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?

  • $\begingroup$ LaTeX suggestion: you should use $\sin$ instead of $sin$, because it looks better $\endgroup$ – Dennis Gulko Oct 30 '12 at 20:20

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}


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

  • $\begingroup$ That identity ought to be better known. I've not encountered it before, so thanks doubly for that. $\endgroup$ – Mark Bennet Oct 30 '12 at 20:48
  • 1
    $\begingroup$ @MarkBennet Go here for more funny identities: math.stackexchange.com/questions/8814/funny-identities $\endgroup$ – user17762 Oct 30 '12 at 20:52

You made an error in each side of the identity. Both sides simplify to

$$\sin^2 \alpha(4\cos^2 \alpha-1)$$

  • $\begingroup$ What errors have I done? $\endgroup$ – Alon Gubkin Oct 30 '12 at 20:27
  • $\begingroup$ @wnvl : Standard $\TeX$ usage: I changed $sin^2 \alpha(4cos^2 \alpha-1)$ to $\sin^2 \alpha(4\cos^2 \alpha-1)$. When you write a\sin b, you get $a\sin b$, with proper spacing before and after $\sin$, and $\sin$ is not italicized. $\endgroup$ – Michael Hardy Oct 30 '12 at 20:28
  • $\begingroup$ See answer Marvis for errors. $\endgroup$ – wnvl Oct 30 '12 at 20:36

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