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化简y=sin2x/1%Cosx

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

已经是最简了。

>> clear >> x=-10:0.01:10; >> y=sin(2*x).*cos(x); >> plot(x,y,'color','r','linewidth',3); >> grid on

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

求y=cosx/cos2x的二阶导数 ,并求二阶导数的零点 解:dy/dx=(-cos2xsinx+2cosxsin2x)/cos²2x 令d²y/dx²=[cos²2x(2sin2xsinx-cos2xcosx-2sinxsin2x+4cosxcos2x)+4cos2xsin2x(-cos2xsinx+2cosxsin2x)]/cos⁴(2x)=0 约分...

变形=2sinxcosx/[cosx+(1-cos2x)/2]dx =-2cosx[cosx+(1-cos2x)/2]d(cosx) 令t=cosx

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

先画出cosx图像 再向下平移3单位得到y=cosx-3 先画出sinx图像 将图像缩小1/2得到sin2x 如果你认可我的回答,请点击“采纳为满意答案”,祝学习进步!

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