Prove all $2\times2$ real matrices with eigenvalues $1$ and $-1$ can be represented as My exercise asks:
Prove that all $2\times2$ real matrices with eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ can be represented as
\begin{equation}
\begin{bmatrix}
\cos\theta & a\sin\theta \\
\frac{1}{a}\sin\theta & -\cos\theta
\end{bmatrix}
\end{equation}
Starting like
\begin{equation}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\end{equation}
Using the fact that $\operatorname{Tr}(A) = a+d =\lambda_1+\lambda_2 = 0 \Rightarrow a=-d$. And using the fact that $\det A = ad-bc=\lambda_1\lambda_2=-1 \Rightarrow-a^2-bc=-1$
\begin{equation}
a^2+bc=1
\end{equation}
How can I complete the proof?

$a,b,c,d$ are fixed numbers and we have to show that they can be represented as $\cos\theta, k\sin\theta, \frac1k \sin\theta, -\cos\theta$
Let $\cos \theta=a \implies d = -a = -\cos \theta$
This can be done because there always exists a $\theta \in \mathbb{C}$ such that $\cos\theta = a$
Now,
$$ a^2 + bc=1$$
$$ bc = 1-a^2 = \sin^2 \theta$$
You will always be able to choose a $k$  such that $b=k\sin\theta$ 
This is because
$$ \sin^2\theta + \cos^2\theta=1$$
$$ \dfrac{b^2}{k^2} + a^2 = 1 $$
$$ k = \pm\dfrac{b}{\sqrt{1-a^2}} $$
This $k$ will also satisfy $c = \dfrac1k \sin\theta$ because $c = \dfrac{\sin^2\theta}{b}$
That completes the proof.
