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-a1×a5=?请问一下为什么最后把这些式子化到10a3-cosa3(1+根号2+根号下根号2+2)后,因为an是以π÷8为等差的数列,所以cosa3
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![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-a1×a5=?请问一下为什么最后把这些式子化到10a3-cosa3(1+根号2+根号下根号2+2)后,因为an是以π÷8为等差的数列,所以cosa3](/uploads/image/z/3724727-23-7.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%BA%CF%80%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%3D5%CF%80%2C%E5%88%99%5Bf%28a3%29%5D%5E2-a1%C3%97a5%3D%3F%E8%AF%B7%E9%97%AE%E4%B8%80%E4%B8%8B%E4%B8%BA%E4%BB%80%E4%B9%88%E6%9C%80%E5%90%8E%E6%8A%8A%E8%BF%99%E4%BA%9B%E5%BC%8F%E5%AD%90%E5%8C%96%E5%88%B010a3-cosa3%EF%BC%881%2B%E6%A0%B9%E5%8F%B72%2B%E6%A0%B9%E5%8F%B7%E4%B8%8B%E6%A0%B9%E5%8F%B72%2B2%EF%BC%89%E5%90%8E%2C%E5%9B%A0%E4%B8%BAan%E6%98%AF%E4%BB%A5%CF%80%C3%B78%E4%B8%BA%E7%AD%89%E5%B7%AE%E7%9A%84%E6%95%B0%E5%88%97%2C%E6%89%80%E4%BB%A5cosa3)
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-a1×a5=?请问一下为什么最后把这些式子化到10a3-cosa3(1+根号2+根号下根号2+2)后,因为an是以π÷8为等差的数列,所以cosa3
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-a1×a5=?
请问一下为什么最后把这些式子化到10a3-cosa3(1+根号2+根号下根号2+2)后,因为an是以π÷8为等差的数列,所以cosa3(1+根号2+根号下根号2+2)是不含π的式子?
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-a1×a5=?请问一下为什么最后把这些式子化到10a3-cosa3(1+根号2+根号下根号2+2)后,因为an是以π÷8为等差的数列,所以cosa3
∵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