Quod Erat Demonstrandum

2012/03/18

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

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

又談舊題,見諒。

Prove that

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

由歐拉公式(Euler’s formula about complex analysis):

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

\theta 變做 -\theta,得

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

把上兩式加和減,得

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)

從而有

\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}

由 (1) 得

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

由 (2) 得

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)

於是

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 ………. (*)

其中 a=x+x^3+x^4+x^5+x^9b=x^2+x^6+x^7+x^8+x^{10}

考慮

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})

利用 calc101 網,得

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)

由 (3),(4),得 a,b 乃下式的根(roots)

y^2+y+3=0

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

由 (*),及 \tan \frac{3\pi}{11}+4\sin \frac{2\pi}{11} > 0

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)

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