Members of SO(3) are a secret thing that are very difficult to find anywhere. So far, I haven't found them from anywhere else than from my own notes

Here they are.
<br />
\theta=(\theta_1,\theta_2,\theta_3) = |\theta|(n_1,n_2,n_3)\in\mathbb{R}^3<br />
<br />
\exp\Big(\left[\begin{array}{ccc}<br />
0 & -\theta_3 & \theta_2 \\<br />
\theta_3 & 0 & -\theta_1 \\<br />
-\theta_2 & \theta_1 & 0 \\<br />
\end{array}\right]}\Big)<br />
=\sum_{k=0}^{\infty}\frac{1}{k!}<br />
\left[\begin{array}{ccc}<br />
0 & -\theta_3 & \theta_2 \\<br />
\theta_3 & 0 & -\theta_1 \\<br />
-\theta_2 & \theta_1 & 0 \\<br />
\end{array}\right]^k<br />
<br />
=\left[\begin{array}{ccc}<br />
1 & 0 & 0 \\<br />
0 & 1 & 0 \\<br />
0 & 0 & 1 \\<br />
\end{array}\right]<br />
\;+\;\left[\begin{array}{ccc}<br />
0 & -n_3 & n_2 \\<br />
n_3 & 0 & -n_1 \\<br />
-n_2 & n_1 & 0 \\<br />
\end{array}\right] \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}|\theta|^{2k+1}<br />
\;+\;\left[\begin{array}{ccc}<br />
n_1^2 -1 & n_1n_2 & n_1n_3 \\<br />
n_1n_2 & n_2^2 - 1& n_2n_3 \\<br />
n_1n_3 & n_2n_3 & n_3^2 - 1 \\<br />
\end{array}\right] \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k)!}|\theta|^{2k}<br />
<br />
=\left[\begin{array}{ccc}<br />
n_1^2(1-\cos|\theta|) + \cos|\theta| & n_1n_2(1-\cos|\theta|) - n_3\sin|\theta|<br />
& n_1n_3(1-\cos|\theta|) + n_2\sin|\theta| \\<br />
n_1n_2(1-\cos|\theta|) + n_3\sin|\theta| & n_2^2(1-\cos|\theta|) + \cos|\theta|<br />
& n_2n_3(1-\cos|\theta|) - n_1\sin|\theta| \\<br />
n_1n_3(1-\cos|\theta|) - n_2\sin|\theta| & n_2n_3(1-\cos|\theta|) + n_1\sin|\theta|<br />
& n_3^2(1-\cos|\theta|) + \cos|\theta| \\<br />
\end{array}\right]<br />
It is also possible to interpret the operator
<br />
e^{\theta\times}<br />
as an member of SO(3). First convince yourself with some geometric arguments, that the mapping
<br />
x\mapsto x + ((x\cdot n)n - x)(1-\cos|\theta|) + (n\times x)\sin|\theta|<br />
is the rotation of x around the angle theta, and then verify that the series
<br />
e^{\theta\times}x = x \;+\; \theta\times x \;+\; \frac{1}{2}\theta\times(\theta\times x) \;+\; \frac{1}{3!}\theta\times(\theta\times(\theta\times x)) \;+\; \cdots<br />
converges towards this. This is the same thing as the matrix calculation, in fact.