作业帮10、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）=5π10、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-a1a5=______．A 0 B 1/16*π2
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![10、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）=5π10、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-a1a5=______．A 0 B 1/16*π2](/uploads/image/z/5208341-5-1.jpg?t=10%E3%80%81%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2x-cosx%2C%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BA%CF%80%2F8%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cf%EF%BC%88a1%EF%BC%89%2Bf%EF%BC%88a2%EF%BC%89%2B%E2%80%A6%2Bf%EF%BC%88a5%EF%BC%89%3D5%CF%8010%E3%80%81%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2x-cosx%2C%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BA%CF%80%2F8%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cf%EF%BC%88a1%EF%BC%89%2Bf%EF%BC%88a2%EF%BC%89%2B%E2%80%A6%2Bf%EF%BC%88a5%EF%BC%89%3D5%CF%80%2C%E5%88%99%5Bf%28a3%29%5D2-a1a5%3D______%EF%BC%8EA+0+B+1%2F16%2A%CF%802)
10、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）=5π10、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-a1a5=______．A 0 B 1/16*π2
10、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
10、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-a1a5=______．
A 0 B 1/16*π2 C 1/8*π2 D 13/16*π2
选D
∵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
我的疑问是这解题中的“∴(cosa1+cosa2+……+cosa5)=0“这步怎么来的?
10、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）=5π10、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-a1a5=______．A 0 B 1/16*π2
f(x)=2x-cosx
所以f(a1）=2a1-cosa1
f(a2)=2a2-cosa2
.
整理得
因为f(a1)+f(a2)+……+f(a5)=5π
2(a1+a2+……+a5)-(cosa1+cosa2+……+cosa5)=5π
注意这里
因为a5=a1+4*π/8=a1+π/2
cosa1+cosa2+cosa3+cosa4+cosa5
=[cos(a3-π/4)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+π/4)
展开化简即可.