Question: Can anyone elaborate on my friend's comments below concerning symmetry and how this more or less proves the result(s) concerning the number of leaves?
I know $r=a\sin n\theta$ and $r=a\cos n\theta$ are $n$-leaved if $n$ is odd and $2n$-leaved if $n$ is even, but I don't really know explicitly why: it is clear with graphing and experimenting, but I was hoping for an algebraic proof, more or less.
Talking with a friend, he said there's a symmetry, which is true of course. That $-r$ is the same as adding odd multiples of $\pi$, which I also know why that is true. That cosine and sine shift signs every $\pi$ (presumably meaning $\cos(x)=-\cos(x+\pi)$ and $\sin(x)=-\sin(x+\pi)$). So scaling the argument by an odd number still gives odd $\pi$. So symmetry. Even number converts to even $\pi$'s. So lose the symmetry.
Progress: It looks as though I can write $\sin(n\theta)=-\sin(n(\theta+\pi))$ when $n$ is odd and $\sin(nx)=\sin(n(x+\pi))$ when $n$ is even (and likewise for cosine). My question, then (I suppose), is how to verbalize that the graph produces $2n$ unique leaves when $n$ is even and the graph is on the interval $\theta\in[0,2\pi)$.