prove that the rose (in the polar plane) has $2n$ "petals" when $n$ is even

Question

prove that the rose $r=\cos(n\theta)$ (in the polar plane) has $2n$ "petals" when $n$ is even. How can I start this demonstration? I would appreciate your help

Answer 1

You should first define a "petal". That alone will make the question much easier to think about. I am guessing the definition will lead you to consider rays from the origin which don't intersect your curve (except at origin), but don't take my word for it...

Answer 2

a) We can define a petal in polar curves $\sin n\theta$ and $\cos n\theta$ as a curve that begin from origin and return to it while it wend one time.

b) $\displaystyle\cos n(2k+1)\frac{\pi}{2n}=0$ for even $n$ and $k\in\Bbb Z$.

c) Every petal for $\displaystyle\cos n(2k+1)\frac{\pi}{2n}=0$ is in interval $\displaystyle \Big[(2k_1+1)\frac{\pi}{2n},(2k_2+1)\frac{\pi}{2n}\Big]$ for $k_1,k_2\in\Bbb Z$.

d) Total of petals will be $$(2k_1+1)\frac{\pi}{2n}=2\pi+(2k_2+1)\frac{\pi}{2n}$$ so $k_2-k_1=2n$ is total petals.

Answer 3

Given $r=\cos(n\theta)$, we have $$ \begin{align} x(\theta)&=\cos(n\theta)\cos(\theta)\\ y(\theta)&=\cos(n\theta)\sin(\theta) \end{align} $$ Note that since $n$ is even, $$ \begin{align} \frac{\mathrm{Im}\left((x+iy)^{n/2}\right)}{\mathrm{Re}\left((x+iy)^{n/2}\right)} &=\frac{\cos^{n/2}(n\theta)\,\mathrm{Im}\left(\cos\left(\frac n2\theta\right)+i\sin\left(\frac n2\theta\right)\right)}{\cos^{n/2}(n\theta)\,\mathrm{Re}\left(\cos\left(\frac n2\theta\right)+i\sin\left(\frac n2\theta\right)\right)}\\ &=\tan\left(\frac n2\theta\right) \end{align} $$ is a function of $x$ and $y$. Likewise, $$ \begin{align} \cos(n\theta) &=\frac{1-\tan^2\left(\frac n2\theta\right)}{1+\tan^2\left(\frac n2\theta\right)}\\ &=\frac{\mathrm{Re}\left((x+iy)^{n/2}\right)^2-\mathrm{Im}\left((x+iy)^{n/2}\right)^2}{|x+iy|^n} \end{align} $$ is also a function of $x$ and $y$, unless $|x+iy|=|\cos(n\theta)|=0$. Thus, $$ \begin{align} \tan\left(\frac\theta2\right) &=\frac{\sin(\theta)}{1+\cos(\theta)}\\ &=\frac{y}{\cos(n\theta)+x} \end{align} $$ is also a function of $x$ and $y$. That is, $\theta\mapsto(x,y)$ is injective from $[0,2\pi)$ except where $\cos(n\theta)=0$.

The "tip" of a petal would be when $\frac{\mathrm{d}r}{\mathrm{d}\theta}=0$, that is when $n\sin(n\theta)=0$. This happens at each multiple of $\frac\pi n$. Since $\cos(n\theta)=\pm1$ at each of these $2n$ points, there must be $2n$ petals.