Question

急！设函数f(x)=2x-cosx,{An}是公差为TT/8的等差数列，f(a1)+f(a2)+…f(a5)=5TT，则 f[(a3)]^2-a1a3=

Answer 1

f(a1)+f(a2)+f(a3)+f(a4)+f(a5)=2(a1+a2+a3+a4+a5)-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-[cos(a3-2π/8)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+2π/8)]
=5π
10a3-5π=[cos(a3-2π/8)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+2π/8)]
=[cos(a3-2π/8)+cos(a3+2π/8)]+cosa3+[cos(a3-π/8)+cos(a3+π/8)]
=2cosa3cos(π/4)+cosa3+2cosa3cos(π/8)
=[1+2cos(π/4)+2cos(π/8)]cosa3
=[1+√2+√(2+√2)]cosa3
设g(x)=-[1+√2+√(2+√2)]cosx+10x-5π
g'(x)=[1+√2+√(2+√2)]sinx+10＞0
g(x)没有拐点，单调递增，最多有1个解。
g‘’(x)=-[1+√2+√(2+√2)]cosx
g'(x)在x=kπ+π/2处有拐点，

f[(a3)]^2-a1a3=(2a3-cosa3)^2-a1a3
=[2(a1+π/4)-cos(a1+π/4)]^2-a1(a1+π/4)

=4(a1+π/4)^2+[cos(a1+π/4)]^2-4(a1+π/4)cos(a1+π/4)-a1(a1+π/4)

Answer 2

设a2=a1+π/8,a3=a1+π/4,a4=a1+3π/8,a5=a1+π/2
则f(a1)+f(a2)+f(a3)+f(a4)+f(a5)=2(a1+a2+a3+a4+a5)-(cosa1+cosa2+cosa3+cosa4+cosa5)
=2(5a1+5π/4)-(cosa1+cosa2+cosa3+cosa4+cosa5)
所以cosa1+cosa2+cosa3+cosa4+cosa5=10a1-5π/2
当a1=π/4时，左边=0，右边=0
故a1=π/4,a2=3π/8,a3=π/2,a4=5π/8,a5=3π/4
[f(a3)]^2-a1a3=π^2-(π/4)(π/2)=7/8π^2

追问

a1=TT/4是怎么算出来的，如果左右边都不等于0是等于其他数字？

追答

这个怎么说呢，就有点碰运气的感觉了
首先左边必须保证是简单的数字（0，1等等）
因为如果不是，左边将成为十分复杂的数，无法计算
而右边的a1需要是关于π的数
综合以上考虑，a1=π/4是唯一的