Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. 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

In the book Young and Freedman 13th edition, the wave equation is

$y(x, t) = A\,\text{cos}(kx-\omega t)$

The problem is, I find it hard to console with the fact that

$y(x, t) = A\,\text{sin}(\omega t-kx)$.

How to derive $A\,\text{sin}(\omega t-kx)$ from $A\,\text{cos}(kx-\omega t)$?

share|cite|improve this question
Minor notational comment: Usual angular frequency is denoted with a Greek omega $\omega$, not a double-u $w$. – Qmechanic Mar 12 '13 at 16:03
They both describe waves. For given $k$ they propagate in the opposite direction, and they are out of phase, but they are both correct descriptions of waves. Nor are they the only ways, so get used to seeing different forms. – dmckee Mar 12 '13 at 16:23
up vote 2 down vote accepted

If you look at the graphs for the sine and cosine functions, and know about the relation between the two:

$\sin(x) = \cos{\left(\frac{\pi}{2}-x\right)}$

you should be able to understand what happened. The expressions aren't completely equivalent, but both are solutions to the wave equation.

enter image description here

share|cite|improve this answer
Thanks completely makes sense, I was thinking of <b>x-pi/2</b> all these times that I have failed to use visualization faculty of my brain lol. Thanks again. – Joey Arnold Andres Mar 12 '13 at 15:52

Note that $\sin x \ne \cos (- x)$, rather $\sin(x) = \cos{\left(\frac{\pi}{2}-x\right)}$. In other words,

$$A\,\text{cos}(kx-\omega t) \ne A\,\text{sin}(\omega t-kx)$$ They really are two different equations, you can't derive one from the other.

However, $A\,\text{sin}(\omega t-kx)=-A\,\text{sin}(kx-\omega t)$. The two equations you gave really are two different wave equations:

$$y_1(x,t)=A\,\text{cos}(kx-\omega t)$$ $$y_2(x,t)=-A\,\text{sin}(kx-\omega t)$$

enter image description here

(Image from Google)

If they were the same wave, the two graphs would overlap. Since the graphs of $\cos x$ and $\sin x$ don't overlap, we know that $y_1(x,t)$ and $y_2(x,t)$ just don't describe the same wave.

share|cite|improve this answer

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.