How do you convert $r =4sin(2θ)$ to rectangular form?

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Answer 1 · 2016-09-25T05:17:27

$(x^2+y^2)^3=64x^2y^2$

Use the conversion equation $r(cos theta, sin theta)=(x, y)$ that

gives $r = sqrt(x^2+y^2) >=0$ , for the principal square root and

$sin 2theta = 2 sin theta cos theta = (2xy)/r^2=(2xy)/(x^2+y^2)$ .

So, after rationalization, the given polar equation becomes

$(x^2+y^2)^3=64x^2y^2$

Some nuances for the interested readers:

If r is allowed to be negative, there are four petals in the four

quadrants, from the center r = 0, for $theta in$ [0, pi]#.

For strictly $r >=0,$ there are just two petals, in $Q_1 and Q_3$ ..

Interestingly, as the period of r is $pi$ , the second is

contributed by this periodicity. .A Table for $theta in [0, 2pi]$ would

reveal all this.

I welcome a graphical depiction of these aspects.

How do you convert r =4sin(2θ)r =4sin(2θ) to rectangular form?