f(x)=cos²2x-sin²2x+sin4x=cos4x+sin4x
=√2[(√2/2)cos4x+(√2/2)sin4x]=√2sin(4x+π/4)
所以,最小正周期T=2π/4=π/2
当sin(4x+π/4)=-1时,最小值为-√2