(方法一)
cos(π/7)+cos(3π/7)+cos(5π/7)
=2sin(π/7)(cos(π/7)+cos(3π/7)+cos(5π/7))/2sin(π/7)
=(sin(2π/7)+sin(4π/7)-sin(2π/7)+sin(6π/7)-sin(4π/7))/2sin(
π/7)
=sin(6π/7)/2sin(π/7)
=1/2
类似地可以证明另一个式子
(方法二)
设z=cos(π/(2n+1))+isin(π/(2n+1))
则 z^(2n+1)=cosπ+isinπ=-1(棣莫佛公式)
则z+z^3+z^5+...+z^(2n-1)=z(1-Z^2n)/(1-z^2)
=(z+1)/(1-z^2)=1/(z+1)
将z=cos(π/(2n+1))+isin(π/(2n+1))代入得：
1/(z+1)=1/(1+cos(π/(2n+1))+isin(π/(2n+1)))
=(1+cos(π/(2n+1))-isin(π/(2n+1))/(2+2cos(π/(2n+1))
=1/2-isin(π/(2n+1))/(2+2cos(π/(2n+1))
得用复数相等的定义：
可得：cos[pai/(2n+1)]+cos[pai*3/(2n+1)]+.
+cos[pai*(2n-1)/*(2n+1)]=0.5
将n=3代入即可得到
cos(pai/7)+cos(pai*3/7)+cos(pai*5/7)=0.5