Question

设函数f(x)=(sinωx+cosωx)^2+2cosωx^2-2 (ω＞2)的最小正周期为2π/3 （1）求ω的值

（2）若把函数y=f(x)的图象向右平移π/2个单位长度，得到了函数y=g(x)的图象，求函数y=g(x),x属于[-π/3,π/12]的值域

Answer 1

1. f(x)=(sinωx+cosωx)^2+2cosωx^2-2 (ω＞2)
= 1+ sin(2ωx) + cos(2ωx) - 1
= √2 sin(2ωx + π /4) 2π / (2ω) = 2π/3 => ω = 3/2
2. g(x) =√ 2 sin(2ωx +π/4 - π/2) =√ 2 sin(3x -π/4 )
x∈ [-π/3,π/12], 3x-π/4∈ [ -5π/4, 0],
g(x) ∈[ 1, -√ 2 ]