Map Two Polar Disks To Four Dimensional Spherical Coordinates ($\left(\mathbb{R}^2\right)^2\to\mathbb{R}^4$) We have two points $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$ with polar parameterizations $(r_1, \theta_1), (r_2, \theta_2)$ (i.e., $x_1=r_1 cos(\theta)$ and so on.) How would the point $(x_1, y_1, x_2, y_2) \in \mathbb{R}^4$ be represented with four-dimensional spherical coordinates $(r, \phi_1, \phi_2, \phi_3)$?
Please correct my proposed answer if I have made a mistake or upvote it if you concur.
Finally, how would one inductively generalize equivalency to $\mathbb{R}^{2^n}$ or $\mathbb{R}^{2n}$?

We begin by equating the components of the two numbers in polar coordinates and the single number in spherical coordinates. Here, $r_3 = \sqrt{r_1^2 + r_2^2}$.
$$r_1 \cos \left( \theta_1 \right) = r_3 \cos \left( \phi_1 \right)$$
$$r_1 \sin \left( \theta_1 \right) = r_3 \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)$$
$$r_2 \cos \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right)$$
$$r_2 \sin \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \sin \left( \phi_3 \right)$$
Next, the standard definition of $arccos$ does not include the full input plane, so we form a continuous extension to $\mathbb{R}^2 \setminus \mathbb{R}^+$. (We exclude the point intersecting $\mathbb{R}^+$ with $\theta = 0$ or $\theta = 2 \pi$). With an abuse of notation, we will still refer to this continuous extension as $arccos$. An interesting application of this definition of $arccos$ is that $\mathbb{R}^3 \setminus 0$ is not homologically trivial, which is an exercise in Munkre's Analysis on Manifolds.
$$\phi_1 = \arccos \left( r_1 r_3^{-1} \cos \left( \theta_1 \right) \right)$$
$$\phi_2 = \arccos \left( r_1 r_3^{-1}  \frac{ sin (\theta_1) }{sin (\phi_1)} \right)$$
$$\phi_3 = \arccos \left( r_2 r_3^{-1} \frac{ \cos \left( \theta \right) }{ \sin \left( \phi_1 \right) \sin \left( \phi_2 \right)   }  \right)$$
Fortunately, the other direction is simpler to compute.
$$\theta_{1} = \arccos(r_3 r_1^{-1} \cos \left( \phi_1 \right))=\arcsin \left( r_3 r_1^{-1} \sin \left( \phi_1 \right) \cos \left( \phi_2 \right) \right)$$
$$\theta_2 = \arccos \left( r_3 r_2^{-1} \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right) \right)=...$$
$$\frac{ r_1 \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } = \frac{ r_1 \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)   }$$
$r_1 = \left|\frac{ \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } - \frac{ \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)   }\right|^{-1}$ provided that $cos(\phi_1)\neq0$, $cos(\phi_2)\neq 0$, and $sin(\phi_1)\neq 0$.
$r_2 = \left|\frac{1}{sin(\phi_1) sin(\phi_2)} \left( \frac{cos (\theta_2) }{cos(\phi_3)} - \frac{sin(\theta_2)}{sin(\phi_3)} \right)\right| ^{-1}$ provided that $sin(\phi_1)\neq 0$, $sin (\phi_2) \neq 0$, $cos(\phi_3)\neq 0$, and $sin(\phi_3) \neq 0$.
I wonder if there would be an easier formula for $r_1$ and $r_2$ by taking the components of $r_3$ along two axes. (In this case, $r_1$, $r_2$, and $r_3$ would all be tangent vectors on the same plane, though finding a parameterization or normal vector for the plane is not trivial; one cannot simply compute $\vec{r_1} \times \vec{r_2}$.)
By stipulating $r_3>0$ and $\theta_1, \theta_2, \theta_3 \in [0, 2 \pi)$, proof of injection (being a one-to-one map) should follow immediately. Likewise, every pair of $(x_1, y_1)=(r_1, \theta_1)$ and $(x_2, y_2) = (r_2, \theta_2)$ should produce a value for $(r, \phi_1, \phi_2, \phi_3)$, allowing surjection to follow as well. $f(r_1, \theta_1, r_2, \theta_2)= (r, \phi_1, \phi_2, \phi_3)$ should then satisfy the criteria of an isomorphic map. If feel that I have missed something. Please correct me in the comments or my answer directly if you have sufficient reputation.
