Working on this problem on linear invariant systems in signal processing, but unsure if I've got the right answer:
Show that for a real impulse response function of $H(\omega)$, the response to a sine input of $\sin(\omega_0 t)$ is the imaginary part of $H(\omega_0)e^{i\omega_0 t}$.
My answer:
Let's assume the input $f(t)$ is equal to $\sin(\omega_0 t)$. Expressing $\sin(\omega_0 t)$ in an exponential form, we have:
$$\sin(\omega_0 t)=\dfrac{e^{i\omega_0 t}-e^{-i\omega_0 t}}{2i}.$$
So the output will be as follows: $$\dfrac1{2i}(H(\omega_0)e^{i\omega_0 t}-H(-\omega_0)e^{-i\omega_0 t}),$$ where we have used the fact that the output of $e^{i\omega_0 t}$ is $H(\omega_0)e^{i\omega_0 t}$.
Using the identity $H(\omega_0)=H^*(-\omega_0)$ (since real IRF), we end up with output as: $$\dfrac{1}{i}(\mbox{Im}(H(\omega_0)e^{i\omega_0 t})$$
I think I'm being a bit silly, but is that it?
I checked your steps, they are correct except the last one. I think the last step should be:
$$y(t)=\frac{1}{2i}[H(\omega_o)e^{i\omega_o t}-H(-\omega_o)e^{-i\omega_ot}]$$
For $h(t)$ is real then $H(t)$ has conjugate symmetry property: $H(\omega_o)=H^*(-\omega_o)$. So
$$y(t)=\frac{1}{2i}[H(\omega_o)e^{i\omega_o t}-H^*(\omega_o)e^{-i\omega_ot}] =\frac{1}{2i}[2i Im\{H(\omega_o)e^{i\omega_ot}\}] =Im\{H(\omega_o)e^{i\omega_ot}\}$$