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

Apr 16, 2015

Using the angle sum and difference formula of the function sinus on the first member:

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

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

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

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

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

that is the second member.