abreviatsjw

2021-12-30

We have:
$\left\{\begin{array}{c}2x=y\mathrm{tan}\theta +\mathrm{sin}\theta \\ 2y=x\mathrm{cot}\theta +\mathrm{cos}\theta \end{array}$
And want to prove ${x}^{2}+{y}^{2}=1$
My works:
I multiplied first equation by $\mathrm{cos}\theta$ and second one by $\mathrm{sin}\theta$ and get:
$\left\{\begin{array}{c}2x\mathrm{cos}\theta =y\mathrm{sin}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \\ 2y\mathrm{sin}\theta =x\mathrm{cos}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \end{array}$
By extracting $\mathrm{sin}\theta \mathrm{cos}\theta$ we get:
$2x\mathrm{cos}\theta -y\mathrm{sin}\theta =2y\mathrm{sin}\theta -x\mathrm{cos}\theta$
$x\mathrm{cos}\theta =y\mathrm{sin}\theta$

Edward Patten

The system
$\left\{\begin{array}{c}2x\mathrm{cos}\theta =y\mathrm{sin}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \\ 2y\mathrm{sin}\theta =x\mathrm{cos}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \end{array}$
is a linear system of two (independent) equations in two variables, and it is readly checked that is a solution. Therefore it is the unique solution. Now ${x}^{2}+{y}^{2}=1$ follows.

sirpsta3u

You got $x=y\mathrm{tan}\theta$. Now substitute back into the original equations to get

Vasquez

We have from your last step $3x\mathrm{cos}\left(\theta \right)=3y\mathrm{sin}\left(\theta \right)$ thus $x=y\mathrm{tan}\left(\theta \right)$ putting this in (1) we get
$2y\mathrm{tan}\left(\theta \right)=y\mathrm{tan}\left(\theta \right)+\mathrm{sin}\left(\theta \right)⇒y=\mathrm{cos}\left(\theta \right)$
from this we get $x=\mathrm{sin}\left(\theta \right)$ now note the identity that ${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=1$

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