Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How might one go about calculating the time average of the product of two identical waves with different phases? For example, what would be the time average of:

$$ \cos(k x-w t) \cdot \cos(kx-wt+a) $$

and how would you get it? Thanks!

share|cite|improve this question

Your function is fully periodic, which means that your average integral is just an average integral over the fundamental domain.

$$ \begin{align*} \lim_n \frac{\int_0^n cos(kx-wt)cos(kx-wt+a) dt}{n} &= \lim_n \frac{\int_{-n}^n cos(kx-wt)cos(kx-wt+a) dt}{2n}\\ &= \frac{\int_{0}^{2 \pi/w} cos(kx-wt)cos(kx-wt+a) dt}{\frac{2\pi}{w}} \, \end{align*} $$

share|cite|improve this answer
Re-writing the ingtegrand via $\cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$ leads to a useful simplification with the answer depending only on $\cos(a)$. – Dilip Sarwate Oct 18 '11 at 21:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.