cosπ/7x cos2/7πx cos4/7π等于多少?
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[cos(π/7)]x[ cos(2π/7)]x[ cos(4π/7)]
=1/[8sin(π/7)]*4[sin(2π/7)][cos(2π/7)][cos(4π/7)]
=1/[8sin(π/7)]*2[sin(4π/7)][cos(4π/7)]
=sin(8π/7)/[8sin(π/7)]
=-1/8.
=1/[8sin(π/7)]*4[sin(2π/7)][cos(2π/7)][cos(4π/7)]
=1/[8sin(π/7)]*2[sin(4π/7)][cos(4π/7)]
=sin(8π/7)/[8sin(π/7)]
=-1/8.
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