# 十七边形

17

4.
a2cotπ.
17

22.735491898417a2

17..
o = 15814.
17
o
158.82352941176°

## 作圖方法

### 作圖

1796年高斯证明了可以用尺規作圖作出正十七邊形，同時發現了可作圖多邊形的條件。正十七邊形其中一个作圖方法如下：

$\operatorname{cos}{2\pi\over17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}.$

### 證明

$16 \alpha = 360^\circ - \alpha$

$\sin 16\alpha = - \sin \alpha$，而

\begin{align} \sin 16\alpha & = 2\sin 8\alpha \cos 8\alpha \\ & = 2^2\sin 4\alpha \cos 4\alpha \cos 8\alpha \\ & = 2^4 \sin \alpha \cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha \\ \end{align}

$16 \cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha = -1$

$2(\cos \alpha + \cos 2\alpha + \cdots + \cos 8\alpha) = -1$

$x = \cos \alpha + \cos 2\alpha + \cos 4\alpha + \cos 8\alpha$

$y = \cos 3\alpha + \cos 5\alpha + \cos 6\alpha + \cos 7\alpha$

$x + y = - \frac {1}{2}$

\begin{align} xy & = (\cos \alpha + \cos 2\alpha + \cos 4\alpha + \cos 8\alpha)(\cos 3\alpha + \cos 5\alpha + \cos 6\alpha + \cos 7\alpha) \\ & = \frac {1}{2} (\cos 2\alpha + \cos 4\alpha + \cos 4\alpha + \cos 6\alpha + \cdots + \cos \alpha + \cos 15\alpha) \\ & = -1 \\ \end{align}

$x = \frac {-1 + \sqrt{17}}{4}$

$y = \frac {-1 - \sqrt{17}}{4}$

$x_1 = \cos \alpha + \cos 4\alpha$$x_2 = \cos 2\alpha + \cos 8\alpha$

$y_1 = \cos 3\alpha + \cos 5\alpha$$y_2 = \cos 6\alpha + \cos 7\alpha$

$x_1 + x_2 = \frac {-1 + \sqrt{17}}{4}$

$y_1 + y_2 = \frac {-1 - \sqrt{17}}{4}$

$\cos \alpha + \cos 4\alpha = x_1$

$\cos \alpha \cos 4\alpha = \frac{y_1}{2}$

$\cos \alpha=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$

Q.E.D