化简函数f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx 0分
问题描述：
化简函数f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx 0分
答
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)