<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] You may manipulate the relations $$x(t) = 2\cos(t) - \sin(2t)  \\ 
   y(t) = 2\sin(t) - \cos(2t)$$ many times to get the right answer, but plotting it by Maple:

and having the whole shape in my mind I could find the right connecting relation here. Note that we need a proper transformation to make the plot skew. 
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