input is $f_2(t) = Acos(w_0 t + \phi)$
output $y_2(t)$ is $\frac{A}{2} e^{j \phi} H(jw_0) e^{jw_0 t} + \frac{A}{2} e^{-j \phi} H(-jw_0) e^{jw_0 t} $
okay so I need the step in between, how were the terms in the Re{} achieved?
Like, what happened to the negatives? $e^{-j \phi}, H(-jw_0)$ and so on?
It's almost like $e^{2t} + e^{-2t} = \frac{e^{4x} + 1}{e^{2t}}$
yet the answer doesn't look like that at all
I'm confused
Also, it's related to Linear Time Invariant systems, and the terms are supposed to be "conjugate pairs".
For convenience, denote $$ z = Ae^{j\phi}H(j \omega_0)e^{j \omega_0t} $$ First, note that $y_2(t)$ can be rewritten as $\frac 12 (z + \overline{z})$, which is simply $\operatorname{Re}\{z\}$. In particular: note that for any real number $x$, we have $\overline{e^{jx}} = e^{-jx}$. Moreover, we can rewrite $$ z = Ae^{j \phi}[H(j \omega_0)]e^{j \omega_0t} = A \left[|H(j\omega_0)| e^{j\angle H(j \omega_0)}\right] e^{j \phi} e^{j \omega_0 t} = A |H(j\omega_0)| e^{j[\angle H(j \omega_0) + \phi + \omega_0 t]} $$ So, we have $$ y_2(t) = \operatorname{Re}\{z\} = \operatorname{Re}\left\{ \left[A |H(j\omega_0)|\right] e^{j[\angle H(j \omega_0) + \phi + \omega_0 t]} \right\} =\\ A |H(j\omega_0)| \operatorname{Re}\left\{ e^{j[\angle H(j \omega_0) + \phi + \omega_0 t]} \right\} = \\ A |H(j\omega_0)| \cos \left[\angle H(j \omega_0) + \phi + \omega_0 t \right] $$ as desired.