Finding value of $\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ 
Finding value of $$\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$$

Try: let $$z^{2n}=1\Rightarrow z=(1)^{\frac{1}{2n}}=e^{\frac{k\pi}{n}}$$
So $$z^{2n}-1= \prod^{2n}_{k=1}(z-e^{\frac{k\pi}{n}})=(z^2-1)\prod^{n-1}_{k=1}(z-e^{\frac{k\pi}{n}})\cdot (z-e^{\frac{-k\pi}{n}})$$
So $$z^{2n}-1=(z^2-1)\prod^{n-1}_{k=1}(z^2-2z\cos(k\pi/n)+1$$
Could some help me how to prove my original formula, Thanks

You are pretty close to derive the answer yourself.
Let $S$ be the product at hand, we have
$$\begin{align}S^2 &= \prod_{k=1}^{n-1} \cos^2\frac{k\pi}{2n} =
\prod_{k=1}^{n-1}\frac12\left(1+\cos\frac{k\pi}{n}\right) =
4^{1-n}\prod_{k=1}^{n-1}\left(2+2\cos\frac{k\pi}{n}\right)\\
&= 4^{1-n}\lim_{z\to-1}\prod_{k=1}^{n-1}\left(1+z^2-2z\cos\frac{k\pi}{n}\right)
= 4^{1-n}\lim_{z\to-1}\frac{z^{2n}-1}{z^2-1}
= \frac{n}{4^{n-1}}
\end{align}
$$
Since $S$ is clearly positive, taking square root give us $S = \frac{\sqrt{n}}{2^{n-1}}$.
