# 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 $r^2$.

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Try two things: (1) Use mathjax to format your questions. (2) Multiply out the terms in your identity. –  copper.hat Jul 17 '12 at 22:29
hint: put $r$ outside of the parenthesis $\to r^2\$! –  Raymond Manzoni Jul 17 '12 at 22:30
$(a b )^2 = a^2 b^2$. Ulfric you need to do a little work, at least. –  copper.hat Jul 17 '12 at 22:30
I shall but unfortunately I am not sure how to use mathjax –  jose Jul 17 '12 at 22:34
Ok I am not sure if I did this correctly.I did r(4sin^2Acos^2A)+r^2(cos2A)(cos2A) –  jose Jul 17 '12 at 22:35

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.

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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 $$4\sin^2 A\cos^2 A+(\cos^2 A-\sin^2 A)^2=(\cos^2 A+\sin^2 A)^2=1.$$