數學白板/YuLM/2008-09-24-0030

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$\left\{ \begin{align} & x+\frac{1}{x}=2\cos \theta \\ & 0\le \theta \le \pi \\ \end{align} \right.\Rightarrow x^{n}+\frac{1}{x^{n}}=2\cos \left( n\theta \right)$

$x+\frac{1}{x}=2\cos \theta \Rightarrow x^{2}+1=2\cos \theta \cdot x$ 註：兩邊乘以 $x\,$
$\Rightarrow x^{2}-2\cos \theta \cdot x+1=0$
$\Rightarrow x=\frac{2\cos \theta \pm \sqrt{\left( -2\cos \theta \right)^{2}-4}}{2}=\frac{2\cos \theta \pm \sqrt{4\cos ^{2}\theta -4}}{2}$ 註：一元二次方程式公式解
$=\frac{2\cos \theta \pm 2\sqrt{\cos ^{2}\theta -1}}{2}=\frac{2\cos \theta \pm 2\sqrt{-\sin ^{2}\theta }}{2}=\cos \theta \pm \sqrt{-\sin ^{2}\theta }$
$\because 0\le \theta \le \pi\Rightarrow \sin \theta \ge 0\therefore \sqrt{-\sin ^{2}\theta }=i\sin \theta$
$\therefore x=\cos \theta \pm \sqrt{-\sin ^{2}\theta }=\cos \theta \pm i\sin \theta$
$\Rightarrow \frac{1}{x}=\frac{1}{\cos \theta \pm i\sin \theta }=\left( \cos \theta \pm i\sin \theta \right)^{-1}=\cos \left( -\theta \right)\pm i\sin \left( -\theta \right)$
$=\cos \theta \mp i\sin \theta$
$\left\{ \begin{align} & x=\cos \theta \pm i\sin \theta \Rightarrow x^{n}=\cos \left( n\theta \right)\pm i\sin \left( n\theta \right) \\ & \frac{1}{x}=\cos \theta \mp i\sin \theta \Rightarrow \frac{1}{x^{n}}=\cos \left( n\theta \right)\mp i\sin \left( n\theta \right) \\ \end{align} \right.$
$\therefore x^{n}+\frac{1}{x^{n}}=\left[ \cos \left( n\theta \right)\pm i\sin \left( n\theta \right) \right]+\left[ \cos \left( n\theta \right)\mp i\sin \left( n\theta \right) \right]$
$=2\cos \left( n\theta \right)\#$