Question

已知函数f（x）=sin(π-ωx)cosωx+cos²(ω>0）的最小正周期为π 1.求ω的值

Answer 1

已知函数f（x）=sin(π-ωx)cosωx+cos²ωx
(ω>0）的最小正周期为π
,
(1).求ω的值
(2).将函数y=f（x）的图像上各点的横坐标缩短到原来的1/2，纵坐标不变,得到函数y=g(x)在区间[0，π/6]上的最小值
解:(1)f(x)=sinωxcosωx+cos²ωx=(1/2)sin2ωx+(1/2)cos2ωx+1/2
=[(√2)/2]sin(2ωx+π/4)+1/2
T=2π/2ω=π,故ω=1.
(2)
f(x)=[(√2)/2]sin(2x+π/4)+1/2=[(√2)/2]sin[2(x+π/8)+1/2
故g(x)=[(√2)/2]sin[4(x+π/8)]+1/2=[(√2)/2]sin(4x+π/2)+1/2
在[0,
π/6]上,mixg(x)=g(π/6)=[(√2)/2]sin(2π/3+π/2)+1/2=[(√2)/2]cos(2π/3)+1/2
=-[(√2)/2]cos(π/3)+1/2=-[(√2)/2](1/2)+1/2=-(√2)/4+1/2