y=ycosy-siny

∵cos(x+y)cosy+sin(x+y)siny=0==>cos[(x+y)-y]=0(应用余弦差角公式cos(A-B)=cosAcosB+sinAsinB)==>cosx=0∴cosx=0.

z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3.此时z=3

[sin(2x+y)/sinx]-2cos(x+y)={[sin(x+y)cosx+cos(x+y)sinx]/sinx}-2cos(x+y)={[sin(x+y)cosx+cos(x+y)sinx-

再问：大哥，你题目看错了。。。再答：哪里有错？再问：第一条等式就错了。。是sin(x+y)=sinx+siny。后面是cos（x+y）·（1+y'）=cosx+cosy·y'？再答：OK，那我改下

两边求导：cos(x+y)*(1+y')=cosx+cosy*y'y'=(cosx-cos(x+y))/(cos(x+y)-cosy)e^x+1=e^y*y'+y'y'=(e^x+1)/(e^y+1)

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

dy=dx+dsiny=dx+cosydy即y'=dy/dx=1/(1-cosy)对x求导y''=-1/(1-cosy)²*(1-cosy)'=-siny*y'/(1-cosy)²

设A=(X+Y)/2,B=(X-Y)/2X=A+B,Y=A-BSINX=SIN(A+B)=SINACOSB+COSASINBSINY=SIN(A-B)=SINACOSB-COSASINBSINX+SI

∫e^ysinydy=-∫e^yd(cosy)=-[e^y*cosy-∫cosyd(e^y)]=∫cosy*e^ydy-e^ycosy=∫e^yd(siny)-e^ycosy=e^ysiny-∫sin

1+y'=cosy*y'y'=1/(cosy-1)dy/dx=1/(cosy-1)

令a=x+y,则条件变为3sin(a-x)=sin(a+x),展开得3sinacosx-3cosasinx=sinacosx+cosasinx,移项2sinacosx=4cosasinxtana=2t

解arcsiny=x中y是自变量,x是因变量∴(arcsiny)'=x'=1/√(1-y^2)≠1例如y=sinx,(sinx)'=y'≠1

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)

x*e^y+siny=0e^y+x*e^y*y'+cosy*y'=0=>y'=-e^y/[xe^y+cosy]再问：你好！我数学太烂。。能不能补充一下完整的答案。。。再答：x*e^y+siny=0两边

sin[(x+y)+x]=5sin[(x+y)-x]sin(x+y)·cosx＋cos(x+y)·sinx＝5·sin(x+y)·cosx－5·cos(x+y)·sinx4·sin(x+y)·cosx

y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx

dy/dx=(ycos(y/x)-x)/(xcos(y/x))=y/x-sec(y/x)设u=y/x,y=ux,dy/dx=u+u'x即u'x=-secucosudu=-dxsinu=-x+C即通解为

交换积分顺序,

两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)