TITLE: Showing $\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}$
QUESTION [19 upvotes]: How can one show :
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$

REPLY [4 votes]: It's my very old problem (from 1979). 
$$\sum_{k=1}^6\cos\frac{2k\pi}{13}=\frac{\sum\limits_{k=1}^62\sin\frac{\pi}{13}\cos\frac{2k\pi}{13}}{2\sin\frac{\pi}{13}}=$$
$$=\frac{\sum\limits_{k=1}^6\left(\sin\frac{(2k+1)\pi}{13}-\sin\frac{(2k-1)\pi}{13}\right)}{2\sin\frac{\pi}{13}}=\frac{\sin\frac{13\pi}{13}-\sin\frac{\pi}{13}}{2\sin\frac{\pi}{13}}=-\frac{1}{2}.$$
Let $$\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=x$$ and
$$\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=y.$$
Thus, $x+y=-\frac{1}{2}$ and
$$xy=\frac{1}{2}(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{12\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}+$$
$$+\cos\frac{2\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+$$
$$+\cos\frac{4\pi}{13}+\cos\frac{12\pi}{13}+\cos\frac{2\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13})=$$
$$=\frac{3}{2}\sum_{k=1}^6\cos\frac{2k\pi}{13}=-\frac{3}{4}.$$
Since $x>0$ and $y<0$, we obtain: $x=\frac{\sqrt{13}-1}{4}$ and $y=\frac{-\sqrt{13}-1}{4}$.
Now, $$\cos\frac{2\pi}{13}\cos\frac{6\pi}{13}\cos\frac{8\pi}{13}=$$
$$=\frac{1}{2}\left(\cos\frac{4\pi}{13}+\cos\frac{8\pi}{13}\right)\cos\frac{8\pi}{13}=\frac{1}{4}\left(\cos\frac{4\pi}{13}+\cos\frac{12\pi}{13}+1+\cos\frac{10\pi}{13}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1+\sqrt{13}}{4}\right)=\frac{3-\sqrt{13}}{16}.$$
By another hand, 
$$\cos\frac{4\pi}{13}\cos\frac{10\pi}{13}+\cos\frac{4\pi}{13}\cos\frac{12\pi}{13}+\cos\frac{10\pi}{13}\cos\frac{12\pi}{13}=$$
$$=\frac{1}{2}\sum_{k=1}^6\cos\frac{2k\pi}{13}=-\frac{1}{4}$$ and
$$\cos\frac{4\pi}{13}\cos\frac{10\pi}{13}\cos\frac{12\pi}{13}=$$
$$=\frac{1}{2}\left(\cos\frac{6\pi}{13}+\cos\frac{12\pi}{13}\right)\cos\frac{12\pi}{13}=\frac{1}{4}\left(\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+1+\cos\frac{2\pi}{13}\right)=$$
$$=\frac{1}{4}\left(1+\frac{\sqrt{13}-1}{4}\right)=\frac{3+\sqrt{13}}{16}.$$
Thus,
$$\sin\frac{2\pi}{13}\sin\frac{6\pi}{13}\sin\frac{8\pi}{13}=\sqrt{\frac{\left(1-\cos\frac{4\pi}{13}\right)\left(1-\cos\frac{12\pi}{13}\right)\left(1-\cos\frac{10\pi}{13}\right)}{8}}=$$
$$=\sqrt{\frac{1+\frac{1+\sqrt{13}}{4}-\frac{1}{4}-\frac{3+\sqrt{13}}{16}}{8}}=\sqrt{\frac{13+3\sqrt{13}}{128}}.$$
Id est,
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=-\tan\frac{2\pi}{13}\tan\frac{6\pi}{13}\tan\frac{8\pi}{13}=$$
$$=-\frac{\sqrt{\frac{13+3\sqrt{13}}{128}}}{\frac{3-\sqrt{13}}{16}}=\sqrt{\frac{(13+3\sqrt{13})(3+\sqrt{13})^2}{8}}=\sqrt{65+18\sqrt{13}}.$$
Done!