General form of matrices in $ \mathrm{O}_{3}(\mathbb{R}) $ It can be shown that every matrix in $ \mathrm{O}_{2}(\mathbb{R}) $ is of the form
$$ \begin{pmatrix} \cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi \end{pmatrix} $$
or
$$ \begin{pmatrix} \cos \varphi & \sin \varphi \\ \sin \varphi & -\cos \varphi \end{pmatrix} $$
for some angle $ \varphi $ where the former is a counter-clockwise rotation by $ \varphi $ and the latter is a reflection about the line which passes through the origin and $ (\cos \varphi/2, \sin \varphi/2) $.
My question is: can a similar "general" form can be found for matrices in $ \mathrm{O}_{3}(\mathbb{R}) $? 
So far, I attempted this by showing that the first column corresponds to a point on the unit sphere in $ \mathbb{R}^3 $. Therefore, it must be of the form $ (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta) $ for some choice of angles $ \theta $ and $ \varphi $. However, I am unable to determine the other two columns from this.
 A: I'm a little rusty, but here goes.  Given a rotation angle $\theta$, we have rotations by $\theta$ radians
$$
R_z(\theta) = \begin{pmatrix} \cos{\theta} & -\sin{\theta} & 0 \\
                                                            \sin{\theta} & \cos{\theta} & 0 \\
                                                            0 & 0 & 1 \\\end{pmatrix}
$$ 
$$
R_y(\theta) = \begin{pmatrix} \cos{\theta} & 0 & -\sin{\theta} \\
                                                                     0 & 1 & 0 \\
                                                                     \sin{\theta} & 0 & \cos{\theta} \\\end{pmatrix}
$$
$$
R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\
                                                                     0 & \cos{\theta} & -\sin{\theta} \\
                                                                     0 & \sin{\theta} & \cos{\theta} \\\end{pmatrix}
$$
in the $xy$-, $xz$-, and $yz$-planes respectively (subscript denotes the axis of rotation).  You can always use these to generate any rotation in $SO(3)$ by moving the axis first (then moving it back).
Let $A$ be a non-identity element of $SO(3)$.  Then $A$ has a 1-diml eigenspace (the axis of rotation), about which we rotate by an angle $\theta$.  This eigenspace intersects the unit sphere somewhere, so write this point of intersection as
$$
\mathbf{n} = \langle \cos{\alpha}\sin{\beta},\sin{\alpha}\sin{\beta},\cos{\beta}\rangle^T
$$
with spherical coordinates (and as a column vector). 
Note that $R_y(\beta)R_z(\alpha)^{T}\mathbf{n} = \langle 0,0,1\rangle^T$ so we have moved the axis so that this rotation occurs in the $xy$-plane.  We now rotate by our original angle $\theta$ and then rotate our axis back to its original position.  You can do the math and check that
$$
A = R_z(\alpha)R_y(\beta)^TR_z(\theta)R_y(\beta)R_z(\alpha)^{T}
$$
accomplishes this.
