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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, $[x] = \{ y \in \mathbb{R} \mid \sin(x) = \sin(y) \}$).

enter image description here

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!


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 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 at 2:25
@André Nicolas Thank you for the clarity! –  Michael Zalla Jun 5 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 at 3:00

2 Answers 2

Hint: By basic trigonometry, $\sin(x)=\sin(x+2\pi)$.

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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, $$\Bigl[\frac{\pi}{6}\Bigr] =\Bigl\{x\in\mathbb{R}\mid\sin x=\sin\Bigl(\frac{\pi}{6}\Bigr)\Bigr\} =\Bigl\{\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi\mid k\in\mathbb{Z}\Bigr\}\ .$$

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Thank you for getting me a step closer to the solution. Should the concept of defining my equivalence classes to be associated with some fixed number tell me that I should be defining one class for every possible value of sin? Given that we are dealing with the set of all real numbers, wouldn't that imply that there are an infinite number of classes, each associated with some real value in the range [-1, 1]? (Just as, in your example, the equivalence class is associated with the value Pi/6). –  Michael Zalla Jun 5 at 3:15
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 at 3:38
David, I've gone ahead and accepted your answer. Purely out of curiosity, I would still appreciate any additional direction (see my edit), such that I can define a series of disjoint equivalence classes that encompass all real numbers. –  Michael Zalla Jun 5 at 7:58
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 at 8:02

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