帮我化简一个三角函数

f(x)
=2cosx[(1/2)sinx+(√3/2)cosx]-√3(sinx)"+sinxcosx
=2sinxcosx +√3cos2x
=sin2x+√3cos2x
=2sin[(π/3)+2x]

解，依题：
f(x)=2cosx*[1/2sinx+根号3/2cosx]-根号3×sin^2x+sinxcosx
=sinxcosx+根号3cos^2x-根号3sin^2x+sinxcosx
=2sinxcosx+根号3×(cos^2x-sin^2x)
=sin2x+根号3×cos2x
=2sin(2x+pai/3)

f(x)=2cosx(sinxcos∏/3+cosxsin∏/3)-(√3)*sin^2(x)+sinxcosx
=sinxcosx+(√3)*cos^2(x)-(√3)*sin^2(x)+sinxcosx
=2sinxcosx+(√3)*cos2x
=sin2x+(√3)*cos2x
=2sin(2x+∏/3)