How to derive spherical trigonometry of sphere of radius $r$? Does anyone know how to derive spherical trigonometry of sphere of radius $r$?
I know that it is as follow if $r=1$:
$$\cos a = \cos b \cos c + \sin b \sin c \cos A$$
where $a,b,c$ are sides of triangle and $A$ is the angle between $b$ and $c$.
 A: Here is a way to do it with vector algebra. Define the vectors $\vec{P}$,$\vec{Q}$ and $\vec{R}$ as the unit length vectors from the center of the sphere to a corner of the triangle such that $\vec{P}$,$\vec{Q}$ subtend the arc with length $c$, $\vec{R}$,$\vec{Q}$ subtend the arc with length $a$ and $\vec{P}$,$\vec{R}$ subtend the arc with length $b$. Then, by this definition the following scalar products can be computeed
$$\begin{eqnarray}\vec{P}\cdot\vec{Q} & = & \cos c \\ 
\vec{R}\cdot\vec{Q} & = & \cos a \\ \vec{P}\cdot\vec{R} & = & \cos b \end{eqnarray}$$  
Then, to find out a formula for the angle $A$, notice that we can construct a vector tangent to the arc with length $b$ in the point $\vec{P}$ by considering a vector orthogonal to $\vec{P}$ and the normal of the plane containing $\vec{P}$ and $\vec{R}$. This can be done quickly with the cross product as follows:
$$\vec{P}\times(\vec{P}\times\vec{R})=\vec{P}(\vec{P}\cdot\vec{R}) - \vec{R}(\vec{P}\cdot\vec{P})= \vec{P}\cos b - \vec{R} $$
To arrive at the result in the middle, I used an identity for the cross product. Applying it to obtain a vector tangent to the arc with length $c$ in the point $\vec{P}$ we obtain likewise
$$\vec{P}\times(\vec{P}\times\vec{Q})=\vec{P}(\vec{P}\cdot\vec{Q}) - \vec{Q}(\vec{P}\cdot\vec{P})= \vec{P}\cos c - \vec{Q} $$
By construction, the scalar product of thos two vectors is equal to $\cos A$ times the lengths of these vectors
$$(\vec{P}\cos b - \vec{R})\cdot(\vec{P}\cos c - \vec{Q}) = \cos A \; \|\vec{P}\cos b - \vec{R}\| \; \|\vec{P}\cos c - \vec{Q}\|$$
Working out the left hand side gives $-\cos b \cos c + \cos a$, while the lengths og the vectors are easily worked out to be $\sin b$ and $\sin c$ respectively, leading to
$$-\cos b \cos c + \cos a = \cos A \sin b \sin c$$
or after rearranging
$$\cos a = \cos b \cos c + \sin b \sin c \cos A $$
As a final comment, I have to specify that I have shoved under the rug a few points concerning for instance the orientation of the vectors I created with the cross product. If the orientation had been different for one of the vectors, we would actually be calculating $\cos (\pi-A)=-\cos A$. Likewise, the $\sin b$ might be negative whereas the lengths of the vectors are always positive. The goal is to check that for all the types of triangles you can have, these sign issues cancel out nicely and you still get the formula above. As it stands, the proof is not complete, but it gives you the essential part.
