Why are the graphs of $y=\frac12\sin(2\sin^{-1}(x+1))$ and $2\sin^{-1}(x+1)=\sin^{-1}(2y)$ not the same?

Question

Is there a valid reason why they are not the same? Because if you rearrange one equation then you get the other equation.

Answer 1

To reduce this to its simplest form, draw the graphs of $y=\sin x$ and $x=\sin^{-1}y$ for a suitably large range of $x$ (say, $x\in[-\pi,\pi]$). You will see that the second graph is just a partial copy of the first graph.

Answer 2

It is true that for all real $\ x,\quad \sin(\arcsin x) = x. $

However, $\ \arcsin(\sin x) = x\quad $ is only true for $\ x\in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].\ $

So whenever you have an equation and you take $\ \arcsin\ $ of both sides, you sometimes have to restrict the values of $x$ and $y$ (compared to before you took $\ \arcsin\ $ of both sides) so that the equation remains valid.

However, if you take $\ \sin\ $ of both sides of an equation, you do not have to restrict the values of $x$ and $y$ (compared to before you took $\ \sin\ $ of both sides).