f(1)=sinπ/3=√3/2
f(2)=sin2π/3=√3/2
f(3)=sin3π/3=0
f(4)=sin4π/3=-√3/2
f(5)=sin5π/3=-√3/2
f(6)=sin6π/3=0
f(7)=sin7π/3=sinπ/3=f(1)
所以是6个一循环
且f(1)+f(2)+……+f(6)=0
2007÷6余数是3
所以原式=f(1)+f(2)+f(3)=√3
f(1)=sinπ/3=√3/2
f(2)=sin2π/3=√3/2
f(3)=sin3π/3=0
f(4)=sin4π/3=-√3/2
f(5)=sin5π/3=-√3/2
f(6)=sin6π/3=0
f(7)=sin7π/3=sinπ/3=f(1)
所以是6个一循环
且f(1)+f(2)+……+f(6)=0
2007÷6余数是3
所以原式=f(1)+f(2)+f(3)=√3