Question

已知函数f(x)=2sinwx 乘coswx+2又根号3乘(coswx)^2-根号3,且函数f(x)的最小正周期为π

1.求w的值
2.将函数图像向右平移4分π个单位长度，再将图像各点横坐标缩小为原来的2分一倍得到函数g(x)图像求g(x)在区间负六分π，24分π上的单调区间

Answer 1

f(x)=2sinwx coswx+2√3cos^2wx-√3,
=2sinwx coswx+√3(2cos^2wx-1)
=sin2wx +√3cos2wx
=2(1/2sin2wx +√3/2cos2wx)
=2sin(2wx +π/3)
最小正周期为 ,π,2π/2w=π,w=1
f(x)=2sin(2x +π/3)
2,f(x)=2sin(4x +π/12)
3.g(x)在[-π/6，π/24]的单调区间
4x +π/12在[-2π/3,π/6]
f(x)在[2Kπ-2π/3,2Kπ-π/2]上是减函数,
在[2Kπ-π/2,2Kπ+π/4]上是增函数

Answer 2

f(x)=2sinwx *coswx+2√3*(coswx)^2-√3=sin2wx+√3*[cos2wx+1]-√3
=2{sin2wx*cos(π/3)+cos2wx*sin(π/3)}=2sin(2wx+π/3)
T=2π/2w=π w=1
f(x)=2sin(2x+π/3)
g(x)=2sin[2(x-π/4)+π/3]*(1/2)=sin(2x+π/12)
当-π/6<x<π/24
-π/4<2x+π/12<π/6 g(x)单调递增

追问

不是纵坐标缩小是横坐标

追答

将函数图像向右平移4分π个单位长度得h(x)=2sin[2(x-π/4)+π/3]=2sin(2x-π/6)
g(x)=2sin[2x*2-π/6]=2sin(4x-π/6)
当-π/6≤x≤π/24
-π<-5π/6≤4x-π/6≤0
由-π/2≤4x-π/6≤0 得：-π/12≤x≤π/24
g(x)的单调增区间为[-π/12,π/24]，单调减区间为[-π/6,-π/12]