f(n)=sin nπ/6,n∈Z,求f(1)+f(2)+f(3)+……+f(2010)

题目

f(n)=sin nπ/6,n∈Z,求f(1)+f(2)+f(3)+……+f(2010)

答案

f(1)=sin π/6=1/2
f(2)=sin π/3=√3/2
f(3)=sin π/2=1
f(4)=sin 2π/3=√3/2
f(5)=sin 5π/6=1/2
f(6)=sin π=0
f(7)=sin 7π/6=-1/2
f(8)=sin 4π/3=-√3/2
f(9)=sin 3π/2=-1
f(10)=sin 5π/3=-√3/2
f(11)=sin 11π/6=-1/2
f(12)=sin 2π=0
f(n)是关于12个数循环、12个数的和为：0
2010/12=167.6
f(1)+f(2)+f(3)+……+f(2010)
=0+0+0.+0+1/2+√3/2+1+√3/2+1/2+0
=1/2+√3/2+1+√3/2+1/2
=1/2+1/2+√3/2+√3/2+1
=1+√3+1
=2+√3