Vector Parameterisation of Ellipse How would you show that, if
$$\mathbf r = \mathbf A \sin \theta + \mathbf B \cos \theta$$
where $\mathbf A$ and $\mathbf B$ are arbitrary constant vectors in $\Bbb R^2$,
then $\mathbf r$ may be written as,
$$\mathbf r = \mathbf a \sin (\theta + \alpha) + \mathbf b \cos (\theta + \alpha)$$
for some constant orthogonal vectors $\mathbf a$ and $\mathbf b$ in $\Bbb R^2$, and real constant $\alpha$.
[If it helps: Note that an explicit expression for $\mathbf a$ and $\mathbf b$ is not necessarily needed, just proof of their existence.]

Let $\boldsymbol{x}=
\begin{pmatrix}
  \cos \theta \\ \sin \theta
\end{pmatrix} \in S^1$
\begin{align*}
  \boldsymbol{x}^T \boldsymbol{x} &= 1 \\
  \boldsymbol{y} &=
\begin{pmatrix}
  \boldsymbol{a} & \boldsymbol{b}
\end{pmatrix}
\begin{pmatrix}
  \cos \theta \\ \sin \theta
\end{pmatrix} \\
  \boldsymbol{y}^T \boldsymbol{y} &=
  \boldsymbol{x}^T
  \begin{pmatrix}
    \boldsymbol{a} \cdot \boldsymbol{a} &
    \boldsymbol{a} \cdot \boldsymbol{b} \\
    \boldsymbol{b} \cdot \boldsymbol{a} &
    \boldsymbol{b} \cdot \boldsymbol{b}
  \end{pmatrix}
  \boldsymbol{x}
\end{align*}
Eigenvalues:
$$\lambda_{min} \le \boldsymbol{y}^T \boldsymbol{y} \le \lambda_{max}$$
$$
\frac{a^2+b^2-\sqrt{(a^2-b^2)^2+4(\boldsymbol{a} \cdot \boldsymbol{b})^2}}{2}
\le \boldsymbol{y}^T \boldsymbol{y} \le
\frac{a^2+b^2+\sqrt{(a^2-b^2)^2+4(\boldsymbol{a} \cdot \boldsymbol{b})^2}}{2}$$
Eigenvectors:
$$\boldsymbol{\alpha}=
\boldsymbol{a} \cos
\left(
  \frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)+
\boldsymbol{b} \sin
\left(
  \frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)$$
$$\boldsymbol{\beta}=
\boldsymbol{b} \cos
\left(
  \frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)-
\boldsymbol{a} \sin
\left(
  \frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)$$
In general, the angle will be distorted and the sense of rotation may not be preserved:
$$\boldsymbol{y}=
\boldsymbol{\alpha} \cos (\phi+\phi_0) \pm
\boldsymbol{\beta} \sin (\phi+\phi_0)$$
See another answer here.

