二十八角形

正二十八角形においては、中心角と外角は12.857…°で、内角は167.142…°となる。一辺の長さが a の正二十八角形の面積 S は

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

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

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

別の表し方として

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

関係式

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

三次方程式の係数を求めると

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

解と係数の関係より

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

変数変換

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

整理すると

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

三角関数、逆三角関数を用いた解は

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

平方根、立方根で表すと

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

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

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

• 七角形
• 十四角形
• 二十一角形