f(1)=sin(π/2)+cos(π/2)=1f(2)=sin(π)+cos(π)=-1f(3)=sin(3π/2)+cos(3π/2)=-1f(4)=sin(2π)+cos(2π)=1f(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