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倒数第二个例题 Cos3xCos2x 怎么变到Cosx+Cos5x的?

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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/...

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...

cos2xcos3x =cos3xcos2x =(1/2){cos(3x+2x)+cos(3x-2x)} =1/2(cos5x+cosx)

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...

证明: cos3x=4(cosx)^3-3cosx 4(cosx)^3=cos3x+3cosx 16(cosx)^6=(cos3x)^2+6cos3xcosx+9(cosx)^2 =(1+cos6x)/2+3(cos4x+cos2x)+9/2*(1+cos2x) 32(cosx)^6=1+cos6x+6cos4x+6cos2x+9+9cos2x 64(cosx)^7=2cos6xcosx+12cos4xcosx+30cos2xcosx+20cos...

cos5x=cos(3x+2x)=cos3xcos2x-sin3xsin2x cosx=cos(3x-2x)=cos3xcos2x+sin3xsin2x cosx+cos5x=2cos3xcos2x cos3xcos2x=1/2(cosx+cos5x) ∫cos3xcos2xdx =∫1/2(cosx+cos5x)dx =1/2∫cosxdx+1/2∫cos5xdx =1/2∫cosxdx+1/10∫cos5xd5x =1/2sinx+1/10sin...

积化和差公式

cos4a =cos²2a-sin²2a =2cos²2a-1 =2[cos2a]²-1 =2[2cos²a-1]²-1 =2[4cos^4a-4cos²a+1]-1 =8cos^4a-8cos²a+1

用复数比较方便 e^(xi)=cosx+isinx e^(2xi)=cos2x+isin2x ... e^(nxi)=cosnx+isinnx 相加得 e^(xi)+e^(2xi)+...+e^(nxi)=(cosx+cos2x+...+cosnx)+i(sinx+sin2x+...+sinnx) e^(xi)+e^(2xi)+...+e^(nxi), 用等比数列求和 =e^(xi) [1-e^(nxi)]/(1-...

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