y=sin(x)*cos(5x)用matlab-搜答案-大师作文网

sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sin(x+y+x-y)=sin2x再问：sin(x+y)cos(x-y)+cos(x+y)sin(x-y)为什么=sin(x+y+

y'=[-sinx(1-sinx)-cosx(-cosx)/(1-sinx)²=[-sinx+sin²x+cos²x]/(1-sinx)²=1/(1-sinx)

sin(x+y)sin(x-y)=[sinxcosy+sinycosx][sinxcosy-cosxsiny]=(sinxcosy)^2-(cosxsiny)^2=(1-cos^2y)cos^2y-c

COS（X+Y）COS（X-Y）=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

整理方程,得y=1+2sinxcosx+2(cosx)^2利用降幂公式和二倍角公式,得y=sin2x+cos2x+2再利用辅助角公式,得y=根号2*sin(2x+π/4)+2所以当2x+π/4属于[2

y'=[(1+sin²x)'*cosx²-(1+sin²x)*(cosx²)']/cos²(x²)=[2sinxcosx*cos(x&sup

symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)

复合函数应该用链式法则求导：若y=g(u),u=f(x),则dy/dx=dy/du*du/dxy=sin^5xdy/dx=dsin^5x/dsinx*dsinx/dx=5sin⁴xcosx

这个方程无解取区间[-π,π]-1

再答：希望采纳！谢谢

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

y=1+sinxcoswx=1+1/2[sin(x+wx)+sin(x-wx)]你确定有w么?hou'yi'b后一步用到了积化和差的公式

cos(x+y)cosy+sin(x+y)siny=cos((x+y)-y)=cosx=4/5sinx=正负3/5tanx=正负3/4

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

正确方式：x=0:1:40;y=sin(x).*cos(x);plot(x,y)原因：注意多个数值做乘除运算时要用点乘（.*）,直接用乘（*）则报错

令sinx+cosx=t两边平方：1+2sinxcosx=t^2sinxcosx=(t^2-1)/2所以y=(t^2-1)/2+t=t^2/2+t-1/2因为t=sinx+cosx=√2sin(x+π

y=1-sin^2x+sinx-2令sinx=ty=-(t^2-t+1/4)-3/4y=-(t-1/2)^2-3/4

y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2

cos（x+y）=cosxcosy-sinxsinysin(x+y)=sinxcosy+cosxsiny