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Math Help - The product of sines and the product of cosines of (k\pi)/(2n+1) for 0<k<n+1

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    The product of sines and the product of cosines of (k\pi)/(2n+1) for 0<k<n+1

    I have to prove the following identities:

    \prod_{k=1}^n \cos\frac{k\pi}{2n+1}=\frac{1}{2^n}

    \prod_{k=1}^n \cos\frac{k\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^n}

    The suggestion is to prove and use this:

    \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right ) for z\in\mathbb{C}

    I can prove this last one. The roots of the polynomial on the left-hand side are \cos\frac{2k\pi}{2n+1}\pm i\sin\frac{2k\pi}{2n+1}. They are also the roots of the quadratic expressions on the right-hand side. I don't know how to use it though. The only way I have found to see \cos\frac{k\pi}{2n+1} and \sin\frac{k\pi}{2n+1} here is this:

    \cos\frac{2k\pi}{2n+1}\pm i\sin\frac{2k\pi}{2n+1}=\left (\cos\frac{k\pi}{2n+1}\pm i\sin\frac{k\pi}{2n+1}\right )^2.

    Now I know that if W(x):=\sum_{k=0}^{2n}z^n, and z_k^{\pm}:=\cos\frac{k\pi}{2n+1}\pm i\sin\frac{k\pi}{2n+1}, then W\left ((z_k^{\pm})^2\right )=0 for every k\in\{1,2,...,n\} and whether we take the plus or the minus. This is 2n equations but I really have no idea what to do with them.

    Of course, if you have any ways of solving this without this suggestion, I will appreciate them to.

    Edit: z^n corrected to z^k in the third formula. Sorry.
    Last edited by ymar; October 17th 2011 at 03:50 PM.
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  2. #2
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    Re: The product of sines and the product of cosines of (k\pi)/(2n+1) for 0<k<n+1

    Quote Originally Posted by ymar View Post
    I have to prove the following identities:

    \prod_{k=1}^n \cos\frac{k\pi}{2n+1}=\frac{1}{2^n}

    \prod_{k=1}^n {\color{red}\sin}\frac{k\pi}{2n+1}=\frac{\sqrt{2n+  1}}{2^n}

    The suggestion is to prove and use this:

    \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right ) for z\in\mathbb{C}
    In the identity \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right ), have you tried putting z=1 and z=1?
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    Re: The product of sines and the product of cosines of (k\pi)/(2n+1) for 0<k<n+1

    Quote Originally Posted by Opalg View Post
    In the identity \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right ), have you tried putting z=1 and z=1?
    Thanks! I haven't. I only tried z=1/\sqrt 2. I wish I were able to see such things -- I really shouldn't study maths.
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  4. #4
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    Re: The product of sines and the product of cosines of (k\pi)/(2n+1) for 0<k<n+1

    Quote Originally Posted by ymar View Post
    I only tried z=1/\sqrt 2. I wish I were able to see such things -- I really shouldn't study maths.
    If you tried z=1/\sqrt 2 then you were obviously thinking along the right lines. Don't be discouraged!
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