二十六角形
α = 2 cos  2 π 26 + 2 cos  6 π 26 + 2 cos  18 π 26 = 1 + 13 2 β = 2 cos  14 π 26 + 2 cos  10 π 26 + 2 cos  22 π 26 = 1 − 13 2 {\displaystyle {\begin{aligned}&\alpha =2\cos {\frac {2\pi }{26}}+2\cos {\frac {6\pi }{26}}+2\cos {\frac {18\pi }{26}}={\frac {1+{\sqrt {13}}}{2}}\\&\beta =2\cos {\frac {14\pi }{26}}+2\cos {\frac {10\pi }{26}}+2\cos {\frac {22\pi }{26}}={\frac {1-{\sqrt {13}}}{2}}\\\end{aligned}}}
三次方程式の係数を求めると
2 cos  2 π 26 ⋅ 2 cos  6 π 26 + 2 cos  6 π 26 ⋅ 2 cos  18 π 26 + 2 cos  18 π 26 ⋅ 2 cos  2 π 26 = − 1 2 cos  2 π 26 ⋅ 2 cos  6 π 26 ⋅ 2 cos  18 π 26 = β − 2 {\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}+2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}+2\cos {\frac {18\pi }{26}}\cdot 2\cos {\frac {2\pi }{26}}=-1\\&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}=\beta -2\end{aligned}}}
解と係数の関係より
x 3 − α x 2 − x − ( β − 2 ) = 0 {\displaystyle x^{3}-\alpha x^{2}-x-(\beta -2)=0}
変数変換
x = y + α / 3 {\displaystyle x=y+\alpha /3}
整理すると
y 3 − 13 + 13 6 y + 26 + 5 13 27 = 0 {\displaystyle y^{3}-{\frac {13+{\sqrt {13}}}{6}}y+{\frac {26+5{\sqrt {13}}}{27}}=0}
三角関数、逆三角関数を用いて解は
x = 1 + 13 6 + 2 3 13 + 13 2 cos  ( 1 3 arccos  − ( 26 + 5 13 ) 2 ( 13 + 13 2 ) 3 2 ) {\displaystyle x={\frac {1+{\sqrt {13}}}{6}}+{\frac {2}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}\cos \left({\frac {1}{3}}\arccos {\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}\right)}