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

Solution 1:

Let $u=\cos\theta$ so $x=3u-2u^3$ and $y=-2u\pm2\sqrt{1-u^2}+2u^3$.

Let $z=x+y$ so $4(1-u^2)=(y+2u-2u^3)^2=(z-u)^2$ giving $$5u^2-2zu+z^2-4=0\implies5u=z\pm2\sqrt{5-z^2}.$$ Apply the quadratic equality repeatedly to get \begin{align}x&=3u-2u\cdot5^{-1}(2zu-z^2+4)=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-1}zu^2\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-2}z(2zu-z^2+4)\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1}-8\cdot5^{-2}z^2)5^{-1}(z\pm2\sqrt{5-z^2})+4\cdot5^{-2}(z^2-4).\end{align} Simplification leads to $$125x=22(x+y)^3-45(x+y)\pm2(35+2(x+y)^2)\sqrt{5-(x+y)^2}.$$

Solution 2:

Indulging the possibility/likelihood of a typographic error in the source material (which was the source of the problem in OP's previous question), consider the system $$\begin{align} x &= \cos\theta\, ( 2 - \cos2\theta ) \\ y &= \sin\theta\, ( 2 - \color{red}{\cos2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\sin2\theta$}} \end{align}$$ From here, we easily have, defining $r:=2-\cos2\theta$, $$\begin{align} x^2+y^2 &= (2-\cos2\theta)^2\phantom{\,\cos2\theta} = r^2 \\ x^2-y^2 &= (2-\cos2\theta)^2\cos2\theta = r^2(2-r) = (x^2+y^2)(2-r) \end{align}$$ whence $r(x^2+y^2) = x^2+3y^2$, so that

$$(x^2+y^2)^3 = ( x^2 + 3y^2 )^2 \tag{$\star$}$$

Alternatively, if we had $$\begin{align} x &= \cos\theta\, ( 2 - \color{red}{\sin2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\cos2\theta$}}\\ y &= \sin\theta\, ( 2 - \sin2\theta ) \end{align}$$ then, with $r:=2-\sin2\theta$, $$\begin{align} x^2+y^2 &= (2-\sin2\theta)^2 \phantom{\sin2\theta\,} = r^2\\ 2 x y &= ( 2 - \sin2\theta)^2\sin2\theta = (x^2+y^2)(2-r) \end{align}$$ whence

$$(x^2 + y^2)^3 = 4 (x^2 - x y + y^2)^2 \tag{$\star\star$}$$

Of the two, I favor $(\star)$ as the correction of the ostensible typo. Either way, these solutions are more in keeping with the spirit of similar exercises in the source than the (perfectly valid) ones given in other answers to this question as stated.