1.6 Trigonometric Functions:
The Unit circle
The Unit Circle
A circle with radius of 1
Equation x2 + y2 = 1
0,1
 1,0

0,1
cos , sin  
1,0
Do you remember 30º, 60º, 90º
triangles?
Do you remember 45º, 45º, 90º
triangles?
Do you remember 45º, 45º, 90º
triangles?
When the hypotenuse is 1
The legs are 2
2
1
2
2
2
2
Let's pick a point on the circle. We'll choose a point where the x is 1/2. If the x is
1/2, what is the y value?
x  y 1
2
1
2

y
1
 
2
3
2
y 
4
(-1,0)
3
y
2
2
You can see there are two
y values. They can be
found by putting 1/2 into
the equation for x and
solving for y.
2
x = 1/2
(0,1)
1 3
 ,

2 2 


(1,0)
1
3
 ,


2 
(0,-1)  2
We'll look at a larger
version of this and
make a right
triangle.
We know all of the sides of this triangle. The bottom leg is just the x value of the
point, the other leg is just the y value and the hypotenuse is always 1 because it is
a radius of the circle.
(0,1)
(-1,0)
1
1 3
 ,

2 2 


3
2

sin  
(1,0)
1
2
tan  
(0,-1)
cos
3
2  3
1
2
1
21
 1 2
3
2  3
1
2
Notice the sine is just the y value of the unit circle point and the cosine is just the x
value.
So if I want a trig function for  whose terminal side contains a point on the unit
circle, the y value is the sine, the x value is the cosine and y/x is the tangent.
(0,1)
 2 2


,
 2 2 


(-1,0)

1
3
 ,

2 2 


sin  
(1,0)
tan 
(0,-1)
1
3
 ,

2
2 


cos
2
2
2

2
2
 2  1
2

2
We divide the unit circle into various pieces and learn the point values so we can
then from memory find trig functions.
Here is the unit circle divided into 8 pieces. Can you figure out how many degrees
are in each division?
These are
easy to
0,1
memorize
 2 2
 2 2
90°
since they all




 2 , 2  135°
 2 , 2 



 have the same
value with
45°
2
different signs

depending on
2
the quadrant.
180°
45°
 
sin 225 
 1,0
0°
1,0
225°
 2
2


 2 , 2 


315°  2
2


270°
,

 2
2 

We can label this all the way around with how many degrees an angle would be and the
0,1
point on the unit circle that corresponds with the terminal side of the angle. We could
then find any of the trig functions.


Can you figure out what these angles would be in radians?
0,1
 2 2


 2 , 2  135°



 1,0
7
sin

4
2
2
90°

3
4
2

4
180°
5
4225°
 2
2


,

 2
2 

 2 2


 2 , 2 


45°
0°
3
2
270°
7
4 315° 
2

0,1
 2

,
1,0
2

2 
The circle is 2 all the way around so half way is . The upper half is divided into 4
pieces so each piece is /4.
Here is the unit circle divided into 12 pieces. Can you figure out how many
degrees are in each division?
You'll need to
 1 3
memorize
1 3
0
,
1



,


3
, 
 2 2 
these too but



2
2
90°


2
you can see
120°
60°
the pattern.
 3 1
  ,  150°
 3 1
 2 2
 , 


 2 2

30° 
 
cos 330 
 1,0
180°
30°
0°
1,0
210°
 3 1


 2 , 2 


240°
270°
330°  3 1 
 , 
 2 2


300°
1
 1
3
3
  , 
 , 
2 2 
 2 2 
0,1




We can again label the points on the circle and the sine is the y value, the cosine is the x
value and the tangent is y over x.
sin 240  
3
2


Can you figure out what the angles would be in radians?
 
 1 3
0,1
 , 
 2 2 


90°
120°
 3 1
 , 
 2 2


 1,0
1 3
 , 
2 2 


60°
150°

30°6
180°
 3 1
 , 
 2 2

30° 
0°
We'll see
them all put
together on
the unit
circle on
the next
screen.
1,0
210°
 3 1


 2 , 2 


240°
270°
330°  3 1 
 , 
 2 2


300°
1
3
 1
3


,

  , 

2 2 
 2 2 
0
,

1



It is still  halfway around the circle and the upper half is divided into 6 pieces so each
piece is /6.


You should
memorize this.
This is a great
reference
because you
can figure out
the trig
functions of all
these angles
quickly.
1
3
 ,

2
2 

The Unit Circle with Radian Measures
2
The Six Trig functions
b
adjacent

c hypotenuse
a
opposite
Sin   
c hypotenuse
a opposite
Tan  
b adjacent
Cos 
Sin 
Tan 
Cos

Reciprocal
Identities
1
Cos ®
= Sec
cos
1
Sin ®
= Csc
sin
1
Tan ®
= Cot
tan
Lets find the six trig functions if
2

3
Think where this angle is on the unit circle.
 2
Cos
 3
 1 3 
 ,

 2 2 


2

3
 1

 2
3
 2 
Sin 

 3  2
3
 2 
Tan
 2  3
 3  1
2
Sin 
Tan 
Cos
Find the six trig functions
2
of   3
Think where this angle is on the unit circle.
 2
Cos
 3
 1

 2
 2   2
 2
Sec

 3  1
 2
Sin 
 3
3



 2
2 3
 2  2

Csc

3
3
 3 
3
 2 
Tan
 2  3
 3  1
2
 2   1  3

Cot 

3
3
 3 
How about


4
 2 2


 2 , 2 


2
  
Cos

 4  2
 
Sin
 4


4
 2  2


 2 , 2 


  2

2

2
  
Tan
  2  1
 4   2
2
There are times when Tan or Cot
does not exist.
At what angles would this happen?
tan 90° = UD
tan 270° = UD
cot180° = UD
cot 360° = UD
Ex 1: Find the values of the sine and cosine functions of an angle in standard
position with measure θ if the point (3,4) lies on it’s terminal side.
Ex 2: If the point (5,12) lies on its terminal side.
Ex 3: Find the sin θ when cos θ = and the terminal side of θ is in the 1st
quadrant.
Ex 4: Find the sin θ when cos θ = and the terminal side of θ is in the 1st
quadrant.
7.) The terminal side of an angle θ in standard position contains the point with
coordinates (8,-15). Find the value of all six trig functions.
8.) contains the point (-3,-4)
9.) If csc θ = -2 and θ lies in Quad III, find the values of the five trig functions.
10.) If sec θ = 2 and θ lies in Quad IV:
YOUR TURN!!!
FILL IN A BLANK UNIT
CIRCLE!!
YAAAHHHHH!!!