*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 x*^{2} + y^{2}
= r^{2} 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 = r^{2}
*
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
x*^{2} + y^{2} = r^{2}
*x*^{2} + 144 = 169
*x*^{2} = 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
90*^{o} and Cos 90^{o}
* *
*
Solution:
The terminal side of a 90*^{o} angle is on the y-axis (0, y)
*
x = 0, y = y and r = y*
*
Thus, the Sin 90*^{o} = y/y = 1 and Cos 90^{o} = 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
180*^{o} and Cos 180^{o}
* *
*
Solution:
The terminal side of a 180*^{o} angle is on the negative
x-axis.
Choose the point (-1, 0) (See note above)
*
x = -1, y = 0, r = 1*
*
Thus, Sin 180*^{o} = 0/1 = 0 and Cos 180^{o} = -1/1 = -1
* *
*3) Find the
Sin
540*^{o} and Cos 540^{o}
* *
*
Solution: Since the angle 540*^{o} has the same terminal
side
as 180^{o}, 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 360*^{o} or 2p
* Conclusion:*
*Sin (q+
360*^{o}) = Sin q
*Cos (q+
360*^{o}) = Cos q
*Sin ( q+
2p) = Sin q*
*Cos ( q+
2p) = Cos q*

* *