函数f(x)=asin(wx+π/4)+b(a,w>0)的最小正周期为π,最大值为2根号2,最小值为0

题目

函数f(x)=asin(wx+π/4)+b(a,w>0)的最小正周期为π,最大值为2根号2,最小值为0
1,写出f（x）的解析式,画出f（x）在一个周期的图像
2,设F（x）=f（x）-f（x-π/4）,若F（x）=6/5,求sinx的值

答案

1.
依题意
最小正周期为π
所以w=2
最大值为2根号2=a+b=2根号2
最小值为0=b-a=0
所以a=b=根号2
所以f(x)=根号2sin(2x+π/4)+根号2
图像自己画画
2.
F(x)=f(x)-f(x-π/4)=根号2sin(2x+π/4)+根号2-[根号2sin(2x-π/4)+根号2]
=根号2sin(2x+π/4)-根号2sin(2x-π/4)
=根号2sin(2x+π/4)+根号2cos(2x+π/4)
=2sin(2x+π/4+π/4)
F(x)=6/5
2sin(2x+π/4+π/4)=6/5
sin(2x+π/2)=3/5
cos2x=3/5
1-2(sinx)^2=3/5
(sinx)^2=1/5
所以sinx=±根号5/5