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!
Answer
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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