# 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.

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It might be useful to derive $X^2+Y^2 = 5 - 4 \sin 3t$ using the sine addition formula. For the $\theta$, I see no nice expansion. However, for $\theta$ you need to pay attention to the signs of $X, Y$. – Lord_Farin Apr 24 '13 at 14:23

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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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\rm X}\pars{t} = 2\cos\pars{t} - \sin\pars{2t}\,,\quad {\rm Y}\pars{t} = 2\sin\pars{t} - \cos\pars{2t}}$

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

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

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.
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Some problems of careless signs when taking square roots notwithstanding, how does this answer the query for a parametrization in polar coordinates? – Did Dec 1 '13 at 21:30
Signs already checked !!! – Felix Marin Dec 1 '13 at 23:15
"Already checked" meaning "modified or added since last comment"? – Did Dec 2 '13 at 6:24