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Cos2xCos4x值为多少?

y=sin^4x+cos^4x =sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x =(sin^2 2x+cos^2 2x)^2-(1/2)(2sinxcosx)^2 =1-(1/2)(sin2x)^2 =1-(1/2)(1-cos^2 2x) =1-1/2*(1-2/9) =1-7/18 =11/18

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 + si...

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/...

积化和差公式: sinαsinβ=-[cos(α+β)-cos(α-β)]/2 cosαcosβ= [cos(α+β)+cos(α-β)]/2 sinαcosβ= [sin(α+β)+sin(α-β)]/2 cosαsinβ= [sin(α+β)-sin(α-β)]/2 sin2x*sin4x=-(cos6x-cos2x)/2 sin2x*cos4x= (sin6x-sin2x)/2 cos2x*sin4x= (sin6x+sin2x)/2

原式乘以sinx,把原式化为sin8x/8 再除以sinx,即为cotπ/9/8

回答:一般不等,在特殊情况下相等 解析: 1、首先看倍角公式:cos4x=(cos2x)^2-(sin2x)^2=2(cos2x)^2-1 2、于是:如果说cos4x=2cos2x 就有2(cos2x)^2-1=2cos2x 相当于2x^2-2x-1=0要在(-1,1)区间内有解 能够解得cos2x的一个负数解, 3、总结...

分区间去绝对值符号后积分,当然也可以换元后利用奇偶性去绝对值符号

因式分解就可以了。

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...

据题,有 y=sinx?cosx?cos2x?cos4x,=12sin2xcos2xcos4x=14sin4xcos4x=18sin8x,所以y′=18cos8x?(8x)′=cos8x.

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