十三角形
2 cos  8 π 13 + 2 cos  12 π 13 = − 2 + − 260 − 156 3 i 3 ω 2 + − 260 + 156 3 i 3 ω 6 = 1 3 ( − 1 + 13 ⋅ − 5 − 3 3 i 2 13 3 ω 2 + 13 ⋅ − 5 + 3 3 i 2 13 3 ω ) {\displaystyle 2\cos {\frac {8\pi }{13}}+2\cos {\frac {12\pi }{13}}={\frac {-2+{\sqrt[{3}]{-260-156{\sqrt {3}}i}}\omega ^{2}+{\sqrt[{3}]{-260+156{\sqrt {3}}i}}\omega }{6}}={\frac {1}{3}}\left(-1+{\sqrt {13}}\cdot {\sqrt[{3}]{\frac {-5-3{\sqrt {3}}i}{2{\sqrt {13}}}}}\omega ^{2}+{\sqrt {13}}\cdot {\sqrt[{3}]{\frac {-5+3{\sqrt {3}}i}{2{\sqrt {13}}}}}\omega \right)}