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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 $\langle f(t)\rangle = 0$?

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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 $$ \left<f\right>= \frac{1}{T}\int_0^Tf(t)dt $$ 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 $f(t)=cos(\omega t)$, $$ \left<f\right>= \frac{1}{T}\int_0^T cos(\omega t)dt = \frac{sin(\omega T)}{\omega T} $$ that goes to zero for $T \rightarrow \infty$.

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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.

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