Central angle between independent uniform sphere points ($f_{\Theta }(\theta )=\frac{1}{2}\sin \theta ,0<\theta <\pi$)

For two independent uniform points on the surface of a sphere, rotational invariance permits fixing the first at the north pole. The surface area of a latitude band of angular width $d\theta$ is $2\pi R^2\sin \theta d\theta$, yielding the displayed density for their central angle. Equivalently their unit position vectors have a dot product uniformly distributed on $[-1,1]$. The law is independent of sphere radius.

Join three independent uniform points on a sphere when their central angle is acute. Given angle $\gamma$ between the first pair, the third point's two favorable hemispheres overlap in a spherical lune of area fraction $(\pi -\gamma )/(2\pi )$. If $\gamma <\pi /2$, connection requires membership in their union, of probability $1/2+\gamma /(2\pi )$. If $\gamma >\pi /2$, it requires membership in their intersection, of probability $1/2-\gamma /(2\pi )$. Integrating these against the central-angle density $\sin \gamma /2$ proves the displayed probability. Connectivity is weaker than all three pairwise angles being acute.

For three independent uniform points on a sphere, condition on the central angle $\theta$ between the first two. The third point makes acute angles with both exactly when it lies in the intersection of their positive hemispheres. That intersection is a spherical lune of opening $\pi -\theta$, hence has probability $(\pi -\theta )/(2\pi )$. The first two points also need $\theta <\pi /2$. Integrating against the central angle between independent uniform sphere points density gives $(4\pi {)}^{-1}{\int }_0^{\pi /2}(\pi -\theta )\sin \theta d\theta =(\pi -1)/(4\pi )$.