不乖灵
春芽
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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
1年前
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将军2号
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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?
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不乖灵
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))