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How do I convert the Polar Equation $r=\sin(k \theta)$ to Cartesian Equation?

I understand that $r^2=x^2+y^2$ and that $x=r\cos\theta$ and $y=r\sin\theta$, but no matter how I try to arrange them it seems that I can never cancel out both r and $\theta$.

I've looked at Writing a Polar Equation for the Graph of an Implicit Cartesian Equation and several mathematics sites on the internet and some videos by PatrickJMT on this, but my knowledge of trigonometry is limited, and I haven't been able to find any sort of a way to get a general cartesian equation for a polar rose.

Also, the Parametric equations for polar rose such that $r=\cos(k\theta)$ are $x=\cos(k t)\sin t$ and $y=\cos(k t)\cos t$.

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4 Answers 4

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Each one may be solved individually: Polar to cartesian form of $ r = \sin(2\theta)$

$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$ $$r^3 = 2(r\sin\theta)(r\cos\theta)$$

$$x = r\cos\theta$$ $$y = r\sin\theta$$

$$r^3 =2xy$$

$$r = (x^2 + y^2)^{\frac 12}$$

$$(x^2+y^2)^{\frac 32} =2xy$$ $$(x^2+y^2)^3=4x^2y^2$$

See Also: Polar to cartesian form of r=cos(2θ)

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If you expect to write it as $y=f(x)$, it will not happen as the graph is certainly not a function, neither on $x$ nor on $y$.

What you can do is express the equation parametrically. As $$ x=r\cos\theta,\ \ y=r\sin\theta, $$ you can write $$ (x(t),y(t))=(\sin kt\,\cos t,\ \sin kt\,\sin t), \ \ t\in[0,2\pi] $$

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  • $\begingroup$ I understand that it isn't a function. Is there a way to write it as multiple functions? $\endgroup$
    – martinjacobd
    Nov 19, 2012 at 18:33
  • $\begingroup$ You could. You would have two functions for each petal; $2k$ functions in total. Not sure why you would want them, though. $\endgroup$
    – Martin Argerami
    Nov 19, 2012 at 18:43
  • $\begingroup$ What would those functions look like? $\endgroup$
    – martinjacobd
    Nov 19, 2012 at 19:27
  • $\begingroup$ That's the thing. You need to solve $x=\sin kt\,\sin t$. Here: wolframalpha.com/input/… is what it looks like for the case $k=2$. $\endgroup$
    – Martin Argerami
    Nov 19, 2012 at 19:30
  • $\begingroup$ I see. You would have to restrict the domain somehow. Thanks for all of your help. $\endgroup$
    – martinjacobd
    Nov 19, 2012 at 19:40
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For k=2 squaring,

$$ r^2 = x^2 + y^2 = \sin^2 (2 \theta ) = 4 \,\sin^2 \theta \cos^2 \theta = 4 \,x^2 \, y^2 / (x^2 + y^2) $$.

For integral k such conversion using De Moivre expansion is possible . For non-integral k an infinite series results. The domain is restricted due to circular trig functions.

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The polar rose $r=\sin\left(a\theta\right)$ for odd $a$ values, can be written as $\sqrt{x^2+y^2}=\sin\left(a\left(\arcsin\left(\frac{y}{\sqrt{x^2+y^2}}\right)\right)\right)$. For even $a$ values, it can be written as $\sqrt{x^2+y^2}=±\sin\left(a\left(\arcsin\left(\frac{y}{\sqrt{x^2+y^2}}\right)\right)\right)$.

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