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cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程

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cos(2π/7)cos(4π/7)cos(6π/7)怎么解·要过程
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cos(2π/7)cos(4π/7)cos(6π/7)
=cos(2π/7)cos(4π/7)[-cos(π-6π/7)]
=-cos(π/7)cos(2π/7)cos(4π/7)
=-2sin(π/7)cos(π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-sin(2π/7)cos(2π/7)cos(4π/7)/[2sin(π/7)]
=-2sin(2π/7)cos(2π/7)cos(4π/7)/[4sin(π/7)]
=-sin(4π/7)cos(4π/7)/[4sin(π/7)]
=-2sin(4π/7)cos(4π/7)/[8sin(π/7)]
=-sin(8π/7)/[8sin(π/7)]
=-sin(π+π/7)/[8sin(π/7)]
=sin(π/7)/[8sin(π/7)]
=1/8