a,b属于[0,pai/3],===>π/3≥a-b≥-π/3===>1≥cos(a-b)≥1/2===>2π/3≥a+b≥0===>1≥cos(a+b)≥-1/2===>当a-b=a+b=π/3时y最小===>a=π/3,b=0时最小===>y=2cos(a-b)+cos(a+b)的最小值为2*1/2+1/2=3/2...