# How do i show: \sin^2(x+y)-\sin^2(x-y)\equiv\sin(2x)\sin(2y)

How do i show:
${\mathrm{sin}}^{2}\left(x+y\right)-{\mathrm{sin}}^{2}\left(x-y\right)\equiv \mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)$
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Malaika Ridley
${\left(\mathrm{sin}\left(x+y\right)\right)}^{2}-{\left(\mathrm{sin}\left(x-y\right)\right)}^{2}={\left(\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y\right)}^{2}-{\left(\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y\right)}^{2}$
$=\left({\mathrm{sin}}^{2}{\mathrm{cos}}^{2}y+2\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y\right)-\left({\mathrm{sin}}^{2}{\mathrm{cos}}^{2}y-2\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y\right)$
$=4\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y$
$=\left(2\mathrm{sin}x\mathrm{cos}x\right)\left(2\mathrm{sin}y\mathrm{cos}y\right)$
$=\mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)$
###### Not exactly what you’re looking for?
meizhen85ulg
Known Identity
${\mathrm{sin}}^{2}a=\frac{1-\mathrm{cos}2a}{2}$
$\mathrm{cos}x-\mathrm{cos}y=-2\mathrm{sin}\left(\frac{x-y}{2}\right)\mathrm{sin}\left(\frac{x+y}{2}\right)$
Given
${\mathrm{sin}}^{2}\left(x+y\right)-{\mathrm{sin}}^{2}\left(x-y\right)$
$=\frac{1-\mathrm{cos}2\left(x+y\right)-\left(1-\mathrm{cos}2\left(x-y\right)\right)}{2}$
$=\frac{\mathrm{cos}2\left(x-y\right)-\mathrm{cos}2\left(x+y\right)}{2}$
$=\mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)$