Question

cosπ/7-cos2π/7+cos3π/7怎么算，过程详细点

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Answer 1

左边=cosπ/7cos2π/7sin(π/2-3π/7)
=sin(π/14)cosπ/7cos2π/7
=2sin(π/14)cos(π/14)cosπ/7cos2π/7/[2cos(π/14)]
=sin(π/7)cos(π/7)cos(2π/7)/[2cos(π/14)]
=2sin(π/7)cos(π/7)cos(2π/7)/[4cos(π/14)]
=sin(2π/7)cos(2π/7)/[4cos(π/14)]
=2sin(2π/7)cos(2π/7)/[8cos(π/14)]
=sin(4π/7)/[8cos(π/14)]
=cos(4π/7-π/2)/[8cos(π/14)]
=cos(π/14)/[8cos(π/14)]
=1/8

Answer 2

引用果香热红酒的回答：
左边=cosπ/7cos2π/7sin(π/2-3π/7)
=sin(π/14)cosπ/7cos2π/7
=2sin(π/14)cos(π/14)cosπ/7cos2π/7/[2cos(π/14)]
=sin(π/7)cos(π/7)cos(2π/7)/[2cos(π/14)]
=2sin(π/7)cos(π/7)cos(2π/7)/[4cos(π/14)]
=sin(2π/7)cos(2π/7)/[4cos(π/14)]
=2sin(2π/7)cos(2π/7)/[8cos(π/14)]
=sin(4π/7)/[8cos(π/14)]
=cos(4π/7-π/2)/[8cos(π/14)]
=cos(π/14)/[8cos(π/14)]
=1/8

看清楚题目
利用积化和差，先构造积
=2sinπ/7(cosπ/7 - cos2π/7+cos3π/7) / 2sinπ/7
=(2sinπ/7cosπ/7 - 2sinπ/7cos2π/7 + 2sinπ/7cos3π/7）/ 2sinπ/7
=sin2π/7 - (sin3π/7-sinπ/7)+(sin4π/7-sin2π/7) ]/ 2sinπ/7
= (sinπ/7+sin4π/7-sin3π/7)/ 2sinπ/7
=1/2 (sin4π/7-sin3π/7=0)