For $f(x)=\sin(x^2-y^2)$, sketch a picture showing regions in $\mathbb{R}^2$ where the expression is positive or negative.

Question

For $f(x)=\sin(x^2-y^2)$, sketch a picture showing regions in $\mathbb{R}^2$ where the expression is positive or negative, zero, or not defined.

$z=\sin(x^2-y^2)$

$z=0\to x^2 = y^2$, its a $\times$ at the origin, two crossed lines

If $z\neq 0$, then we get the form $\sin(x^2-y^2)=c$, where $c$ is either positive or negative. Notice that $-1\leq c \leq 1$.

So then we get that $x^2-y^2=\sin^{-1}(c)$

If the $\sin^{-1}(c)>0$, then we have a hyperbola facing sideways, and if $\sin^{-1}(c)<0$, we have hyperbola facing up/down.

So our contour plot is something like this.

$z=0$ when we are in those "crossed lines at the origin"

What about when $z>0, z<0$, this how do I find out?

How do I show the values where $\sin^{-1}c > 0$, and $\sin^{-1}c < 0$?

Answer 1

$\sin(z)$ equals zero if $z\in \pi \mathbb{Z}$, is positive if $z\in(0,\pi)+2\pi\mathbb{Z}$ and is negative if $z\in(-\pi,0)+2\pi\mathbb{Z}$. It follows that $\sin(x^2-y^2)$ equals zero on the lines $y=\pm x$ and on the (equilateral) hyperbolas $x^2-y^2 = \pm \pi n$ with $n\in\mathbb{N}^+$.

Then the pattern of signs is fairly easy to understand, without even invoking the $\arcsin$ function.