函数f(x)=sin(π-ωx)cosωx+cos^2ωx最小正周期为π,求ω

问题描述:

函数f(x)=sin(π-ωx)cosωx+cos^2ωx最小正周期为π,求ω

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+1/2(1+cos2ωx)
=1/2sin(2ωx)+1/2cos2ωx+1/2
=1/2+=√2/2sin(2ωx+π/4)
最小正周期为π,所以最小正周期为2π/2ω=π
因此 ω=1