化简函数f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx

楼上的复杂了啊!
很简单啊!
把cos(x-π/6)化开,f(x)=√3(cos)^2-√3(sinx)^2+sin2x=√3cos2x+sin2x=2sin(2x+π/3)

f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx
=[cos(2x-π/6)+cosπ/6]-√3/2 (1-cos2x)+1/2 sin2x
=cos(2x-π/6)-√3/2 cos2x+1/2 sin2x
=cos(2x-π/6)-cos(2x-π/6)
=0.

f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx
=2cosx*(cosx*cosπ/6+sinx*sinπ/6)-√3(sinx)^2+sinx*cosx
=2cosx*(√3/2*cosx+1/2*sinx)-√3(sinx)^2+sinx*cosx
=√3(cosx)^2+sinx*cosx-√3(sinx)^2+sinx*cosx
=√3((cosx)^2-(sinx)^2)+2sinx*cosx
=√3cos2x+sin2x
=2sin(2x+π/3)