What happens to a cosine when sending it through a LTI system?
This question came up in my last signal processing class. The professor elegantly answered that it simply gets scaled and shifted. I tried to verify this at home but the result didn't confirm what he said:
The input-signal is defined as $x(t)=cos(\omega t)$. Furthermore it can be shown that the output of a LTI-system is calculated by $ y(t)=\int_{-\infty}^{+\infty} x(t-\tau)h(\tau)d\tau$.
\begin{align*} y(t)&=\int_{-\infty}^{+\infty} cos(\omega(t-\tau))h(\tau)d\tau\\ &=\int_{-\infty}^{+\infty} (\frac{1}{2}e^{-i\omega (t-\tau)}+\frac{1}{2}e^{i\omega (t-\tau)})h(\tau)d\tau\\ &=\frac{1}{2}e^{-i\omega t}\int_{-\infty}^{+\infty}e^{i\omega \tau}h(\tau)d\tau + \frac{1}{2}e^{i\omega t}\int_{-\infty}^{+\infty}e^{-i\omega \tau}h(\tau)d\tau\\ &=\frac{1}{2}e^{-i\omega t}H(-j\omega)+\frac{1}{2}e^{i\omega t}H(j\omega) \end{align*}
But a scaled and shifted cosine would actually look like this: $$ a*cos(t+b)=\frac{1}{2}ae^{-ib-it}+\frac{1}{2}ae^{ib+it} $$
Is my professors statement right? Did I make a mistake in my calculations / is my aprroach wrong?