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Math Help - Trigonometric proof(requires use of complex numbers)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Trigonometric proof(requires use of complex numbers)?

    Prove that if m, n\ \mbox{and}\ p are arbitrary integers then the expression:

    \sin \frac{m\pi}{p}\sin \frac{n\pi}{p}\ +\ \sin \frac{2m\pi}{p}\sin \frac{2n\pi}{p}\ +\ \sin \frac{3m\pi}{p}\sin \frac{3n\pi}{p}\ + \mbox{...}\ +\ \sin \frac{(p - 1)m\pi}{p}\sin \frac{(p - 1)n\pi}{p} is equal to:

    \blacksquare\ -\frac{p}{2} when (m + n) is divisible by 2p and (m - n) is not,

    \blacksquare\ \frac{p}{2} when (m - n) is divisible by 2p and (m + n) is not,

    \blacksquare\ 0 when both (m + n) and (m - n) are divisible by 2p or not divisible by 2p.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Prove that if m, n\ \mbox{and}\ p are arbitrary integers then the expression:

    \sin \frac{m\pi}{p}\sin \frac{n\pi}{p}\ +\ \sin \frac{2m\pi}{p}\sin \frac{2n\pi}{p}\ +\ \sin \frac{3m\pi}{p}\sin \frac{3n\pi}{p}\ + \mbox{...}\ +\ \sin \frac{(p - 1)m\pi}{p}\sin \frac{(p - 1)n\pi}{p} is equal to:

    \blacksquare\ -\frac{p}{2} when (m + n) is divisible by 2p and (m - n) is not,

    \blacksquare\ \frac{p}{2} when (m - n) is divisible by 2p and (m + n) is not,

    \blacksquare\ 0 when both (m + n) and (m - n) are divisible by 2p or not divisible by 2p.
    Start by using the trig formula \sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr). That gives

    \sum_{r=1}^{p-1}\sin\tfrac{mr\pi}p\sin\tfrac{nr\pi}p = \frac12\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p - \cos\tfrac{(m+n)r\pi}p\bigr).

    You can sum the cosine series using \sum_{r=1}^{p-1}\cos\tfrac{kr\pi}p = \text{Re}\sum_{r=1}^{p-1}e^{kr\pi i/p} (geometric series).

    I'll leave it to you to take it from there. I haven't tried to complete the calculation, but my gut feeling is to be suspicious of the claimed result. I would expect the answer to depend on whether m\pm n and 2p have a nontrivial common divisor, rather than whether they are divisible by 2p. But I may well be wrong about that.
    Last edited by Opalg; May 25th 2009 at 12:08 AM.
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  3. #3
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    Observation...

    Notice that if 2p|m+n then \frac{2m+2n}{2p}=2k_+ and equivalently, if 2p|m-n then \frac{2m-2n}{2p}=2k_-

    From the equation: S=\frac12\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p - \cos\tfrac{(m+n)r\pi}p\bigr)

    It is immediately apparent that condition 3 holds, as both cosine terms are actually \cos(2kr\pi) for some k, which is zero regardless of r.

    So, condition 2 translates to: \sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m+n)r\pi}p\bigr)=-p for all m,n,p for 2p\not| m+n

    Likewise condition 1 becomes: \sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p\bigr)=-p for all m,n,p for 2p\not| m-n

    This should greatly simplify the geometric series step.
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