I'm looking over some class notes, and I'm sure that my professor has made an error. Consider the following equation:

$$ e^{-q} \sin \left(kx - \omega{t} \right) - e^{q} \sin \left(kx + \omega{t} \right)$$

In my notes, this can be simplified as:

$$ e^{-q} \left(\sin\left(kx - \omega{t}\right) - \sin \left(kx + \omega{t}\right)\right)- \left(e^{q} - e^{-q}\right) \sin \left(kx - \omega{t}\right)$$

However, I'm not convinced this is correct. I thought that the simplification for the left hand terms should be

$$\sin\left(kx - \omega{t}\right) = \sin(kx)\cos(\omega{t}) - \cos(kx) \sin(\omega{t})$$

which is different from his solution. Also, I'm not sure what is happening with the terms on the right hand side. For example, why is $e^q$ simplified to $e^q - e^{-q}$

Any advice would be appreciated

  • $\begingroup$ I think it may have to do with the nature of problem you are dealing with, for example see the damped oscillator equation and how it is formed and solved and converted into a form which is physically meaningful. Also, I am assuming that this has to do something with physics because of the $\sin(kx-wt)$ resembles so. However, i am not sure , if it not related to physics then i really don't know why that simplification to be exact. $\endgroup$ – Mann Dec 31 '14 at 11:33
  • $\begingroup$ So, if this is a physics example, which it is, I don't see how the solution provided is correct. $\endgroup$ – Emma Tebbs Jan 1 '15 at 10:16
  • $\begingroup$ Yea , it wasn't correct. I didn't see it there was a error of + ,- as answered by the other users ! $\endgroup$ – Mann Jan 2 '15 at 7:27

It should be $$ e^{-q} \left(\sin\left(kx - \omega{t}\right) - \sin \left(kx + \omega{t}\right)\right)- \left(e^{q} - e^{-q}\right) \sin \left(kx \mathop{\color{red}{+}} \omega{t}\right) . $$ Your professor has just added and subtracted $e^{-q} \sin(kx+\omega t)$.

  • $\begingroup$ Wow, i didn't even look at this. I feel dumb enough now. ^^ $\endgroup$ – Mann Dec 31 '14 at 11:38
  • $\begingroup$ why would you add and subtract that? $\endgroup$ – Emma Tebbs Dec 31 '14 at 18:22
  • $\begingroup$ There's too little context here to answer that question. You'll need to look at what happens later in the computation, and then it should hopefully become clear what the purpose was. (Or just ask your professor!) $\endgroup$ – Hans Lundmark Jan 1 '15 at 10:24

I would agree if the result was $$ \mathrm{e}^{-q}\left(\sin(kx-\omega t) -\sin(kx+\omega t)\right) -\left(\mathrm{e}^{q}-\mathrm{e}^{-q}\right)\sin(kx+\omega t)\tag{*} $$ As you simply add $$ \mathrm{e}^{-q}\left(\sin(kx+\omega t)-\sin(kx+\omega t)\right) $$ To the original equation. So if there was a minus sign in yours/his notes for the last term in Eq (*)


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