How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
 A: You can find the solution in this page:


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*http://natto.2ch.net/math/kako/1002/10029/1002903143.html
Translation of the page into  English.
$I = \tan (3π/11) +4 \sin (2π/11)$
and $t = 3π/11 $
　$11t = 3π$
　⇔ $6t = 3π-5t$
　⇒ $\sin (6t) = \sin (3π-5t)$ taking sin of both sides
　⇔ $2\sin (3t) \cos (3t) = \sin(5t)$ double angle formula
  ⇔ $[3\sin(t)-4 \sin^3 (t)] [4 \cos^3 (t)-3\cos(t)] = 16 \sin^5(t) -20 \sin^3(t) +5 \sin(t)$
　⇔ $[3-4 \sin^2 t ] [4 \cos^3 t -3\cos t] = 16 \sin^4 t - 20 \sin^2 t  +5$ dividing by  $\sin t ≠ 0$
　⇔ $32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0$, where $\sin^2 t  = 1 - \cos^2 t$, $x = \cos t$    
Thus $x = \cos (3π/11)$ is a solution of 
$ 32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1 = 0 $ 
Since $(2π/11) = [1 - (9 / 11)] π = (π-3t)$, so
$I = \tan (3π/11) +4 \sin (2π/11)$
$　= \tan t +4 sin (π-3t)$
$　= \tan t +4 \sin (3t)$
$　= (\sin t / \cos t) +4 [3\sin t-4 \sin^3 t ]$
$　= (\sin t / \cos t) [16 \cos^3 t- 4 \cos t +1]$  
$I ^ 2 = (\sin t / \cos t) ^ 2 [16 \cos^3 t -4 \cos t +1]^2$
$　= [(1 - \cos^2 t) / \cos^2 t] [16 \cos^3 t -4 \cos t +1] ^ 2$
$　= [(1-x^2) (16x^3-4x +1)^2]/x^2$, where $x = \cos t$ 
Molecule {(1-x ^ 2) (16x ^ 3-4x +1) ^ 2} a 
{32x ^ 5-16x ^ 4-32x ^ 3 +12 x ^ 2 +6 t-1} is divided by 
← 2 11x ^ quotient remainder omitted 
A: A slightly more general one is
$$ (\tan 3x+4\sin 2x)^{2}= 11-\frac{\cos 8x(\tan 8x+\tan 3x)}{\sin x\cos 3x}.$$ The proof is similar, see e.g. on Mathlinks here or the attached file on the bottom of this post.
A: 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!
A: This is a famous problem! 
A proof, which I got from just googling, appears as a solution Problem 218 in the College Mathematics Journal.
Snapshot:

You should be able to find a couple of different proofs more and references here: http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf
A: Similar to the proof from the College Mathematics Journal, but structured slightly differently.
Let $\omega=e^{i\pi /11}$. Then we get $\sin\dfrac{k\pi}{11}=\dfrac{\omega^{2k}-1}{2i\omega^k}$ and $\tan\dfrac{k\pi}{11}=\dfrac{\omega^{2k}-1}{i(\omega^{2k}+1)}$
Substitution followed by some algebraic manipulations should lead to $\displaystyle\sum_{i=0}^{10}\omega^{2i}=0$, which is certainly true.
A: Another way to solve it using the following theorem found here (author B.Sury): 

Let $p$ be an odd prime, $p\equiv -1 \pmod 4$ and let $Q$ be the set of squares in  $\mathbb{Z}_p^*$. Then, $$\sum_{a\in Q}\sin\left(\frac{2a\pi}{p}\right)=\frac{\sqrt{p}}{2}$$

You may also need to use $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.
