# How do you graph r=2sin2x?

Jul 19, 2018

See graph and explanation.

#### Explanation:

The graph of $r = a \sin \left(n \theta - \alpha\right)$, n = 1, 2, 3, 4, ...# shows

n equal and symmetrical loops, around the pole r = 0.

The graph of $r = a \sin \left(n \theta - \alpha\right)$ is obtained by rotating

anticlockwise graph of $r = a \sin \theta$, about $\theta = 0$,

through $\alpha$.

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

$\sin 2 \theta = 2 \sin \theta \cos \theta$ and get the Cartesian form of

$r = 2 \sin 2 \theta$ as

${\left({x}^{2} + {y}^{2}\right)}^{1.5} - 4 x y = 0$.

Now, the Socratic graph is immediate.
graph{ (x^2 + y^2 )^1.5 - 4 xy = 0[-4 4 -2 2 ]}

For anticlockwise rotation, through $\alpha = \frac{p}{2}$, use

$r = 2 \sin \left(2 \left(\theta - \frac{\pi}{2}\right)\right) = - 2 \sin 2 \theta$

The graph is immediate.
graph{ (x^2 + y^2 )^1.5 + 4 xy = 0[-4 4 -2 2 ]}

For clockwise rotation, through $\alpha = \frac{\pi}{4}$, use

$r = 2 \sin \left(2 \left(\theta + \frac{\pi}{4}\right)\right) = 2 \cos 2 \theta$.

See the graph, using

$\cos 2 \theta = \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) = \frac{{x}^{2} - {y}^{2}}{r} ^ 2$

graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-4 4 -2 2]}