# How do you prove Sin^2x-sin^2y=sin(x+y)sin(x-y)?

May 24, 2016

#### Explanation:

$\sin \left(x + y\right) \sin \left(x - y\right)$

= $\left(\sin x \cos y + \cos x \sin y\right) \left(\sin x \cos y - \cos x \sin y\right)$

= ${\sin}^{2} x {\cos}^{2} y - {\cos}^{2} x {\sin}^{2} y$

= ${\sin}^{2} x \left(1 - {\sin}^{2} y\right) - \left(1 - {\sin}^{2} x\right) {\sin}^{2} y$

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

= ${\sin}^{2} x - \cancel{{\sin}^{2} x {\sin}^{2} y} - {\sin}^{2} y + \cancel{{\sin}^{2} x {\sin}^{2} y}$

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