 Solution

For a strong $h$ term, maxima lie where $S(h)\cos (2\pi hx)=+1$. At either possible set of such maxima, $\cos (4\pi hx)=+1$: doubling a phase of either $0$ or $\pi$ gives zero modulo $2\pi$. A strong $2h$ term reinforces those positive maxima when $S(2h)=+1$. The direct-methods pair relationship is thereforeA negative strong second harmonic would oppose the maxima favored by the first and tend to create implausible negative or flattened density elsewhere. The relationship is a strong-reflection inference, not a sign identity for arbitrary weak coefficients.