Euler Angles

Before talking about the Euler angles and the rigid body motion, we first briefly review the rotations of a point in a single fixed world frame. Given a point $\mathbf{V}\equiv [x, y, z]^T$, if we rotate it along the z-axis by an angle $\theta$ to a new position $\mathbf{V'}\equiv [x', y', z']^T$, then we have the relation \begin{equation} \mathbf{V'}= R_Z(\theta)\mathbf{V}\,,\quad R_Z(\theta)\equiv\left[\begin{array}[ccc]& \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{array}\right]\,.\tag{1}\end{equation} If we continue to rotate $\mathbf{V'}$ along the x-axis by another angle $\phi$, the point ends with the coordinates $\mathbf{V''}\equiv [x'', y'', z'']^T$ with relation \begin{equation} \mathbf{V''}= R_X(\phi)\mathbf{V'}\,,\quad R_X(\phi)\equiv\left[\begin{array}[ccc]& 1 & 0 & 0 \\ 0 & \cos(\phi) & -\sin(\phi) \\ 0 & \sin(\phi) & \cos(\phi)\end{array}\right]\,.\tag{2}\end{equation} Combining two rotations togther, we have \begin{equation} \mathbf{V''}= R_X(\phi)R_Z(\theta)\mathbf{V}\,.\tag{3}\end{equation} Note that during the two rotations, the world frame is fixed and there are no changes of the x and z axis. We simply rotate a point twice in this fixed frame. 

When talking about the rigid body motion, there are two frames: one is the fixed world frame $\{ \mathbf{i}, \mathbf{j}, \mathbf{k}\}$ and the other one is the body frame which keeps changing as the rigid body rotates. Suppose initially the body frame aligns perfectly with the world frame, if we rotate the rigid body along its z-axis by an angle $\theta$, the axis of the new body frame $\{ \mathbf{i'}, \mathbf{j'}, \mathbf{k'}\}$ becomes \begin{eqnarray} \mathbf{i'} &=& \cos(\theta) \,\mathbf{i} + \sin(\theta) \,\mathbf{j}\,,\\ \mathbf{j'} &=& -\sin(\theta) \,\mathbf{i} + \cos(\theta) \,\mathbf{j}\,,\\ \mathbf{k'}& =& \mathbf{k} \,,\end{eqnarray} where $\mathbf{i'}, \mathbf{j'}, \mathbf{k'}$ and $\mathbf{i}, \mathbf{j}, \mathbf{k}$ represent the world frame coordinates (column) of the unit vectors along body frame's x, y, z axis  before and after the rotation. Writing in the matrix form, we have \begin{equation} \left[\mathbf{i'} \quad  \mathbf{j'}\quad \mathbf{k'}\right] = \left[\mathbf{i}\quad \mathbf{j}\quad \mathbf{k}\right] \,R_Z(\theta) \,.\tag{4}\end{equation} For those who are less familiar with linear algebra, a common MISTAKE is to write the matrix form as \begin{equation} \left[\begin{array}[c] & \mathbf{i'} \\ \mathbf{j'} \\ \mathbf{k'}\end{array}\right] = R_Z^T(\theta) \left[\begin{array}[c] & \mathbf{i} \\ \mathbf{j} \\ \mathbf{k}\end{array}\right]\end{equation} which has mismatched matrix dimensions. Similarly, if we continue to rotate the rigid body along its new x-axis ($\mathbf{i'}$) by another angle $\phi$ to a new pose $\{ \mathbf{i''}, \mathbf{j''}, \mathbf{k''}\}$, we have the equation \begin{equation} \left[\mathbf{i''} \quad  \mathbf{j''}\quad \mathbf{k''}\right] = \left[\mathbf{i'}\quad \mathbf{j'}\quad \mathbf{k'}\right] \,R_X(\phi) \,.\tag{5}\end{equation} Indeed, $\mathbf{i''}=\mathbf{i'}$. In sum, if we rotate the rigid body by its z-axis and then by its NEW x-axis, we have the relation  \begin{equation} \left[\mathbf{i''} \quad  \mathbf{j''}\quad \mathbf{k''}\right] = \left[\mathbf{i}\quad \mathbf{j}\quad \mathbf{k}\right] R_Z(\theta) R_X(\phi) \,,\tag{6}\end{equation} by combining Eq. (4) and (5).

The two rotations discussed above for the rigid body is called intrinsic rotations, since they are along the axis of the body frame. In contrast, there are also extrinsic rotations that are rotations along the axis of the fixed world frames. What is the corresponding rigid body pose $\left\{\mathbf{i''},\mathbf{j''}, \mathbf{k''}\right\}$ in the world frame if we rotate it along the fixed world frame z-axis by an angle $\theta$ and then rotate again along the fixed world frame x-axis ($\mathbf{i}$) by an angle $\phi$? Applying Eq. (3) with $\mathbf{V} = \mathbf{i}, \mathbf{j}, \mathbf{k}$ leads to the solution: \begin{equation} \mathbf{i''}= R_X(\phi)R_Z(\theta)\mathbf{i}\,,\quad \mathbf{j''}= R_X(\phi)R_Z(\theta)\mathbf{j}\,,\quad \mathbf{k''}= R_X(\phi)R_Z(\theta)\mathbf{k}\,.\tag{7} \end{equation}

Note that $\left[\mathbf{i}\quad \mathbf{j}\quad \mathbf{k}\right]=I_{3\times 3}$, from Eq. (6) and (7), the rigid body pose $\left[\mathbf{i''} \quad  \mathbf{j''}\quad \mathbf{k''}\right]$ is equal to $R_Z(\theta) R_X(\phi)$ and $R_X(\phi)R_Z(\theta)$ after the corresponding intrinsic rotations and extrinsic rotations. 

Therefore, we have proved the following statement in the wiki: "Any extrinsic rotation is equivalent to an intrinsic rotation by the same angles but with inverted order of elemental rotations, and vice versa." Actually, there is an even simple proof without so many linear algebra as above. The key insight is the relativity of motion: the intrinsic rotations of rigid body in the world frame (active view) are equivalent to the corresponding extrinsic rotations of the entire world in the still body frame (passive view). The underlying math is nothing but $(AB)^{-1}=B^{-1}A^{-1}$.

A few more comments on Euler angles:

1. To describe the rigid body motion, we need three Euler angles. Since the two consecutive rotations along the same axis can be combined into a single rotation, there are only $3\times 2\times 2=12$ combinations of three rotation axis for the three Euler angles.  Proper Euler angles are the 6 cases in which the first and third rotation axis are the same while Tait-Bryan angles have three distinct rotation axis.
2. Both Proper Euler angles and Tait-Bryan angles can be extrinsic or intrinsic. The differences are in the product order of the elementary rotation matrices, as proved above.
3. In theoretical mechanics such as Landau's mechanics book, Euler angles are intrinsic and of the rotating order $\mathbf{k}-\mathbf{i'}-\mathbf{k''}$.