TITLE: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
QUESTION [36 upvotes]: How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$

REPLY [6 votes]: Since $\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}>0$, it's enough to prove that
$$\left(\sin\frac{3\pi}{11}+4\sin\frac{2\pi}{11}\cos\frac{3\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$\left(\sin\frac{3\pi}{11}+2\sin\frac{5\pi}{11}-2\sin\frac{\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$1-\cos\frac{6\pi}{11}+4-4\cos\frac{10\pi}{11}+4-4\cos\frac{2\pi}{11}+4\cos\frac{2\pi}{11}-4\cos\frac{8\pi}{11}-$$
$$-4\cos\frac{2\pi}{11}+4\cos\frac{4\pi}{11}-8\cos\frac{4\pi}{11}+8\cos\frac{6\pi}{11}=11+11\cos\frac{6\pi}{11}$$ or
$$\sum_{k=1}^5\cos\frac{2k\pi}{11}=-\frac{1}{2}$$ or
$$\sum_{k=1}^52\sin\frac{\pi}{11}\cos\frac{2k\pi}{11}=-\sin\frac{\pi}{11}$$ or
$$\sum_{k=1}^5\left(\sin\frac{(2k+1)\pi}{11}-\sin\frac{(2k-1)\pi}{11}\right)=-\sin\frac{\pi}{11}$$ or
$$\sin\frac{11\pi}{11}-\sin\frac{\pi}{11}=-\sin\frac{\pi}{11}.$$
Done!