Determine whether $R$ is an equivalence relation: $xRy$ if $\cos(x)^2+\sin(y)^2=1$

Question

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.

Answer 1

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 $2$.

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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.

Answer 2

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 $x=y$.

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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 ...

Answer 3

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 $\sin^2(x)=\sin^2(y)$.