= Solution
Rewrite the relation from part (c) as
$$
(\gamma-1)\sin\beta\cos(\beta-\theta)
=(\gamma+1)\cos\beta\sin(\beta-\theta).
$$
The angle-addition identities reduce this to
$$
\sin(2\beta-\theta)=\gamma\sin\theta.
$$
For $0<\theta<\arcsin(1/\gamma)$, the right side lies strictly between zero and one. There are two physically admissible <strong oblique-shock angle branches>:
$$
\boxed{\begin{aligned}
\beta_{\rm weak}&=\frac12\left[\theta+\arcsin(\gamma\sin\theta)\right],\\
\beta_{\rm strong}&=\frac12\left[\theta+\pi-\arcsin(\gamma\sin\theta)\right].
\end{aligned}}
$$
Both lie between $\theta$ and $\pi/2$. They merge when $\gamma\sin\theta=1$, proving the <maximum deflection through a strong perfect-gas shock>,
$$
\boxed{\theta_{\max}=\arcsin(1/\gamma),\qquad
\beta_* =\frac\pi4+\frac{\theta_{\max}}2.}
$$
For small positive $\theta$, the formal strong-shock-limit asymptotics are
$$
\boxed{\beta_{\rm weak}\sim\frac{\gamma+1}{2}\theta,\qquad
\beta_{\rm strong}\sim\frac\pi2-\frac{\gamma-1}{2}\theta.}
$$
The weak branch is normally observed for an unconfined supersonic stream turning through a modest compression corner. It has a smaller normal component of upstream <velocity>, hence a smaller downstream <pressure> increase; for small turns the downstream flow remains supersonic. The nearly normal strong branch can be selected by suitable downstream <pressure> conditions, so it is not mathematically forbidden.
The two limits must be ordered honestly. At finite $M$, the weak <oblique shock> approaches the <Mach angle> $\arcsin(1/M)$ as $\theta\to0$, rather than zero. The displayed weak-branch asymptotic applies when the large-$M$ limit is taken first, or in the overlapping range $1/M\ll\theta\ll1$, where $M\sin\beta\gg1$ remains valid.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2021-q4-shock-angle-branches.png]
{title=Calculated weak and strong shock-angle branches in the infinite-Mach-number limit, for a monatomic gas}
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