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 $r^2$.
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$\begingroup$Don't worry about the "$r$"'s. Let them hang along for the ride. Think of $r$ as just a specific number, if you like.
Let's start with the left hand side of your identity: $$\tag{1} (\color{maroon}2r\color{maroon}{\sin A\cos A})^2+r^2(\color{darkgreen}{\cos^2 A-\sin^2 A})^2 $$ and apply the double angle identities you've noted. Replacing, as we may, in $(1)$ the expression $\color{maroon}{2\sin A\cos A}$ with $\sin (2A)$ and the expression $\color{darkgreen}{\cos^2A-\sin^2A}$ with $\cos( 2A)$ we obtain $$\tag{2} \bigl(r\sin( 2A)\bigr)^2 +r^2\bigl(\cos(2A)\bigr)^2 $$
Using the rule $(ab)^2=a^2b^2$ and the notations $(\sin x)^2=\sin^2x$ and $(\cos x)^2=\cos^2 x$, we can write $(2)$ as $$ \tag{3} r^2\sin^2(2A)+r^2\cos^2(2A). $$ Next, let's factor the $r^2$ term out. We can write $(3)$ as $$\tag{4} r^2\bigl(\sin^2 (2A)+\cos^2(2A)\bigr). $$ But Pythagoras tell us that $\sin^2 (2A)+\cos^2 (2A)=1$; thus the expression in $(4)$ is simply $r^2$, which is what you wanted to show.
$\endgroup$ $\begingroup$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 $$4\sin^2 A\cos^2 A+(\cos^2 A-\sin^2 A)^2=(\cos^2 A+\sin^2 A)^2=1.$$
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