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Trigonometric Functions
Trigonometric Functions

Let point P with coordinates (x, y) be any
point that lies on the terminal side of θ.

θ is a position angle of point P

Suppose P’s distance to the origin is “r”
units.

“r” is known as the radius vector of point
P and is always considered positive
Trigonometric Functions
P(x, y)
hypotenuse  r
O
θ
x
y  opposite θ
W
 adjacent θ

By using Pythagoras Theorem, we can
see that r2 = x2 + y2.

By taking any two of the three values for r,
x, and y, we can form 6 different ratios
Trigonometric Functions

There are 6 trigonometric functions
Primary Ratios
Sine θ = opp  sin 
hyp
adj
 cos 
Cosine θ =
hyp
Tangent θ =
opp
 tan 
adj
Reciprocal Ratios
Cosecant θ = hyp  csc 
opp
Secant θ = hyp  sec
adj
Cotangent θ =
adj
 cot 
opp
Trigonometric Functions
Example 1: If θ is the position angle of the point
P(3, 4), find the values of the six trigonometric
functions of θ.
Solution: To determine the values of the six
trigonometric functions, we first need:
1)
The values of x, y (the coordinates of a point on
the terminal side of θ
2)
The value of r (the distance of the point from
the origin)
Trigonometric Functions

Since P(3, 4) lies on the terminal side of θ, we
know that x = 3, and y = 4.
Since r2 = x2 + y2  r2 = (3)2 + (4)2
r2 = 9 + 16
r2 = 25  r = 5
Thus:
3
4
4

sin  
5
5
csc  
4
cos  
sec  
5
5
3
tan  
3
3
cot  
4
Trigonometric Functions
5
Example 2: If tan  
and θ is a third quadrant
12
angle, find the value of the other trigonometric
functions of θ.
Solution: Because θ is in the 3rd quadrant, we
know that the values of x and y are both negative.
 r2 = x2 + y2  r2 = (12)2 + (5)2
r2 = 169  r = 13
Trigonometric Functions

Therefore, the other trigonometric
functions are:
5
sin   
13
13
csc   
5
12
cos   
13
13
sec   
12
cot  
12
5
Homework

Do # 1 – 15 odd numbers only on page 231
from Section 7.3 for Monday June 8th 
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