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# 五十角形

## 正五十角形

${\displaystyle S={\frac {50}{4}}a^{2}\cot {\frac {\pi }{50}}\simeq 198.68181a^{2}}$

${\displaystyle \cos(2\pi /50)}$を冪根で表すと

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{50}}=&\cos {\frac {\pi }{25}}\\=&{\frac {1}{2}}{\sqrt {2+2\cos {\frac {2\pi }{25}}}}\\=&{\frac {1}{2}}{\sqrt {2+{\sqrt[{5}]{{\frac {{\sqrt {5}}-1}{4}}+{\frac {\sqrt {10+2{\sqrt {5}}}}{4}}i}}+{\sqrt[{5}]{{\frac {{\sqrt {5}}-1}{4}}-{\frac {\sqrt {10+2{\sqrt {5}}}}{4}}i}}}}\end{aligned}}}

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{50}}=&\cos {\frac {\pi }{25}}={\frac {1}{2}}\left({\sqrt[{5}]{\cos {\frac {\pi }{5}}+i\cdot \sin {\frac {\pi }{5}}}}+{\sqrt[{5}]{\cos {\frac {\pi }{5}}-i\cdot \sin {\frac {\pi }{5}}}}\right)\\=&{\frac {1}{2}}\left({\sqrt[{5}]{\cos {\frac {2\pi }{10}}+i\cdot \sin {\frac {2\pi }{10}}}}+{\sqrt[{5}]{\cos {\frac {2\pi }{10}}-i\cdot \sin {\frac {2\pi }{10}}}}\right)\\=&{\frac {1}{2}}\left({\sqrt[{5}]{{\frac {{\sqrt {5}}+1}{4}}+i\cdot {\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}}+{\sqrt[{5}]{{\frac {{\sqrt {5}}+1}{4}}-i\cdot {\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}}\right)\\\end{aligned}}}