Question

Prove that $si{n}^2\alpha +si{n}^2\beta -si{n}^2\gamma =2sin\alpha sin\beta sin\gamma ,where\alpha +\beta +\gamma =\pi$

Answer 1

$LHS=si{n}^2\alpha +si{n}^2\beta -sin2\gamma =si{n}^2\alpha +(si{n}^2\beta -si{n}^2\gamma )$
$=si{n}^2\alpha +sin(\beta +\gamma )sin(\beta -\gamma )$
$=si{n}^2\alpha +sin(\pi -\alpha )sin(\beta -\gamma )[\because \alpha +\beta +\gamma =\pi ]$
$=si{n}^2\alpha +sin\alpha sin(\beta -\gamma )$
$=sin\alpha [sin\alpha +sin(\beta -\gamma )]$
$=sin\alpha [sin(\pi -(\beta +\gamma ))+sin(\beta -\gamma )]$
$=sin\alpha [sin(\beta +\gamma )+sin(\beta -\gamma )]$
$=sin\alpha [2sin\beta cos\gamma ]=2sin\alpha sin\beta cos\gamma =RHS$