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prove this identity : $$\sin(x+y)\sin(x-y)=\sin^2 x - \sin^2 y$$ I tried solving it with additional formulas but I can't get the right answer. I get $$\sin^2 x \cos^2 y-\cos^2 x \sin^2 y$$

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marked as duplicate by lab bhattacharjee, 6005, user127.0.0.1, AlexR, Umberto P. Feb 6 '14 at 18:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you want a geometrical proof or just a proof using some common facts like $\sin^2+\cos^2=1$? – Ragnar Feb 6 '14 at 9:27
    
I think common facts – dona12 Feb 6 '14 at 9:28
    
Do you know the formulas for $\sin(x\pm y)$? – Martín-Blas Pérez Pinilla Feb 6 '14 at 9:29
up vote 2 down vote accepted

$\sin^2 x \cos^2 y-\cos^2 x \sin^2 y=\sin^2 x(1-\sin^2 y) -(1-\sin^2 x) \sin ^2 y$

$=\sin^2 x -\sin^2 x\sin^2y -\sin ^2 y + \sin^2x\sin^2y$

$=\sin^2 x - \sin ^2 y$

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Use the identity $$\sin(x\pm y)=\sin x\cos y\pm \sin y\cos x$$ and we can get \begin{align*} LHS&=\sin(x+y)\sin(x-y)\\ &=(\sin x\cos y+\cos x\sin y)(\sin x\cos y-\cos x\sin y)\\ &=\sin^2x\cos^2y-\cos^2x\sin^2y \\ \end{align*}

See if you can take it from here using the identity $$\sin^2x+\cos^2x=1.$$

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