If you try and find out how you compute the distance between points on a sphere, you’ll get a bunch of sites which offer to calculate it for you if you enter the coordinates. If you search harder, you can even get the formula. But no one seems to be offering a derivation. So here you go.

First, some preliminaries. The half angle formulas give us 2 sin² ½ψ = 1−cos ψ. (Easily derivable from cos 2α = cos² α − sin² α.) The straight-line distance (cutting through the inside of the sphere) between two points on a unit sphere with an angle ψ between them is 2 sin ½ψ. (Bisect the triangle formed by the two points and the center of the sphere.) Together, these mean that the square of the straight-line distance is 2 − 2 cos ψ.

Consider A and B to be our two points we want to get the distance between, with latitudes φ_A and φ_B, and longitudes which differ by Θ. We’ll operate mostly on the disk formed by the parallel through B. The point where it intersects the pole is D, and the projection of A onto it is C. Let EB be perpendicular to DC.

The radius of the parallel through A is r_A = cos φ_A = DC; the one through B is r_B = cos φ_B = DB. Angle CDB is Θ, so EB = r_B sin Θ and ED = r_B cos Θ.
EC = CD − ED = r_A − r_B cos Θ.
BC² = EC² − EB² = r_A² − 2 r_A r_B cos Θ + r_B² cos² Θ + r_B² sin² Θ = r_A² + r_B² − 2 r_A r_B cos Θ.

AC = |sin φ_A − sin φ_B|.
AB² = BC² + AC² = r_A² + r_B² − 2 r_A r_B cos Θ + sin² φ_B − 2 sin φ_A sin φ_B
   = cos² φ_A + sin² φ_A + cos² φ_B + sin² φ_A + sin² φ_B − 2 r_A r_B cos Θ − 2 sin φ_A sin φ_B
   = 2 − 2 r_A r_B cos Θ − 2 sin φ_A sin φ_B.

If ψ is the angle AOB (which is also the along-surface distance between A and B), then
AB² = 2 − 2 cos ψ, from the “preliminaries”.
So, 1 − r_A r_B cos Θ − sin φ_A sin φ_B = 1 − cos ψ.
ψ = arccos(cos φ_A cos φ_B cos Θ + sin φ_A sin φ_B).