十四角形

正十四角形

${\displaystyle S={\frac {14}{4}}a^{2}\cot {\frac {\pi }{14}}\simeq 15.3345a^{2}}$

となる。

${\displaystyle \cos(2\pi /14)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{14}}=\cos {\frac {\pi }{7}}={\frac {1}{6}}\left({\sqrt[{3}]{{\frac {7}{2}}\left(-1+3{\sqrt {3}}\cdot i\right)}}+{\sqrt[{3}]{{\frac {7}{2}}\left(-1-3{\sqrt {3}}\cdot i\right)}}+1\right)={\frac {1}{6}}\left({\sqrt {7}}\cdot {\sqrt[{3}]{\frac {-1+3{\sqrt {3}}\cdot i}{2{\sqrt {7}}}}}+{\sqrt {7}}\cdot {\sqrt[{3}]{\frac {-1-3{\sqrt {3}}\cdot i}{2{\sqrt {7}}}}}+1\right)=0.9009688...}$
${\displaystyle \sin {\frac {2\pi }{14}}={\frac {\sqrt {3\left(28-2{\sqrt[{3}]{28-84i{\sqrt {3}}}}-2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}}$
${\displaystyle \cos {\frac {2\pi }{14}}={\frac {\sqrt {3\left(20+2{\sqrt[{3}]{28-84i{\sqrt {3}}}}+2{\sqrt[{3}]{28+84i{\sqrt {3}}}}\right)}}{12}}}$

脚注

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関連項目

• Weisstein, Eric W. "Tetradecagon". MathWorld (英語).
• algebra precalculus - Roots of $8x^3-4x^2-4x+1$ - Mathematics Stack Exchange