The steady state part can be expressed in the form
$\displaystyle v_{steady}=A\sin (\omega t + \alpha) = A\sin \omega t \cos \alpha + A \cos \omega t \sin \alpha,$
where $\displaystyle A $ is the amplitude.
Equate coefficients of $\displaystyle \sin \omega t$ and $\displaystyle \cos \omega t$ with the corresponding expressions in your formula for $\displaystyle v_{steady},$ and eliminate the $\displaystyle \cos \alpha \text{ and } \ sin \alpha$ by squaring and adding.
If $\displaystyle A\sin \theta = p \text{ and } A\cos \theta = q,$
then ,squaring and adding,
$\displaystyle A^{2}(\sin ^{2}\theta + \cos ^{2}\theta) = p^{2} + q^{2},$
so
$\displaystyle A = \sqrt {p^{2} + q^{2}.$
And sorry, but I know absolutely nothing about the filters mentioned in the question.
I didn't derive anything in post 2, I simply stated what should (at your stage) be a well-known trig identity, and what allows the the expression for $\displaystyle v_{steady}$ given in the question to be written in an alternative form.
Another trig identity, with which you should be familiar, is used later. $\displaystyle \cos^{2}\theta+\sin^{2}\theta=1.$