问题: 关于三角函数的
若f（n）=sin（nπ／6）,n∈N+,则f(1)+f(2)+L+f(2008)=
解答:
若f(n)=sin(nπ/6),n∈N+,则f(1)+f(2)+...+f(2008)=
f(1)+f(11) = sin(π/6)+sin(11π/6) = sin(π/6)-sin(π/6) = 0
f(2)+f(10) = sin(2π/6)+sin(10π/6) =sin(2π/6)-sin(2π/6)=0
...
f(5)+f(7) = sin(5π/6)+sin(7π/6) = sin(5π/6)-sin(5π/6) = 0
f(6)+f(12) = sin(π)+sin(2π) = 0
--->f(1)+f(2)+...+f(12) = 0
∵f(13)+f(14)+...+f(24) = f(1)+f(2)+...+f(12) = 0 ...
且2008÷12=167...4
--->f(1)+f(2)+...+f(2008) = 167*0 + f(1)+f(2)+f(3)+f(4)
= sin(π/6)+sin(2π/6)+sin(3π/6)+sin(4π/6)
= 1/2 + √3/2 + 1 + √3/2
= 3/2 + √3
版权及免责声明
1、欢迎转载本网原创文章，转载敬请注明出处：侨谊留学（www.goesnet.org）；
2、本网转载媒体稿件旨在传播更多有益信息，并不代表同意该观点，本网不承担稿件侵权行为的连带责任；
3、在本网博客/论坛发表言论者，文责自负。
