# How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$

How can we prove the following trigonometric identity?

$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$

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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

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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

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Ask Wolfram Alpha for the value of tan(3 pi/11) + 4 sin (2 pi/11). Look at the first alternate form shown below the main answer.

Just how Alpha knows this is an interesting question in itself.

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"how Alpha knows this" is the question. – The Chaz 2.0 May 14 '11 at 1:54
@The Chaz: To be more accurate it is isomorphic to this question :-) – Asaf Karagila May 14 '11 at 11:24
@Asaf: your comment notwithstanding, I still plan on running for "MSE 2011 Funniest User"! – The Chaz 2.0 May 14 '11 at 16:58
@Asaf & Chaz: I suppose this is how a rivalry develops... :) – J. M. May 14 '11 at 17:35

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 $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.

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Cute: this is a variant on the quadratic Gauss sum. (Added: as Sury points out in his article, to say the least.) – Pete L. Clark May 13 '11 at 18:49

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.

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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.

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How did you reach the $\sum_i \omega^{2i}=0$? – ziyuang Jul 2 '14 at 17:13