# How do you prove sin(alpha+beta)sin(alpha-beta)=sin^2alpha-sin^2beta?

Aug 19, 2016

$= {\sin}^{2} \left(\alpha\right) - {\sin}^{2} \left(\beta\right)$

#### Explanation:

$\sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) =$
$\sin \left(\alpha\right) \cos \left(\beta\right) + \cos \left(\alpha\right) \sin \left(\beta\right) \cdot \sin \left(\alpha\right) \cos \left(\beta\right) - \cos \left(\alpha\right) \sin \left(\beta\right)$

now multiply
${\left(\sin \left(\alpha\right) \cos \left(\beta\right)\right)}^{2} + \cancel{\cos \left(\alpha\right) \sin \left(\beta\right) \sin \left(\alpha\right) \cos \left(\beta\right)} - \cancel{\cos \left(\alpha\right) \sin \left(\beta\right) \sin \left(\alpha\right) \cos \left(\beta\right)} - {\left(\cos \left(\alpha\right) \sin \left(\beta\right)\right)}^{2}$

then replace the cosine
${\sin}^{2} \left(\alpha\right) \left(1 - {\sin}^{2} \left(\beta\right)\right) - {\sin}^{2} \left(\beta\right) \left(1 - {\sin}^{2} \left(\alpha\right)\right)$
$= {\sin}^{2} \left(\alpha\right) \cancel{- {\sin}^{2} \left(\alpha\right) {\sin}^{2} \left(\beta\right)} - {\sin}^{2} \left(\beta\right) + \cancel{{\sin}^{2} \left(\alpha\right) {\sin}^{2} \left(\beta\right)}$
$= {\sin}^{2} \left(\alpha\right) - {\sin}^{2} \left(\beta\right)$