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Thread: 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 $\displaystyle m, n\ \mbox{and}\ p$ are arbitrary integers then the expression:

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

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

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

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

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

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

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

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

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

    You can sum the cosine series using $\displaystyle \sum_{r=1}^{p-1}\cos\tfrac{kr\pi}p = \text{Re}\sum_{r=1}^{p-1}e^{kr\pi i/p}$ (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 $\displaystyle m\pm n$ 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 $\displaystyle 2p|m+n$ then $\displaystyle \frac{2m+2n}{2p}=2k_+$ and equivalently, if $\displaystyle 2p|m-n$ then $\displaystyle \frac{2m-2n}{2p}=2k_-$

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

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

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

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

    This should greatly simplify the geometric series step.
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