# How do you graph r^2=2sin(3θ)?

May 10, 2016

Thanks to Sam's updating, it is a 3-petal curve, without using non-negative r, For the three loops, choose the ranges $\theta \in \left[0 , \frac{\pi}{3}\right] , \left[\frac{2 \pi}{3} , \pi\right] \mathmr{and} \left[\frac{4 \pi}{3} , \frac{5 \pi}{3}\right]$..

#### Explanation:

Upon insertion of the graph by Sam, I have duly revised my answer,

to make the answer almost perfect.. ${r}^{2} = 2 \sin 3 \theta \ge 0$. Maximum r = 2.

Choose the range $\theta \in \left[0 , \frac{\pi}{3}\right]$,

so that $3 \theta \in \left[0 , \pi\right]$ is in the first two quadrants, where sine is

positive. It is similar, for the the other two petals, in the respective

ranges for $\theta$....

For every $\sin 3 \theta , r = \pm \sqrt{2 \sin 3 \theta}$

(If negative r is allowed, there would be six loops altogether, keeping $2 \sin 3 \theta \ge 0$. For the range $\theta \in \left[0 , \frac{\pi}{3}\right]$, one loop traced by the (+r)-hand, is in the first quadrant, and its mirror image about the pole, traced with the (-r)-hand, is in the third quadrant. If the convention is to treat r as modulus (length) of a vector, r$\ge$ 0, the mirror images are not done.)

As the period of $2 \sin 3 \theta$ is $\frac{2 \pi}{3}$ the petals are retraced

when the ends of the intervals for $\theta$ are increased by integer

multiple $n \left(\frac{2 \pi}{3}\right)$ of $\frac{2 \pi}{3}$...