Polar roses and a symmetric proof 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 $\theta\in[0,2\pi)$.
I know $=\sin \theta$ and $=\cos \theta$ are $$-leaved if $n$ is odd and $2$-leaved if $$ 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 $−$ 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 $\cos()=−\cos(+\pi)$
 and $\sin()=−\sin(+\pi)$. 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 $\sin(\theta)=−\sin((\theta+\pi))$ when $n$ is odd and $\sin()=\sin((+\pi))$ when $n$ is even (and likewise for cosine). My question, then (I suppose), is how to verbalize that the graph produces $2$ unique leaves when $$ is even and the graph is on the interval $\theta∈[0,2\pi)$.
I know $=\sin \theta$ and $=\cos \theta$ are $$-leaved if $n$ is odd and $2$-leaved if $$ 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 $−$ 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 $\cos()=−\cos(+\pi)$
 and $\sin()=−\sin(+\pi)$. 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 $\sin(\theta)=−\sin((\theta+\pi))$ when $n$ is odd and $\sin()=\sin((+\pi))$ when $n$ is even (and likewise for cosine). My question, then (I suppose), is how to verbalize that the graph produces $2$ unique leaves when $$ is even and the graph is on the interval $\theta∈[0,2\pi)$.