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!
 A: Suppose first that $A$ is diagonal.  Then, since you consider only the traceless case (the formula clearly fails for, say, the identity matrix), we have $A = \begin{pmatrix} a & 0 \\ 0 & -a \end{pmatrix} = a \begin{pmatrix} 1 & \\ & -1 \end{pmatrix}$ for some $a\in A$.  It will be convenient to use instead $\alpha = -ai$.
Then $\exp(A) = \begin{pmatrix} e^{i\alpha} & 0 \\ 0 & e^{-i\alpha} \end{pmatrix}$.  Of course $\sqrt{\det A} = \pm \alpha$.  It is ambiguous by a sign,  but that won't matter, as $\cos x$ and $\frac{\sin x}x$ are both even functions.  Note also that the latter should be taken to be $1$ at $x=0$.  The RHS then evaluates to:
$$ \cos\left( \alpha \right)\begin{pmatrix} 1 & \\ & 1\end{pmatrix} + i\sin\left(\alpha\right) \begin{pmatrix} 1 & \\ & -1\end{pmatrix} $$
The comparison of the two sides is then the well-known Euler formula.
Second, suppose that $A$ is diagonalizable.  Note that for any analytic function $f$ and any invertible $B$, we have $f(BAB^{-1}) = Bf(A)B^{-1}$, and that conjugating a traceless matrix leaves it traceless.  Thus any formula for analytic functions of a matrix variable which holds for diagonal matrices holds for diagonalizable matrices.  Diagonalizing also does not change the determinant.
Third and finally, the diagonalizable matrices are well-known to be dense among all matrices.  But both sides of the equation you care about are continuous.  This completes the proof.
It is worth stating again the main takeaway of the above argument: to check any conjugation-invariant continuous formula of a matrix variable (and in particular any formula consisting of analytic functions of a real variable evaluated at matrices), it suffices to check just the diagonal inputs.
A: 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 $\lambda\neq 0$, 
$$\exp(A)=\cos(\lambda)I_2+\frac{1}{\lambda}\sin(\lambda)A.$$
A: By Cayley-Hamilton theorem, $A^2$ is a constant or degree-$1$ polynomial in $A$. It follows that $\exp(A)=pI+qA$ for some $p,q\in\mathbb{C}$. Since $A$ is traceless, its two eigenvalues are $iz$ and $-iz$ for some $z\in\mathbb{C}$ with $\det(A)=z^2$. Without loss of generality, we take $\color{red}{\sqrt{\det(A)}=z}$. Take the traces of both sides of $\exp(A)=pI+qA$, we get $e^{iz}+e^{-iz}=2p$, i.e. $p=\frac12\left(e^{iz}+e^{-iz}\right)=\cos z$. To find $q$, consider an eigenvector of $A$ corresponding to the eigenvalue $iz$. We have $e^{iz}v=\exp(A)v=(pI+qA)v=\cos(z)v+iqzv$. Since $e^{iz}=\cos z + i\sin z$, we get $q=\frac{\sin z}{z}$.
A: A possible method, if you so choose to accept it, is to do the following: Define the matrix valued functions 
$$
u(t) = \exp(tA)
$$
and
$$
v(t) = \cos\left(\sqrt{\det(tA)}\right)I + \frac{\sin\left(\sqrt{\det(tA)}\right)}{\sqrt{\det(tA)}}tA
$$
Then show that $u(0)=v(0)$ and $u'(t) = v'(t)$. This will imply that $u(t)=v(t)$ for all $t \geq 0$, in particular for $t=1$.
Along the way it might help to remember that $\sqrt{\det(tA)} = t \sqrt{\det(A)}$ (since $A$ is $2 \times 2$). 
