Show $\exp(A)=\cos(\sqrt{\det(A)})I+\frac{\sin(\sqrt{\det(A)})}{\sqrt{\det(A)}}A,A\in M(2,\mathbb{C})$ 
Show $$\exp(A)=\cos(\sqrt{\det(A)})I+\frac{\sin(\sqrt{\det(A)})}{\sqrt{\det(A)}}A$$
  for $A\in M(2,\mathbb{C})$. In addition, $\operatorname{trace}(A)=0$.

Can anyone give me a hint how this can connect with cosine and sine? Thanks!

The caracteristic polynomial of $2\times 2$ matrix $A$ is 
$$X^2-\mathrm{Tr}(A)X+\mathrm{det}(A),$$
so that a trace $0$ matrix satisfies the equation
$$A^2=-\mathrm{det}(A)I_2.$$
Let $\lambda\in\Bbb C$ be a square root of $\det(A)$. It follows from the equation above that for every integer $p$
$$A^{2p}=(-1)^p\lambda^{2p}I_2\qquad\text{and}\qquad A^{2p+1}=(-1)^p\lambda^{2p}A$$
From then definition of the exponential function, 
$$\begin{align}\exp(A)&=\sum_{n\in\Bbb N}\frac{A^n}{n!}\\
&=\sum_{p=0}^{\infty}\frac{A^{2p}}{(2p)!}+\sum_{p=0}^{\infty}\frac{A^{2p+1}}{(2p+1)!}\\
&=\sum_{p=0}^{\infty}\frac{(-1)^p\lambda^{2p}}{(2p)!}I_2+\sum_{p=0}^{\infty}\frac{(-1)^p\lambda^{2p}}{(2p+1)!}A.
\end{align}$$
Thus, if $\lambda\neq 0$, 
$$\exp(A)=\cos(\lambda)I_2+\frac{1}{\lambda}\sin(\lambda)A.$$
