# Equivalence classes of a relation

Define on R the relation $$xTy$$ if and only if $$cos^2(x) + sin^2 (y) = 1$$. Prove that this is an equivalence relation and find R/T

About that second part, what do the equivalence classes look like? I found that, for every real number a, $$[a]_T$$ = {$$y: (y = 2kπ +-a)$$ or $$(y = 2kπ + (π-a))$$ or $$(y = 2kπ + (π+a))$$, for all k in Z}, but I cannot describe this partition properly.

• Don't worry about what they look like. Just prove it's an equivalence relation via the three conditions. – Don Thousand May 14 '19 at 2:10
• Oh this I proved easily, that's why I did not include it... my question was more about R/T, fellow yugioh fan – JBuck May 14 '19 at 2:11
• Haha didn't realize anyone would recognize my name. Anyways, for any given $y$, we know the problem becomes finding $x$ such that $\cos^2(x)=1-\sin^2(x)$. So, let $k\in[0,1]$ be arbitrary. The problem becomes finding $x$ such that $\cos^2(x)=k$. Can you solve from here? – Don Thousand May 14 '19 at 2:15
• I get x = +-$arccos(sqrt(k))$... does this lead anywhere? – JBuck May 14 '19 at 2:20
• You forgot the $+2\pi k$ term, but yes, that's right. – Don Thousand May 14 '19 at 2:22

You have already (mostly) established that \begin{align}T&=\{\langle x,y\rangle\in\Bbb R^2\mid\cos^2(x)+\sin^2(y)=1\}\\&=\{\langle x,k\pi\pm x\rangle\mid x\in\Bbb R, k\in\Bbb Z\}\\[2ex]{[x]}_T &=\{y\in\Bbb R\mid\exists k\in\Bbb Z: y=k\pi\pm x\}&\text{for all }x\in\Bbb R \\ &=\{k\pi\pm x\mid k\in\Bbb Z\}\end{align}
Now, we know $$\Bbb R/T = \{{[x]}_T: x\in\Bbb R\}$$ by definition of Quotient Set, so...