Question

设函数f(x)=2x-cosx,{An}是公差为π/8的等差数列，f(a1)+f(a2)+…f(a5)=5π，则[f(a3)]^2-a1×a3为？

为什么cosa1+cosa2+...+cosa5=0???

Answer 1

{An}是公差为π/8的等差数列
a1=x-2π/8,a2=x-π/8,a3=x,a4=x+π/8,a5=x+2π/8
f(a1)=2(x-2π/8)-cos(x-2π/8)
f(a2)=2(x-π/8)-cos(x-π/8)
f(a3)=2x-cosx
f(a4)=2(x+π/8)-cos(x+π/8)
f(a5)=2(x+2π/8)-cos(x+2π/8)
f(a1)+f(a2)+…f(a5)=5π
2(x-2π/8)-cos(x-2π/8)+2(x-π/8)-cos(x-π/8)+2x-cosx+2(x+π/8)-cos(x+π/8)+2(x+2π/8)-cos(x+2π/8)=5π
cos(x-2π/8)+cos(x-π/8)+cosx+cos(x+π/8)+cos(x+2π/8)=10x-5π
2cosxcos(2π/8)+2cosxcos(π/8)+cosx=10x-5π
(2cosπ/4+2cosπ/8+1)cosx=10x-5π

[f(a3)]^2-a1×a3
=(2x-cosx)^2-(x-2π/8)×x

看来题目有问题