Verify $(2r\sin A\cos A)^2+r^2(\cos^2 A-\sin^2 A)^2=r^2$ How would I verify this confounding identity:
$$(2r\sin A\cos A)^2+r^2(\cos^2 A-\sin^2 A)^2=r^2.$$
I know that
$$\sin 2\theta = 2\sin \theta \cos \theta$$
and that
$$\cos^2 \theta - \sin^2 \theta = \cos 2\theta,$$
but in my problem there is an $r$ variable, so I am not sure how to proceed and make the left side equal $r^2$.

Double angle identities do shorten the calculation, but are unnecessary.  Note that we are looking at 
$$r^2\left(4\sin^2 A\cos^2 A +(\cos^2A-\sin^2 A)^2\right).\tag{$1$}$$
The following non-trigonometric identity is easy to verify, and quite useful:
$$(x-y)^2+4xy=(x+y)^2.\tag{$2$}$$
Let $x=\cos^2 A$ and $y=\sin^2 A$. 
We get
$$4\sin^2 A\cos^2 A+(\cos^2 A-\sin^2 A)^2=(\cos^2 A+\sin^2 A)^2=1.$$
