Cosine Rule on a Sphere

Let \(\textbf{u}\), \(\textbf{v}\) and \(\textbf{w}\) be unit vectors from the middle of the sphere to the points \(u\), \(v\) and \(w\) of our trianlge. The lengths of the sides are then give by the dot product:

\[\cos(a) = \textbf{u}\cdot\textbf{v},\]

\[\cos(b) = \textbf{u}\cdot\textbf{w},\]

\[\cos(c) = \textbf{v}\cdot\textbf{w}.\]

The angle C is the inverse cosine of two tangents, say \(\textbf{t}_a\) and \(\textbf{t}_b\), which are along the sides \(a\) and \(b\), i.e.

\[C=\cos^{-1}(\textbf{t}_a\cdot\textbf{t}_b),\]

\[\cos(C)=\textbf{t}_a\cdot\textbf{t}_b.\]

But we know more information about these tangent vectors. For instance we know that the they are perpendicular to \(\textbf{u}\) which is given by the component of \(\textbf{v}\) perpendiculat to \(\textbf{u}\).

This gives (once normalized):

\[\textbf{t}_a=\frac{\textbf{v}-\textbf{u}(\textbf{u}\cdot\textbf{v})}{|\textbf{v}-\textbf{u}(\textbf{u}\cdot\textbf{v})|} = \frac{\textbf{v}-\textbf{u}\cos(a)}{\sin(a)}.\]

Similarly,

\[\textbf{t}_b=\frac{\textbf{w}-\textbf{u}(\textbf{u}\cdot\textbf{w})}{|\textbf{w}-\textbf{u}(\textbf{u}\cdot\textbf{w})|} = \frac{\textbf{w}-\textbf{u}\cos(b)}{\sin(b)}.\]

So substituting these values into the above equation we get:

\[\cos(C)=\frac{\textbf{v}-\textbf{u}\cos(a)}{\sin(a)}\cdot \frac{\textbf{w}-\textbf{u}\cos(b)}{\sin(b)},\]

\[\cos(C)=\frac{\textbf{v}\cdot\textbf{w}-(\textbf{v}\cdot\textbf{u})\cos(b)-(\textbf{u}\cdot\textbf{w})\cos(a)+(\textbf{u}\cdot\textbf{u})\cos(a)\cos(b)}{\sin(a)\sin(b)},\]

\[\cos(C)=\frac{\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}.\]

Which when rearranged gives the required cosine rule:

\[\cos(c)=\cos(a)\cos(b)+\sin(a)\sin(b)\cos(C).\]