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:

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

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

The suggestion is to prove and use this:

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

I can prove this last one. The roots of the polynomial on the left-hand side are $\displaystyle \cos\frac{2k\pi}{2n+1}\pm i\sin\frac{2k\pi}{2n+1}.$ 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 $\displaystyle \cos\frac{k\pi}{2n+1}$ and $\displaystyle \sin\frac{k\pi}{2n+1}$ here is this:

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

Now I know that if $\displaystyle W(x):=\sum_{k=0}^{2n}z^n,$ and $\displaystyle z_k^{\pm}:=\cos\frac{k\pi}{2n+1}\pm i\sin\frac{k\pi}{2n+1},$ then $\displaystyle W\left ((z_k^{\pm})^2\right )=0$ for every $\displaystyle k\in\{1,2,...,n\}$ and whether we take the plus or the minus. This is $\displaystyle 2n$ 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.

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

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Originally Posted by

**ymar** I have to prove the following identities:

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

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

The suggestion is to prove and use this:

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

In the identity $\displaystyle \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right )$, have you tried putting z=–1 and z=1?

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

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Originally Posted by

**Opalg** In the identity $\displaystyle \sum_{k=0}^{2n}z^k=\prod_{k=1}^n \left (1-2z\cos\frac{2k\pi}{2n+1}+z^2\right )$, have you tried putting z=–1 and z=1?

Thanks! I haven't. I only tried $\displaystyle z=1/\sqrt 2$. I wish I were able to see such things -- I really shouldn't study maths. :(

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** I only tried $\displaystyle z=1/\sqrt 2$. I wish I were able to see such things -- I really shouldn't study maths. :(

If you tried $\displaystyle z=1/\sqrt 2$ then you were obviously thinking along the right lines. Don't be discouraged! (Nod)