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化简y=sin2x/1%Cosx

y=(sin2xsinx)/(1-cosx)=2sinxcosxsinx/(1-cosx)=2sin^2xcosx/(1-cosx)=2(1-cos^2x)sinx/(1-cosx) y=(sin2xsinx)/(1-cosx)=2(1+cosx)cosx=2cosx+2cos^2x=2(cosx+1/2)^2-1/2 所以值域为-1/2=<y<=4

y=sin2x+3cosx =2sinxcosx +3cosx =cosx (2sinx+3)

令t=sinx+cosx,原式变为y=(t^2-1)/(1-t+t^2-1)=1+1/t. 因为t∈[-√2,√2] 所以值域为y>=1+√2/2,y<=1-√2/2 望采纳~

(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)]/(sin2x-cos2x)=sinx+cosx

你好:y=(sin2x)/(1-cosx)的函数图像如下:(x的单位是:弧度)

y=sin2x/(1+cosx)=2cosx*sinx/(1+cosx)=2cosxtan(x/2)y'=2tan(x/2)*(-sinx)+2cosx*sec(x/2)*1/2=cosxsec(x/2)-2sinxtan(x/2)=2cosx-2/(1+cosx)微分:dy=[2cosx-2/(1+cosx)]dxx=tcost,y=tsintdx/dt=cost+t(-sint)=cost-tsintdy/dt=sint+tcost=sint+tcostdy/dx=(dy/dt)/(dx/dt)=(sint+tcost)/(cost-tsint)

y=(sin2x)/2

根据公式sin2x=2sinxcosx可知:y=sinxcosx可化简为y=1/2sin2x(二分之一倍的sin2x)

y=3/2sin2x-cosxcosx-1/2=y=3/2sin2x-1/2sin2x-1/2=sin2x-1/2,因为sin2x=2sinxcosx真诚相助望给予满意采纳!

解:∵y''=sin(2x)-cosx ==>y'=-cos(2x)/2-sinx+C1 (C1是积分常数) ==>y=-sin(2x)/4+cosx+C1x+C2 (C2是积分常数) ∴原方程的通解是y=-sin(2x)/4+cosx+C1x+C2 (C1,C2是积分常数).

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