Both sin and cos are considered sinusoidal waveforms. As a practical matter, sin and cos are essentially the same thing, just offset by 90 degrees.Since "time 0" is arbitrary, the distinction between sin and cos only matters if you are comparing phase against another signal.
Since Euler's equation works out to cos for the real component and sin for the imaginary component, it's just more useful to use the cos than the sin for real voltage.
Superimposition is OK, but you made a mistake: \$I_0\$ is a DC component, so it cannot pass through a capacitor (as you say, we are analyzing steady-state). Therefore there is no point in computing the impedance, because it is apparent that the capacitor won't contribute to it (at DC it has infinite impedance, so in the parallel with R is completely negligible, if you want to use this POV).
Therefore \$I_0\$ will pass through R alone, that's why your book computes \$v_u'\$ simply as \$R I_0\$.
At the end you will have a DC component in \$v_u\$ due to \$I_0\$ and a sinusoidal component with frequency \$f\$ due to \$I_s\$, so your output voltage will have this form:
\$
v_u(t) = V_0 + A \cos(2 \pi f t + \phi)
\$
EDIT
(in response to a comment)
The impedance of a capacitor in the phasor domain is
\$
Z = \dfrac{1}{j\omega C} = j \, \dfrac{-1}{\omega C} = j X_C
\$
where \$X_C=\dfrac{-1}{\omega C}\$ is called the reactance of the capacitor.
If you want to make a comparison with what happens in DC circuits, i.e. with resistance, you should take the modulus of the impedance \$|Z|\$, which expresses intuitively how much the current flow is impeded when trying to flow in the capacitor.
A you can see:
\$
|Z| = \dfrac {1}{\omega C}
\rightarrow \infty \quad\textrm{as}\quad \omega \rightarrow 0
\$
And since \$I=\dfrac V Z \;\Rightarrow\; |I|= \dfrac{|V|}{|Z|} = 0 \$ when \$\omega=0\$
Best Answer
No, \$\omega\$ and \$\omega t\$ and are not the same, and they don't have the same units. \$\omega t\$ has units of radians, which makes it possible to add the phase angle \$\phi\$ to it and get something sensible.
In the equation you provide, the values of \$V_m\$, \$\omega\$, and \$\phi\$ are constants. So, if we want to find the voltage as a function of time, then there must be a variable in the equation that represents the specific value of time in question. That variable is \$t\$.