I've found many formulas for the shortest distance between two points on a sphere, but all of them using coordinate geometry. I wanted a formula in terms of radius and angle subtended at the centre, since I feel those terms are much more intuitive. Heres my attempt to do so:
Consider points A and B on a sphere of radius $R$, subtending angle $\theta$ at the centre. A plane passing through these two points cuts the sphere in a circle of radius $r$. A and B subtend angle $\beta$ at the centre of this circle. Let $l$ be the length of the arc AB of the circle
Now, $\ d(A,B) = 2R \sin \frac \theta2 = 2r \sin \frac \beta2$
$\therefore l = r \beta = R\beta \frac {\sin \frac \theta2}{\sin \frac \beta2}$
For minima, $\frac {dl}{d\beta} = 0$
$\therefore \frac {d}{d\beta} (\frac {\beta}{\sin \frac \beta2})= 0$
$\therefore \sin \frac \beta2 - (\frac \beta2) \cos \frac \beta2 = 0$
$\therefore \tan \frac \beta2 = \frac \beta2$
$\therefore \sin \frac \beta2 = \frac {\beta}{\sqrt {\beta ^2 + 4}}$
$\therefore l= \sqrt {\beta ^2 + 4} * R \sin \frac \theta2$
Now, to get the value of $\beta$, I know that for $\theta = \pi$, shortest distance is going to be $\pi R$
Substituting values of $l$ and $\theta$ in the final equation, I get $\beta \approx 2.42$
However, this value of $\beta$ is not consistent with my previous equation, $\tan \frac \beta2 = \frac \beta2$
Where did I go wrong?
I figured out what was wrong, and thought I'd just post the answer instead of deleting the question. Who knows, maybe it will help somebody else.
For the cut circle to actually be on the sphere, the limits ($\theta \leq\beta \leq \pi$) need to be in place. However there were no such limits while calculating minima, and thus it gave the answer as $\beta=0$
In fact, after graphing $l$ against $\beta$ on a graphing calculator, I saw that the only minima is at $\beta=0$, and $l$ increases with increasing $\beta$ (for within the limits we are concerned with). Hence for $\theta \leq\beta \leq \pi$, the minima is at $\beta = \theta$ !
Therefore minimum $l=R\theta$