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=

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)