[Math] Finding the equivalence classes of a trigonometric relation

equivalence-relationsrelations

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

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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)?

Best Answer

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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