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Graphs of Sine and Cosine FunctionsDate: 02/28/2003 at 17:22:35 From: Nati Subject: Relation between the sine and the cosine function How is the graph of y = cos(x) linked to the graph of y = sin(x)? The Pythagoran Identity, meaning sin^2 x + cos^2 x = 1, is what I found out about the relation concerning sine and cosine. But aren't there any general explanations other than the fact that the sine is just a displaced cosine curve along the x-axis ? Date: 02/28/2003 at 17:31:21 From: Doctor Tom Subject: Re: Relation between the sine and the cosine function Hi Nati, Basically you are right - the only difference between the sine and the cosine is that they are shifted versions of each other. In a right triangle, you have the right angle and two other angles A and B. Since A and B together make 90 degrees, A is the complement of B and vice-versa. The (co)sine of an angle is is the sine of the (co)mplement. In other words, the sine of A is the cosine of B and the sine of B is the cosine of A. The same idea holds for the other trigonometric functions: the cotangent is the tangent of the complement and the cosecant is the secant of the complement. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/
Date: 02/28/2003 at 22:57:05
From: Doctor Dotty
Subject: Re: Relation between the sine and the cosine function
Hi Nati,
Thanks for the question!
Lets go back to basics. Triangles:
|.
| . c
b|_ .
|_|_ _ _ _ _x_(_ _.
a
The following equations are true:
b
sin x = -
c
a
cos x = -
c
But, suppose we put on another triangle:
_ _ _ _a_ _ _ _ _
|. |_|
| . |
b|_ c . y |b
|_|_ _ _ _ _x_(_ .|
a
Where y is the angle between c and the right hand b.
x + y = 90
So x = 90 - y.
We know that:
b
sin x = -
c
If you look at the diagram, this gives:
a
sin (90 - x) = -
c
Which is cos x.
So:
sin (90 - x) = cos x
and therefore
cos (90 - x) = sin x
This is why the graph of cos x is the graph of sin x displaced by 90
degrees.
Note that if you are working in radians, 90 degrees is (Pi / 2).
Does that all make sense?
If I can help any more with this problem or any other, please write
back.
- Doctor Dotty, The Math Forum
http://mathforum.org/dr.math/
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