百四十四角形

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正百四十四角形

${\displaystyle S=36a^{2}\cot {\frac {\pi }{144}}}$

{\displaystyle {\begin{aligned}&x_{1}=2\cos {\frac {2\pi }{144}}+2\cos {\frac {98\pi }{144}}+2\cos {\frac {94\pi }{144}}=0\\&x_{2}=2\cos {\frac {14\pi }{144}}+2\cos {\frac {110\pi }{144}}+2\cos {\frac {82\pi }{144}}=0\\&x_{3}=2\cos {\frac {10\pi }{144}}+2\cos {\frac {86\pi }{144}}+2\cos {\frac {106\pi }{144}}=0\\&x_{4}=2\cos {\frac {70\pi }{144}}+2\cos {\frac {26\pi }{144}}+2\cos {\frac {122\pi }{144}}=0\\&x_{5}=2\cos {\frac {50\pi }{144}}+2\cos {\frac {142\pi }{144}}+2\cos {\frac {46\pi }{144}}=0\\&x_{6}=2\cos {\frac {62\pi }{144}}+2\cos {\frac {130\pi }{144}}+2\cos {\frac {34\pi }{144}}=0\\&x_{7}=2\cos {\frac {38\pi }{144}}+2\cos {\frac {134\pi }{144}}+2\cos {\frac {58\pi }{144}}=0\\&x_{8}=2\cos {\frac {22\pi }{144}}+2\cos {\frac {74\pi }{144}}+2\cos {\frac {118\pi }{144}}=0\\\end{aligned}}}

{\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{144}}\cdot 2\cos {\frac {98\pi }{144}}+2\cos {\frac {98\pi }{144}}\cdot 2\cos {\frac {94\pi }{144}}+2\cos {\frac {94\pi }{144}}\cdot 2\cos {\frac {2\pi }{144}}=-3\\&2\cos {\frac {2\pi }{144}}\cdot 2\cos {\frac {98\pi }{144}}\cdot 2\cos {\frac {94\pi }{144}}=2\cos {\frac {2\pi }{48}}\\\end{aligned}}}

${\displaystyle u^{3}-3u-2\cos {\frac {2\pi }{48}}=0}$

{\displaystyle {\begin{aligned}u_{1}=2\cos {\frac {2\pi }{144}}=&{\sqrt[{3}]{\cos {\frac {2\pi }{48}}+i\sin {\frac {2\pi }{48}}}}+{\sqrt[{3}]{\cos {\frac {2\pi }{48}}-i\sin {\frac {2\pi }{48}}}}\\4\cos {\frac {2\pi }{144}}=&{\sqrt[{3}]{8\cos {\frac {2\pi }{48}}+i8\sin {\frac {2\pi }{48}}}}+{\sqrt[{3}]{8\cos {\frac {2\pi }{48}}-i8\sin {\frac {2\pi }{48}}}}\\4\cos {\frac {2\pi }{144}}=&{\sqrt[{3}]{4{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}+i\cdot 4{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}}}+{\sqrt[{3}]{4{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}-i\cdot 4{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}}}\\\end{aligned}}}

${\displaystyle \cos(2\pi /144)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{144}}={\frac {1}{4}}{\sqrt[{3}]{4{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}+i\cdot 4{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}}}+{\frac {1}{4}}{\sqrt[{3}]{4{\sqrt {2+{\sqrt {2+{\sqrt {3}}}}}}-i\cdot 4{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}}}}$

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