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Trigonometry
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how do you soulve identity sin(x+y)sin(x-y) = sin^2x - sin^2y?
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Write \(\sin(x+y)=A+B\) and \(\sin(x-y)=A-B\) where \(A=\sin x\cos y\) and \(B=\cos x\sin y\). Then \[\begin{align*}\Large \sin(x+y)\sin(x-y)&=A^2-B^2\\ &=\sin^2x\cos^2y-\cos^2x\sin^2y\\ &=(\sin^2x\cos^2y+\sin^2x\sin^2y)-(\sin^2x\sin^2y+\cos^2x\sin^2y)\\ &=\sin^2x-\sin^2y. \end{align*}\]
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