Orthogonality of Cosine and Sine Functions: A Proof by Explicit Integration

[Jack21222]

Homework Statement

This would be easier if I knew how to use LaTeX, but I'll give it a shot.

Show by explicit integration that the following is true:

Integral from 0 to a of sin(m*pi*x/a)*cos(n*pi*x/a)=0

Homework Equations

The Attempt at a Solution

I've used an identity to change sin(m*pi*x/a)*cos(n*pi*x/a) into sin[(m+n)*pi*x/a]+sin[(m-n)*pi*x/a] and integrated each side. When I do this, m+n=even (including m=n), the answer is zero, but the answer is not zero for m+n=odd. Instead, I get -a*m/[(m^2-n^2)*pi]

So, I decided to check with my TI-89. When I enter the original problem into my TI-89, I get the same thing.

Mathematica tells me it equals -((a (-m + m Cos[m \[Pi]] Cos[n \[Pi]] +
n Sin[m \[Pi]] Sin[n \[Pi]]))/((m^2 - n^2) \[Pi])), which ends up reducing to the same thing. I tested Mathematica with m = 2 and n = 3, and it gave me -4a/(5*pi), which is the same as my TI-89 and the same as the answer I got by hand.

Is the question wrong?

 

[Stephen Tashi]

Are the arguments of the functions really $$\frac{m \pi x}{a}$$?

Perhaps it should say $$\frac{m 2 \pi x}{a}$$

 

[Jack21222]

Are the arguments of the functions really $$\frac{m \pi x}{a}$$?

Perhaps it should say $$\frac{m 2 \pi x}{a}$$

I emailed the professor after posting this, and he came back with the correction that yes, it should have been 2*pi, not pi. He forgot the factor of 2.

Everything's correct now, thanks. If I had hair, I would have pulled it out over this.