What is the domain of $z=\arcsin\dfrac{x}{y}$?

Question

I get that it should be $|y|>|x|$ and in the Wolfram it looks like this. But when I graph it by hand is that it should be only the "upper" part of intersection and not the "bottom" part as well, since function values are smaller there for both inequalites: $y>x$ and $y>-x$

Answer 1

Hint: Let $$z=\arcsin\left(\dfrac{x}{y}\right)=\arcsin(w)$$ Obviously $$-1 \le w \le 1\Rightarrow -1 \le \dfrac{x}{y} \le 1\Rightarrow $$ \begin{cases} -1 \le \dfrac{x}{y}\\ \dfrac{x}{y} \le 1 \end{cases} \begin{cases} 0 \le \dfrac{x+y}{y}\\ \dfrac{x-y}{y} \le 0 \end{cases} \begin{cases} y\gt 0 \Rightarrow x+y\ge0,\qquad x-y \le 0 \\ y \lt 0 \Rightarrow x+y\le0,\qquad x-y \ge 0 \end{cases} \begin{cases} y\gt 0 \Rightarrow -y \le x \le y\\ y \lt 0 \Rightarrow y \le x \le -y \end{cases} Thus the solution is $$|x| \le |y|,\qquad y\neq0$$

Answer 2

$f(x,y)=\arcsin\dfrac{x}{y} = g(h(x,y))$, where $g(t)=\arcsin t$ and $h(x,y)=\dfrac{x}{y}$.

Now, $\hbox{dom }f= \hbox{dom }g\circ h = \hbox{dom }h \cap h^{-1}(\hbox{dom }g)$.

You should be able to write $\hbox{dom }f$ easily from $$\hbox{dom }g = \{ t \in \mathbb R : |t|\le 1 \}$$ $$\hbox{dom }h = \{ (x,y) \in \mathbb R^2 : y\ne 0 \}$$

Answer 3

You must have $$ -1\le \frac{x}{y}\le 1 $$ that is $$ \frac{|x|}{|y|}\le 1 $$ which means $$ |x|\le|y|,\quad y\ne0. $$