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.
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
Hint: multiply by sin(x)
The standard way of doing problems like these is to look at the coefficients of the Chebyshev polynomials. The polynomial $T_n$ of degree $n$ such that $T_n(2 \cos \theta) = 2 \cos n \theta$ has leading term $1$, and we want to compute something like the fourth root of the product of the roots of $T_{2k+1}(x)^2 = 4$. Vieta's formulas and some reflection identities should handle it from here.
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 $z$.
