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As title describes, I was wondering how I would put this into Cartesian form, from polar.

All I have is $ r = \sin(2\theta)$.

I'm not really sure what to do, I've been trying to find similar problems on the internet to no avail (at least with an explanation), nor can I figure it out myself. Any help would be great.

sardobi
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Mark
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  • If you don't know that $\sin(2\theta)=2\sin\theta\cos\theta$, then you need to review trigonometry. – Michael Hardy May 07 '14 at 01:21
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    @MichaelHardy Really? Don't you think that was a little harsh? There are nicer ways to suggest something. – homegrown May 07 '14 at 01:28
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    Suggesting reviewing trigonometry is harsh? Or saying it appears that someone doesn't know a certain fact is harsh? It does appear likely that the poster didn't know that, since otherwise that would be among the first things to come to mind. – Michael Hardy May 07 '14 at 01:56
  • Related: http://math.stackexchange.com/questions/1929330/polar-to-cartesian-form-of-r-cos2%CE%B8/1929340?noredirect=1#comment3960822_1929340 – User3910 Sep 16 '16 at 15:34
  • Inverse Question: http://math.stackexchange.com/questions/104163/writing-a-polar-equation-for-the-graph-of-an-implicit-cartesian-equation – User3910 Sep 16 '16 at 19:51
  • r=sin(4θ): http://math.stackexchange.com/questions/1930325/polar-to-cartesian-form-of-r-sin4%CE%B8 – User3910 Sep 17 '16 at 15:02

2 Answers2

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$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta \to r^3 = 2(r\sin\theta)(r\cos\theta)$. Then use:

$x = r\cos\theta$, and $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$ to finish.

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

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

$$r^3 =2xy$$

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

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

User3910
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