How to prove that $ \prod_{k=1}^{m} (\cos(\frac{(2k-1)\pi}{2m+1} )+1) = \frac{2m+1}{2^{m}} $

Question

$$ \mbox{How to prove that }\quad\prod_{k = 1}^{m}\left\{% \cos\left(\left[2k - 1\right]\pi \over 2m+1 \right)+1\right\} = {2m + 1 \over 2^{m}}\quad {\Large ?} $$ where $m$ is an integer and $1 < m < 90$.

I try to use the identity that $\cos\left(2\theta\right) + 1 = 2\cos^{2}\left(\theta\right)$, but I'm stuck with $\prod_{k = 1}^{m}\cos^{2}\left(\left[2k - 1\right]\pi \over 4m + 2\right)$ part.

Please give me an advice, I appreciate it and thank you for every reply.

Answer 1

If $\cos(2m+1)x=0,$

$(2m+1)x=(2n+1)\dfrac\pi2$ where $n$ is any integer

$x=\dfrac{(2n+1)\pi}{2(2m+1)}$ where $0\le n\le2m$

Now using this

$\cos(2m+1)x=2^{2m}\cos^{2m+1}x+\cdots+(-1)^m(2m+1)\cos x$

So, the roots of $$2^{2m}c^{2m+1}+\cdots+(-1)^m(2m+1)c=0$$ are $\cos\dfrac{2(2n+1)\pi}{(2m+1)},0\le n\le2m$

If $\cos x=0,n=m$

So, the roots of $$2^{2m}c^{2m}+\cdots+(-1)^m(2m+1)=0$$ are $\cos\dfrac{(2n+1)\pi}{2(2m+1)},0\le n\le2m, m\ne n$

$\implies2^{2m}\prod_{n=0,n\ne m}^{2m}\cos\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$

As $\pi-\dfrac{(2n+1)\pi}{2(2m+1)}=\dfrac{\pi(1+2(2m-n))}{2(2m+1)}$

$\implies2^{2m}\prod_{n=0}^{m-1}(-1)^m\cos^2\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$

Can you take it from here?

Answer 2

The numbers $\cos\left(\frac{\pi\left(k+\frac{1}{2}\right)}{n}\right)$, for $k\in\{0,1,\ldots,n-1\}$, are the roots of the Chebyshev polynomial $T_n(x)$. By picking $n=2m+1$ we have that $$ \prod_{k=1}^{m}\left(\cos\frac{\pi(2k-1)}{2m+1}+1\right)=2^m\prod_{k=0}^{m-1}\cos^2\frac{\pi\left(k+\frac{1}{2}\right)}{2m+1} $$ equals the absolute value of $2^m$ times the product of the non-zero roots of $T_{2m+1}(x)$, i.e. $$ 2^m \left|\frac{T_{2m+1}'(0)}{[x^{2m+1}]T_{2m+1}(x)}\right|=2^m\frac{2m+1}{2^{2m}}=\color{red}{\frac{2m+1}{2^m}}. $$