作业帮设函数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=
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![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=](/uploads/image/z/3724729-25-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2x-cosx%2C%7BAn%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BATT%2F8%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cf%28a1%29%2Bf%28a2%29%2B%E2%80%A6f%28a5%29%3D5TT%2C%E5%88%99+f%5B%28a3%29%5D%5E2-a1a3%3D)
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=
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=
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)
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+...
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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
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