Define equivalence classes for $\tan(x)+k=\tan(y)$

Question

Let $ x,y \in \mathbb{R}, x\sim y \iff \exists k \in \mathbb{Z}\ \tan(x) + k = \tan(y)$ be equivalence relation. Define equivalence classes. I understand that equivalence classes for $x$ are such $y$ that $\tan(y) - k = \tan(x)$. But is there a way to express $x$, and classes for it (I don't think that $\cot$ has to to something with it here)?

Answer 1

An equivalence class is a set $S\subseteq \mathbb R$ where $a=b$ if and only if $a,b \in S$.

Often, it's useful to identify the classes via a representative element. For example, the smallest non-negative member, if that exists.

Then, you can try and define the other members constructively from that one. For example, via a function like $f(x,i,\dots)$. $i, j, \dots$ that parameterize the class. Ideally, there should be a 1-1 relationship between those parameters and members -- each member is uniquely identified.

The definition you are given is implicit: it's not a function that gives you other members of the class.

$k$ seems like a useful parameter to start with. So, you find all solutions $y$ in terms of $x$ and $k$:$$ y=f(x,k,n)=\arctan (\tan x+k) + n\pi $$

The "smallest non negative member" way of identifying a representative member seems reasonable: we can use $x \in [0, \tfrac \pi 4)$. Because $\tan \tfrac \pi 4=1$, we can see that $\tfrac \pi 4$ is in the same class as $0$.

Because $\tan$ is periodic, $n$ emerges as an additional parameter. Because of the limited range of $\arctan$, and its monotonicity, you can easily see that no two $(k,n)$ produce the same $y$. Since the domain of $\arctan$ is $\mathbb R$, all $k$ lead to a valid result.

So an equivalence class and its members are nicely parametrized by two integers, given $x\in[0, \tfrac \pi 4)$.

Answer 2

Each element of a specific equivalence class will be congruent to $(\theta)$, where $\theta$ is some real number such that $0 \leq \theta < \pi/4$.

This is because $\tan(0) = 0$ and $\tan(\pi/4) = 1.$

That is, as $\theta$ ranges from $0$ (inclusive) to $\pi/4$ exclusive, $\tan(\theta)$ takes on all values between $0$ inclusive and $1$ (exclusive). So, for any real number $y$, there will have to be a unique $\theta$ such that $0 \leq \theta < \pi/4$ and the fractional part of $\tan(y)$ equals $\tan(\theta).$

Note:
If (for example) $\displaystyle ~\tan(y) = -\frac{7}{3} = -3 + \frac{2}{3},$
then I am interpreting the fractional part of $~\tan(y)~$ to be $~\dfrac{2}{3}.$

So, you can let $\{\theta ~: 0 \leq \theta < \pi/4\}$ represent the set of all equivalence classes.