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:


*

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

*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.

*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)?
 A: 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,
$$\Bigl[\frac{\pi}{6}\Bigr]
  =\Bigl\{x\in\mathbb{R}\mid\sin x=\sin\Bigl(\frac{\pi}{6}\Bigr)\Bigr\}
  =\Bigl\{\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi\mid k\in\mathbb{Z}\Bigr\}\ .$$
A: Hint: By basic trigonometry, $\sin(x)=\sin(x+2\pi)$. 
A: Let $P(\alpha)=(x,y)$ be the point on the unit circle that corresponds to the angle $\alpha$. Then $\sin \alpha = y$. So if $\sin \alpha = \sin \beta$, then
$$\text{$\alpha = \beta + 2m\pi$ or $\alpha = \pi - \beta + 2n\pi$ for some $m,n \in \mathbb Z$}$$.
So $$[\theta] = (\theta + 2\pi m) \cup (\pi - \theta +2\pi n)$$ for some $m,n \in \mathbb{Z}.$
A: All the distinct equivalence classes are ${[x]:x\in[0,\pi /2] \cup (\pi ,3\pi /2]}$
A: Each element $x \in \left [-\frac{\pi}{2}, \frac{\pi}{2} \right ]$ defines a unique equivalence class. An element $x$ is a member of a given equivalence class $\left [x_{i}  \right ]$ such that: $x \equiv x_{i} \pmod {2 \pi}$.
