spherical trigonometry

In the following we deduce the cosine law for a spherical trihedron.

Let $\mathbf{e_{1},e_{2},e_{3}}$ be the vertex unitary vectors as shown in the figure.

The cosine of the angle $\alpha$ formed by the plane defined by $\mathbf{e_{1},e_{2}}$ and the plane defined by $\mathbf{e_{1},e_{3}}$ is:

\cos\alpha=\frac{(\mathbf{e_{1}}\times\mathbf{e_{3}})\cdot(\mathbf{e_{1}}% \times\mathbf{e_{2}})}{\left\|\mathbf{e_{1}}\times\mathbf{e_{3}}\right\|\left% \|\mathbf{e_{1}}\times\mathbf{e_{2}}\right\|}=\frac{(\mathbf{e_{1}}\times% \mathbf{e_{3}})\cdot(\mathbf{e_{1}}\times\mathbf{e_{2}})}{\sin b\sin c}

Now, using the cyclic property of the triple vector product and Lagrange’s formula (http://planetmath.org/TripleCrossProduct), we can write:

\cos\alpha=\frac{\mathbf{e_{1}}\cdot(\mathbf{e_{3}}\times(\mathbf{e_{1}}\times% \mathbf{e_{2}}))}{\sin b\sin c}=\frac{\mathbf{e_{1}}\cdot((\mathbf{e_{3}}\cdot% \mathbf{e_{2}})\mathbf{e_{1}}-(\mathbf{e_{3}}\cdot\mathbf{e_{1}})\mathbf{e_{2}% })}{\sin b\sin c}=\frac{\cos a-\cos b\cos c}{\sin b\sin c}

Hence:

\cos a=\cos b\cos c+\sin b\sin c\cos\alpha