Is this a relation of equivalence? $R := \{(x,y) \in \mathbb{R} \times \mathbb{R} ; \sin(x)=\sin(y)\}$

Question

We have to prove reflexivity, symmetry, and transitivity.

$1.)$ Reflexivity: If our relation is reflexive we know that if $(x,y)\in R$ then $(x,x)\in R$

$(x,x)\in R \to \sin(x)=\sin(x)$ which is true. Out relation is reflexive.

$2.)$ Symmetry: If our relation is symmetric we know that if $(x,y)\in R$ then $(y,x)\in R$

$(y,x)\in R \to \sin(y)=\sin(x)$ which is the same as our original relation: $\sin(x)=\sin(y)$

$3.)$ Transitivity: If our relation is symmetric we know that if $(x,y) \in R \land (y,z) \in R$ then $(x,z) \in R$

$(x,y) \in R \to \sin(x)=\sin(y)$

$(y,z) \in R \to \sin(y)=\sin(z)$

$\sin(x)=\sin(y) \land \sin(y)=\sin(z) \to \sin(x)=\sin(z)$ Which ios what we wanted to prove.

So $R$ is an equivalent relation on $\mathbb{R} \times \mathbb{R}$

Answer 1

It's correct.

Actually it holds for an arbitrary function in place of $\sin$, moreover, using the quotient construction one can prove that a relation $R\subseteq X\times X$ is an equivalence relation if and only if there is a function $f:X\to Y$ such that $$R=\{(x,x'):f(x)=f(x')\}\,.$$