sin3x-cos3x-sinx+cosx
=3sina-4sin^3a-（4cos^3a-3cosa）-sinx+cosx
=2sina-4sin^3a-4cos^3a+4cosx
=2sin(1-2sin^2x)+4cos(1-cos^2x)
=2sincos2x+4cossin^2x
=2sin(cos2x+sin2x)
（復(fù)習(xí)三角公式)
上面用到的公式：
sin(3x)=sin(x+2x)=sin2xcosx+cos2xsinx=2sinx(1-sin²x)+(1-2sin²x)sinx=3sinx-4sin^3x
cos3x=cos(2x+x)=cos2xcosx-sin2xsinx=(2cos²x-1)cosx-2(1-cos^x)cosx=4cos^3x-3cosx
sin2x=2sinx·cosx
cos=cos^2(x)-sin^2(x) =2cos^2(x)-1 =1-2sin^2(x)sin3x-cos3x-sinx+cosx=2sinx怎么解？解 sin3x-cos3x-sinx+cosx=2sinx 3sinx-4sin^3x-（4cos^3x-3cosx）-sinx+cosx=2sinx-4sin^3x-4cos^3x+4cosx=0sin^3x+cos^3x-cosx=0sin^3x+cosx(cos^2x-1)=0sin^3x+cosx(-sin^2x)=0sin^2x(sinx-cosx)=0解得x=π/4+nπ 或x=nπ （n=0.1.2.....)解畢。