等差数列AP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-a1a3=?

Question

等差数列AP
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-a1a3=?

Answer 1

要用到和差化积,cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]
f(a1)+f(a2)+...+f(a5)
=10a3-(cosa1+cosa5)-(cosa2+cosa4)-cosa3
=10a3-2cos(a1+a5)/2*cos(a1-a5)/2-2cos(a2+a4)/2*cos(a2-a4)/2-cosa3
=10a3-2cosa3*cosπ/4-2cosa3*cosπ/8-cosa3
=10a3-cosa3(2cosπ/4+2cosπ/8+1)
记g(x)=10x-cosx(2cosπ/4+2cosπ/8+1)
g'(x)=10+(2cosπ/4+2cosπ/8+1)sinx>0
所以g(x)单调递增
而g(π/2)=5π
所以a3=π/2是唯一的
从而[f(a3)]²-a1a3
=π²-π²/8=7π²/8