Baker–Campbell–Hausdorff formula for generators of $SO(3)$ By noting ${e_1 = (1, 0, 0), e_2 = (0, 1, 0), e_3 = (0, 0, 1)}$ and $E_i = [e_i]_\times$ the generators of $SO(3)$, then we have the following commutator properties :
$$
\begin{align}
  E_1 &= [E_2, E_3] \\
  E_2 &= [E_3, E_1] \\
  E_3 &= [E_1, E_2]
\end{align}
$$
Is it possible to find an explicit expansion of the Baker–Campbell–Hausdorff formula for these generators, such as $\log(e^{E_1} e^{E_2}) $ ?
 A: Of course it's possible. Rodrigues solved the problem in 1840.
In your somewhat idiosyncratic notation, $E_1=L_x$ and $E_2=L_y$ physicists learn in Sophomore year. I will, predictably, choose easy angles $\theta= \pi/2$, instead of your freaky $\theta=1$.
You then trivially have a roll and a pitch,
$$
\exp \left ({\pi\over 2} E_1 \right ) = \begin{bmatrix}
 1 &  0 & 0 \\
 0 &  0 & -1 \\
 0          &   1         & 0 
\end{bmatrix}    , \qquad \exp  \left({\pi\over 2}  E_2  \right) 
= \begin{bmatrix}
 0 &  0 & 1 \\
 0 &  1 & 0 \\
 -1          &   0          & 0  
\end{bmatrix}  ~~~\leadsto \\
\exp  \left({\pi\over 2}  E_1  \right)  \exp  \left({\pi\over 2}  E_2  \right) = 
\begin{bmatrix}
 0 &  0 & 1 \\
 1 &  0 & 0 \\
 0          &   1         & 0  
\end{bmatrix}  \\  = \exp\left (  {2\pi\over 3}    (E_1+E_2+E_3)/\sqrt 3\right ),
$$
Where we have used this formulation of the  Rodrigues formula, for $\theta=2\pi /3$,
$$M\equiv E_1+E_2+E_3\equiv \sqrt{3} K=\begin{bmatrix}
 0 &  -1 & 1 \\
 1 &  0 & -1 \\
-1  &   1 & 0  
\end{bmatrix} , \leadsto\\
\exp (\theta K)= I+\sin\theta ~K+ (1-\cos\theta) K^2 , ~~~\leadsto \\
\exp \left ( {2\pi\over 3} M/\sqrt{3}\right ) =I +{1\over 2} M +{1+1/2\over 3} M^2.
 $$
For general angle formulas, you should use the first link, the Rodrigues-Gibbs formula. (After you've appreciated its music, the Original Rodrigues paper is also instructive.)
