Derivation of Spherical Law of Cosines I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown. 
The Wikipedia page says that for unit vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ : 
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
The unit vector $\mathbf{t}_a$ is defined as the unit vector perpendicular to $\mathbf{u}$ in the u-v plane, whose direction is given by the component of v perpendicular to u. Wikipedia explains that
$$
\begin{align}
\mathbf{t}_a = 
{
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\over
{
\left\lVert 
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\right\rVert
}}
=
{{{\mathbf{v}-\mathbf{u}\cos(a)}}
\over
{\sin(a)}}
\end{align}
$$
Similarly,
$$
\begin{align}
{\mathbf{t}_b} 
=
{{{\mathbf{w}-\mathbf{u}\cos(b)}}
\over
{\sin(b)}}
\end{align}
$$
Without any further justification, their proof ends by the claim:
$$
\begin{align}
{\mathbf{t}_a} \cdot {\mathbf{t}_b}
=
{{\cos(c)-\cos(a)\cos(b)}
\over
{\sin(a)\sin(b)}}
\end{align}
$$
It is the final step, which yeilds the formula for $\mathbf{t}_a\cdot\mathbf{t}_b$, that I do not comprehend. Can somebody please explain to me the justification for this expression?
 A: $$\mathbf{t}_a \cdot \mathbf{t}_b = \frac{\mathbf{v} - \mathbf{u} \cos(a)}{\sin(a)} \cdot \frac{\mathbf{w} - \mathbf{u} \cos(b)}{\sin(b)} \\ = \frac{\mathbf{v} \cdot \mathbf{w} - \mathbf{v} \cdot \mathbf{u} \cos(b) - \mathbf{u} \cdot \mathbf{w} \cos(a) + \mathbf{u} \cdot \mathbf{u}\cos(a)\cos(b)}{\sin(a)\sin(b)} \\ = \frac{\cos(c) - \cos(a)\cos(b) - \cos(a)\cos(b) + \cos(a)\cos(b)}{\sin(a)\sin(b)} = \frac{\cos(c) - \cos(a)\cos(b)}{\sin(a)\sin(b)}.$$
Note that $\mathbf{u} \cdot \mathbf{u} = 1$ since it's a unit vector.
A: It looks like you just follow the angle definitions given and distribute.
You have:
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
$$
\begin{align}
\mathbf{t}_a = 
{{{\mathbf{v}-\mathbf{u}\cos(a)}}
\over
{\sin(a)}}
\end{align}
$$
$$
\begin{align}
{\mathbf{t}_b} 
=
{{{\mathbf{w}-\mathbf{u}\cos(b)}}
\over
{\sin(b)}}
\end{align}
$$
So:
$$
\begin{align}
{\mathbf{t}_a} \cdot {\mathbf{t}_b}
=
\frac{1}
{\sin(a)\sin(b)}(\mathbf{v}\cdot\mathbf{w}-\mathbf{u}\cdot\mathbf{w}\cos(a)-\mathbf{v}\cdot\mathbf{u}\cos(b)+\mathbf{u}\cdot\mathbf{u}\cos(a)\cos(b))
\end{align}
$$
And simplify.
