Convert $$$r = \tan{\left(3 \theta \right)}$$$ to rectangular coordinates

Convert $$$r = \tan{\left(3 \theta \right)}$$$ to rectangular coordinates.

Apply the formula $$$\tan{\left(3 \alpha \right)} = \frac{3 \tan{\left(\alpha \right)} - \tan^{3}{\left(\alpha \right)}}{1 - 3 \tan^{2}{\left(\alpha \right)}}$$$ with $$$\alpha = \theta$$$: $$$r = \frac{- \tan^{3}{\left(\theta \right)} + 3 \tan{\left(\theta \right)}}{1 - 3 \tan^{2}{\left(\theta \right)}}$$$.

From $$$x = r \cos{\left(\theta \right)}$$$ and $$$y = r \sin{\left(\theta \right)}$$$, we have that $$$\cos{\left(\theta \right)} = \frac{x}{r}$$$, $$$\sin{\left(\theta \right)} = \frac{y}{r}$$$, $$$\tan{\left(\theta \right)} = \frac{y}{x}$$$, and $$$\cot{\left(\theta \right)} = \frac{x}{y}$$$.

The input becomes $$$r = \frac{\frac{3 y}{x} - \frac{y^{3}}{x^{3}}}{1 - \frac{3 y^{2}}{x^{2}}}$$$.

Simplify: the input now takes the form $$$x^{3} \left(r x \left(x^{2} - 3 y^{2}\right) + y \left(- 3 x^{2} + y^{2}\right)\right) = 0$$$.

In rectangular coordinates, $$$r = \sqrt{x^{2} + y^{2}}$$$ and $$$\theta = \operatorname{atan}{\left(\frac{y}{x} \right)}$$$.

Thus, the input can be rewritten as $$$x^{3} \left(x \left(x^{2} - 3 y^{2}\right) \sqrt{x^{2} + y^{2}} + y \left(- 3 x^{2} + y^{2}\right)\right) = 0$$$.

Simplify: the input now takes the form $$$x^{3} \left(x \left(x^{2} - 3 y^{2}\right) \sqrt{x^{2} + y^{2}} - y \left(3 x^{2} - y^{2}\right)\right) = 0$$$.

$$$r = \tan{\left(3 \theta \right)}$$$A in rectangular coordinates is $$$x^{3} \left(x \left(x^{2} - 3 y^{2}\right) \sqrt{x^{2} + y^{2}} - y \left(3 x^{2} - y^{2}\right)\right) = 0$$$A.