- 已知：f(n)=cos(nπ/6),求f(1)+f(2)+f(3?
- 已知：f(n)=c(nπ/6),求f(1)+f(2)+f(3)+···+f(960)的值。
写下过程！
谢谢帮助！
- 解：f(n+12)=cos[2π+(nπ/6)]=cos(nπ/6)=f(n)，即f(n)周期为12
f(1)=cos(π/6)=√3/2，f(2)=cos(π/3)=1/2，f(3)=cos(π/2)=0
f(4)=cos(2π/3)=-1/2，f(5)=cos(5π/6)=-√3/2，f(6)=cosπ=-1
f(7)=cos(7π/6)=-√3/2，f(8)=cos(4π/3)=-1/2，
f(9)=cos(3π/2)=0，f(10)=cos(5π/3)=1/2，
f(11)=cos(11π/6)=√3/2，f(12)=cos(2π)=1
计算得f(1)+f(2)+…+f(12)=0，960÷12=80，故
f(1)+f(2)+f(3)+…+f(960)=80[f(1)+f(2)+…+f(12)]=0