Frequency response of an input with a phase (steady state)
Your input is a pure sine wave.   Let's call it $$x(t)=A\sin(\omega t)$$

Lets further say that input \$x(t)\$ will produce output \$y(t)\$.

Now, let's change the input to $$x^{'}(t)=A\sin(\omega t +\phi)$$
Because this is a pure sine wave of the same frequency of \$x(t)\$, you can represent the new input as a time delay of the original input.  Thus,
$$x^{'}(t) = x(t-t_{0})$$

Time invariance now says that the output will be a time delayed version of the original output:
$$y^{'}(t) = y(t-t_{0})$$