Number of Petals of Odd Index Rhodonea Curve

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Theorem

Let $n$ be an odd positive integer.

Let $R$ be a rhodonea curve defined by one of the polar equations:

\(\ds r\) \(=\) \(\ds a \cos n \theta\)
\(\ds r\) \(=\) \(\ds a \sin n \theta\)


Then $R$ has $n$ petals.


Proof

Let $R$ be defined by $r = a \cos n \theta$.

The tips of each of the petals of $R$ occur when $\cos n \theta = \pm 1$.

This happens whenever $n \theta \in \set {0, \pi}$.

Thus for $0 \le \theta < 2 \pi$, the tips occur at:

$\theta \in \set {\dfrac {2 k \pi} {2 n}: k \in \set {0, 1, \ldots, 2 n - 1} }$


Let $k < n$.

Then each of $\dfrac {2 k \pi} {2 n}$ is a distinct value between $0$ and $\pi$.

Hence there are at least $n$ petals, one for each $k \in \set {0, 1, \ldots, n - 1}$.


Now consider $\theta \in \hointr \pi {2 \pi}$.

From Cosine of Angle plus Straight Angle, we have that:

$\cos \theta = -\map \cos {\theta - \pi}$

But as $\theta$ is on the opposite side of the circle to $\theta - \pi$, the points they describe are exactly the same.

Hence the petals for $\theta \ge \pi$ are in the same place as that for $\theta < \pi$.


So for $\pi \le \theta < 2 \pi$ the pattern will be traced out again.

Hence the result.

$\blacksquare$


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Sources

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