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

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The discussion centers on proving that the definite integral from 0 to 2π of sin(nx) * cos(mx) equals 0 for arbitrary integers n and m, as stated in Tom Apostol's "Calculus" (Volume 1). The key insight is that both sin(nx) and cos(mx) integrate to zero over the interval [0, 2π]. The user initially struggled with the proof but ultimately realized that using the identity sin(nx) * cos(mx) = 0.5 * [sin((n+m)x) - sin((n-m)x)] simplifies the problem, leading to the conclusion that the integral evaluates to zero.

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uman
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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.
 
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If you know

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

then what is

\sin nx \cos mx + \cos nx \sin mx\;?
 
sorry my original post had an error.

I meant to say that I had proven that sin nx * cos mx = -cos mx * sin nx
 
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 :-)
 
uman said:
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.
 
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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