Unit Circle and Reference Angles
 1 3

− ,
 2 2 


2π
3

2 2

−
 2 , 2 
 3π

4
1 3

 ,
2 2 


(0,1)
120○
π
π
2
3
90
π
60○
○
(− 1,0)

3 1

−
 2 ,− 2 


π
7π
6

2
2

−
,
−

 2
2


45○
Numerator is
always one
less than
denominator
times π
150○
180
○
225
○
5π
4
30
○
6
 3 1


 2 , 2


sine value
Numerator is
always one
more than
denominator
times π
Numerator is
always one less
than twice
denominator
times π 315○
240○
300○
3π
2
(0,−1)
5π
3
0
0
360○
2π
330○
7π
4
270○
4π
3
 1
3

 − ,−

 2
2


Some hints when
dealing with radians.
π
Numerator is
always π
○
210
cosine
value
4
135○

3 1

−
 2 , 2
 5π

6
 2 2


 2 , 2 


11π
6
(1,0)
 3 1


 2 ,− 2 


 2
2


,
−

 2
2


1
3

 ,−

2
2


Value of Denominator
?
6
?
4
?
3
Reference Angle
30○
45○
60○
A. Reference Angles:
Example:
π 
Draw 60°   in Quadrant II.
3
1. (a) We are in the Quadrant II with a reference angle of 60 degrees. The angle in radians has
a denominator of 3. Thus, the angle in Quadrant II is 120 degrees or 2π/3.
(b)Formulas for Reference Angles:
Note: Let θ be the reference angle given.
Quadrant II
180 - θ
Quadrant I
θ
Quadrant II
θ-180
Quadrant II
360- θ
In this example, we in are in Quadrant II. Thus, the angle is 180 degrees-60 degrees =
120 degrees.
Solution:
120º
(2π/3)
60º
Note: Knowing reference angles speeds up things considerably.
Positive and Negative Quadrants
S
Quadrant II
(+)
sine and cosecant
A
Quadrant I
All trigonometric
functions are
positive
(-)
cosine secant
tangent cotangent
T
C
Quadrant III
(+)
tangent and cotangent
Quadrant IV
(+)
cosine and secant
(-)
sine cosecant
cosine secant
(-)
sine cosecant
tangent cotangent
All trigonometric functions are positive in Quadrant I
Sine and cosecant are positive in Quadrant II
Tangent and cotangent are positive in Quadrant III
Cosine and secant are positive in Quadrant IV
*Note: This information is used in conjunction with reference angles.
B. Using Reference Angles To Solve Trigonometric Equations.
Example: Solve :
sin 2θ = −
2
.
2
To solve this equation, take the inverse sine (arcsin) of both sides.

2

sin -1 (sin 2θ ) = sin −1  −

2


Note: 2θ means you have to make two revolutions around the unit circle. nθ determines
the number of revolutions.
Steps:
1. Determine which quadrants your desired angles lie in. In this example, sine is negative in
Quadrants III and IV.
S
A
T
C
2. Find corresponding angles in those quadrants. In this example, we need an angle in Quadrant
III and an angle in Quadrant IV which has a sine of
−
2
. The reference angle is 45 degrees which means the angles have a denominator
2
of four.
First time around:
--Quadrant III (angle in radians is one more than denominator)
5π
4
--Quadrant IV (angle in radians is 1 less and twice denominator)
7π
4
Second time around (just add 2π to the previous angles):
--Quadrant III (angle in radians is one more than denominator)
5π
13π
+ 2π =
4
4
--Quadrant IV (angle in radians is 1 less and twice denominator)
7π
15π
+ 2π =
4
4
5π 7π 13π 15π
,
,
,
Solution:
4 4 4
4