cosxcos2xcos3xcos4xcos5x，求值，怎么算，算不出来把，x换成排/11。求值。

Question

cosxcos2xcos3xcos4xcos5x，求值，怎么算，算不出来把，x换成排/11。求值。谢谢。

Answer 1

cosx*cos2x*cos4x = 2 sinx*cosx*cos2x*cos4x / (2sinx)
= sin2x * cos2x *cos4x /(2sinx) = sin8x / (8sinx)
cos3x*cos5x =(1/2) ( cos8x +cos2x)
原式= (1/16) (1/sinx) [ sin8x cos8x + sin8xcos2x ]
= (1/32) (1/sinx) [ sin16x + sin10x +sin6x ]

Answer 2

cos(3π/11)
=cos(π-8π/11)
=-cos(8π/11)

cos(5π/11)
=cos(16π/11-π)
=-cos(16π/11)

所以，
cos(π/11)·cos(2π/11)·cos(3π/11)
·cos(4π/11)·cos(5π/11)
=cos(π/11)·cos(2π/11)·cos(4π/11)
·cos(8π/11)·cos(16π/11)
=32sin(π/11)·cos(π/11)·cos(2π/11)
·cos(4π/11)·cos(8π/11)·cos(16π/11)
÷32sin(π/11)
=16sin(2π/11)·cos(2π/11)·cos(4π/11)
·cos(8π/11)·cos(16π/11)÷32sin(π/11)
=8sin(4π/11)·cos(4π/11)·cos(8π/11)
·cos(16π/11)÷32sin(π/11)
=4sin(8π/11)·cos(8π/11)·cos(16π/11)
÷32sin(π/11)
=2sin(16π/11)·cos(16π/11)÷32sin(π/11)
=sin(32π/11)÷32sin(π/11)
=sin(3π-π/11)÷32sin(π/11)
=sin(π/11)÷32sin(π/11)
=1/32本回答被提问者和网友采纳