y=f(x)的最大值为2故A/2+A/2=2 得A=2其图像相邻两对称轴间的距离为2故最小正周期T=4即2π/(2w)=4得w=π/4故f(x）=1-cos(π/2x+2ψ)带入点（1,2）得cos(π/2+2ψ)=-1又0≤ψ≤π/2得ψ=π/4 f(1)=2f(2)=1-cos(π+π/2)=1f(3)=1-cos(3π/2+π/2)=0f(4)=1-cos(2π+π/2)=1故f(1)+f(2)+f(3)+f(4)=4故f(1)+f(2)+``````f(2008)=502[f(1)+f(2)+f(3)+f(4)]=2008