How to calculate $e^{tA}$ with $tr(A)=0$ and invertible matrix $A\in{M_{2}(\mathbb{R})}$?
By Cayley-Hamilton, we can get $$A^2+|A|I=0$$ Then we calculate the non-zero eigenvalues $$\left\{ \begin{array}{} \pm\sqrt{\det{}A}i & |A|>0\\ \pm\sqrt{-\det{}A} & |A|<0\\ \end{array}\right.$$ so it exists an invertible matrix $P$ for which $$P^{-1}\exp{(tA)}P=\left\{ \begin{array}{} {\rm diag}(e^{t\sqrt{\det{}A}i},e^{-t\sqrt{\det{}A}i}) & |A|>0\\ {\rm diag}(e^{t\sqrt{-\det{}A}},e^{-t\sqrt{-\det{}A}}) & |A|<0\\ \end{array}\right.$$
but how to calculate $e^{tA}$ with the result $$e^{tA}=\left\{ \begin{array}{} \cos(t\sqrt{\det {A}})I+\frac{\sin(t\sqrt{\det{}A})}{\sqrt{\det{A}}}A & |A|>0\\ \cosh(t\sqrt{-\det {A}})I+\frac{\sinh(t\sqrt{-\det{}A})}{\sqrt{-\det{A}}}A & |A|<0\\ \end{array}\right.$$
Let $d=\sqrt{|\det A|}$.
You have $$ e^{tdi}=\cos td+i \sin td. $$ So \begin{align} e^{tA}&=P\,\begin{bmatrix}\cos td+i \sin td&0\\0& \cos td-i \sin td\end{bmatrix}\,P^{-1}\\[0.2cm] &=P\,\begin{bmatrix}\cos td&0\\0& \cos td\end{bmatrix}\,P^{-1} +i P\begin{bmatrix}\sin td&0\\0& -\sin td\end{bmatrix}\,P^{-1}\\[0.2cm] &=\cos td\,I+i\,\frac{\sin td}{d}\,P\begin{bmatrix}d&0\\0&-d \end{bmatrix}\,P^{-1}\\[0.2cm] &=\cos td\,I+i\,\frac{\sin td}{d}\,A. \end{align} The case with $e^{td}$ is similar, now with the hyperbolic sine and cosine.