Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ Recently, I ran across a product that seems interesting.
Does anyone know how to get to the closed form:
$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$
I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.
This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even. 
So, I tried taking that into account, but must have approached it wrong. 
How can this be shown? Thanks everyone.

The idea of using $\cos x=\frac{1}{2}\frac{\sin 2x}{\sin x}$ is a nice one, and works quickly.   We should not use this when $x=\frac{n\pi}{n}$,because of the $\frac{0}{0}$ issue. Also, as you observe, the product is $0$ if $n$ is even, so we needn't bother. So let $n$ be odd. 
Look at the product from $k=1$ to $k=n-1$. As $k$ ranges over these values, the numbers $2k$ range, modulo $n$, over all numbers from $1$ to $n-1$.  So the $\cos(2k\pi/n)$ (apart from sign) range in some order over the $\cos(\pi j/n)$. Thus, apart from sign, there is cancellation and the product has absolute value $\frac{1}{2^{n-1}}$.  
There is no sign issue. If $n\equiv 1\pmod {4}$, the product from $1$ to $n-1$ has an even number of negative terms. The term $\cos(\pi n/n)$ then gives us an extra $-1$, and the product is negative. For the same reason, if $n\equiv 3\pmod{4}$ then the product is positive. The $-\sin(n\pi/2)$ term captures these sign facts, and also produces the right answer of $0$ when $n$ is even.
