# Finding the equivalence classes of a trigonometric relation

I have been asked to respond to the following:

Define a binary relation R on $\mathbb{R}$ as ${\{(x, y) \in \mathbb{R} \times \mathbb{R} \mid \sin(x) = \sin(y)\}}$. Prove that R is an equivalence relation. What are its equivalence classes?

Given that the relation R is based on equality, the first part of the question is rather simple:

1. Is R reflexive?

Let $a \in \mathbb{R}$, then $(a, a) \in R$ because $\sin(a) = \sin(a)$.

2. Is R symmetric?

Let $a, b \in \mathbb{R} \mid (a,b) \in R$. Then $(b,a) \in R$ by the symmetric property of equality.

3. Is R transitive?

Let $(a,b) \in R$; thus, $\sin(a) = \sin(b)$.
Let $(b,c) \in R$; thus, $\sin(b) = \sin(c)$.
Thus, $(a,c) \in R$, as $\sin(a) = \sin(c)$.

However, I am having trouble following a process to find the equivalence classes for R. As we have demonstrated that R is an equivalence relation, we know that we can decompose R into a series of equivalence classes such that, for any $x \in \mathbb{R}$, then $x \in [x]$, and that $[x] = \{ y \in \mathbb{R} \mid (x,y) \in R \}$ (or, more specifically, $[x] = \{ y \in \mathbb{R} \mid \sin(x) = \sin(y) \}$).

Examining the unit circle, we know that, for any value $x \in \mathbb{R}$, the value of $\sin(x)$ will be equivalent to that of any value $x_1 \in \mathbb{R}$ such that $x_1 = x * 2\pi k$, where $k \in \mathbb{Z}$ (after all, sin is a periodic function).

We can also see that $\sin(x) = \sin(\pi - x)$ for any x in the range $[0, \pi / 2]$; similarly, we know that $\sin(\pi + x)$ and $\sin(2\pi - x)$ are both equal to $-\sin(x)$. Along a single period of $\sin(\theta)$, there are exactly two values for $\theta$ (in that domain) for which $\sin(\theta)$ will be equal.

I don't know which of the above information is relevant to the task at hand, and I'm unsure about how to proceed. Any helpful explanations or clues would be appreciated!

EDIT:

I can begin to define some of the equivalence classes of R:

$[0] = \{ k\pi \mid k \in \mathbb{Z} \}$
$[1] = \{ \pi/2 + 2k\pi \mid k \in \mathbb{Z} \}$
$[-1] = \{ 3\pi/2 + 2k\pi \mid k \in \mathbb{Z} \}$

$[\pi/6] = \{ \pi/6 + 2k\pi, 5\pi/6 + 2k\pi \mid k \in \mathbb{Z} \}$
$[-\pi/6] = \{ -\pi/6 + 2k\pi, 7\pi/6 + 2k\pi \mid k \in \mathbb{Z} \}$

How can I generalize this to include all possible equivalence classes (accounting for all possible values of sin x)?

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The information you listed is all relevant. The equivalence class of $x$ consists of all numbers $x+2n\pi$ and $(2n+1)\pi -x$, where $n$ ranges over the integers, positive, negative, and $0$. –  André Nicolas Jun 5 '14 at 2:20
In your point 2, "commutative" is not the right terminology - this refers to an operation with the property that, for example, $x+y=y+x$. A better way to say it would be that it is due to the symmetric property of equality. In other words, you assume that $\sin a=\sin b$ and from this deduce that $\sin b=\sin a$. –  David Jun 5 '14 at 2:25
@André Nicolas Thank you for the clarity! –  Michael Zalla Jun 5 '14 at 2:59
@David Ah, I see. This is the first class in mathematics that I have taken in several years, so I appreciate you offering an explanation of the distinction here. I'll remember that for future proofs! –  Michael Zalla Jun 5 '14 at 3:00

Hint: By basic trigonometry, $\sin(x)=\sin(x+2\pi)$.
Start with the definition you have stated: $$[x] = \{ y \in \mathbb{R} \mid \sin(x) = \sin(y) \}\ .$$ I'm going to slightly change the notation: $$[a] = \{ x \in \mathbb{R} \mid \sin x = \sin a \}\ .$$ The reason I have done this is to emphasize what you have to do: given a fixed number $a$, find all $x$ which satisfy the equation. You should be able to do this by basic trigonometric methods, and the diagram and graph in your question ought to help. To give one example, $$\Bigl[\frac{\pi}{6}\Bigr] =\Bigl\{x\in\mathbb{R}\mid\sin x=\sin\Bigl(\frac{\pi}{6}\Bigr)\Bigr\} =\Bigl\{\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi\mid k\in\mathbb{Z}\Bigr\}\ .$$
There will be an equivalence class for every element of $\Bbb R$ (that is, every value of $a$), but they will not all be different. For example, in your question we have $[\frac{\pi}{6}]=[\frac{5\pi}{6}]$. In this case there will be an infinite number of equivalence classes, but this is not always true. Define a relation on $\Bbb R$ by saying $x\mathrel{R}y$ iff $x,y$ are both positive, both negative or both zero. This is an equivalence relation but it will only have three equivalence classes. –  David Jun 5 '14 at 3:38
Thanks Michael. Not keen on giving complete solutions to a homework problem, but let me offer another hint: if $a$ is given, can you solve the equation $\sin x=\sin a$ in the same way that I have solved $\sin x=\sin\frac{\pi}{6}$? If you look at the graph you posted (taking $a$ where you have written $\theta$), this pretty much shows you the answer. –  David Jun 5 '14 at 8:02