试证明：cos(4π/n)+cos(8π/n)+...+cos(4(n-1)π/n)+cos(4nπ/n ) = 0

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) = 0.1，2应该是检验单独成立吧。。。用欧拉公式似乎也可以。n = 1时是cos(4π) = 1,n = 2时是cos(2π)+cos(4π) = 2.Euler公式确实可以.e^(4π/n)+e^(8π/n)+...+e^(4nπ/n)= e^(4π/n)·(e^(4nπ/n)-1)/(e^(4π/n)-1) (等比数列求和, 这里要求e^(4π/n) ≠ 1)= 0.取实部即得结论.