Transformations from n-sphere coordinates to cartesian coordinates. I was wondering how one would proceed to convert between coordinate systems in $ \mathbb R^n $.
For $ \mathbb R^2 $ the conversion is easy and just basic trigonometry.
Given $(r, \theta)$ we can convert these polar coordinates to cartesian coordinates to $(x=r\cos\theta, y=r\sin\theta)$
The same follows for $ \mathbb R^3 $.
Given $(r,\theta_1,\theta_2)$ in spherical coordinates we can convert the spherical coordinates to cartesian coordinates with the following transformation $(x=r\cos\theta_2\cos\theta_1,y=r\cos\theta_2\sin\theta_1, z=r\sin\theta_2)$, just so my point of reference is clear:$\theta_1$is the angle lying in the x-y plane, and  $\theta_2$ is the angle the vector makes with the x-y plane.
So what about for four dimensions and up? For the sake of simplicity we can define the n-sphere vector as $(r,\theta_1,\theta_2,...,\theta_n)$ and the cartesian vectors $(x_1,x_2,...,x_n)$ and maintain the same reference of $\theta_1,\theta_2$ as before.
How do we define a transformation in $ \mathbb R^4 $?
A look into this wikipedia page under the spherical coordinates section gave me something that I don't believe is quite right.
As according to the section
$(x_1=r\cos\theta_1,x_2=r\sin\theta_1 \cos\theta_2,x_3=\sin\theta_1 \sin\theta_2\cos\theta_3,...)$ which is already in contradiction to my statement about $ \mathbb R^3$ transformations. 
What is the general transformation rule? Am I missing something that I should know from linear algebra that I might be forgetting because my linear algebra is rusty? If so, please let me know what I should review or point me in the right direction.
 A: You're misreading the formulas in the Wiki article. As they are written, they only make sense for $n \ge 4$, since the last formula (for $x_n$) breaks the pattern. For $n=3$, the intended meaning is clearly what you have (up to renaming of the variables).
For $n=4$, there are several ways to set up coordinates. For example, as on Wikipedia:
$$
\begin{align*}
x_1 &= r \cos\theta_1 \\
x_2 &= r \sin\theta_1 \cos\theta_2 \\
x_3 &= r \sin\theta_1 \sin\theta_2 \cos\theta_3 \\
x_4 &= r \sin\theta_1 \sin\theta_2 \sin\theta_3
\end{align*}
$$
But the following also works:
$$
\begin{align*}
x_1 &= r \cos\theta_1 \cos\alpha \\
x_2 &= r \sin\theta_1 \cos\alpha \\
x_3 &= r \cos\theta_2 \sin\alpha \\
x_4 &= r \sin\theta_2 \sin\alpha
\end{align*}
$$
And as $n$ grows, so does the number of the possibilities. You could classify them by drawing some kind of tree-like diagrams. (This is done in papers by Kalnins and Miller from the 1980s, where they classify coordinate systems in which the Hamilton–Jacobi equation from classical mechanics may be solved by separation of variables, but their general scheme is more complicated, since it involves not only spherical coordinates but also ellipsoidal and paraboloidal coordinates.)
