Mechanical Waves

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 $A\,\text{cos}(kx-\omega t)$?

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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

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.

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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:

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

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.