一道三角函数求和题cosπ/13+cos3π/13+cos9π/13

这题难了,一般和差化积貌似不行的：
已知(或容易求)：
cos(π/13)+cos(3π/13)+...+cos(11π/13)=1/2
cos(2π/13)+cos(4π/13)+...+cos(12π/13)=-1/2
令：m=cos(π/13)+cos(3π/13)+cos(9π/13)
n=cos(5π/13)+cos(7π/13)+cos(11π/13)
则：m+n=1/2
经过复杂但不难理解的推导可得：
m*n=(3/2)(cos(2π/13)+cos(4π/13)+...+cos(12π/13))=-3/4
解方程组：m+n=1/2和m*n=-3/4
得：m=(1+√13)/4
还有一个m=(1-√13)/4要舍去经过复杂但不难理解的推导可得：m*n=(3/2)(cos(2π/13)+cos(4π/13)+...+cos(12π/13))=-3/4什么意思呀？呵呵，就是推导很复杂，却不难做，只要保证不算错：m*n=(cos(π/13)+cos(3π/13)+cos(9π/13))*(cos(5π/13)+cos(7π/13)+cos(11π/13))=(1/2)(cos(6π/13)+cos(4π/13))+(1/2)(cos(8π/13)+cos(6π/13))+(1/2)(cos(12π/13)+cos(10π/13))+(1/2)(cos(8π/13)+cos(2π/13))+(1/2)(cos(10π/13)+cos(4π/13))+(1/2)(cos(14π/13)+cos(8π/13))+(1/2)(cos(14π/13)+cos(4π/13))+(1/2)(cos(16π/13)+cos(2π/13))+(1/2)(cos(18π/13)+cos(2π/13))一共9大项，即18小项的和，cos(2π/13)、cos(4π/13)、cos(6π/13)cos(8π/13)、cos(10π/13)、cos(12π/13)每项有3个即：m*n=(3/2)(cos(2π/13)+cos(4π/13)+...+cos(12π/13))