Polar to cartesian form of $ r = \sin(2\theta)$
By David Jones •
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.
$\endgroup$ 62 Answers
$\begingroup$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta \to r^3 = 2(r\sin\theta)(r\cos\theta)$. Then use:
$x = r\cos\theta$, and $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$ to finish.
$\endgroup$ 1 $\begingroup$$$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$$
$\endgroup$ 1More in general
"Zoraya ter Beek, age 29, just died by assisted suicide in the Netherlands. She was physically healthy, but psychologically depressed. It's an abomination that an entire society would actively facilitate, even encourage, someone ending their own life because they had no hope. Th…"
