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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.

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# Problem from Apostol book

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