Start by using the substitution \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\(r^2 = 2 \sin 2\theta\)
\(r^2 = 2(2 \sin \theta \cos \theta)\)
\(r^2 = 4 \sin \theta \cos \theta\)
Then, multiply both sides by \(r^2\):
\(r^4 = 4r^2 \sin \theta \cos \theta\)
\((r^2)^2 = 4(r \sin \theta)(r \cos \theta)\)
Since \(r^2 = x^2 + y^2\), \(r \cos \theta = x\), and \(r \sin \theta = y\), we have that
\(\bf (x^2 + y^2)^2 = 4xy\)
