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# 二十二角形

## 正二十二角形

${\displaystyle S={\frac {22}{4}}a^{2}\cot {\frac {\pi }{22}}\simeq 38.25334a^{2}}$

${\displaystyle \cos(2\pi /22)}$の値は冪根を用いて以下となる。正十一角形も参照のこと。

{\displaystyle {\begin{aligned}\cos {\frac {2\pi }{22}}=&\cos {\frac {\pi }{11}}=\cos \left(\pi -{\frac {10\pi }{11}}\right)=-\cos {\frac {10\pi }{11}}\\=&{\frac {1}{10}}\\&-{\frac {-1-{\sqrt {5}}+i{\sqrt {10-2{\sqrt {5}}}}}{40}}\left\lbrace {\sqrt[{5}]{-{\frac {11}{4}}\left\lbrace 89+25{\sqrt {5}}+\left(45{\sqrt {5-2{\sqrt {5}}}}-5{\sqrt {5+2{\sqrt {5}}}}\right)i\right\rbrace }}+{\sqrt[{5}]{-{\frac {11}{4}}\left\lbrace 89-25{\sqrt {5}}-\left(5{\sqrt {5-2{\sqrt {5}}}}+45{\sqrt {5+2{\sqrt {5}}}}\right)i\right\rbrace }}\right\rbrace \\&-{\frac {-1-{\sqrt {5}}-i{\sqrt {10-2{\sqrt {5}}}}}{40}}\left\lbrace {\sqrt[{5}]{-{\frac {11}{4}}\left\lbrace 89+25{\sqrt {5}}-\left(45{\sqrt {5-2{\sqrt {5}}}}-5{\sqrt {5+2{\sqrt {5}}}}\right)i\right\rbrace }}+{\sqrt[{5}]{-{\frac {11}{4}}\left\lbrace 89-25{\sqrt {5}}+\left(5{\sqrt {5-2{\sqrt {5}}}}+45{\sqrt {5+2{\sqrt {5}}}}\right)i\right\rbrace }}\right\rbrace \\=&0.959492...\end{aligned}}}