已知函数f(x)=sin(π-ωx)cosωx+cosωx(ω>0)的最小正周期为π 帮忙解一下

(1) f(x)=sin(π-ωx)cosωx+cosωx =sinωxcosωx+cosωx =(1/2)2sinωxcosωx+cosωx-(1/2)+(1/2) =(1/2)sin2ωx+(1/2)(2cosωx-1)+(1/2) =(1/2)sin2ωx+(1/2)cos2ωx+(1/2) =[(√2)/2]sin(2ωx+π/4)+1/2 2ω=2π/T=2，ω=1. (2)g(x)=[(√2