# Proving $\cos^2x+\sin^2y=1$ is reflexive, symmetric, transitive.

I want to make sure that I got the hang of the following relations.

For reflexivity, if $$x=y,\cos^2x+\sin^2y=\cos^2x+\sin^2x=1 \implies xRx$$, then it is reflexive.

For symmetry, $$xRy\implies\cos^2x+\sin^2y=1$$ and $$yRx\implies\cos^2 y+\sin^2x=1 \implies$$ symmetry.

For transitivity, $$\cos^2x+\sin^2y=1$$ and $$\cos^2y+\sin^2z=1$$ then $$\cos^2x+\sin^2y+\cos^2y+\sin^2z=2 \implies \cos^2x+\sin^2z=2-1=1$$ so transitivity holds.

Is this enough to prove the symmetric property? Anti-symmetric is easy because I only need to prove $$x=y$$ but symmetry needs to be for all $$x,y$$ but I can't list all possibilities in all questions.

• You may use \sin instead of sin... May 7, 2019 at 15:37
• If you use that $1 - sin(y)^2 = cos(y)^2$, this is a special case of the fact that $x \sim y$ iff $f(x) = f(y)$ defines an equivalence relation for any function $f$. May 7, 2019 at 16:22

For symmetric:

$$\sin^2 x + cos ^2 y = 1$$

We need to show that:

$$\sin^2 y + \cos^2 x = 1$$

Given: $$\sin^2 x + cos ^2 y = 1$$ $$\implies 1-\cos^2 x + 1- sin ^2 y = 1$$

$$\implies \cos^2 x + \sin^2 y = 1$$

Reflexive

$$cos^2x+sin^2x=1$$

symmetric

Suppose that $$cos^2x+sin^2y=1$$, $$cos^2y+sin^2x=1-sin^2y+1-cos^2x=2-(cos^2x+sin^2y)=2-1=1$$

Transitive

$$cos^2x+sin^2y=1, cos^2y+sin^2z=1$$

$$cos^2x+sin^2z=cos^2x+1-cos^2y=cos^2x+sin^2y=1$$