cos(π/7)×cos(2π/7)×cos(3π/7)=?过程
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cos(π/7)×cos(2π/7)×cos(3π/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)
=-sin(4π/7)cos(4π/7)/4sin(π/7)
=-sin(8π/7)/8sin(π/7)
=sin(π/7)/8sin(π/7)
=1/8
=-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)
=-sin(4π/7)cos(4π/7)/4sin(π/7)
=-sin(8π/7)/8sin(π/7)
=sin(π/7)/8sin(π/7)
=1/8
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