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.



An identity for the cosine function
Let
$S(x)=\prod\_{j=1}^k \text{sin}(jx)$
and
$C(x)=\prod\_{j=1}^k \text{cos}(jx)$.
Let x = $\frac{\pi}{2k+1}$.
Then $S(2x) = S(x)$ (from $\text{sin}(\pi-x)=\text{sin}(x)$), and $S(2x)=2^kS(x)C(x)$ (from $\text{sin}(2x)=2\text{sin}(x)\text{cos}(x)$), from which the result follows.


Steve


