kramtus51

Answered

2021-12-29

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

Answer & Explanation

sukljama2

Expert

2021-12-30Added 32 answers

Let $u=\mathrm{cos}\theta \text{}so\text{}x=3u-2{u}^{3}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}y=-2u\pm 2\sqrt{1-{u}^{2}}+2{u}^{3}$

Let$z=x+y\text{}so\text{}4(1-{u}^{2})={(y+2u-2{u}^{3})}^{2}={(z-u)}^{2}$ giving

$5{u}^{2}-zu+{z}^{2}-4=0\Rightarrow 5u=z\pm 2\sqrt{5-{z}^{2}}$

Apply the quadratic equality repeatedly to get

$x=3u-2u\cdot {5}^{-1}(2zu-{z}^{2}+4)=(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1})u-4\cdot {5}^{-1}z{u}^{2}$

$=(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1})u-4\cdot {5}^{-2}z(2zu-{z}^{2}+4)$

$=(3+2\cdot {5}^{-1}{z}^{2}-8\cdot {5}^{-1}-8\cdot {5}^{-2}{z}^{2}){5}^{-1}(z\pm 2\sqrt{5-{z}^{2}})+4\cdot {5}^{-2}({z}^{2}-4)$

Simplification leads to

$125x=22{(x+y)}^{3}-45(x+y)\pm 2(35+2{(x+y)}^{2})\sqrt{5-{(x+y)}^{2}}$

Let

Apply the quadratic equality repeatedly to get

Simplification leads to

Jonathan Burroughs

Expert

2021-12-31Added 37 answers

Indulging the possibility/likelihood of a typographic error in the source material , consider the system

From here, we easily have, defining

whence

Alternatively, if we had

then, with

whence

nick1337

Expert

2022-01-08Added 573 answers

(Observe

Adding two equations.

From the first equation we get,

By squaring,

Substituting the first result,

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