How do you graph  4cos(3theta)?

Aug 14, 2018

See explanation and graph.

Explanation:

$0 \le 4 \sqrt{{x}^{2} + {y}^{2}} = 4 \cos 3 \theta \in \left[0 , 4\right]$

The period for this cosine function $= \frac{2 \pi}{3}$.

$r \ge 0$ for $\theta \in \left[0 , \frac{\pi}{6}\right] , \left[\frac{\pi}{2} , \frac{2}{3} \pi\right] , \left[\frac{7}{6} \pi , \frac{3}{2} \pi\right] \mathmr{and}$

$\left[\frac{11}{6} \pi , 2 \pi\right]$, in one revolution $\theta \in \left[0 , 2 \pi\right]$. The first

and the last are halves, from the same loop. For subsequent

revolutions, these three three loops are redrawn.

For Socratic graphic utility that keeps off r-negative pixels, the

Cartesian form.

$\sqrt{{x}^{2} + {y}^{2}}$

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

$= 4 \left({x}^{3} / {r}^{3} - \frac{3 x {y}^{2}}{r} ^ 3\right)$, giving

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

is used. See graph.
graph{( x^2 + y^2 )^2 - 4 ( x^3 - 3 xy^2 )=0}