Question

简化。。cos(π/7)cos(2π/7)cos(3π/7)

cos(π/7)cos(2π/7)cos(3π/7)

答案是1/8
求过程

Answer 1

cos(π/7)cos(2π/7)cos(3π/7)
=sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/sin(π/7)
=1/2sin(2π/7)cos(2π/7)cos(3π/7)/sin(π/7)
=1/4sin(4π/7)cos(3π/7)/sin(π/7)
=1/4sin(3π/7)cos(3π/7)/sin(π/7) (sina=sin(π-a))
=1/8sin(6π/7)/sin(π/7) (sina=sin(π-a))
=1/8sin(π/7)/sin(π/7)
=1/8

Answer 2

cos(π/7)cos(2π/7)cos(3π/7)=sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/sin(π/7)
=sin(2π/7)cos(2π/7)cos(3π/7)/2sin(π/7)
=sin(4π/7)cos(3π/7)/4sin(π/7)
=sin(3π/7)cos(3π/7)/4sin(π/7)
=sin(6π/7)/8sin(π/7)
=sin(π/7)/8sin(π/7)
=1/8