Why is a matrix $A\in \operatorname{SL}(2,\mathbb{R})$ with $|\operatorname{tr}(A)|<2$ conjugate to a matrix of the following form?

The trace $\operatorname{tr}(A)$ of a matrix $A$ is the sum of its diagonal entries. Apparently if $A\in \operatorname{SL}(2,\mathbb{R})$ and $|\operatorname{tr}(A)|<2$, then $A$ is conjugate in $\operatorname{SL}(2,\mathbb{R})$ to a matrix of the form

$$\left(\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right).$$

Why is this? I seem to have forgotten my linear algebra.

When we assume that $|tr(A)|<2$ for some $A\in SL(2,\mathbb{R})$, then the roots of the characteristic polynomial $$ 0=\det(A-tI_2)=t^2-tr(A) t+\det A=t^2-tr(A) t+1 $$ are complex conjugates of each other, and hence on the unit circle, so of the form $e^{\pm i\theta}$. This already implies that $A$ would be conjugate to that rotation matrix in $SL(2,\mathbb{C})$. To show that they are conjugate also in $SL(2,\mathbb{R})$ requires a bit more.

Check theorem 3.1 from this document: Decomposing $SL_2(R)$