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

#### ymar

I have to prove the following identities:

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

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

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.

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#### Opalg

MHF Hall of Honor
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?

ymar

#### ymar

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.

#### Opalg

MHF Hall of Honor
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)

ymar