关于三角函数的

显然f(1)+f(2)+...+f(12)=0，这是由于正弦函数的对称性同时由于f(n)周期为12那么考虑2008/12=167.333而167*12=2004,那么前2004项和为零原式=f(2005)+...+f(2008)=f(12*167+1)+...+f(12*167+4)=f(1)+...+f(4)=Sin（π／6）+Sin（π／3）+Sin（π／2）+Sin（2π／3)=1/2+√3/2+1-√3/2=2 收起

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) = 0f(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) = 0f(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 收起