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

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 
