Question

已知函数f（x）=cos^2（wx）+根3sinwxcoswx（w>0）的最小正周期为π.

（1）求f（2π/3）的值
（2）求函数f（x）的单调区间及其图像的对称轴方程.

Answer 1

解：f(x)=(1+cos2wx)/2+(√3/2)sin2wx
=sin2wx*cos(π/6)+cos2wx*sin(π/6)+1/2
=sin(2wx+π/6)+1/2
T =π=2π/2w
所以 w=1
f(x)=sin(2x+π/6)+1/2
(1)f(2π/3)=sin(4π/3+π/6)+1/2=sin(7π/6)+1/2=-1/2+1/2=0
(2) 增：
2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
增区间为 【kπ-π/3，kπ+π/6】，k∈Z
减：
2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+π/3≤2x≤2kπ+4π/3
kπ+π/6≤x≤kπ+2π/3
减区间为 【kπ+π/6，kπ+2π/3】，k∈Z

2x+π/6=kπ+π/2
2x=kπ+π/3
对称轴方程 x=kπ/2+π/6,k∈Z

追问

f（2π/3）为什么我算的等于-1/2 是我错了么？？

追答

我算错了，抱歉
f(2π/3)=sin(4π/3+π/6)+1/2=sin(3π/2)+1/2=-1+1/2=-1/2

Answer 2

(1)、f（x）=cos^2（wx）+根3sinwxcoswx
=1/2[2cos^2(wx)-1]+√3/2 (2sinwxcoswx)+1/2
=1/2cos(2wx)+√3/2sin(2wx)+1/2
=sin(2wx+π/6)+1/2
最小正周期为π,则有T=2π/2w=π 即：w=1
所以可得：f(x)=sin(2x+π/6)+1/2
即：f(2π/3)=sin(4π/3+π/6)+1/2=sin(3π/2)+1/2=-1/2

(2)、正弦函数的单调递增区间为：[2kπ-π/2,2kπ+π/2]
所以有：2kπ-π/2 ≤2x+π/6≤2kπ+π/2
解得：kπ-π/3 ≤x ≤kπ+π/6
所以此函数的递增区间为:[kπ-π/3 ,kπ+π/6]
正弦函数的单调递减区间为：[2kπ+π/2,2kπ+3π/2]
所以有：2kπ+π/2 ≤2x+π/6≤2kπ+3π/2
解得：kπ+π/6 ≤x ≤kπ+π/6
所以此函数的递减区间为:[kπ-π/6 ,kπ+2π/3]

正弦函数对称轴为：x=kπ+π/2
所以有：2x+π/6=kπ+π/2
即：x=kπ/2+π/6

Answer 3

f（x）=cos^2（wx）+根3sinwxcoswx=cos^2（wx）+根3/2sin2wx
=2cos(2wx-π/3)
T=π,所以w=1
（1）f（2π/3）=2cosπ=-2
(2)f(x)=2cos(2x-π/3)
单调递减区间： 2kπ<2x-π/3<2kπ+π
2kπ +π/6 <x<kπ+2/3π
单调递增区间：2kπ-π<2x-π/3<2kπ
kπ-π/3 <x<kπ+π/6
k为整数
对称轴方程|f(x)|=1 x=π/6+kπ/2