是求值吗? 若是这样,则方法如下:
原式=cos(2π/14)cos(4π/14)cos(6π/14)
=2sin(2π/14)cos(2π/14)cos(4π/14)cos(6π/14)/[2sin(2π/14)]
=sin(4π/14)cos(4π/14)cos(6π/14)/[2sin(π/7)]
=2sin(4π/14)cos(4π/14)cos(6π/14)/[4sin(π/7)]
=sin(8π/14)cos(6π/14)/[4sin(π/7)]
=sin(π/2+π/14)cos(π/2-π/14)/[4sin(π/7)]
=2cos(π/14)sin(π/14)/[8sin(π/7)]
=sin(π/7)/[8sin(π/7)]
=1/8
注:若你的意图不是我所猜测的那样,则请你补充说明.