Closed formula from product of sine Someone knows a formula for expression below?
$$\prod_{k=2}^{n}\sin(\pi/k)$$
 A: Partial Solution: As $k\to+\infty$, $\frac{\pi}k\to0$ so we can use the small-angle approximation: $\sin\left(\frac{\pi}k\right)\approx\frac{\pi}k$. We can also ignore $k=2$ and start from $k=3$, since $\sin(\pi/2)=1$.
$$\begin{aligned}
P(n)&=\prod_{k=3}^n\sin\left(\frac\pi k\right)\\
&=\prod_{k=3}^n\left[\left(\frac\pi k\right)+\frac1{3!}\left(\frac\pi k\right)^3+\frac1{5!}\left(\frac\pi k\right)^5+\ldots\right]\\
&\approx\frac{\pi^{n-2}}{n!/2}\\
\end{aligned}$$
This fits reasonably well and we get a maximum absolute error of $0.2101$ (at $x=4$) and the approximation seems to have a maximum relative error of around $1.0$ (although I've only managed to check this up to $n=224$, due to the factorial term). The approximation (black) and true value (green) of $P(n)$ are shown in the figure below.

You could also consider using the fact that:
$$\begin{aligned}
\ln P&=\ln\left[\prod_{k=2}^n\sin\left(\frac\pi k\right)\right]\\
&=\sum_{k=2}^n\ln\left[\sin\left(\frac\pi k\right)\right]\\
\end{aligned}$$
So if you can find a closed form for $\ln\sin(\pi/k)$, you'd have an solution. 
