Question

设f(x)=sin，则f(1)+f(2)+f(3)+…+f(2 005)=_________________.

Answer 1

解析:∵sin=sin(x+2π)=sin，

∵f(1)=sin=，

f(2)=sin=，

f(3)=sinπ=0，f(4)=sin=-,f(5)=sin=-,f(6)=sin2π=0.

又f(2 005)=sin=sin(334×2π+)=sin=

∴原式=［f(1)+f(2)+…+f(6)］×334+f(2 005)

=0×334+

=.

答案: