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

kramtus51

kramtus51

Answered

2021-12-29

Eliminating θ from x=cosθ(2cos2θ) and y=sinθ(2sin2θ)

Answer & Explanation

sukljama2

sukljama2

Expert

2021-12-30Added 32 answers

Let u=cosθ so x=3u2u3 and y=2u±21u2+2u3
Let z=x+y so 4(1u2)=(y+2u2u3)2=(zu)2 giving
5u2zu+z24=05u=z±25z2
Apply the quadratic equality repeatedly to get
x=3u2u51(2zuz2+4)=(3+251z2851)u451zu2
=(3+251z2851)u452z(2zuz2+4)
=(3+251z2851852z2)51(z±25z2)+452(z24)
Simplification leads to
125x=22(x+y)345(x+y)±2(35+2(x+y)2)5(x+y)2
Jonathan Burroughs

Jonathan Burroughs

Expert

2021-12-31Added 37 answers

Indulging the possibility/likelihood of a typographic error in the source material , consider the system
x=cosθ(2cos2θ)
y=sinθ(2cos2θ) instead of s2θ
From here, we easily have, defining r=2cos2θ
x2+y2=(2cos2θ)2=r2
x2y2=(2cos2θ)2cos2θ=r2(2r)=(x2+y2)(2r)
whence r(x2+y2)=x2+3y2, so that
Alternatively, if we had
x=cosθ(2sin2θ) instead of cos2θ
y=sinθ(2sin2θ)
then, with r=2sin2θ
x2+y2=(2sin2θ)2=r2
2xy=(2sin2θ)2sin2θ=(x2+y2)(2r)
whence
(x2+y2)3=4(x2xy+y2)2

nick1337

nick1337

Expert

2022-01-08Added 573 answers

x=cosθcosθcos2θ
y=sinθsinθsin2θ
(Observe x=2cosθcosθcos2θ and y=2sinθsinθsin2θ)
Adding two equations.
x+y=cosθ+sinθ(cosθcos2θ+sinθsin2θ)cosθ=sinθ
From the first equation we get,
x=cosθ(1cos2θ)=2sin2θcosθ
By squaring,
x2=4sin4θcos2θ=4sin4θ(1sin2θ)
Substituting the first result,
x2=4(x+y)4(1(x+y)2)

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