# How do you create a graph of r = sin ((4theta)/3)?

Jul 5, 2018

See the 4-loop graph and the crisscrossing of loops. Also, see the idiosyncratic 5-loop graph of $r = \sin \left(\left(\frac{5}{4}\right) \theta\right)$.

#### Explanation:

Use astutely

$\left(x , y\right) = r \left(\cos \theta , \sin \theta\right)$,

$r = \sqrt{{x}^{2} + {y}^{2}} = \sin \left(\frac{4 \theta}{3}\right)$

$\sin 4 \theta = 4 \sin \theta \cos \theta \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)$

$= \sin \left(3 \left(\frac{4}{3}\right) \theta\right)$

$= 3 {\cos}^{2} \left(\left(\frac{4}{3}\right) \theta\right) \sin \left(\left(\frac{4}{3}\right) \theta\right) - {\sin}^{3} \left(\left(\frac{4}{3}\right) \theta\right)$

and arrive at the Cartesian form

$4 x y \left({x}^{2} - {y}^{2}\right) = {\left({x}^{2} + {y}^{2}\right)}^{2.5} \left(3 - 4 \left({x}^{2} + {y}^{2}\right)\right)$.

The Socratic graph:.

graph{4xy(x^2-y^2)-(x^2+y^2)^2.5(3-4(x^2+y^2))=0[-2 2 -1 1]}

Similar astute approach gives the graphs of $r = \cos \left(\left(\frac{3}{4}\right) \theta\right)$ and # r = sin ((5/4)theta).

Graph of $r = \cos \left(\left(\frac{3}{4}\right) \theta\right)$:
graph{x^3-3xy^2-(x^2+y^2)^1.5(8(x^2+y^2)^2-8(x^2+y^2)+1)=0[-2 2 -1 1]}
Idiosyncratic graph of $r = \sin \left(\left(\frac{5}{4}\right) \theta\right)$:
graph{5x^4y-10x^2y^3+y^5-4(x^2+y^2)^3(1-x^2-y^2)^0.5(1-2(x^2+y^2))=0[-4 4 -2 2]}