f(x)=2√2sinwxcos(wx+π/4)
=2√2sinwx(coswxcosπ/4-sinwxsinπ/4)
=2√2sinwx*√2/2(coswx-sinwx)
=2sinwxcoswx-2sin^2 wx
=sin2wx +cos2wx-1
=√2sin(2w+π/4) -1
∴T最小=2π/2w=π
w=1
f(x)=2√2sinωxcos(ωx+π/4)
=2√2sinωx(√2/2cosωx-√2/2sinωx)
=2sinωx(cosωx+sinωx)
=2sinωxcosωx+2sinωx
=sin2ωx-cos2ωx+1
=√2sin(2ωx-π/4)+1
因为T=2π/ω，所以2ω=2π/T=2，所以ω=1
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