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Prove sin(mx)cosn(nx) is odd

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following formula
    [itex]
    \int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx = 0\\
    (m, n = \pm 1, \pm 2, \pm 3, ...)
    [/itex]
    2. Relevant equations
    [itex]
    \sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)]
    [/itex]

    3. The attempt at a solution

    [itex]
    \int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx\\
    \int_{-\pi}^{\pi} \sin(mx-nx)+\sin(mx+nx)\,dx\\
    \int_{-\pi}^{\pi} \sin((m-n)x)+\sin((m+n)x)\,dx
    [/itex]
    Edit
    [itex]
    \int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx
    [/itex]
    the function, sine of any constant times x, is an odd function, therefore when it is integraded from -a to a, its value is 0.
    Therefore, we get
    [itex]
    \int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx = 0 + 0 = 0
    [/itex]
    Would that be sufficient in the proof?

    Thanks guys!
     
    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 3, 2012 #2

    vela

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    Just do the integrals, taking care to treat the case where m=n separately. You know how to integrate sin kx, right?
     
  4. Oct 3, 2012 #3

    LCKurtz

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    What do you know about odd and even functions and integrals? Are sines and cosines odd or even? Can you draw any conclusions about those integrands?
     
  5. Oct 3, 2012 #4
    Ok well of course -sin(x) = sin(-x) but does that hold true when there is a constant in front of x?
     
  6. Oct 3, 2012 #5

    LCKurtz

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    x can be anything in that identity, can't it? Like, for example, cx.
     
  7. Oct 3, 2012 #6

    HallsofIvy

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    a(-x)= -(ax).
     
  8. Oct 3, 2012 #7
    Ok thanks guys! I think I got it, could you tell me if my attempted answer to the question appears to be sufficient?
     
  9. Oct 3, 2012 #8

    vela

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    Yeah, that's sufficient. But can you see how could have done the whole problem in one step?
     
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