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  1. #1
    Moo
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    Prove a trigonometric identity

    Hi !

    (I know a proof, so it's like a challenge question)

    Prove that \sin \tfrac \pi m \cdot \sin \tfrac{2 \pi}{m} \cdots \sin \tfrac{(m-1) \pi}{m}=\tfrac{m}{2^{m-1}}, where m \in \mathbb{N} and m \geqslant 2
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    See this blog post of mine from last February for a proof of this, as a corrolary of a geometric proof using complex numbers.

    --Kevin C.
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    Quote Originally Posted by Moo View Post
    Hi !

    (I know a proof, so it's like a challenge question)

    Prove that \sin \tfrac \pi m \cdot \sin \tfrac{2 \pi}{m} \cdots \sin \tfrac{(m-1) \pi}{m}=\tfrac{m}{2^{m-1}}, where m \in \mathbb{N} and m \geqslant 2
    I posted a proof a long time ago here.

    Another interesting one (from number theory) is if p\equiv 1(\bmod 4) then:
    \cos \frac{1^2\pi}{p} + \cos \frac{2^2\pi}{p} + \cos \frac{3^2\pi}{p} + ... + \cos \frac{(p-1)^2\pi}{p} = \sqrt{p}

    (I do not think this is solvable by ordinary trigonometry techinques it is a special result from number theory).
    It generalizes to when p is any positive integer integer.
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    Quote Originally Posted by ThePerfectHacker View Post

    Another interesting one (from number theory) is if p\equiv 1(\bmod 4) then:
    \cos \frac{1^2\pi}{p} + \cos \frac{2^2\pi}{p} + \cos \frac{3^2\pi}{p} + ... + \cos \frac{(p-1)^2\pi}{p} = \sqrt{p}

    (I do not think this is solvable by ordinary trigonometry techinques it is a special result from number theory).
    It generalizes to when p is any positive integer integer.
    are you sure you wrote the sum correctly?! i think if p is any odd integer, then S_p=\sum_{j=1}^{p-1} \cos \left(\frac{j^2 \pi}{p} \right)=0. the reason is that S_p=\sum_{j=1}^{\frac{p-1}{2}}\left[\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right) \right],

    and it's easy to see that \cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right)=0.
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    Quote Originally Posted by NonCommAlg View Post
    are you sure you wrote the sum correctly?! i think if p is any odd integer, then S_p=\sum_{j=1}^{p-1} \cos \left(\frac{j^2 \pi}{p} \right)=0. the reason is that S_p=\sum_{j=1}^{\frac{p-1}{2}}\left[\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right) \right],

    and it's easy to see that \cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right)=0.
    Sorry , \sum_{t=0}^{p-1} \cos \frac{2\pi t^2}{p} = \sqrt{p}.

    There is a similar one with \sum_{t=0}^{p-1}\cos \frac{2\pi t^3}{p} but this one was only recently solved, p\equiv 1(\bmod 3)
    (Known as Kummer's problem)
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