How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityHow can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
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:
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 $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.
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.
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
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!
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.
$$x=\tan(\frac{3\pi}{11})+4\sin(\frac{2\pi}{11})$$
For simpliying equation, I used $$u=\frac {\pi}{11}$$ and $$11u=\pi$$ transformations,
Hence,
$$x=\tan3u+4\sin2u$$
$$2\cos3u*x=2\cos3u*\tan3u+8\cos3u*\sin2u$$
Hence,
$$2\cos3u*x=2\sin3u+8\cos3u*\sin2u$$
After squaring both sides,
$$(2\cos3u*x)^2=(2\sin3u+8\cos3u*\sin2u)^2$$
$$4(\cos3u)^2*x^2=4(\sin3u)^2+32\sin3u*\cos3u*\sin2u+64(\cos3u)^2*(\sin2u)^2$$
$$=2*(1-\cos6u)+16\sin6u*\sin2u+16*(1+\cos6u)*(1-\cos4u)$$
$$=2-2\cos6u+8\cos4u-8\cos8u+16*(1+\cos6u-\cos4u-\cos6u*\cos4u)$$
$$=2-2\cos6u+8\cos4u-8\cos8u+16+16\cos6u-16\cos4u-16\cos6u*\cos4u$$
$$=18+14\cos6u-8\cos4u-8\cos8u-8*(\cos10u+\cos2u)$$
$$=18+14\cos6u-8\cos4u-8\cos8u-8\cos10u-8\cos2u$$
After multiplying both sides with $\sin u$,
$$4(\cos3u)^2*\sin u*x^2=18\sin u+14\cos6u*\sin u-8\cos4u*\sin u-8\cos8u*\sin u-8\cos10u*\sin u-8\cos2u*\sin u$$
$$=18\sin u+7\sin7u-7\sin5u-(4\sin5u-4\sin3u)-(4\sin9u-4\sin7u)-(4\sin11u-4\sin9u)-(4\sin3u-4\sin u)$$
$$=18\sin u+7\sin7u-7\sin5u-4\sin5u+4\sin3u-4\sin9u+4\sin7u-4\sin11u+4\sin9u-4\sin3u+4\sin u$$
$$=22\sin u+11\sin7u-11\sin5u-4\sin11u$$
$$=22\sin u+11*(\sin7u-\sin5u)-4*\sin(\pi)$$
$$=22\sin u+22\cos6u*\sin u-4*0$$
Thus,
$$4(\cos3u)^2*\sin u*x^2=22\sin u*(1+\cos6u)$$
$$4(\cos3u)^2*x^2=22*(1+\cos6u)$$
$$4(\cos3u)^2*x^2=44*(\cos3u)^2$$
$$x^2=11$$
$$x=\sqrt11$$