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I believe this question had been asked two years ago. Nevertheless, I will be inquiring again because every polar equation has a different cartesian equivalence. So, what is the exact cartesian counterpart of this specific polar equation?

$$r=sin\left[\frac{10}{4.3}\theta\right]$$

Justin Gabriel
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  • If by "exact Cartesian counterpart" you mean a single function $y=f(x)$ having the same graph as your polar equation, you're well out of luck. Have you tried graphing the polar equation to see what it looks like? If you do, you will understand how hard it will be to make sense of "exact Cartesian counterpart". – Gerry Myerson Jun 29 '20 at 09:53
  • @GerryMyerson so are saying it's impossible to convert that polar equation into cartesian equation? – Justin Gabriel Jun 29 '20 at 10:21
  • I'm saying, graph it and see for yourself. – Gerry Myerson Jun 29 '20 at 12:40
  • @GerryMyerson I already did, I don't see the benefits of graphing it again as you have told me in order for clarity to submerge into my consciousness. I went here in hopes of knowing its cartesian equation/conversion. So, short answer, is it possible or not? – Justin Gabriel Jun 29 '20 at 17:11
  • What did it look like, when you graphed it? – Gerry Myerson Jun 30 '20 at 03:08
  • @GerryMyerson it looks like a flower but more complex than the standard one. – Justin Gabriel Jun 30 '20 at 08:08
  • I think https://math.stackexchange.com/questions/240721/general-cartesian-rectangular-equation-for-polar-rose-r-sink-theta should be helpful. You can always use $r=\sqrt{x^2+y^2}$ and $\theta=\arcsin(y/\sqrt{x^2+y^2})$ to get $$\sqrt{x^2+y^2}=\sin((10/4.3)\arcsin(y/\sqrt{x^2+y^2}))$$ but I don't know whether you'll be happy with that. – Gerry Myerson Jul 01 '20 at 07:19
  • That will do @GerryMyerson. Thank you for all. – Justin Gabriel Jul 02 '20 at 01:39

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