How do I find 2x2 orthonormal diagonalizing matrices using only trigonometry? I have a matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where all values are known), and I eventually want to diagonalize it into:
$$
A=UDV^T
$$
for orthonormal U and V. If I represent U and V as:
$$
U=\begin{bmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{bmatrix}
,
V=\begin{bmatrix} cos(\phi) & sin(\phi) \\ -sin(\phi) & cos(\phi) \end{bmatrix}
$$
Is there a geometric way to quickly find $\theta$ and $\phi$ using $a$, $b$, $c$, and $d$?
I realize that I could find the eigenvalues and work from there algebraically, but I'm wondering if there's a clean visual way of "seeing" $\theta$ and $\phi$, and then going from there.
I was thinking they would be related to the angles of triangles made up of the values of $A$, but I can't seem to figure out which angles they are.
Thanks!
EDIT: I realized I miswrote $U$ and $V$. $U$ and $V$ need to be rotation matrices for this to work. Any ideas?

Problem statement and restrictions
The problem you propose is this
$$
\begin{align}
\mathbf{A} &= \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \\
% A
\left[
\begin{array}{cc}
 a & b \\
 c & d \\
\end{array}
\right]
%
&=
%
% U
\left[
\begin{array}{cr}
 \cos \theta & -\sin \theta \\
 \sin \theta & \cos \theta \\
\end{array}
\right]
% S
\left[
\begin{array}{cc}
 \sigma_{1} & 0 \\
 0 & \sigma_{2}\\
\end{array}
\right]
% V*
\left[
\begin{array}{rr}
 \cos \phi & \sin \phi \\
 -\sin \phi & \cos \phi \\
\end{array}
\right] \\[3pt]
%
&=
% svd
\left[
\begin{array}{cc}
 \sigma_{1}\cos \theta \cos \phi+\sigma_{2}\sin \theta \sin \phi & \sigma_{1}\cos \theta \sin \phi-\sigma_{2}\cos \phi
   \sin \theta \\
 \sigma_{1}\cos \phi \sin \theta-\sigma_{2} \cos \theta \sin \phi & \sigma_{2}\cos \theta \cos \phi+\sigma_{1}\sin \theta
   \sin \phi \\
\end{array}
\right]
%
\end{align}
$$
The task is to connect the Roman alphabetics to the Greeks. Note this restricts us to class of problems where the domain matrices are rotation matrices. This excludes reflection matrices and matrices which are compositions with reflections.
To proceed, require 


*

*rank $\left( \mathbf{A} \right)$ = 2 $ \quad \Rightarrow \quad$ $\det \mathbf{A} \ne 0$.

*The variables $a$, $b$, $c$, and $d$ are real.

Linear regression example
Consider a closely related example which has been solved before. The problem is from linear regression
$$
 y(x) = a_{0} + a_{1} x
$$
Given a sequence of $m$ measurements of the form $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, one can assemble to systems matrix $\mathbf{A}\in\mathbb{C}^{m\times 2}_{2}$ and the data vector $y\in\mathbb{C}^{m}$ which is not in the null space: $y\notin\mathcal{N}\left( \mathbf{A}^{*} \right)$. 
The linear system is
$$
\begin{align}
  \mathbf{A} a & = y\\
% A
\left[
\begin{array}{cc}
 1 & x_{1} \\
 \vdots & \vdots \\
 1 & x_{m} \\
\end{array}
\right]
% a
\left[
\begin{array}{c}
 a_{0} \\
 a_{1}
\end{array}
\right]
%
& =
%
\left[
\begin{array}{c}
 y_{1} \\
 \vdots \\
 y_{m}
\end{array}
\right]
%
\end{align}
$$
The dimensions of this matrix $\mathbf{A}$ here do not match those of your example, there are too many rows. But this may not be a problem, because once we compute the matrices $\mathbf{S}$ and $\mathbf{V}$, we can construct the matrix $\mathbf{U}$ via
$$
 \mathbf{S}^{-1} \mathbf{A} \, \mathbf{V} = \mathbf{U}.
$$
Eigensystem
The first step in decomposition is to construct the product matrix $\mathbf{A}^{*}\, \mathbf{A}$ and resolve the eigensystem.
$$
 \mathbf{A}^{*}\, \mathbf{A} =
\left[
\begin{array}{cc}
 \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
 x \cdot \mathbf{1}  & x \cdot x
\end{array}
\right]
$$
The characteristic polynomial is 
$$ 
\begin{align}
p(\lambda) &= \lambda^{2} - \lambda \, \text{tr}
\left( \mathbf{A}^{*} \mathbf{A}\right) + \det 
\left( \mathbf{A}^{*} \mathbf{A}\right) \\
\end{align}
$$
The roots $p\left( \lambda_{\pm} \right)$ are the eigenvalues. The square root of the eigenvalues produces the singular values
$$
\sigma = \sqrt{\frac{1}{2} 
\left( \mathbf{1}\cdot\mathbf{1} + x \cdot x \pm \sqrt{4
\left(
\mathbf{1}\cdot x
\right)^{2} - 
\left( \mathbf{1}\cdot\mathbf{1} - x \cdot x
\right)^{2}}
\right)}
$$
Introduce rotation matrix
Onto computing the domain matrix, and this is where your insight will help. Use the relationship
$$
\begin{align}
 \mathbf{A}^{*} \mathbf{A} &= \mathbf{V} \, \mathbf{S}^{2} \, \mathbf{V}^{*} \\
%
&=
% V
\left[
\begin{array}{cr}
 \cos \phi & \sin -\phi \\
 \sin \phi & \cos \phi \\
\end{array}
\right]
%
\mathbf{S}^{2}
% V*
\left[
\begin{array}{rc}
 \cos \phi & \sin \phi \\
 -\sin \phi & \cos \phi \\
\end{array}
\right] \\[3pt]
%
\left[
\begin{array}{cc}
 \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
 x \cdot \mathbf{1}  & x \cdot x
\end{array}
\right]
%
&=
%
\left[
\begin{array}{cc}
 \sigma_{1}^{2} \cos^{2} \phi + \sigma_{2}^{2} \sin^{2} \phi & 
 \left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi \\
 \left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi  & 
 \sigma_{2}^{2} \cos^{2} \phi + \sigma_{1}^{2} \sin^{2} \phi
\end{array}
\right]
%
\end{align}
$$
Solution
We find solutions like 
$$
 \cos \phi = \sqrt{\frac{ 
\left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{1}^{2} - 
\left( x \cdot x \right) \sigma_{2}^{2}}
{\sigma_{1}^{4} - \sigma_{2}^{4}}}, \qquad
%
 \sin \phi = \sqrt{\frac{ 
\left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{2}^{2} - 
\left( x \cdot x \right) \sigma_{1}^{2}}
{\sigma_{1}^{4} - \sigma_{2}^{4}}}
$$
This constructs the angles in the rotation matrix for the domain interms of the data.
