Trying to find the inverse of $$ 2*\arcsin\left(\frac{1}{1+x}\right) $$
My approach: $$ 2*\arcsin\left(\frac{1}{1+x}\right) = y \iff \arcsin\left(\frac{1}{1+x}\right) = \frac{y}{2}$$
$$ \iff \frac{1}{1+x} = \sin\left(\frac{y}{2}\right) \iff 1+x= \frac{1}{\sin\left(\frac{y}{2}\right)} \iff x = \frac{1}{\sin\left(\frac{y}{2}\right)} – 1 $$
For domain and range of the inverse I have $$\mathbf {D} = \{x \in \Bbb {R}: -\pi \le x \le \pi\} $$
$$\mathbf {R} = \{y \in \Bbb {R}: -2 \le x \le 0, x\ne1\} $$
Checking the inverse and plot in Wolframalpha, it doesn't look at all like this. Is the problem here the multiplication by $(1+x)$ without knowing whether it is positive or negative? How to do this?
Best Answer
You are right and the problem you seem to have interpreting your answer is the following fact: consider the function $$f(x)=\frac{1}{1+x}$$ You have $f(-2)=-1$, $f(0)=1$ and $|f(x)|\gt 1$ for $-2\lt x\lt 0$ hence arcosin $f(x)$ is not defined in this domain despite your answer-function it is (actually the domain of your answer is all $\Bbb R$ excepting the points in which $\sin \frac y2=0$)