n = 1,2的时候不成立,应该有限制n ≥ 3.
此时sin(2π/n) ≠ 0.
而2sin(2π/n)·(cos(4π/n)+cos(8π/n)+...+cos(4nπ/n))
= 2sin(2π/n)·cos(4π/n)+2sin(2π/n)·cos(8π/n)+...+2sin(2π/n)·cos(4nπ/n)
= (sin(6π/n)-sin(2π/n))+(sin(10π/n)-sin(6π/n))+...+(sin((4n+2)π/n)-sin((4n-2)π/n))
= sin((4n+2)π/n)-sin(2π/n)
= sin(4π+2π/n)-sin(2π/n)
= sin(2π/n)-sin(2π/n)
= 0.
即得cos(4π/n)+cos(8π/n)+...+cos(4nπ/n) = 0.