Determine whether $R$ is an equivalence relation: $xRy$ if $\cos(x)^2+\sin(y)^2=1$ I'm having troubles with this question. I understand that for a relation to be equivalent, it needs to be reflexive, symmetric, and transitive.
So far I've split this problem into 3 sections, one to check each requirement.
But am I overthinking it?  Can I just state that:
Using Pythagoras' theorem, $((\cos(x))^2)+((\sin(x))^2)=1$, so we see that $x = y$. So $x$ is related to $y$ by equality, therefore this relation is an equivalence relation.
 A: You cannot say that, because $x,y$ may be different values; that is to say, we need not have
$$\cos^2(x) + \sin^2(y) = 1$$
As an example, take $x = 0, y = \pi/2$. Then $\cos(x) = \sin(y) = 1$ and thus the sum of their squares is $2$.

Granted, this argument does work if $x=y$, i.e. if we're showing $xRx$. But you cannot conclude $R$ is an equivalence relation from that alone.
A: You are correct that the rule is important, but that is not what it says.


*

*Witness that $\cos(0)^2+\sin(\pi)^2=1$, so $\cos(x)^2+\sin(y)^2=1$ does not entail that $x=y$.



Now Pythagoras's Rule does say that for all $x$, $\cos^2(x)+\sin^2(x)=1$, so you can conclude $R$ is reflexive.
Now for symmetry, 


*

*Consider $\cos^2(x)+\sin^2(y)=(1-\sin^2(x))+(1-\cos^2(y))$. 

*So for any $x,y$, if $\cos^2(x)+\sin^2(y)=1$, then ...
Likewise for transitivity. 


*

*For any $x,y,z$, if $\cos^2(x)+\sin^2(y)=1$ and $\cos^2(y)+\sin^2(z)=1$, then ...

A: Given any non-empty set $X$ and some function $f\colon X\to Y$ the relation $R_f$ defined by $x R_f y \iff f(x)=f(y)$ is an equivalence relation. Your case is $X=Y=\mathbb{R}$ and $f:=\sin^2$ since as already mentioned $\cos^2(x)+\sin^2(x)=1$ and thus $\cos(x)^2+\sin(y)^2=1$ is equivalent to $\sin^2(x)=\sin^2(y)$.
