设函数f（x）=2x-cosx，{an}是公差为π8的等差数列，f（a1）+f（a2）+…+f（a5）=5π，则[f（a3）]2-a1a5

设函数f（x）=2x-cosx，{an}是公差为π8的等差数列，f（a1）+f（a2）+…+f（a5）=5π，则[f（a3）]2-a1a5=______．

∵f（x）=2x-cosx，
∴可令g（x）=2x+sinx，∵{an}是公差为
π
8 的等差数列，f（a1）+f（a2）+…+f（a5）=5π
∴g（a1-
π
2 ）+g（a2-
π
2 ）+…+g（a5-
π
2 ）=0，则a3=
π
2 ，a1=
π
4 ，a5=
3π
4
∴[f（a3）]2-a1a5=π2-
π
4 ?
3π
4 =
13π2
16 ，
故答案为：
13π2
16

∵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