1. Now lets add a circle through the point and with the origin as its center.
Lets start with a point.
P(x,y) r
What does a point really mean well it means that it is x units left or right from the origin and y units up or down from the origin. Y ø X
Therefore we can find the trig ratios of any point from the origin (reference angle) just by its x and y.
Sin ø = y/r csc ø = r/y
Cos ø = x/r sec ø = r/x
tan ø = y/x cot ø = x/y
In the unit circle the radius is 1 so
sin = y csc = 1/y
cos = x sec = 1/x
tan = y/x cot = x/y

2. r y x
x2 + y2 = r x and y represent the coordinates of the point on the unit circle and the radius is always 1 (unit circle)

3. Lets look at some types of problems that will use all the information about trig ratios and the unit circle.
Finding all the trig functions exact values given a point on the unit circle:
Example: ( ½ , -√3/2)
Sin ø = y/r csc ø = r/y
Cos ø = x/r sec ø = r/x
tan ø = y/x cot ø = x/y
Sin ø = -√3/2
Cos ø = ½
Now use the triangle ratios to find the rest:
Tan ø = y/x or sin/cos
-√3/2 ÷ ½ = -√3
csc ø = 1/sin = -2/√3
sec ø = 1/cos = 2
cot ø = 1/tan = -1/√3

4. Finding all the trig functions exact values given a point on the unit circle:
Practice #15 ( -2/5 , -√21/5)
Remember from Friday the x value is the value of cosine and the y value is the value of sine so…
Sin ø =
Cos ø =
Now use the triangle ratios to find the rest:
Tan ø = y/x or sin/cos
csc ø = 1/sin =
sec ø = 1/cos =
cot ø = 1/tan =