问题
填空题
设函数f(x)=2x-cosx,{an}是公差为
|
答案
∵f(x)=2x-cosx,
∴f(a1)+f(a2)+…+f(a5)=2(a1+a2+…+a5)-(cosa1+cosa2+…+cosa5),
∵{an}是公差为
的等差数列,π 8
∴a1+a2+…+a5=5a3,由和差化积公式可得,
cosa1+cosa2+…+cosa5
=(cosa1+cosa5)+(cosa2+cosa4)+cosa3
=[cos(a3-
×2)+cos(a3+π 8
×2)]+[cos(a3-π 8
)+cos(a3+π 8
)]+cosa3π 8
=2cos
cos(a3-
)+(a3+π 4
)π 4 2
+2cos(a3-
)-(a3+π 4
)π 4 2
cos(a3-
)+(a3+π 8
)π 8 2
+cosa3(a3-
)-(a3+π 8
)π 8 2
=2cosa3•
+2cosa3•cos(-2 2
)+cosa3π 8
=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-a1a3=π2-(
-2•π 2
)•π 8
=π2-π 2
=π2 8
,7π2 8
故答案为:7π2 8