# 二十八角形

## 正二十八角形

${\displaystyle S={\frac {28}{4}}a^{2}\cot {\frac {\pi }{28}}\simeq 62.12672a^{2}}$

${\displaystyle \cos(2\pi /28)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{28}}=\cos {\frac {\pi }{14}}={\sqrt {\frac {1+\cos {\frac {2\pi }{14}}}{2}}}={\sqrt {{\frac {1}{2}}\left(1+{\frac {1}{12}}{\sqrt {3\left(20+2{\sqrt[{3}]{28-84i{\sqrt {3}}}}+2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}\right)}}}$

${\displaystyle \cos {\frac {2\pi }{28}}={\frac {1}{2}}{\sqrt {2+{\sqrt {2+2\cdot \cos {\frac {2\pi }{7}}}}}}={\frac {1}{2}}{\sqrt {2+{\sqrt {2+2\cdot {\frac {1}{6}}\left({\sqrt {7}}\cdot {\sqrt[{3}]{\frac {1+3{\sqrt {3}}\cdot i}{2{\sqrt {7}}}}}+{\sqrt {7}}\cdot {\sqrt[{3}]{\frac {1-3{\sqrt {3}}\cdot i}{2{\sqrt {7}}}}}-1\right)}}}}}$

{\displaystyle {\begin{aligned}&\alpha =2\cos {\frac {2\pi }{28}}+2\cos {\frac {6\pi }{28}}+2\cos {\frac {18\pi }{28}}={\sqrt {7}}\\&\beta =2\cos {\frac {10\pi }{28}}+2\cos {\frac {26\pi }{28}}+2\cos {\frac {22\pi }{28}}=-{\sqrt {7}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{28}}\cdot 2\cos {\frac {6\pi }{28}}+2\cos {\frac {6\pi }{28}}\cdot 2\cos {\frac {18\pi }{28}}+2\cos {\frac {18\pi }{28}}\cdot 2\cos {\frac {2\pi }{28}}=0\\&2\cos {\frac {2\pi }{28}}\cdot 2\cos {\frac {6\pi }{28}}\cdot 2\cos {\frac {18\pi }{28}}=-{\sqrt {7}}\\\end{aligned}}}

${\displaystyle x^{3}-{\sqrt {7}}x^{2}+{\sqrt {7}}=0}$

${\displaystyle x=y+{\frac {\sqrt {7}}{3}}}$

${\displaystyle y^{3}-{\frac {7}{3}}y+{\frac {13{\sqrt {7}}}{27}}=0}$

${\displaystyle x={\frac {\sqrt {7}}{3}}+{\frac {2{\sqrt {7}}}{3}}\cos \left({\frac {1}{3}}\arccos {\frac {-13}{14}}\right)}$

${\displaystyle x={\frac {\sqrt {7}}{3}}+{\frac {\sqrt {7}}{3}}{\sqrt[{3}]{{\frac {-13}{14}}+i{\frac {3{\sqrt {3}}}{14}}}}+{\frac {\sqrt {7}}{3}}{\sqrt[{3}]{{\frac {-13}{14}}-i{\frac {3{\sqrt {3}}}{14}}}}}$

${\displaystyle \cos(2\pi /28)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{28}}={\frac {\sqrt {7}}{6}}+{\frac {\sqrt {7}}{6}}{\sqrt[{3}]{{\frac {-13}{14}}+i{\frac {3{\sqrt {3}}}{14}}}}+{\frac {\sqrt {7}}{6}}{\sqrt[{3}]{{\frac {-13}{14}}-i{\frac {3{\sqrt {3}}}{14}}}}}$