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I must represent the domain of the function: $$z=\frac{x-y}{\sin x-\sin y}$$ Therefore, $\sin x\neq\sin y$. So I must plot $\sin x=\sin y$. How do I do this?

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1  
Do you perhaps mean sin instead of sen? –  qaphla Aug 5 '13 at 22:47
3  
Sen(x) is the portuguese, and italian spelling of it. pt.wikipedia.org/wiki/Seno, it.wikipedia.org/wiki/Seno_(matematica). Probably the same in some other languages too. –  Simon Means Aug 5 '13 at 22:50
    
Just draw the graph of $\sin x$ and then draw a horizontal line crossing it. The intersection points are the $x$ values that share the same sine value. You do not what your $x$ and $y$ be any two of these intersection points. How would you summarize this situation? –  Maesumi Aug 5 '13 at 23:05
    
For any given $x$, $\sin x = \sin((2n + 1)\pi - x) = \sin(x + 2n\pi)$ for $n \in \mathbb Z$, and this exhausts all possibilities. So, $\sin x = \sin y$ when $y = x + 2n\pi$ or $y = (2n + 1)\pi -x$. The graph is the union of these infinitely many straight lines. –  Tunococ Aug 5 '13 at 23:07
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Graphically, it's a grid rotated by 45 degrees. –  Tunococ Aug 5 '13 at 23:08

3 Answers 3

$\sin y = \sin x \iff y = n\pi +(-1)^nx,\text{ for some $n\in \mathbb{Z}$}$.

Putting $n=0$, we see that $y = x$ is on the graph.

Putting $n=1$, we see that $y = -x +\pi$ is also on the graph.

So for all even $n$, we have the set of straight lines:

$y = x + -4\pi,\,\quad y = x + -2\pi,\,\quad y = x,\,\quad y = x + 2\pi$ and so on so forth.

Likewise, for all odd $n$:

$y = -x + -3\pi,\,\quad y = -x + -\pi,\,\quad y = -x + \pi,\,\quad y = -x + 3\pi$ and so forth.

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Did you mean to say these are to be excluded from $xy$ planeto get the domain, as opposed to "it is on graph"? –  Maesumi Aug 5 '13 at 23:12
    
@Maesumi Yes, you are right that these have to be excluded from the domain. I believe the OP is trying to plot all the points that are not in the domain of his/her function. These points as he/she concludes, precisely satisfy $\sin x = \sin y$. –  Alraxite Aug 5 '13 at 23:16

$\sin y = \sin x \iff y = x +2k\pi \text{ or } y = \pi-x+2k\pi,\text{ for some $k\in \mathbb{Z}$}$.

The two equations on the RHS of the equivalence correspond to lines.

Therefore, the set $\{(x,y) \in \mathbb{R}^2 \mid \sin x = \sin y\}$ (over which the function is not defined) is a "crisscross" of such lines; the restriction of that set over $[-40,40]^2$ looks as follows:

enter image description here


$\LaTeX$ code:

\documentclass{article}
\usepackage{pgfplots}

\begin{document}
    \begin{tikzpicture}
    	\def\PI{3.1416}
    	\begin{axis}[xlabel=$x$, ylabel=$y$,ymin=-40,ymax=40]
    		\foreach \k in {-15,...,15}{
    			\addplot[red, ultra thin,domain=-40:40] (x,x+2*\PI*\k);
    			\addplot[red, ultra thin,domain=-40:40] (x,\PI-x+2*\PI*\k);
    		}
    	\end{axis}
    \end{tikzpicture}
    \end{document}
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$$\begin{align} \sin(y) &=\sin(x)\\ \implies y&=\arcsin(\sin(x))+2\pi\,n\text{ or }\pi-\arcsin(\sin(x))+2\pi\,n\\ \implies y&=\arcsin(\sin(x))+2\pi\,n\text{ or }\pi-\arcsin(\sin(x))+2\pi\,n\\ \end{align}$$ for some $n\in\mathbb{Z}$.

Now $\arcsin(\sin(x))$ is a sawtooth function with slope $1$ over $(-\pi/2,\pi/2)$ and slope $-1$ over $(\pi/2,3\pi/2)$. Accounting for all vertical translations by an integer multiple of $2\pi$, and then also for all vertical reflections over the $x$-axis followed by a vertical translation by $\pi$, the domain is the complement of a criss-cross lattice with a generating square with vertices $(-\pi/2,-\pi/2),(-3\pi/2,\pi/2),(-\pi/2,3\pi/2),(\pi/2,\pi/2)$.

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