Graphs of Sine and Cosine Functions
Date: 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.