Question

cosπ/15 cos(2π/15)cos(3π/15)cos(4π/15)cos(5π/15)cos(6π/15)cos(7π/15)=1/128

求证

Answer 1

注意到以下两个结论：
（1）cosxcos2x
=4sinxcosxcos2x/[4sinx]
=2sin2xcos2x/[4sinx]
=sin4x/[4sinx]
（2）cosxcos2xcos4xcos8x
=16sinxcosxcos2xcos4xcos8x/[16sinx]
=8sin2xcos2xcos4xcos8x/[16sinx]
=4sin4xcos4xcos8x/[16sinx]
=2sin8xcos8x/[16sinx]
=sin16x/[16sinx]
于是左式
=cos(π/15)cos(2π/15)cos(3π/15)cos(4π/15)cos(5π/15)cos(6π/15)cos(7π/15)
=[cos(π/15)cos(2π/15)cos(4π/15)cos(7π/15)]*[cos(3π/15)cos(5π/15)cos(6π/15)]
注意到cos(7π/15)=-cos(8π/15)
cos(3π/15)=cos(π/5)
cos(6π/15)=cos(2π/5)
cos(5π/15)=cos(π/3)=1/2
于是左式
=-[cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)]*[cos(π/5)cos(2π/5)]*(1/2)
=-{sin(16π/15)/[16sin(π/15)]}*{sin(4π/5)/[4sin(π/5)]}*(1/2)（运用结论（1）（2））
注意到sin(16π/15)=-sin(π/15)
sin(4π/5)=sin(π/5)
所以左式
=(1/16)*(1/4)*(1/2)=1/128=右式。证毕