I'm trying to show that
$$\tan\frac{\pi}{7}\tan\frac{2\pi}{7}\tan\frac{3\pi}{7}=\sqrt 7$$
My attempt: Since $\tan x=\frac{\sin x}{\cos x} $ so immediately the denominator is recognized as $$\prod_{k=1}^{3}\cos\frac{k\pi}{7}=\frac{1}{2^3}=\frac{1}{8}$$ as known since elementary classes. To tackle with sine product I tried as $\sin(x)=\cos\left(\frac{\pi}{2}-x\right)$. But I fail with this ideas.
How do I deal with $\sin x$ product?
Here by @ Darth Geek, in general we have $$S(n)=\prod_{k=1}^{n}\sin\left(\frac{k\pi}{2n+1} \right)=\frac{\sqrt{2n+1}}{2^n}$$ set $n=3$ we have $S(3)=\frac{\sqrt{7}}{8}$ and we are done. Also $$C(n)=\prod_{k=1}^{n}\cos\left(\frac{k \pi}{2n+1} \right)=\frac{1}{2^n}$$.
Alternative approach: We exploit reflection formula of the gamma function ,
$$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi}{ z}}$$ giving us $$\sin\frac{\pi}{7}\sin\frac{3\pi}{7}\sin\frac{3\pi}{7}=\frac{\pi^3}{\Gamma\left(\frac{1}{7}\right)\Gamma\left(\frac{6}{7}\right)\Gamma\left(\frac{2}{7}\right)\Gamma\left(\frac{5}{7}\right)\Gamma\left(\frac{3}{7}\right)\Gamma\left(\frac{4}{7}\right)}=\frac{\sqrt 7 \pi^3}{8\pi^3 } $$ We deduce the latter result using multiplication theorem for gamma function. Now it is just simplification which gives us $\sqrt 7$.
