35
$\begingroup$

How can we prove the following trigonometric identity?

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

7 Answers 7

33

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

14

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

  • 6
    Cute: this is a variant on the quadratic Gauss sum. (Added: as Sury points out in his article, to say the least.)2011-05-13
8

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

7

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.

3

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!

2

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.

  • 0
    it is the sum (resultant) of the $11$ vectors defining a $11$-gon.2017-10-19
0

$x=\tan(\frac{3\pi}{11})+4\sin(\frac{2\pi}{11})$

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,

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

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

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

$x^2=11$

$x=\sqrt11$

  • 0
    I improve formatting but I don't know about using trigonometric functions in sentencess still.2017-10-19