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

How can we prove the following trigonometric identity?

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

Solutions Collecting From Web of "How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$"

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

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

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

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.

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.

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!

x=tan(3pi/11)+4sin(2pi/11)

2cos(3pi/11)*x=2sin(3pi/11)+8cos(3pi/11)*sin(2pi/11)

For simpliying equation, I used u=pi/11 and 11u=pi transformations

Hence,

2cos3ux=2sin3u+8cos3usin2u

(2cos3ux)^2=(2sin3u+8cos3usin2u)^2

4(cos3u)^2*x^2=4(sin3u)^2+32sin3ucos3usin2u+64(cos3u)^2*(sin2u)^2

4(cos3u)^2*x^2=4*(1-cos6u)/2+16sin6usin2u+64(1+cos6u)/2*(1-cos4u)/2

4(cos3u)^2*x^2=2-2cos6u+8cos4u-8cos8u+16*(1+cos6u)*(1-cos4u)

4(cos3u)^2*x^2=2-2cos6u+8cos4u-8cos8u+16+16cos6u-16cos4u-16cos6u*cos4u

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

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

After multiplying both sides with sinu,

4(cos3u)^2*sinux^2=18sinu+14cos6usinu-8cos4usinu-8cos8usinu-8cos10usinu-8cos2usinu

=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

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

Thus x^2=11 and x=Sqrt(11)