四十五角形
2 cos  2 π 45 = cos  2 π 15 + i sin  2 π 15 3 + cos  2 π 15 − i sin  2 π 15 3 4 cos  2 π 45 = 8 cos  2 π 15 + i 8 sin  2 π 15 3 + 8 cos  2 π 15 − i 8 sin  2 π 15 3 4 cos  2 π 45 = 1 + 5 + 30 − 6 5 + i ( 15 + 3 − 10 − 2 5 ) 3 + 1 + 5 + 30 − 6 5 − i ( 15 + 3 − 10 − 2 5 ) 3 cos  2 π 45 = 1 4 ( 1 + 5 + 30 − 6 5 + i ( 15 + 3 − 10 − 2 5 ) 3 + 1 + 5 + 30 − 6 5 − i ( 15 + 3 − 10 − 2 5 ) 3 ) {\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}}}