TITLE: An identity for the cosine function
QUESTION [5 upvotes]: Let $x = \pi/(2k+1)$, for $k>0$.
Prove that
$$
\cos(x)\cos(2x)\cos(3x)\dots\cos(kx) = \frac{1}{2^k}
$$
I've confirmed this numerically for $n$ from $1$ to $30$.
I'm finding it surprisingly difficult using standard trigonometric formula manipulation.
Even for the case $k = 2$, I needed to actually work out $\cos x$ by other methods to get the result.
Please let me know if you have a neat proof.

REPLY [2 votes]: Another proof arises from the identity
$$ \prod_{m=-k}^k\cos\left(t+\frac{m\pi}{2k+1}\right) = 2^{-2k}\cos((2k+1)t),\tag{1}$$
by putting $t=0$ and taking the square root of both sides. We can deduce (1) from 
$$ \prod_{m=-k}^k \left(z\exp\left(\frac{m\pi i}{2k+1}\right)+z^{-1}\exp\left(\frac{-m\pi i}{2k+1}\right)\right)=z^{2k+1}+z^{-2k-1}\tag{2},$$
by putting $z=\exp(it)$. Finally, we can prove (2) by multiplying both sides by $z^{2k+1}$ and examining the roots of the resulting two polynomials in $z$.