Petals in a Rose Curve

Math Trig Algebra

Question

Show there are $n$ petals in a rose curve with odd $n$ and $2n$ petals in a rose curve with even $n$.

Solution

WLOG the rose curve has the form $r=\cos n\theta$. Then there are 2 cases to get a maximum point on a petal: either $\cos n\theta =1$ or $\cos (n(\theta +\pi ))=-1$

Case 1

$$\cos n\theta =1$$

For all integers $i$:

$$n\theta =2i\pi$$ $$\theta =\frac{2i\pi }{n}$$

Case 2

$\cos (n(\theta +\pi ))=-1$ For any arbitrary integer $i$: $n(\theta +\pi )=2i\pi +1$ $\theta =\frac{(2i+1)\pi -n\pi }{n}$ For $\theta \in [0,2\pi )$ both sequences create

Even Case

Factor out $\frac{\pi }{n}$ from both sides. Case 1 has $\frac{\pi }{n}$ multiplied by an even factor, while case 2 has $\frac{\pi }{n}$ multiplied by an odd factor. Therefore the sequences are unique, creating $2n$ unique petals.

Odd Case

Take case 2: $\theta =\frac{(2i+1)\pi -n\pi }{n}$ $\theta =\frac{2i\pi +(1-n)\pi }{n}$ Since $n$ is odd, $1-n$ is even, and therefore the numerator is an even factor of $\pi$. $\theta =\frac{2i\pi }{n}$ Therefore case 2 produces the same $\theta$ as case 1, resulting in $n$ petals.