i believe the problem was suppose to be $\prod_{k=1}^{2m}\cos\frac{k\pi}{2m+1}=\frac{(-1)^m}{4^m}.$ let $w=cis \frac{2\pi}{2m+1}$ we'll get that $w^k=cis \frac{2k\pi}{2m+1}~~(*)$ thus $w,w^2,..,w^{2m}$ are the roots of $f(x)=\sum_{k=0}^{2m} x^k$ $\Rightarrow f(x)=\prod_{k=1}^{2m} (x-w^k)=\sum_{k=0}^{2m} x^k$ consider $ f(-1)=\prod_{k=1}^{2m} (-1-w^k)=\sum_{k=0}^{2m} (-1)^k$ $\Rightarrow \prod_{k=1}^{2m} (-1-w^k)=1\Rightarrow|\prod_{k=1}^{2m} (1+w^k)|=|1|=1$ $\Rightarrow\prod_{k=1}^{2m} |(1+cis\frac{2k\pi}{2m+1})|=1~~;~from~(*)$ $\Rightarrow \prod_{k=1}^{2m} |cis\frac{k\pi}{2m+1}||\frac{1}{cis\frac{k\pi}{2m+1}}+cis\frac{k\pi}{2m+1}|=1$ $\Rightarrow\prod_{k=1}^{2m} |(2\cos\frac{k\pi}{2m+1})|=1\Rightarrow\prod_{k=1}^{2m}|(\cos\frac{k\pi}{2m+1})|=\frac{1}{2^{2m}} =\frac{1}{4^m}$ since $|\cos\frac{k\pi}{2m+1}|=\cos\frac{k\pi}{2m+1},~~~~\forall k\in\{1,2,..,m\}$ and $|\cos\frac{k\pi}{2m+1}|=-\cos\frac{k\pi}{2m+1},~~~~\forall k\in\{m+1,m+2,..,2m\}$ $\therefore ~\prod_{k=1}^{2m}\cos\frac{k\pi}{2m+1}=\frac{(-1)^m}{4^m},~~~\forall m\in\mathbb{N}~\square$