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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.

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    $\begingroup$ You may use \sin instead of sin... $\endgroup$
    – Dr. Mathva
    May 7, 2019 at 15:37
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    $\begingroup$ 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 $f$. $\endgroup$
    – Elle Najt
    May 7, 2019 at 16:22

2 Answers 2

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For symmetric:

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

We need to show that:

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

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

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

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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

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

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

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