cosαsinβ=[sin(α+β)-sin(α-β)]/2 f(x)=(√3cosx-sinx)sin2x/2cosx+1/2=(√3cosx-sinx)sinx+1/2=2cos(x+π/4)sinx+1/2=sin(x+π/4+x)-sin(x+π/4-x)+1/2=sin(2x+π/4)-(√2-1)/2后面的你会的.我不会化简和单调区间f(x)=(√3cosx-sinx)sin2x/2cosx+1/2=(√3cosx-sinx)sinx+1/2=2cos(x+π/6sinx+1/2=sin(x+π/6x)-sin(x+π/6-x)+1/2=sin(2x+π/6)则（1）f(π/3)=sin(2π/3+π/6)=1/2(2)T=2π/2=π单调递减区间为： [kπ+π/6，kπ+7π/6],k∈Z