= 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 therefore
$$
S(h)^2S(2h)\simeq+1,\qquad\hbox{hence }S(2h)\simeq+1.
$$
A 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.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4-phase-signs.png]
{title=How the sign of a second harmonic changes reinforcement of an atomic maximum}
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