Problem from Apostol book

1. Dec 4, 2007

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.

2. Dec 4, 2007

Ben Niehoff

If you know

$$\sin nx \cos mx = \cos nx \sin mx\,,$$

then what is

$$\sin nx \cos mx + \cos nx \sin mx\;?$$

3. Dec 4, 2007

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

4. Dec 4, 2007

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 :-)

5. Dec 4, 2007

ice109

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.

6. Dec 4, 2007

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...

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