f(x)=(a+b)·b=a·b+|b|^2=sinxcosx-1/2+cosx^2+1=sin(2x)/2+cos(2x)/2+1
=(sqrt(2)/2)*sin(2x+π/4)+1,-π/2≤x≤0,即：-π≤2x≤0,即：-3π/4≤2x+π/4≤π/4
故：sin(2x+π/4)∈[-1,sqrt(2)/2],故：f(x)∈[1-sqrt(2)/2,3/2]
即f(x)的最小值是：1-sqrt(2)/2,最大值是：3/2