19
$\begingroup$

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.

  • 1
    It's finite, isn't it?2012-07-20
  • 0
    You should have $sin(n\pi/2)$.2012-07-20
  • 1
    In the argument of the $\sin$, instead of $k$, it should be $n$.2012-07-20
  • 0
    Yes, I know. It is a careless typo. Sorry.2012-07-20
  • 0
    @Cody: [Step $(2)$](http://math.stackexchange.com/a/173310) uses the identity you were thinking about.2012-07-20
  • 2
    This is almost identical to the second part of [this question](http://math.stackexchange.com/questions/70231/how-to-prove-those-curious-identities)2012-07-20

4 Answers 4