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×a5=?

题目

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×a5=?
请问一下为什么最后把这些式子化到10a3-cosa3（1+根号2+根号下根号2+2）后,因为an是以π÷8为等差的数列,所以cosa3（1+根号2+根号下根号2+2）是不含π的式子?

答案

∵数列{an}是公差为π/8的等差数列,
且f(a1)+f(a2)+……+f(a5)=5π
2a1-cosa1+2a2-cosa2+2a3-cosa3+2a4-cosa4+2a5-cosa5=5π
∴2(a1+a2+……+a5)-(cosa1+cosa2+……+cosa5)=5π
∴(cosa1+cosa2+……+cosa5)=0
即2(a1+a2+……+a5)=2×5a3=5π,
a3=π/2,
a1=π/4
a5=3π/4
∴[f(a3)]²-a1a5
=(2a3-cosa3)²-a1a5
=(2*π/2-cosπ/2)²-π/4*3π/4
=π²-3π²/16
=13π²/16