How are arc components of a spherical system derived? I am studying a flight dynamics book (see Flight Dynamics by Stengel) and am rusty on spherical coordinates. Commonly, aerospace coordinates use a North/East/Down right-hand system. So $z=-h$, increasing as altitude decreases. I'm used to a conventional mathematical system and am rusty with spherical coordinates generally. The velocity vectors are defined as:
$v_1=V\cos\gamma\cos\xi\\v_2=V\cos\gamma\sin\xi\\v_3=-V\sin\gamma$
And $V=\sqrt{v_1^2+v_2^2+v_3^2}$
I believe $\xi$ and $\gamma$ are the equivalent of $\theta$ and $\phi$ of a standard system, except that $\gamma$ increases as the vector rotates up rather than down in the direction of the $z$ axis.
In the book, $\xi=\cos^{-1}[\frac{v_1}{\sqrt{v_1^2+v_2^2}}]=\sin^{-1}[\frac{v_2}{\sqrt{v_1^2+v_2^2}}]$. Also, $\gamma=\sin^{-1}(-v_3/V)$.
I've verified the inverse relationship back checking for $\xi$ by setting $v_3=0$, but have gotten stuck working forwards re-deriving them. 
Q: What trig steps take me from the vector system to the inverse $\xi$ relationships?
note: This isn't critical, but will help refresh personal comfort with this spherical reference frame. Thanks, this is my first Stack Exchange question. References to the process as well as steps are equally welcome.
 A: In the diagram, $r = V\cos(-\gamma) = V\cos\gamma$ is the distance from the $v_{3}$-axis. The formulas
\begin{align*}
v_{1} &= r\cos\xi = V\cos\gamma \cos\xi, \\
v_{2} &= r\sin\xi = V\cos\gamma \sin\xi, \\
v_{3} &= V\sin(-\gamma) = -V\sin \gamma &&
\end{align*}
can be read off various right triangles. Specifically, the first two come from the right triangle with legs $v_{1}$ and $v_{2}$, and hypotenuse $r$, in the $(v_{1}, v_{2})$-plane; the third comes from the right triangle with legs $v_{3}$ and $r$, and hypotenuse $V$, in the longitudinal plane containing the $v_{3}$-axis and the point $(v_{1}, v_{2}, v_{3})$.

A: From the eqs. 
$v_1=V\cos\gamma\cos\xi\\v_2=V\cos\gamma\sin\xi\\v_3=-V\sin\gamma$
It follows that $\gamma=-\sin^{-1}\frac{v_3}{V}$ substitute in the eqs above you get $$\cos \xi=\frac{v_1}{V\cos\gamma}=\frac{v_1}{V\cos(-\sin^{-1}\frac{v_3}{V})}=\frac{v_1}{V\cos(\sin^{-1}\frac{v_3}{V})}...............(*)$$
Let $y=\sin^{-1}\frac{v_3}{V}\Rightarrow \sin y =\frac{v_3}{V}$. Using the right triangle to evaluate $\cos y$ you get $\cos y =\cos(\sin^{-1}\frac{v_3}{V})=\frac{\sqrt{V^2-v^2_3}}{V}$. Subsitituting in (*) we have
$$\cos \xi =\frac{v_1}{V\cos(\sin^{-1}\frac{v_3}{V})}=\frac{v_1}{V\frac{\sqrt{V^2-v^2_3}}{V}}=\frac{v_1}{\sqrt{v_1^2+v_2^2+v_3^2-v_3^2}}=\frac{v_1}{\sqrt{v_1^2+v_2^2}}.$$
Similarly we can prove $\sin \xi =\frac{v_2}{\sqrt{v_1^2+v_2^2}}$.
