设f(x)=COS(n/2π+π/4),求f(1)+f(2)+f(3)+...f(2008)的值

Question

设f(x)=COS(n/2π+π/4),求f(1)+f(2)+f(3)+...f(2008)的值

Answer 1

f(x)=COS(n/2π+π/4),f(1)=cos(3π/4)=-√2/2f(2)=-cos(π/4)=-√2/2f(3)=-cos(π/2+π/4)=sinπ/4=√2/2f(4)=cos(π/4)=√2/2所以f(1)+f(2)+f(3)+f(4)=0从前往后每四项的和为0f(1)+f(2)+f(3)+...f(2008)=0...