給你三角函數的表示形式 你再看看要怎麼把它寫成三次根號的樣子

如果要將正七邊形，和一元三次方程式，扯上關係，會讓人想到有一種可能… 首先 a⁶+a⁵+a⁴+a³+a²+a+1 =(a-b)·(a-b²)·(a-b³)·(a-b⁴)·(a-b⁵)·(a-b⁶) b=exp(2πi/7) 因而 =>a⁶+a⁵+a⁴+a³+a²+a+1=0 Re(a⁶+a⁵+a⁴+a³+a²+a+1)=0 cos(2π/7) + cos(4π/7) + cos(6π/7) + cos(8π/7) + cos(10π/7) + cos(12π/7) + 1 = 0 再來因為 cos(2π/7) = cos(12π/7) = -cos(9π/7) cos(4π/7) = cos(10π/7) = -cos(3π/7) cos(6π/7) = cos(8π/7) = -cos(π/7) cos(π/7) = cos(15π/7) 且 2(cos(2π/7)+cos(4π/7)+cos(6π/7))=-1 cos(2π/7)+cos(4π/7)+cos(6π/7)=-½ 所以 cos(π/7)-cos(2π/7)+cos(3π/7) =cos(3π/7)-cos(6π/7)+cos(9π/7) =cos(5π/7)-cos(10π/7)+cos(15π/7) =½ 又因為 cos(2θ)=2·cos²θ-1 =>2·cos(2θ)=4·cos²θ-2 cos(3θ)=4·cos³θ-3cosθ =>2·cos(3θ)=8·cos³θ-6cosθ 於是 f(cosθ) =2·cos(3θ)-2·cos(2θ)+2·cosθ-1 =(8·cos³θ-6cosθ)-(4·cos²θ-2) +2·cosθ-1 =8·cos³θ-4·cos²θ-4·cosθ+1=0 得 =>x³-4x²-4x+1|(x=cosθ)=0 最後 (cos(π/7),cos(3π/7),cos(5π/7)) =(cos(π/n),cos(3π/n),cos(5π/n))|(n=7) 為三根(5π是七邊形的內角和)