# How do you sketch one cycle of y=-3sin4theta?

Jul 22, 2018

One petal graph only, for one cyle $\theta \in \left(0 , \frac{\pi}{2}\right)$.

#### Explanation:

$r = - 3 \sin 4 \theta \ge 0 \Rightarrow \sin 4 \theta < 0$

$\Rightarrow 3 \theta \in {Q}_{1}$or ${Q}_{2} \Rightarrow \theta \in {Q}_{1}$

If cycle means period, here, the period = (2pi)/4 = pi/2.

Considering one cycle $\theta \in \left[0 , \frac{\pi}{2}\right]$,

$r \ge 0 \in \left(0 , \frac{\pi}{4}\right) \mathmr{and} r < 0 \in \left(\frac{\pi}{4} , \frac{\pi}{2}\right)$.

For creating a graph, convert to Cartesian form, usimg

$r = \sqrt{{x}^{2} + {y}^{2}} \ge 0 , r \left(\cos \theta , \sin \theta\right)$ and

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

( x^2 + y^2 )^2.5 + 12 xy( ( x^2 - y^2 ) = 0#.

Now, the Socratic graph is ready.

graph{( x^2 + y^2 )^2.5 + 12 xy ( x^2 - y^2 ) = 0[-0.1 8 -0.1 4] }

Slide the graph $\rightarrow \uparrow$, to view the other three petals.