2,

In general:
3
2
,
  2 n

5
2
,
7
2
,
9
2
,
11
2
,...
For any integer n.
Stewart’s Table: Finding Exact Values of
Trig Functions
R.A.
Sin
Cos
0
0
0
2
1 1

2 2
2
2
1
3
2
1
2

6

4

3

2
3
2
2
2
4 2
 1
2
2
Each time the square root
number goes up by 1
Tan
1. Find the
value of
0
0
the
1
Reference
12
1
1 2
 
 3

Angle.
3 2 2 3
3
3
2 2
1
2 2
3 2
3 2

 
12
2 1
0
Reverse the order of the
values from sine
1

0
tan   
2. Find the
angles
3
quadrant
to figure
out +/-.
sin  
cos  
Example 3
Find the exact value of the following:
Thought process
tan  53 

Reference Angle:
3
Tangent of Reference Angle: tan  3   3
Quadrant of Angle: 5  60  300  Fourth Quadrant
Sign of Tangent in Fourth Quadrant:
 ,  
 negative
positive  Negative
 , 
Therefore:
tan  53    3
 ,  
 , 
The only thing required for a correct
answer (unless the question says explain)