inverse of $\arcsin (\frac{x}{x-1})$

Question

determine the inverse of

a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$

b) $y= \dfrac{1-2e^{-x}}{4}$

I learned you the steps for finding the inverse are 1) get it in a form of $x= \dots$

2) change $x$ and $y$

a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$

is $y=\arcsin \left(\dfrac{x}{x-1}\right)\quad \iff \quad \sin y = \dfrac{x}{x-1}$? and if so what next?

b) $y= \dfrac{1-2e^{-x}}{4} \quad \iff \quad 4y-1=-2e^{-x} \quad\iff\quad -2y + \frac{1}{2}= e^{-x}$ . but then I'm stuck.

I would really appreciate the help!

Answer 1

$$ \sin y = \dfrac{x}{x-1}$$ $$x\sin y-\sin y=x$$ $$x(\sin y-1)=\sin y$$ $$x=\frac{\sin y}{\sin y-1}$$

For the second $$-2y + \frac{1}{2}= e^{-x}$$ $$0.5-2y=e^{-x}$$ take the $\log$ $$\log(0.5-2y)=-x$$ $$x=-\log(0.5-2y)$$ $$x=\log(\frac{1}{0.5-2y})$$

Answer 2

a) \begin{align} \sin y &= \dfrac{x}{x-1}\\ x\sin y -\sin y &=x\\ (\sin y - 1)x&=\sin y\\ x&=\frac{\sin y}{\sin y -1} \end{align} So the inverse function is $x\mapsto \dfrac{\sin x}{\sin x -1}$

And for b) is similar.

Answer 3

$\sin y(x-1)=x\Rightarrow \sin y x-x=siny \Rightarrow x(\sin y -1)=\sin y$ so $$x=\frac{\sin y}{\sin y -1}\Rightarrow y^{-1}=\frac{\sin x}{\sin x -1}$$