Help with Trigonometry homework - prove an identity I need to prove the following identity:
$\sin^2 2\alpha-\sin^2 \alpha = \sin 3\alpha \sin \alpha$
What I have tried, is to work on each side of the identity. I have started with the left side:
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\sin 2\alpha+\sin \alpha)\\
&=(2\sin\alpha \cos\alpha-\sin \alpha)(2\sin \alpha \cos \alpha+\sin \alpha)\\
&=\sin\alpha(2\cos\alpha-1)\sin \alpha(2\cos \alpha+1)\\
&=\sin^2 \alpha(2\cos^2 \alpha-1)\\
&=\sin^2\alpha \cos 2\alpha
\end{align}
Then, I moved on to the right side of the identity:
\begin{align}
\sin 3\alpha \sin \alpha&=\sin(2\alpha+\alpha) \sin \alpha\\
&=\sin \alpha(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha)\\
&=\sin \alpha[2 \sin \alpha \cos^2 \alpha+\sin \alpha(\cos^2 \alpha-\sin ^2 \alpha)]\\
&=\sin^2 \alpha(3\cos^2\alpha-\sin^2\alpha)\\
&=\sin^2\alpha(4\cos^2\alpha+1)
\end{align}
I'm not sure how to continue. Any tips?
 A: There is a funny identity that goes as $$\sin(x+y) \sin(x-y) = \sin^2(x) - \sin^2(y)$$
Take $x = 2\alpha$ and $y = \alpha$ to get what you want.
Proof:
\begin{align}
\sin(x+y) \sin(x-y) & = \left( \sin(x) \cos(y) + \sin(y) \cos(x) \right) \left( \sin(x) \cos(y) - \sin(y) \cos(x) \right)\\
& = \sin^2(x) \cos^2(y) - \sin^2(y) \cos^2(x)\\
& = \sin^2(x) (1-\sin^2(y)) - \sin^2(y) (1-\sin^2(x))\\
& = \sin^2(x) - \sin^2(x) \sin^2(y) - \sin^2(y) + \sin^2(y) \sin^2(x)\\
& = \sin^2(x) - \sin^2(y)
\end{align}
EDIT:
As for the mistakes in your answer, I have highlighted the corrections in red.
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\sin 2\alpha+\sin \alpha)\\
&=(2\sin\alpha \cos\alpha-\sin \alpha)(2\sin \alpha \cos \alpha+\sin \alpha)\\
&=\sin\alpha(2\cos\alpha-1)\sin \alpha(2\cos \alpha+1)\\
&=\sin^2 \alpha( \color{red}{4\cos^2 \alpha-1})
\end{align}
\begin{align}
\sin 3\alpha \sin \alpha&=\sin(2\alpha+\alpha) \sin \alpha\\
&=\sin \alpha(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha)\\
&=\sin \alpha[2 \sin \alpha \cos^2 \alpha+\sin \alpha(\cos^2 \alpha-\sin ^2 \alpha)]\\
&=\sin^2 \alpha(3\cos^2\alpha-\sin^2\alpha)\\
&=\sin^2\alpha(\color{red}{4\cos^2\alpha-1})
\end{align}
EDIT
For more funny identities: Funny identities
A: You made an error in each side of the identity. Both sides simplify to
$$\sin^2 \alpha(4\cos^2 \alpha-1)$$
