# Quod Erat Demonstrandum

## 2017/06/10

### 正多邊形方程

Filed under: Additional / Applied Mathematics,Fun,mathematics,NSS — johnmayhk @ 12:24 下午
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$\displaystyle r=\frac{1}{\cos(\frac{2}{n}\sin^{-1}(\sin(\frac{n}{2}\theta)))}$

P 由 S 至 A，$\theta$ 由 0∘ 至 36∘，則 $r=\displaystyle\frac{1}{\cos\theta}$

P 由 A 至 M，$\theta$ 由 36∘ 至 72∘，則 $r=\displaystyle\frac{1}{\cos(72^o-\theta)}$

P 由 M 至 B，$\theta$ 由 72∘ 至 108∘，則 $r=\displaystyle\frac{1}{\cos(\theta-72^o)}$

$r=\displaystyle\frac{1}{\cos\phi}$

$\displaystyle 0\le \theta \le \frac{\pi}{5}$，則 $\phi=\theta$

$\displaystyle \frac{\pi}{5}\le \theta \le \frac{2\pi}{5}$，則 $\phi=\frac{2\pi}{5}-\theta$

$\displaystyle \frac{2\pi}{5}\le \theta \le \frac{3\pi}{5}$，則 $\phi=\theta-\frac{2\pi}{5}$

$\displaystyle \frac{3\pi}{5}\le \theta \le \frac{4\pi}{5}$，則 $\phi=\frac{4\pi}{5}-\theta$

$\displaystyle \frac{4\pi}{5}\le \theta \le \frac{5\pi}{5}$，則 $\phi=\theta-\frac{4\pi}{5}$

$\sin^{-1}(\sin(x))\equiv x$ 嗎？

$\displaystyle -\frac{\pi}{2}\le x \le \frac{\pi}{2}$，則 $\sin^{-1}(\sin(x))=x$

$\displaystyle \frac{\pi}{2}\le x \le \frac{3\pi}{2}$，則 $\sin^{-1}(\sin(x))=\pi-x$

$\displaystyle \frac{3\pi}{2}\le x \le \frac{5\pi}{2}$，則 $\sin^{-1}(\sin(x))=x-2\pi$

$\displaystyle \frac{5\pi}{2}\le x \le \frac{7\pi}{2}$，則 $\sin^{-1}(\sin(x))=3\pi-x$

$f(x)=\sin^{-1}(\sin(x))$

$\displaystyle r=\frac{1}{\cos\phi}=\frac{1}{\cos(\frac{2}{5}\sin^{-1}(\sin(\frac{5}{2}\theta)))}$

$\displaystyle r=\frac{1}{\cos(\frac{2}{n}\sin^{-1}(\sin(\frac{n}{2}\theta)))}$

$\displaystyle r=\frac{\cos(\frac{\pi}{n})}{\cos(\frac{2}{n}\sin^{-1}(\sin(\frac{n}{2}\theta)))}$

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