Spherical Law of Cosines

Theorem
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.

Then:


 * $\cos a = \cos b \cos c + \sin b \sin c \cos A$

Proof

 * Spherical-Cosine-Formula.png

Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.

By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.

By definition of a great circle, the center of each of these great circles is $O$.

Let $AD$ be the tangent to the great circle $AB$.

Let $AE$ be the tangent to the great circle $AC$.

Thus the radius $OA$ of $S$ is perpendicular to $AD$ and $AE$.

By construction, $AD$ lies in the same plane as $AB$.

Thus when $OB$ is produced, it will intersect $AD$ at $D$, say.

Similarly, $OC$ can be produced to intersect $AE$ at $E$, say.

The spherical angle $\sphericalangle BAC$ is defined as the angle between the tangents $AD$ and $AE$.

Thus:
 * $\sphericalangle BAC = \angle DAE$

or, denoting that spherical angle $\sphericalangle BAC$ as $A$:


 * $A = \angle DAE$

In the (plane) triangle $OAD$, we have that $\angle OAD$ is a right angle.

We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the length of a side of a spherical triangle.

Thus:

and by similar analysis of $\triangle OAE$, we have:

From consideration of $\triangle DAE$:

From consideration of $\triangle DOE$:

Thus:

and the result follows.

Also known as
Some sources refer to this result as just the cosine-formula.