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I'm working through a trigonometry book and was shown this equation being worked out. I don't understand the rules for doing a particular step:

$$\begin{align} A &= A\sin(x-vt) \\ 1 &= \sin(x-vt) \\ x-vt &= {\pi \over 2} \\ x &= {\pi \over 2}+vt \end{align}$$

How are they going from $1=\sin(x-vt)$ to $x-vt = {\pi \over 2}$? Thanks!

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For what value(s) of $A$ is $\sin (A) = 1$? – David Steinberg Feb 15 '14 at 0:15
I think it will all become clear when I tell you that $\arcsin(1) = \frac{\pi}{2}$. To reiterate, just take the $\arcsin$ of both sides. – recursive recursion Feb 15 '14 at 0:15
@recursiverecursion $\arcsin(\sin(x))$ does not always equal $x$! – JiK Feb 15 '14 at 0:29
true, but it would only make sense if it did in this case. This is generally a strategy one should employ when trying to understand a step like this. – recursive recursion Feb 15 '14 at 0:31
It is physics related, it's dealing with the crest of a wave moving at velocity v, x being the crest. – stewsims Feb 15 '14 at 0:50
up vote 4 down vote accepted

Whenever you confront a step like this, notice that the $\sin$ is being taken off one side. That means that the inverse of the $\sin$ function must have been used. From this reasoning, even if you don't know that $\arcsin(1) = \frac{\pi}{2}$, then you can make the logical assumption that it is, and from that you can understand the step. EDIT: Note that $\arcsin(\sin(x))$ does not always equal $x$. I would not use this step in trying to solve an equation, but for something like this, the strategy shown above helps.

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For whatever reason it didn't dawn on me to try and do that. Thanks, totally clear now. – stewsims Feb 15 '14 at 0:32
I think this is a bit dangerous explanation, because it's a common error to "solve" trigonometric equations by using "inverse function" ($\sin$ and $\cos$ don't have inverse functions). – JiK Feb 15 '14 at 0:34
You're right, let me edit it to emphasize that. – recursive recursion Feb 15 '14 at 0:35

To solve the equation $1 = \sin(x-vt)$, you must ask yourself: what angle $\theta$ is such that $\sin\theta = 1$? One such angle is $\frac{\pi}{2}$. (The others are $\frac{\pi}{2} \pm 2n\pi$).

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$\sin(3\pi/2) = -1$; I think you mean $\pi/2 +- 2\pi n$? – Chris K Feb 15 '14 at 0:17
D'oh. Of course I did. Thanks. – rogerl Feb 15 '14 at 0:18
Also, crucially, these are the only angles $\theta$ such that $\sin\theta=1$; this is the fact that allows us to take the equation the 'other direction' and say that if we have $\sin\theta=1$ then $\theta$ must be of this form. – Steven Stadnicki Feb 15 '14 at 0:19
I'm not sure how to mark both as accepted answers. Both of these totally helped but the other one went into more of how I was failing in reasoning through it. Thanks! – stewsims Feb 15 '14 at 0:35

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