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

Feb 2011
147
19
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.
 
Last edited:

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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?
 
  • Like
Reactions: ymar
Feb 2011
147
19
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
Aug 2007
4,039
2,789
Leeds, UK
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)
 
  • Like
Reactions: ymar