I'll use 3-D coordinates with the $z$-axis passing through the poles, so we start at $\alpha:=(\cos \theta, 0, \sin \theta)$. The optimal path $\gamma$ must be a geodesic, so $\gamma$ is an arc of a great circle through $\alpha$. Let $\nu$ be a vector in $\mathbb{R}^3$ orthogonal to $\gamma$. Since $\alpha \in \gamma$, we have $\alpha \cdot \nu = 0$.
Let the end point of this geodesic have longitude $\psi$. Then $\gamma$ must be the shortest path to the line of longitude $(-\sin \psi, \cos \psi, 0)^{\perp}$. Set $\lambda := (-\sin \psi, \cos \psi, 0)$. So $\gamma=\nu^{\perp}$ and $\lambda^{\perp}$ must meet orthogonally, which means that $\lambda \cdot \nu = 0$.
So $\nu$ is a multiple of $$\alpha \times \lambda = (-\cos (\psi) \sin (\theta),-\sin (\psi) \sin (\theta),\cos (\psi) \cos (\theta )).$$
The point of intersection of the geodesic $\nu^{\perp}$ with the line of longitude $\lambda^{\perp}$ is a scalar multiple of $$\lambda \times \nu=-(\lambda \times (\lambda \times \alpha))=(\cos ^2(\psi) \cos (\theta),\sin (\psi) \cos (\psi) \cos (\theta),\sin (\theta)).$$ Now, taking the cross product with $\lambda$ twice is just (up to scalar) orthogonal projection onto $\lambda^{\perp}$, so we can write this more simply as $\alpha - \langle \lambda, \alpha \rangle \lambda$. And the angle between $\alpha$ and the orthogonal projection of $\alpha$ onto $\lambda^{\perp}$ is $\sin^{-1} (\alpha \cdot \lambda)$ (using that $\alpha$ and $\lambda$ are both unit vectors), which is $$\sin^{-1} {\big(} (\sin \phi) (\cos \theta) {\big)}.$$