四十五角形

正四十五角形においては、中心角と外角は8°で、内角は172°となる。一辺の長さが a の正四十五角形の面積 S は

${\displaystyle S={\frac {45}{4}}a^{2}\cot {\frac {\pi }{45}}\simeq 160.8825a^{2}}$

${\displaystyle \cos(2\pi /45)}$を平方根と立方根で表すと、

${\displaystyle {\begin{aligned}\cos {\frac {2\pi }{45}}=&\cos \left({\frac {\pi }{9}}-{\frac {\pi }{15}}\right)\\=&\cos {\frac {\pi }{9}}\cos {\frac {\pi }{15}}+\sin {\frac {\pi }{9}}\sin {\frac {\pi }{15}}\\=&{\frac {{\sqrt[{3}]{4-4i{\sqrt {3}}}}+{\sqrt[{3}]{4+4i{\sqrt {3}}}}}{4}}\cdot {\frac {1}{8}}\left({\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1\right)+{\frac {i\left({\sqrt[{3}]{4-4i{\sqrt {3}}}}-{\sqrt[{3}]{4+4i{\sqrt {3}}}}\right)}{4}}\cdot {\frac {1}{8}}\left({\sqrt {2\left(5+{\sqrt {5}}\right)}}+{\sqrt {3}}-{\sqrt {15}}\right)\\=&{\frac {1}{32}}\left(\left({{\sqrt[{3}]{4-4i{\sqrt {3}}}}+{\sqrt[{3}]{4+4i{\sqrt {3}}}}}\right)\left({\sqrt {6\left(5+{\sqrt {5}}\right)}}+{\sqrt {5}}-1\right)+{i\left({\sqrt[{3}]{4-4i{\sqrt {3}}}}-{\sqrt[{3}]{4+4i{\sqrt {3}}}}\right)}\left({\sqrt {2\left(5+{\sqrt {5}}\right)}}+{\sqrt {3}}-{\sqrt {15}}\right)\right)\end{aligned}}}$

関係式

${\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{45}}+2\cos {\frac {32\pi }{45}}+2\cos {\frac {28\pi }{45}}=0\\2\cos {\frac {4\pi }{45}}+2\cos {\frac {26\pi }{45}}+2\cos {\frac {34\pi }{45}}=0\\2\cos {\frac {8\pi }{45}}+2\cos {\frac {38\pi }{45}}+2\cos {\frac {22\pi }{45}}=0\\2\cos {\frac {16\pi }{45}}+2\cos {\frac {14\pi }{45}}+2\cos {\frac {44\pi }{45}}=0\\\end{aligned}}}$

三次方程式の係数を求めると

${\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{45}}\cdot 2\cos {\frac {32\pi }{45}}+2\cos {\frac {32\pi }{45}}\cdot 2\cos {\frac {28\pi }{45}}+2\cos {\frac {28\pi }{45}}\cdot 2\cos {\frac {2\pi }{45}}=-3\\&2\cos {\frac {2\pi }{45}}\cdot 2\cos {\frac {32\pi }{45}}\cdot 2\cos {\frac {28\pi }{45}}=2\cos {\frac {2\pi }{15}}={\frac {1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}}{4}}\end{aligned}}}$

解と係数の関係より

${\displaystyle u^{3}-3u-2\cos {\frac {2\pi }{15}}=0}$

三次方程式を解いて、整理すると${\displaystyle \cos(2\pi /45)}$が求められる。

${\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{45}}=&{\sqrt[{3}]{\cos {\frac {2\pi }{15}}+i\sin {\frac {2\pi }{15}}}}+{\sqrt[{3}]{\cos {\frac {2\pi }{15}}-i\sin {\frac {2\pi }{15}}}}\\4\cos {\frac {2\pi }{45}}=&{\sqrt[{3}]{8\cos {\frac {2\pi }{15}}+i8\sin {\frac {2\pi }{15}}}}+{\sqrt[{3}]{8\cos {\frac {2\pi }{15}}-i8\sin {\frac {2\pi }{15}}}}\\4\cos {\frac {2\pi }{45}}=&{\sqrt[{3}]{1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}+i\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}+{\sqrt[{3}]{1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}-i\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}\\\cos {\frac {2\pi }{45}}=&{\frac {1}{4}}\left({\sqrt[{3}]{1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}+i\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}+{\sqrt[{3}]{1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}-i\left({\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}\right)}}\right)\\\end{aligned}}}$

• 五角形
• 九角形
• 十五角形
• 十八角形
• 三十角形