Is it also the case that $\langle f(t)\rangle = 0$ for $f(t) = A \cos(\omega t)$? And how does one get that $\langle f(t)\rangle = 0$?

This page discusses time averaging. It says that time averages are often important when considering oscillating waves of the form $$f(t) = A \sin(\omega t)$$, where $$\omega$$ is the angular frequency and $$A$$ is the amplitude. It is then said that the instantaneous value of this wave varies between $$-A$$ and $$A$$, but the time average of this wave over one period is $$\langle f(t)\rangle = 0$$. Is it also the case that $$\langle f(t)\rangle = 0$$ for $$f(t) = A \cos(\omega t)$$? And how does one get that $$\langle f(t)\rangle = 0$$?

Time averages over a finite time span $$T$$ do depend on $$T$$. However, as already noticed in another answer, if $$T$$ coincides with the period the average is zero.
Even more important, since $$\left= \frac{1}{T}\int_0^Tf(t)dt$$ provided the integral on the right-hand side of the previous formula is bounded, the average goes to zero when $$T \rightarrow \infty$$. For instance, in the case of $$f(t)=cos(\omega t)$$, $$\left= \frac{1}{T}\int_0^T cos(\omega t)dt = \frac{sin(\omega T)}{\omega T}$$ that goes to zero for $$T \rightarrow \infty$$.
The time average of $$\sin$$ and $$\cos$$ depends on the time interval you average those functions over. The time average over a period (or multiples of it) is zero. This is because over a period for every positive value of those functions there always is an equal but negative value as well. You don't even need to average actually, they integrate to zero over a period. You can visually see this by plotting those functions.
Another thing that could help you is this. The $$\sin$$ and $$\cos$$ functions really are the same function, just shifted, so they basically have the same properties.