I need to prove that the curve defined by the polar equation $r=\sin(2\theta)$ is the affine variety $\textbf{V}((x^2+y^2)^3-4x^2y^2)$. First I showed that any point on the curve defined by $r=\sin(2\theta)$ is in the affine variety by simple algebra.
The converse seems to be similar, but the book says to be careful since $r$ can be negative in $r=\sin(2\theta)$. I've looked over my solution and can't see why that might complicate things. Here's what I did:
Let $(x,y)\in \textbf{V}((x^2+y^2)^3-4x^2y^2)$. Then
\begin{align*} &(x^2+y^2)^3-4x^2y^2=0\\ &\Rightarrow (r^2)^3-4(r^2\cos^2(\theta))(r^2\sin^2(\theta))=0\\ &\Rightarrow r^6-r^4(2\cos(\theta)\sin(\theta))^2=0\\ &\Rightarrow r^2-\sin(2\theta)=0\\ &\Rightarrow r=\sin(2\theta) \end{align*}
Is what I did valid, or did I miss something?
Thanks!