已知f(x)=sin(xπ/2)+cos(xπ/2),则f(1)+f(2)+．．．+f(2007)+f(2008)等于

f(1)=sin(π/2)+cos(π/2)=1 f(2)=sin(π)+cos(π)=-1 f(3)=sin(3π/2)+cos(3π/2)=-1 f(4)=sin(2π)+cos(2π)=1 f(4k+a)=sin(2π+aπ/2)+cos(2π+aπ/2)=f(a) 而：f(1)+f(2)+f(3)+f(4)=1-1-1+1=0 所以, f(1)+f(2)+．．．+f(2007)+f(2008) =2008/4*(f(1)+f(2)+f(3)+f(4)) =2008/4*0 =0