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How can we prove the following trigonometric identity?

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

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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:

alt text

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

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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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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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!

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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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  • $\begingroup$ How did you reach the $\sum_i \omega^{2i}=0$? $\endgroup$ – ziyuang Jul 2 '14 at 17:13
  • $\begingroup$ it is the sum (resultant) of the $11$ vectors defining a $11$-gon. $\endgroup$ – G Cab Oct 19 '17 at 19:02
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$$x=\tan(\frac{3\pi}{11})+4\sin(\frac{2\pi}{11})$$

For simpliying equation, I used $$u=\frac {\pi}{11}$$ and $$11u=\pi$$ transformations,

Hence,

$$x=\tan3u+4\sin2u$$

$$2cos3u*x=2cos3u*\tan3u+8cos3u*\sin2u$$

Hence,

$$2cos3u*x=2sin3u+8cos3u*sin2u$$

After squaring both sides,

$$(2cos3u*x)^2=(2sin3u+8cos3u*sin2u)^2$$

$$4(cos3u)^2*x^2=4(sin3u)^2+32sin3u*cos3u*sin2u+64(cos3u)^2*(sin2u)^2$$

$$=2*(1-cos6u)+16sin6u*sin2u+16*(1+cos6u)*(1-cos4u)$$

$$=2-2cos6u+8cos4u-8cos8u+16*(1+cos6u-cos4u-cos6u*cos4u)$$

$$=2-2cos6u+8cos4u-8cos8u+16+16cos6u-16cos4u-16cos6u*cos4u$$

$$=18+14cos6u-8cos4u-8cos8u-8*(cos10u+cos2u)$$

$$=18+14cos6u-8cos4u-8cos8u-8cos10u-8cos2u$$

After multiplying both sides with $$sinu$$,

$$4(cos3u)^2*sinu*x^2=18sinu+14cos6u*sinu-8cos4u*sinu-8cos8u*sinu-8cos10u*sinu-8cos2u*sinu$$

$$=18sinu+7sin7u-7sin5u-(4sin5u-4sin3u)-(4sin9u-4sin7u)-(4sin11u-4sin9u)-(4sin3u-4sinu)$$

$$=18sinu+7sin7u-7sin5u-4sin5u+4sin3u-4sin9u+4sin7u-4sin11u+4sin9u-4sin3u+4sinu$$

$$=22sinu+11sin7u-11sin5u-4sin11u$$

$$=22sinu+11*(sin7u-sin5u)-4*sin(pi)$$

$$=22sinu+22cos6u*sinu-4*0$$

Thus,

$$4(cos3u)^2*sinu*x^2=22sinu*(1+cos6u)$$

$$4(cos3u)^2*x^2=22*(1+cos6u)$$

$$4(cos3u)^2*x^2=44*(cos3u)^2$$

$$x^2=11$$

$$x=\sqrt11$$

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    $\begingroup$ Please edit your answer so that it can be more easily read. $\endgroup$ – carmichael561 May 31 '17 at 1:42
  • $\begingroup$ I don't know MathJax. Please you teach me for perfect stranger. Also LHS after multiplying both sides with sinu is "4(cos3u)^2 cross sinu cross x^2" and before is "(2cos3u cross x)^2" and "4(cos3u)^2 cross x^2". $\endgroup$ – user450946 May 31 '17 at 2:11
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    $\begingroup$ I'm not recommending deletion, but you need to take some time to learn MathJax if you want your answers to stay up. $\endgroup$ – DMcMor May 31 '17 at 2:57
  • $\begingroup$ I improve formatting but I don't know about using trigonometric functions in sentencess still. $\endgroup$ – user450946 Oct 19 '17 at 18:38

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