​ ​Prove that 
tan(3pi/11) + 4(sin(2pi/11) = root(11)

Dear Student,

$$\begin{gathered} Let t=\tan \left(\frac{3\mathrm{\pi }}{11}\right)+4 \sin \left(\frac{2\mathrm{\pi }}{11}\right)=\frac{\sin \left(\frac{3\mathrm{\pi }}{11}\right)}{\cos \left(\frac{3\mathrm{\pi }}{11}\right)}+4 \sin \left(\frac{2\mathrm{\pi }}{11}\right) \\ 2\cos \left(\frac{3\mathrm{\pi }}{11}\right) t=2\sin \left(\frac{3\mathrm{\pi }}{11}\right)+8 \sin \left(\frac{2\mathrm{\pi }}{11}\right) \cos \left(\frac{3\mathrm{\pi }}{11}\right) \\ Take k=\frac{\mathrm{\pi }}{11} \\ Thus, 2t \cos 3k=2 \sin 3k+8 \sin 2k \cos 3k \\ Squaring both sides, \\ {\left(2t \cos 3k\right)}^2={\left(2 \sin 3k+8 \sin 2k \cos 3k\right)}^2 \\ 4t^2 {\cos }^{2}3k=4 {\sin }^{2}3k+64 {\sin }^{2}2k {\cos }^{2}3k+32 \sin 2k \cos 3k \sin 3k \\ 4t^2 {\cos }^{2}3k=4\left(\frac{1-\cos 6k}{2}\right)+64\left(\frac{1-\cos 4k}{2}\right)\left(\frac{1+\cos 6k}{2}\right)+16 \sin 2k \sin 6k \\ 4t^2 {\cos }^{2}3k=2\left(1-\cos 6k\right)+16\left(1-\cos 4k\right)\left(1+\cos 6k\right)+8 \cos 4k-8 \cos 8k \\ 4t^2 {\cos }^{2}3k=2-2\cos 6k+16-16\cos 4k+16\cos 6k-16\cos 4k\cos 6k+8 \cos 4k-8 \cos 8k \\ 4t^2 {\cos }^{2}3k=18+14\cos 6k-8\cos 4k-8 \cos 8k-8 \cos 10k-8 \cos 2k \\ Multiplying both sides by \sin k, we get \\ 4t^2 {\cos }^{2}3k \sin k=18\sin k+14\cos 6k \sin k-8\cos 4k \sin k-8 \cos 8k \sin k-8 \cos 10k \sin k-8 \cos 2k \sin k \\ 4t^2 {\cos }^{2}3k \sin k=18\sin k+7 \sin 7k-7\sin 5k-4\sin 5k+4 \sin 3k -4 \sin 9k +4 \sin 7k-4 \sin 11k+4 \sin 9k-4\sin 3k +4\sin k \\ 4t^2 {\cos }^{2}3k \sin k=22\sin k-11\sin 5k+11 \sin 7k-4 \sin 11k \\ 4t^2 {\cos }^{2}3k \sin k=22\sin k+11 \left(\sin 7k-\sin 5k\right)-4 \sin \pi \left(\because k=\frac{\mathrm{\pi }}{11}\right) \\ 4t^2 {\cos }^{2}3k \sin k=22\sin k+11 \left(2\cos 6k \sin k\right)-0 \\ 4t^2 {\cos }^{2}3k \sin k=22\sin k+22 \cos 6k \sin k=22\sin k\left(1+\cos 6k\right) \\ 4t^2 {\cos }^{2}3k =22\left(1+\cos 6k\right)=22 \left(2{\cos }^{2}3k\right) \\ 4t^2 {\cos }^{2}3k =44 {\cos }^{2}3k \\ {t}^2=11 \\ Thus t=\sqrt{11} \\ i.e. \tan \left(\frac{3\mathrm{\pi }}{11}\right)+4 \sin \left(\frac{2\mathrm{\pi }}{11}\right)=\sqrt{11} \end{gathered}$$

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