问下数学！
- 提问者网友：骨子里的高雅
- 2021-04-30 07:45
- 五星知识达人网友：酒安江南
- 2021-04-30 09:16
（1）因为左边=f（n）＝sin（nπ/6）
右边=sin((n+12)π/6)=sin((nπ+12π)/6)=sin(nπ/6+2π)＝sin（nπ/6）=左边
所以原式成立
(2)
f(1)=sin(π/6) f(11)=sin(11π/6)=sin(11π/6-2π)=sin(-π/6)=-sin(π/6)
f(2)=sin(2π/6) f(10)=sin(10π/6)=sin(10π/6-2π)=sin(-2π/6)=-sin(2π/6)
f(3)=sin(3π/6) f(9)=sin(9π/6)=sin(9π/6-2π)=sin(-3π/6)=-sin(3π/6)
f(4)=sin(4π/6) f(8)=sin(8π/6)=sin(8π/6-2π)=sin(-4π/6)=-sin(4π/6)
f(5)=sin(5π/6) f(7)=sin(7π/6)=sin(7π/6-2π)=sin(-5π/6)=-sin(5π/6)
f(6)=sin(6π/6)=sinπ=0 f(6)=sin(6π/6)=sin(6π/6-2π)=sin(-6π/6)=-sinπ=0
可见，由第一项开始，每12项相加就会等于0
因为167*12=2004，所以f(1)＋f（2）＋…＋f（2004）=0
而f(2005)=f(1),f(2006)=f(2),f(2007)=f(3),f(2008)=f(4)
所以f（1）＋f（2）＋…＋f（2008）=f（2005）+f（2006）+f（2007）+f（2008）
=sin(π/6)+sin(2π/6)+sin(3π/6)+sin(4π/6)
=1/2+（根号3）/2+1+（根号3）/2
=3/2+根号3