Rotational Kinematics and Angular Velocity Vector Transformation Say we have a rigid body with a body-fixed coordinate system XYZ and an inertial coordinate system (North - East - Down) XoYoZo. If we use the x-y-z convention and rotate the body fixed frame first around the Zo axis, then the Yo and then Xo we end up with multiplying the three rotations to get a transformation matrix R = CzCyCx (where C is the general rotation matrix around a certain axis). R is the transformation matrix to describe how is the body oriented with regard to the inertial frame. This matrix can also be used for the linear velocity vector transformation.
What I cannot understand is why this matrix cannot be used for describing how the angular velocities result on the body frame with respect to the inertial frame? Instead, the matrix is:
$$\begin{pmatrix} 1 & \sin\phi\tan\theta & \cos\phi\tan\theta \\
0 & \cos\phi & -\sin\phi \\
0 & \sin\phi/\cos\theta & \cos\phi/\cos\theta \end{pmatrix} $$
,where   $\phi$ and $\theta $ describe the orientation of the body fixed frame with respect to the earth fixed frame around the X and Y axes. I have hard time understanding why the angular velocity around the Z axis is not part of the transformation matrix? Also in general, as I stated, why a rotation around the three axes (like the matrix for linear velocity) wouldn't work for the angular velocity vector? 
I am sorry for the so what badly formulated question but I am not sure how to express my perplexity otherwise.
 A: There seems to be confusion between a transformation of coordinates (matrix $\mathbf{R}$) and the Jacobian (matrix $\mathbf{J}$).


*

*The rotation matrix transforms the components of vectors between the body frame and the inertial frame. This happens for all vectors.
$$ \begin{aligned} 
  \boldsymbol{v}_0 & = \mathbf{R} \boldsymbol{v} \\
  \boldsymbol{\omega}_0 & = \mathbf{R} \boldsymbol{\omega} \\
  \boldsymbol{F}_0 & = \mathbf{R} \boldsymbol{F} \\
  \boldsymbol{\tau}_0 & = \mathbf{R} \boldsymbol{\tau} 
\end{aligned} $$

*The Jacobian relates the three joint speeds $(\dot{\phi},\dot{\psi},\dot{\theta})$ to body rotational velocity $\boldsymbol{\omega}_0$. For a sequence of rotations the body to inertial rotation matrix is: $\mathbf{R} = \mathbf{R}_x \mathbf{R}_y \mathbf{R}_z $. Now the body rotational velocity vector is defined as follows
$$ \boldsymbol{\omega}_0 = \boldsymbol{\hat{\imath}} \dot{\phi} + \mathbf{R}_x \left( \boldsymbol{\hat{\jmath}} \dot{\psi} + \mathbf{R}_y \boldsymbol{\hat{k}} \dot{\theta} \right) $$
Do you see the pattern above? See this post as well as this post for more details.
the above is grouped together into the Jacobian as
$$ \boldsymbol{\omega}_0 = \mathbf{J}  \pmatrix{\dot{\phi} \\ \dot{\psi} \\ \dot{\theta} } $$
You see, the list of joint speeds is not a vector because each joint speed is riding on a different reference frame. The columns of the Jacobian contain the orientation of each rotation axis in the inertial system
$$ \mathbf{J} = \Big[ \begin{array}{c|c|c} \boldsymbol{\hat{\imath}} & \mathbf{R}_x \boldsymbol{\hat{\jmath}} & \mathbf{R}_x \mathbf{R}_y \boldsymbol{\hat{k}}  \end{array} \Big] $$

*The matrix you describe in your post is the inverse Jacobian which relates the joint motions to the body rotational velocity
$$\pmatrix{\dot{\phi} \\ \dot{\psi} \\ \dot{\theta} }  = \mathbf{J}^{-1}   \boldsymbol{\omega}_0$$
where the inverse Jacobian evaluates to
$$ \mathbf{J}^{-1} = \begin{pmatrix} 1 & \sin\phi\tan\psi & -\cos\phi\tan\psi \\
0 & \cos\phi & \sin\phi \\
0 & -\sin\phi/\cos\psi & \cos\phi/\cos\psi \end{pmatrix} $$
