<s>[INST] Parametrization of a curve in polar coordinates I'm trying to change this parametrics equations to polar coordinates
$$ X(t) = 2\cos(t) - \sin(2t)  \\ 
   Y(t) = 2\sin(t) - \cos(2t) $$
What i tryed to do was raise the two equations squared, sum  then and make some algebric manipulations.
$$ X^2 = (2\cos(t) - \sin(2t) )^2  =  4\cos^2(t) -4\cos(t)\sin(2t) + \sin^2(2t) \\
   Y^2 = (2\sin(t) - \cos(2t))^2  =  4\sin^2(t) -4\sin(t)\cos(2t) + \cos^2(2t)  \\
X^2 + Y^2 = 4(\sin(t) + \cos(t))^2 -4(\sin(t)\cos(2t) +\sin(2t)\cos(t))+(\sin(2t)+\cos(2t))^2
\\ \to X^2 + Y^2 = 5 - 4[ \sin(t)(2\cos^2(t) -1) +2\sin(t)\cos^2(t) ]
\\ \to X^2 + Y^2 = 5 - 4\sin(t)(4(\cos(t))^2 -1) 
$$
For last, we can obtain:
$$  X^2 + Y^2 = 5 - 4\sin(3t) $$
Considering that $R = \sqrt{X^2 + Y^2} $ and $  \theta(t) = \arctan(\frac{Y}{X}) $ ,
what i can do to replace the right hand side of the equation for polar? 
EDIT:
Using the help of the Lord_Farin, I derivated the main equation and now i'm trying to found a relation between $\frac{X}{Y} $
and $\sin(3t) $ but i don't see a simplification in my equations. 
$$
 \frac{d}{dt} (X^2 + Y^2) = \frac{d}{dt} (5 -4\sin(3t))  \\
 2\dot{X}X + 2\dot{Y}Y = -4(3\cos(3t)) \to \\
 2\dot{X}\frac{X}{Y} + \frac{2\dot{Y}Y}{Y} = \frac{-12\cos(3t)}{Y} \to \\
 \frac{X}{Y} = \left( \frac {-12\cos(3t)}{Y} -2\dot{Y} \right).\frac{1}{2\dot{X}}
$$
where $$
  \dot{X} = -2\sin(t) -2\cos(2t) \\
  \dot{Y}=  +2\cos(t) +2\sin(2t)$$
so
$$ \frac{X}{Y} = \left( \frac {-12\cos(3t)} {2\cos(t) +2\sin(2t)} -2(2\cos(t) +2\sin(2t)) \right).\frac{1}{2(-2\sin(t) -2\cos(2t))} $$
I manipulated the values and didn't found nothing that could be replaced by $\sin(3t)$. I would be grateful if someone find a relationship.
 [/INST] $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{{\rm X}\pars{t} = 2\cos\pars{t} - \sin\pars{2t}\,,\quad 
     {\rm Y}\pars{t} = 2\sin\pars{t} - \cos\pars{2t}}$

$$
{\rm Y}\pars{t} = 2\sin\pars{t} - \cos\pars{2t}
= 2\sin\pars{t} - \bracks{1 - 2\sin^{2}\pars{t}}
= 2\bracks{\sin\pars{t} + \half}^{2} - {3 \over 2}
$$
$$
\sin\pars{t} = \pm\,{\root{2{\rm Y}\pars{t} + 3} - 1 \over 2}\tag{1}
$$

$$
{\rm X}\pars{t} = 2\cos\pars{t} - \sin\pars{2t}
= 2\cos\pars{t} - 2\sin\pars{t}\cos\pars{t}
= 2\cos\pars{t}\bracks{1 - \sin\pars{t}}
$$
$$
\cos\pars{t} = \half\,{{\rm X}\pars{t} \over 1 - \sin\pars{t}}
\quad\mbox{and}\ \pars{~\mbox{see Eq.}\ \pars{1}~}\quad
\left\lbrace%
\begin{array}{rclcl}
1 - \sin\pars{t} &= & {3 - \root{2{\rm Y}\pars{t} + 3} \over 2} & \mbox{if} & +
\\
1 - \sin\pars{t} &= & {1 + \root{2{\rm Y}\pars{t} + 3} \over 2} & \mbox{if} & -
\end{array}\right.
$$
$$
\cos\pars{t}
=
\left\lbrace%
\begin{array}{lcl}
{{\rm X}\pars{t} \over 3 - \root{2{\rm Y}\pars{t} + 3}} & \mbox{if} & +
\\
{{\rm X}\pars{t} \over 1 + \root{2{\rm Y}\pars{t} + 3}} & \mbox{if} & -
\\[2mm]
&&\mbox{See}\ \pars{1}\ \mbox{for the}\ \pm\ \mbox{signs meaning.}
\end{array}\right.
$$

With $\pars{1}$ and $\pars{2}$ and the identity
$\cos^{2}\pars{t} + \sin^{2}\pars{t} = 1$ we get:
$$\color{#0000ff}{\large\left\lbrace%
\begin{array}{lclcl}
\pars{{\rm X} \over 3 - \root{2{\rm Y} + 3}}^{2}
+
\pars{\root{2{\rm Y} + 3} - 1 \over 2}^{2}
& = & 1 & \mbox{if} & +
\\[2mm]
\pars{{\rm X} \over1 + \root{2{\rm Y} + 3}}^{2}
+
\pars{\root{2{\rm Y} + 3} - 1 \over 2}^{2} & = & 1 & \mbox{if} & -
\end{array}\right.}
$$
See $\pars{1}$ for the $\pm$ signs meaning.

</s>