Eliminating θ from x=cos⁡θ(2−cos⁡2θ) and y=sin⁡θ(2−sin⁡2θ)

Question

sukljama2 · Accepted Answer

Let u=cos⁡θ so x=3u−2u3 and y=−2u±21−u2+2u3  Let z=x+y so 4(1−u2)=(y+2u−2u3)2=(z−u)2 giving  5u2−zu+z2−4=0⇒5u=z±25−z2   Apply the quadratic equality repeatedly to get   x=3u−2u⋅5−1(2zu−z2+4)=(3+2⋅5−1z2−8⋅5−1)u−4⋅5−1zu2  =(3+2⋅5−1z2−8⋅5−1)u−4⋅5−2z(2zu−z2+4)   =(3+2⋅5−1z2−8⋅5−1−8⋅5−2z2)5−1(z±25−z2)+4⋅5−2(z2−4)  Simplification leads to  125x=22(x+y)3−45(x+y)±2(35+2(x+y)2)5−(x+y)2

Jonathan Burroughs · Accepted Answer

Indulging the possibility/likelihood of a typographic error in the source material , consider the system
$x = \cos θ (2 - \cos 2 θ)$
$y = \sin θ (2 - \cos 2 θ) \leftarrow$ instead of $s \in 2 θ$
From here, we easily have, defining $r = 2 - \cos 2 θ$
$x^{2} + y^{2} = {(2 - \cos 2 θ)}^{2} = r^{2}$
$x^{2} - y^{2} = {(2 - \cos 2 θ)}^{2} \cos 2 θ = r^{2} (2 - r) = (x^{2} + y^{2}) (2 - r)$
whence $r (x^{2} + y^{2}) = x^{2} + 3 y^{2}$ , so that
Alternatively, if we had
$x = \cos θ (2 - \sin 2 θ) \leftarrow i n s t e a d o f \cos 2 θ$
$y = \sin θ (2 - \sin 2 θ)$
then, with $r = 2 - \sin 2 θ$
$x^{2} + y^{2} = {(2 - \sin 2 θ)}^{2} = r^{2}$
$2 x y = {(2 - \sin 2 θ)}^{2} \sin 2 θ = (x^{2} + y^{2}) (2 - r)$
whence
${(x^{2} + y^{2})}^{3} = 4 {(x^{2} - x y + y^{2})}^{2}$

nick1337 · Accepted Answer

x=cos⁡θ−cos⁡θcos⁡2θ y=sin⁡θ−sin⁡θsin⁡2θ (Observe x=2cos⁡θ−cos⁡θcos⁡2θ and y=2sin⁡θ−sin⁡θsin⁡2θ) Adding two equations. x+y=cos⁡θ+sin⁡θ−(cos⁡θcos⁡2θ+sin⁡θsin⁡2θ)⏟cos⁡θ=sin⁡θ From the first equation we get, x=cos⁡θ(1−cos⁡2θ)=2sin⁡2θcos⁡θ By squaring, x2=4sin⁡4θcos⁡2θ=4sin⁡4θ(1−sin⁡2θ) Substituting the first result, x2=4(x+y)4(1−(x+y)2)

Eliminating \theta from x=\cos \theta(2-\cos 2\theta) and y=\sin \theta(2-\sin 2\theta)

Answered question

Answer & Explanation

New Questions in Trigonometry