三十五角形
2 cos  2 π 35 + 2 cos  32 π 35 + 2 cos  22 π 35 = 1 4 ( 1 + 5 − 14 ( 5 − 5 ) ) = x 1 2 cos  4 π 35 + 2 cos  6 π 35 + 2 cos  26 π 35 = 1 4 ( 1 − 5 + 14 ( 5 + 5 ) ) = x 2 2 cos  8 π 35 + 2 cos  12 π 35 + 2 cos  18 π 35 = 1 4 ( 1 + 5 + 14 ( 5 − 5 ) ) = x 3 2 cos  16 π 35 + 2 cos  24 π 35 + 2 cos  34 π 35 = 1 4 ( 1 − 5 − 14 ( 5 + 5 ) ) = x 4 {\displaystyle {\begin{aligned}2\cos {\frac {2\pi }{35}}+2\cos {\frac {32\pi }{35}}+2\cos {\frac {22\pi }{35}}={\frac {1}{4}}\left(1+{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}\right)=x_{1}\\2\cos {\frac {4\pi }{35}}+2\cos {\frac {6\pi }{35}}+2\cos {\frac {26\pi }{35}}={\frac {1}{4}}\left(1-{\sqrt {5}}+{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)=x_{2}\\2\cos {\frac {8\pi }{35}}+2\cos {\frac {12\pi }{35}}+2\cos {\frac {18\pi }{35}}={\frac {1}{4}}\left(1+{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}\right)=x_{3}\\2\cos {\frac {16\pi }{35}}+2\cos {\frac {24\pi }{35}}+2\cos {\frac {34\pi }{35}}={\frac {1}{4}}\left(1-{\sqrt {5}}-{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)=x_{4}\\\end{aligned}}}
cos  2 π 35 = 1 6 ( 1 + 5 − 14 ( 5 − 5 ) 4 + − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) − 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 + − 7 + 49 5 + 15 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) + 3 3 ( 7 + 7 5 + 14 ( 5 − 5 ) + 2 14 ( 5 + 5 ) ) i 8 3 ) {\displaystyle {\begin{aligned}\cos {\frac {2\pi }{35}}=&{\frac {1}{6}}\left({\tfrac {1+{\sqrt {5}}-{\sqrt {14\left(5-{\sqrt {5}}\right)}}}{4}}+{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}-3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}+{\sqrt[{3}]{\tfrac {{-7+49{\sqrt {5}}+15{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}}+3{\sqrt {3}}\left(7+7{\sqrt {5}}+{\sqrt {14\left(5-{\sqrt {5}}\right)}}+2{\sqrt {14\left(5+{\sqrt {5}}\right)}}\right)i}{8}}}\right)\\\end{aligned}}}
両辺の立方根を取ると