In my textbook, there's this problem:
(i) $\arcsin {2x \sqrt {1 - x^2}} = 2\arcsin {x}, \frac {-1}{\sqrt {2}} \leq x \leq \frac {1}{\sqrt {2}}$(ii) $\arcsin {2x \sqrt {1 - x^2}} = 2\arccos {x}, \frac {1}{\sqrt {2}} \leq x \leq 1$
I know how to solve this question, but my question is how did they know what the minimum and maximum values of $x$ are? And how is that relevant to the particular question?
I tried defining the range of $x$ by taking the inequality:
$$ -1 \leq 2x \sqrt {1 - x^2} \leq 1 $$ as that is the domain of the $\arcsin$ function. Unfortunately, I couldn't procceed.
We know that the range of the arcsine function is $[-\pi/2, \pi/2]$. Thus, in the first equation, we require that $$-\frac{\pi}{2} \leq \arcsin\left(2x\sqrt{1 - x^2}\right) = 2\arcsin x \leq \frac{\pi}{2}$$ Hence, \begin{align*} -\frac{\pi}{4} & \leq \arcsin x \leq \frac{\pi}{4}\\ \sin\left(-\frac{\pi}{4}\right) & \leq x \leq \sin\left(\frac{\pi}{4}\right)\\ -\frac{\sqrt{2}}{2} & \leq x \leq \frac{\sqrt{2}}{2} \end{align*}
We know that the range of the arcsine function $[-\pi/2, \pi/2]$ and that range of the arccosine function is $[0, \pi]$. Therefore, if $x$ satisfies the equation $$\arcsin\left(2x\sqrt{1 - x^2}\right) = 2\arccos x,$$ then $$0 \leq \arcsin\left(2x\sqrt{1 - x^2}\right) = 2\arccos x \leq \frac{\pi}{2}$$ since the solution must lie in the intersection of their domains. Keep in mind that the arccosine function is decreasing. Can you proceed from here?