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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.

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  • $\begingroup$ Don't worry about what they look like. Just prove it's an equivalence relation via the three conditions. $\endgroup$ – Don Thousand May 14 at 2:10
  • $\begingroup$ Oh this I proved easily, that's why I did not include it... my question was more about R/T, fellow yugioh fan $\endgroup$ – JBuck May 14 at 2:11
  • $\begingroup$ 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? $\endgroup$ – Don Thousand May 14 at 2:15
  • $\begingroup$ I get x = +-$arccos(sqrt(k))$... does this lead anywhere? $\endgroup$ – JBuck May 14 at 2:20
  • $\begingroup$ You forgot the $+2\pi k$ term, but yes, that's right. $\endgroup$ – Don Thousand May 14 at 2:22
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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...

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