# Quod Erat Demonstrandum

## 2012/03/18

### tan(3pi/11)+4tan(2pi/11)=?

Filed under: Pure Mathematics — johnmayhk @ 6:48 上午

Prove that

$\tan \frac{3\pi}{11}+4\sin \frac{2\pi}{11}=\sqrt{11}$.

$e^{i\theta}=\cos\theta+i\sin\theta$

$\theta$ 變做 $-\theta$，得

$e^{-i\theta}=\cos\theta-i\sin\theta$

$e^{i\theta}+e^{-i\theta}=2\cos\theta\Rightarrow \cos\theta =\frac{1}{2}(e^{i\theta}+e^{-i\theta})$

$e^{i\theta}-e^{-i\theta}=2i\sin\theta\Rightarrow \sin\theta =\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ ………. (1)

$\tan\theta=\frac{1}{i} \frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}=\frac{1}{i} \frac{e^{2i\theta}-1}{e^{2i\theta}+1}$ ………. (2)

$x=e^{2i\pi/11}$

$4i\sin \frac{2\pi}{11}=2(x-x^{-1})=2x-2x^{10}$$\because x^{11}=1$

$i\tan \frac{3\pi}{11}$

$=\frac{x^3-1}{x^3+1}$

$=\frac{x^3-x^{33}}{x^3+1}$$\because x^{11}=1$

$=x^3-x^6+x^9-x^{12}+\dots-x^{30}$（sum of G.S.）

$=x^3-x^6+x^9-x+x^4-x^7+x^{10}-x^2+x^5-x^8$$\because x^{11}=1$, we have $x^{12}=x^{11}\cdot x=x$ say）

$i\tan \frac{3\pi}{11}+4i\sin \frac{2\pi}{11}$

$=(x^3-x^6+x^9-x+x^4-x^7+x^{10}-x^2+x^5-x^8)+(2x-2x^{10})$

$=x-x^2+x^3+x^4+x^5-x^6-x^7-x^8+x^9-x^{10}$

$=(x+x^3+x^4+x^5+x^9)-(x^2+x^6+x^7+x^8+x^{10})$

$=a-b$ ………. (*)

$1+a+b=1+x+x^2+x^3+\dots +x^9+x^{10}=\frac{1-x^{11}}{1-x}=0$$\because x^{11}=1$

$a+b=-1$ ………. (3)

$ab$

$=(x+x^3+x^4+x^5+x^9)(x^2+x^6+x^7+x^8+x^{10})$

$ab$

$=x^7+x^6+x^5+2x^4+x^3+2x^2+2x+5+2x^{10}+2x^9+x^8+2x^7+x^6+x^5+x^3$

$=5+2(a+b)=3$ ………. (4)

$y^2+y+3=0$

$(a-b)^2=(a+b)^2-4ab=1-4(3)=-11$

$i\tan \frac{3\pi}{11}+4i\sin \frac{2\pi}{11}=a-b=\sqrt{-11}=i\sqrt{11}$

$\tan \frac{3\pi}{11}+4\sin \frac{2\pi}{11}=\sqrt{11}$

（June 1982 V.N. Murty）