If I choose the sequence of my Euler rotations to be $Z\rightarrow Y \rightarrow Z$ or in terms of matrix multiplication $R_x(\phi)R_y(\theta)R_z(\psi)$, a stationary $3$-axis accelerometer can measure a gravitational vector through the following equation
$$
\left[
\begin{array}{@{\;}c@{\;}}
a^i_x \\ a^i_y \\ a^i_z
\end{array}\right]
=
\left[
\begin{array}{@{\;}r@{\;\;\;} @{\;\;\;}r@{\;\;\;} @{\;\;\;}r@{\;}}
%
\mathrm{c}_{\theta} \mathrm{c}_{\psi}
& \mathrm{c}_{\theta} \mathrm{s}_{\psi}
& -\mathrm{s}_{\theta} \\
%
\mathrm{s}_{\phi} \mathrm{s}_{\theta} \mathrm{c}_{\psi} – \mathrm{c}_{\phi}\mathrm{s}_{\psi}
& \mathrm{s}_{\phi} \mathrm{s}_{\theta} \mathrm{s}_{\psi} + \mathrm{c}_{\phi} \mathrm{c}_{\psi}
& \mathrm{s}_{\phi} \mathrm{c}_{\theta} \\
%
\mathrm{c}_{\phi} \mathrm{s}_{\theta} \mathrm{c}_{\psi} + \mathrm{s}_{\phi} \mathrm{s}_{\psi}
& \mathrm{c}_{\phi} \mathrm{s}_{\theta} \mathrm{s}_{\psi} – \mathrm{s}_{\phi} \mathrm{c}_{\psi}
& \mathrm{c}_{\phi} \mathrm{c}_{\theta}
%
\end{array}
\right]
\left[
\begin{array}{@{\;}c@{\;}}
0 \\ 0 \\ g
\end{array}\right]
=
\left[
\begin{array}{@{\;}r@{\;}}
-g\sin({\theta}) \\
g\sin(\phi) \cos({\theta}) \\
g\cos(\phi) \cos(\theta)
%
\end{array}
\right]
$$
I can follow the derivation and see how all $\psi$ terms are zero-ed out in the matrix multiplication, but I don't grasp the qualitative meaning of it. Gravitational vector is just a vector. Any Euler rotation applied to a vector in 3D space has to be specified by a chosen sequence of three Euler angles $\psi,\theta$ and $\phi$.
The equation however suggests that any frame (or vector) can rotate to any orientation with just two Euler angles $\phi$ and $\theta$ instead of three. Visualizing a rotation of any vector confirms this to be true – any vector can be repositioned to a new vector of any coordinate with just two Euler rotations. If we can rotate a vector or the frame with just two Euler angles then why is Euler angle defined by three angles?
[Edit]
A reason I can think of that can make use of a 3rd rotation is when I visualize a vector to be an airplane instead that rolls around its own axis. Other than that, all geometric vectors like gravitational vector are simply a line with an arrow that makes no use of a Third Euler angle.
Best Answer
The vector $\vec{g}$ points in the z-direction, and the last angle $\psi$ in the sequence of rotations is about the z-axis also.
So the angle $\psi$ has no effect on $\vec{g}$. Or in mathematical terms
$$ R_z(\psi) \vec{g}= \vec{g} $$
If you had chosen a different set of Euler angles, such as the last operation is not a rotation about the z-axis, then the angle $\psi$ would not cancel out.
The two angles that orient a vector, are exactly the angles of the spherical coordinate system, similar to the earth's latitude/longitude angles.