you don't have to use infinitesimal expansions... Just use the series form for the exponential:
e^{A} = \sum_{n} \frac{A^{n}}{n!}
I use \hbar =1 although it doesn't matter to state since you can also rename S^{12} \equiv \frac{1}{\hbar} S^{12} and using the appropriate form for S^{12} repr, as I do below - no hbar, all to keep exponential argument dimensionless. Also some expressions might look terrible, but if you are accustomed to these calculations, you can jump them - i just did everything almost explicitly because I don't know you.
e^{-i \theta S^{12}} = \sum_{n} \frac{(-i \theta S^{12})^{n}}{n!}
e^{-i \theta S^{12}} = 1_{4 \times 4} + \sum_{p=1} (-i)^{2p} (S^{12})^{2p} (\frac{\theta}{2p!})^{2p}+\sum_{p=0} (-i)^{2p+1} (S^{12})^{2p+1} (\frac{\theta}{2p+1!})^{2p+1}
Now [(S^{12})^{2}]^{\rho}_{\omega} = \sum_{\sigma} (S^{12})^{\rho}_{\sigma} (S^{12})^{\sigma}_{\omega} = \sum_{\sigma} (-i)^{2} (g^{1 \rho} \delta^{2}_{\sigma}- g^{2 \rho} \delta^{1}_{\sigma}) (g^{1 \sigma} \delta^{2}_{\omega}- g^{2 \sigma} \delta^{1}_{\omega})= - \sum_{\sigma}[ g^{1 \rho} \delta^{2}_{\sigma} g^{1 \sigma} \delta^{2}_{\omega} - g^{1 \rho} \delta^{2}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}- g^{2 \rho} \delta^{1}_{\sigma}g^{1 \sigma} \delta^{2}_{\omega}+ g^{2 \rho} \delta^{1}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}]
Now from this -due to summing over the delta kroenecker and metric is diagonal- the 1st and last term vanish, while the other two survive... So you have [doing the sum of sigma]:
[(S^{12})^{2}]^{\rho}_{\omega} = g^{1 \rho} g^{22} \delta^{1}_{\omega}+ g^{2 \rho} g^{11} \delta^{2}_{\omega}
From here you see that the squared S^{12} is diagonal (try putting \rho \ne \omega ) and has entries 1 at \rho = \omega = 1,2. I'll drop 12 of S since we are only caring about them, and with S I'll mean the matrix of S^{12}
S^{2} = \begin{bmatrix}<br />
0 & 0 & 0 & 0\\<br />
0 & 1 & 0 & 0\\<br />
0 & 0 & 1 & 0\\<br />
0 & 0 & 0 & 0\\<br />
\end{bmatrix}
So every power S^{2k} = S^{2},~ (k>0). This can also be seen without doing any calculations if you could have noticed the similarity of S to the Pauli matrice in a block form. What about the S^{2k+1} ? Well you just have to multiply S^{2} with the S matrix...
<br />
S \cdot S^{2} = -i \begin{bmatrix}<br />
0 & 0 & 0 & 0\\<br />
0 & 0 & 1 & 0\\<br />
0 & -1 & 0 & 0\\<br />
0 & 0 & 0 & 0\\<br />
\end{bmatrix} <br />
<br />
\begin{bmatrix}<br />
0 & 0 & 0 & 0\\<br />
0 & 1 & 0 & 0\\<br />
0 & 0 & 1 & 0\\<br />
0 & 0 & 0 & 0\\<br />
\end{bmatrix} =<br />
<br />
\begin{bmatrix}<br />
0 & 0 & 0 & 0\\<br />
0 & 0 & -i & 0\\<br />
0 & i & 0 & 0\\<br />
0 & 0 & 0 & 0\\<br />
\end{bmatrix} = S<br />
<br />
And finally you can find the exponential:
e^{-i \theta S} = 1_{4 \times 4} + S^{2} \sum_{p=1} (-i)^{2p} (\frac{\theta}{2p!})^{2p}+ S \sum_{p=0} (-i)^{2p+1} (\frac{\theta}{2p+1!})^{2p+1}= 1_{4 \times 4} + S^{2} \sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p}- i S \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1}
which in matrix form is:
e^{-i \theta S} =<br />
\begin{bmatrix}<br />
1 & 0 & 0 & 0\\<br />
0 & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\<br />
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\<br />
0 & 0 & 0 & 1\\<br />
\end{bmatrix} <br />
That's the \Lambda you have... just see the series expansions of cos and sin in your expression...
e^{-i \theta S} =<br />
\begin{bmatrix}<br />
1 & 0 & 0 & 0\\<br />
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\<br />
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\<br />
0 & 0 & 0 & 1\\<br />
\end{bmatrix} <br />
<br />
=<br />
\begin{bmatrix}<br />
1 & 0 & 0 & 0\\<br />
0 & \cos \theta & - \sin \theta & 0\\<br />
0 & \sin \theta & \cos \theta & 0\\<br />
0 & 0 & 0 & 1\\<br />
\end{bmatrix} <br />
<br />
