7-3 The Sine and Cosine Functions

    The six trigonometric functions are the 6 different ratios that you can set up from a right triangle.  To simplify it, we will form the right triangles with a vertex at the origin and a terminal ray in standard position.  Study the following graph:
Click for demo of Sine Function (Manipula Math)
Click for demo of Cosine Function (Manipula Math)
 
     Let the point P(x,y) be a point on the circle  x2 + y2 = r2 and 0 is an angle in standard position.  We define the following:
Sin q = y/r
Cos q = x/r
x and y get their signs from the quadrants they appear in, and r > 0

 Example
1)  If the terminal side of an angle q  in standard position goes through
(-2, -5), find the Sin qand Cos q.
 
First, draw a sketch: 
Calculate r:          (-2)2 + (-5)2 = 29 = r2
     Thus, 
Thus 

2)  If theta is a second quadrant angle and sin q= 12/13, find Cos q.
 
Solution:  Since the angle is in the second quadrant, x must be negative implying Cos must also be negative.  Since the sin is 12/13, this means y = 12 and r = 13.  Find x by using x2 + y2 = r2
x2 + 144 = 169
x2 = 25
x = 5 or -5.  Take -5
Must be in second quadrant, remember?
Thus, Cos q= -5/13

Signs of the Sine and Cosine Functions
Study the following table for the correct signs:
 
  Function     Quad I    Quad II    Quad III    Quad IV
    Sine         +         +        -          -
   Cosine         +         -         -         +
  
Quadrantal points
1)  Find the Sin 90o and Cos 90o
 
    Solution:  The terminal side of a 90o angle is on the y-axis (0, y)
                      x = 0, y = y and r = y
                    Thus, the Sin 90o = y/y = 1 and Cos 90o = 0/y = 0
Note:  It doesn't matter what y value I take for this problem.  From now on I will choose 1 to make the arithmetic easy.  This also goes for points on the x-axis.
 
2)  Find the Sin 180o and Cos 180o
 
    Solution:  The terminal side of a 180o angle is on the negative x-axis.  Choose the point (-1, 0) (See note above)
                                x = -1, y = 0, r = 1
                    Thus, Sin 180o = 0/1 = 0 and Cos 180o = -1/1 = -1
 
3)  Find the Sin 540o and Cos 540o
 
        Solution:  Since the angle 540o has the same terminal side as 180o, the Sine and Cosine functions have the same value as problem # 2.
This leads to the conclusion that the trig functions repeat their value every 360o or 2p
 Conclusion:
Sin (q+ 360o) = Sin q
Cos (q+ 360o) = Cos q
Sin ( q+ 2p) = Sin q
Cos ( q+ 2p) = Cos q