How to Prove Definite Integral of Sin(nx) * Cos(mx) = 0?

[uman]

Hello all,

In exercise 31, page 106 of Calculus by Tom Apostol (volume 1), the reader is asked to prove that the definite integral from zero to 2*pi of sin(nx) * cos(mx) = 0 (where n and m are arbitrary integers). The book hints that the reader should use the fact that the definite integral from 0 to 2*pi of sin(nx) equals zero (and that the same is true of cos(nx)). I haven't been able to get anywhere on this problem, except using the addition formula for sine to prove that sin(nx)cos(mx) = cos(nx)sin(mx). Any help, anyone?

I'm sure the proof is very simple, but I just can't seem to find it...

Sorry for spelling everything out in words. I have no idea how to use TeX.

 

[Ben Niehoff]

If you know

[tex]\sin nx \cos mx = \cos nx \sin mx\,,[/tex]

then what is

[tex]\sin nx \cos mx + \cos nx \sin mx\;?[/tex]

 

[uman]

sorry my original post had an error.

I meant to say that I had proven that sin nx * cos mx = -cos mx * sin nx

 

[uman]

Wow I just figured out how to prove it about two minutes after I posted that

Thanks though.

It was as simple as I thought it would be :-)

 

[ice109]

sorry my original post had an error.

I meant to say that I had proven that sin nx * cos mx = -cos mx * sin nx

maybe you mean the integral of the left side equals the integral of the right side because what you have written there isn't true.

 

[uman]

That's what I meant.

Two errors, at least. Argh. I think if I could use real notation on this site I wouldn't have done that. time to learn how to use latex...