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Problem from Apostol book

  1. Dec 4, 2007 #1
    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. jcsd
  3. Dec 4, 2007 #2

    Ben Niehoff

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    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]
     
  4. Dec 4, 2007 #3
    sorry my original post had an error.

    I meant to say that I had proven that sin nx * cos mx = -cos mx * sin nx
     
  5. Dec 4, 2007 #4
    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 :-)
     
  6. Dec 4, 2007 #5
    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.
     
  7. Dec 4, 2007 #6
    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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