# 十七边形

（重定向自正十七邊形

17

${\displaystyle \approx 22.735491898417a^{2}}$

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：

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

### 證明

${\displaystyle 16\alpha =360^{\circ }-\alpha }$

${\displaystyle \sin 16\alpha =-\sin \alpha }$ ，而

{\displaystyle {\begin{aligned}\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{aligned}}}

${\displaystyle 16\cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha =-1}$

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

${\displaystyle \cos 15\alpha =\cos 2\alpha }$ ${\displaystyle \cos 12\alpha =\cos 5\alpha }$ ，令

${\displaystyle x=\cos \alpha +\cos 2\alpha +\cos 4\alpha +\cos 8\alpha }$

${\displaystyle y=\cos 3\alpha +\cos 5\alpha +\cos 6\alpha +\cos 7\alpha }$

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

{\displaystyle {\begin{aligned}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{aligned}}}

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

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

${\displaystyle x_{1}=\cos \alpha +\cos 4\alpha }$ ${\displaystyle x_{2}=\cos 2\alpha +\cos 8\alpha }$

${\displaystyle y_{1}=\cos 3\alpha +\cos 5\alpha }$ ${\displaystyle y_{2}=\cos 6\alpha +\cos 7\alpha }$

${\displaystyle x_{1}+x_{2}={\frac {-1+{\sqrt {17}}}{4}}}$

${\displaystyle y_{1}+y_{2}={\frac {-1-{\sqrt {17}}}{4}}}$

${\displaystyle \cos \alpha +\cos 4\alpha =x_{1}}$

${\displaystyle \cos \alpha \cos 4\alpha ={\frac {y_{1}}{2}}}$

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