cosxcos2xcos3xcos4xcos5x,求值,怎么算,算不出来把,x换成排/11。求值。
利用 e^(ix)=cosx+isinx;
e^(ix)+e^(i2x)+e^(i3x)+……+e*(inx)=(cosx+cos2x+……+cosnx)+i(sinx+sin2x+……+sinnx)
=[e^(inx+ix) -e^(ix)]/[e^(ix)-1];
将最后一个等号右端分成实部和虚部(分母和分子同乘以 (cosx-1)-isinx),与等号左端实部和虚部对应相等即得;
cos2Xcos3X=二分之cos5X+cosX
可以化成2cos2xcos3x=cos5x+cosx
然后cos5x+cosx可以化成2cos(5x+x)/2 * cos(5x-x)/2
就是2cos(6x)/2 * cos(4X)/2
然后就是2cos3x * cos2x
就出来了
从后往前算就行了。
*:乘号
知道了吧``
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 ]
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
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