# How do you graph r=9cos12theta?

Dec 17, 2016

See explanation to create this fine 12-petal rose.

#### Explanation:

I have no polar graphic facility . Using, the Socratic Cartesian

graphics, I used

$\frac{1}{9} {\left({x}^{2} + {y}^{2}\right)}^{6.5}$

$= {y}^{12} + {x}^{12} - 66 {x}^{2} {y}^{2} \left({x}^{8} + {y}^{8}\right) + 495 {x}^{4} {y}^{4} \left({x}^{4} + {y}^{4}\right) - 924 {x}^{6} {y}^{6}$

obtained by conversion, using

$\cos 12 \theta = {\cos}^{12} \theta - 12 {C}_{2} {\cos}^{2} \theta {\sin}^{2} \theta \left({\cos}^{10} \theta + {\sin}^{10} \theta\right) + 12 {C}_{4} {\cos}^{4} \theta {\sin}^{4} \theta \left({\cos}^{8} \theta + {\sin}^{8} \theta\right) - 12 {C}_{6} {\cos}^{6} \theta {\sin}^{6} \theta + {\sin}^{12} \theta$

Of course, the 12-petal rose should have equi-sized petals that are

9 units long.

The period of $r \left(\theta\right) = 9 \cos 12 \theta$ is $\frac{2 \pi}{12} = \frac{\pi}{6} \mathmr{and}$

$2 \pi = 12 X \left(\frac{\pi}{6}\right)$, giving 12 petals.

$\theta$ size for each petal is $\frac{\pi}{12}$. For the other half of the period, $r < 0$.

graph{(1/9)(x^2+y^2)^6.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6=0[-20 20 -10 10] }

For wider loops, change r/9 to ${r}^{3} / 729$.

graph{(1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6=0[-20 20 -10 10] }

For the use by artists, I added r-negative loops, to get 24 loops.

This is not to be mistaken for 24 r-positive loops of

$r = \cos 24 \theta$.

graph{((1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6)(-(1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6)=0[-20 20 -10 10] }