kramtus51

2021-12-29

Eliminating $\theta$ from $x=\mathrm{cos}\theta \left(2-\mathrm{cos}2\theta \right)$ and $y=\mathrm{sin}\theta \left(2-\mathrm{sin}2\theta \right)$

sukljama2

Expert

Let
Let giving
$5{u}^{2}-zu+{z}^{2}-4=0⇒5u=z±2\sqrt{5-{z}^{2}}$
Apply the quadratic equality repeatedly to get
$x=3u-2u\cdot {5}^{-1}\left(2zu-{z}^{2}+4\right)=\left(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1}\right)u-4\cdot {5}^{-1}z{u}^{2}$
$=\left(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1}\right)u-4\cdot {5}^{-2}z\left(2zu-{z}^{2}+4\right)$
$=\left(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1}-8\cdot {5}^{-2}{z}^{2}\right){5}^{-1}\left(z±2\sqrt{5-{z}^{2}}\right)+4\cdot {5}^{-2}\left({z}^{2}-4\right)$
$125x=22{\left(x+y\right)}^{3}-45\left(x+y\right)±2\left(35+2{\left(x+y\right)}^{2}\right)\sqrt{5-{\left(x+y\right)}^{2}}$

Jonathan Burroughs

Expert

Indulging the possibility/likelihood of a typographic error in the source material , consider the system
$x=\mathrm{cos}\theta \left(2-\mathrm{cos}2\theta \right)$
$y=\mathrm{sin}\theta \left(2-\mathrm{cos}2\theta \right)←$ instead of $s\in 2\theta$
From here, we easily have, defining $r\phantom{\rule{0.222em}{0ex}}=2-\mathrm{cos}2\theta$
${x}^{2}+{y}^{2}={\left(2-\mathrm{cos}2\theta \right)}^{2}={r}^{2}$
${x}^{2}-{y}^{2}={\left(2-\mathrm{cos}2\theta \right)}^{2}\mathrm{cos}2\theta ={r}^{2}\left(2-r\right)=\left({x}^{2}+{y}^{2}\right)\left(2-r\right)$
whence $r\left({x}^{2}+{y}^{2}\right)={x}^{2}+3{y}^{2}$, so that

$y=\mathrm{sin}\theta \left(2-\mathrm{sin}2\theta \right)$
then, with $r\phantom{\rule{0.222em}{0ex}}=2-\mathrm{sin}2\theta$
${x}^{2}+{y}^{2}={\left(2-\mathrm{sin}2\theta \right)}^{2}={r}^{2}$
$2xy={\left(2-\mathrm{sin}2\theta \right)}^{2}\mathrm{sin}2\theta =\left({x}^{2}+{y}^{2}\right)\left(2-r\right)$
whence
${\left({x}^{2}+{y}^{2}\right)}^{3}=4{\left({x}^{2}-xy+{y}^{2}\right)}^{2}$

nick1337

Expert

$x=\mathrm{cos}\theta -\mathrm{cos}\theta \mathrm{cos}2\theta$
$y=\mathrm{sin}\theta -\mathrm{sin}\theta \mathrm{sin}2\theta$
(Observe $x=2\mathrm{cos}\theta -\mathrm{cos}\theta \mathrm{cos}2\theta$ and $y=2\mathrm{sin}\theta -\mathrm{sin}\theta \mathrm{sin}2\theta$)
$x+y=\mathrm{cos}\theta +\mathrm{sin}\theta -\underset{\mathrm{cos}\theta }{\underset{⏟}{\left(\mathrm{cos}\theta \mathrm{cos}2\theta +\mathrm{sin}\theta \mathrm{sin}2\theta \right)}}=\mathrm{sin}\theta$
From the first equation we get,
$x=\mathrm{cos}\theta \left(1-\mathrm{cos}2\theta \right)=2\mathrm{sin}2\theta \mathrm{cos}\theta$
By squaring,
${x}^{2}=4\mathrm{sin}4\theta \mathrm{cos}2\theta =4\mathrm{sin}4\theta \left(1-\mathrm{sin}2\theta \right)$
Substituting the first result,
${x}^{2}=4\left(x+y{\right)}^{4}\left(1-\left(x+y{\right)}^{2}\right)$

Do you have a similar question?