f(x)=cos（X+π/6）-cos（X-π/6）+ 根号三*cosx
=cosxcosπ/6-sinxsinπ/6-(cosxcosπ/6+sinxsinπ/6)+√3cosx
=-sinx+√3cosx
=-2(1/2sinx-√3/2cosx)
=-2sin(x-π/6)
因为x∈【-π/2,0】
所以x-π/6∈【-π/2-π/6,0-π/6】=【-2π/3,-π/6】
所以最大值=2;
最小值=2*1/2=1.怎么到第二步的？=-sinx+√3cosxcosxcosπ/6-sinxsinπ/6-(cosxcosπ/6+sinxsinπ/6)+√3cosx=cosxcosπ/6-sinxsinπ/6-cosxcosπ/6-sinxsinπ/6+√3cosx=-2sinxsinπ/6+√3cosx(sinπ/6=1/2)=-sinx+√3cosx