问题 填空题
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=______.
答案

∵f(x)=2x-cosx,

∴f(a1)+f(a2)+…+f(a5)=2(a1+a2+…+a5)-(cosa1+cosa2+…+cosa5),

∵{an}是公差为

π
8
的等差数列,

∴a1+a2+…+a5=5a3,由和差化积公式可得,

cosa1+cosa2+…+cosa5

=(cosa1+cosa5)+(cosa2+cosa4)+cosa3

=[cos(a3-

π
8
×2)+cos(a3+
π
8
×2)]+[cos(a3-
π
8
)+cos(a3+
π
8
)]+cosa3

=2cos

(a3-
π
4
)+(a3+
π
4
)
2
cos
(a3-
π
4
)-(a3+
π
4
)
2
+2cos
(a3-
π
8
)+(a3+
π
8
)
2
cos
(a3-
π
8
)-(a3+
π
8
)
2
+cosa3

=2cosa3

2
2
+2cosa3•cos(-
π
8
)+cosa3

=cosa3(1+

2
+
2+
2

则cosa1+cosa2+…+cosa5的结果不含π,

又∵f(a1)+f(a2)+…+f(a5)=5π,

∴cosa3=0,故a3=

π
2

[f(a3)]2-a1a32-(

π
2
-2•
π
8
π
2
2-
π2
8
=
7π2
8

故答案为:

7π2
8

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