# What is the type of curve represented by the equation r=sin6theta?

Nov 21, 2016

$r \ge 0 \to \theta \in \left[0 , \frac{\pi}{6}\right] , \left[\frac{\pi}{3} , \frac{\pi}{2}\right] , \left[\frac{2}{3} \pi , \frac{5}{6} \pi\right] , \left[\pi , \frac{7}{6} \pi\right] , \left[\frac{4}{3} \pi , \frac{3}{2} \pi\right] \mathmr{and} \left[\frac{5}{3} \pi , \frac{11}{6} \pi\right]$ Graph is in the Cartesian frame. See explanation.

#### Explanation:

graph{(x^2+y^2)^3.5-6xy(x^4+y^4)+20x^3y^3=0 [-2 2 -1 1]} $r = \sin 6 \theta \ge 0 \to \theta \in \left[0 , \frac{\pi}{6}\right]$

The period = $\frac{2 \pi}{6} = \frac{\pi}{3}$. Yet, for $\theta \in \left(\frac{\pi}{6} , \frac{\pi}{3}\right) , r < 0.$

There is just one loop: $r \in \left[0 , 1\right]$, for $\theta \in \left[0 , \frac{\pi}{6}\right]$.

For this loop, $\theta = \frac{\pi}{12}$ is the line of symmetry

For conversion, I have used

$r = \sin 6 \theta$

$= 6 {\cos}^{5} \theta \sin \theta - 20 {\cos}^{3} \theta$

${\sin}^{3} \theta + 6 \cos \theta {\sin}^{5} \theta$

$= \frac{6 x y \left({y}^{4} + {x}^{4}\right) - 20 {x}^{3} {y}^{3}}{r} ^ 6$, where $r = \sqrt{{x}^{2} + {y}^{2}}$.

Likewise, in the same Cartesian frame, the graph is created for the

seldom cared yet grand

$r = \sin \left(\frac{\theta}{6}\right)$, with $r \ge 0$ half-period $6 \pi$.

The shape is on uniform scale. The intruded arc over the zenith is

not a part of the graph, for one period. Please ignore it. Sans this

arc, the graph is OK.
graph{y-(x^2+y^2)(1-x^2-y^2)^0.5(6-32(x^2+y^2)(1-x^2-y^2))=0[-3 3 -1.5 1.5]}