Question

Given $\alpha +\beta +{\rm Y}=\pi$, prove that ${\sin }^2\alpha +{\sin }^2\beta -{\sin }^2{\rm Y}=2\sin \alpha \sin \beta \cos {\rm Y}$

Answer 1

$=1-{\cos }^2\alpha +{\sin }^2\beta -{\sin }^2\gamma$

$=1-[{\cos }^2\alpha -{\sin }^2\beta ]-{\sin }^2\gamma$

$=1-\cos (\alpha +\beta ).\cos (\alpha -\beta )-{\sin }^2\gamma$

$=1-\cos (\pi -\gamma ).\cos (\alpha -\beta )-1+{\cos }^2\gamma$

$=\cos \gamma .\cos (\alpha -\beta )+{\cos }^2\gamma$

$=\cos \gamma \left(\cos \right(\alpha -\beta )+\cos (\pi -(\alpha +\beta ))$

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

$=\cos \gamma (2\sin \alpha .\sin \beta )$

$=2\sin \alpha .\sin \beta .\cos \gamma$

Here use formula

$1-{\cos }^{2}a-{\sin }^{2}b=\cos (a+b).\cos (a-b)$

$2-\cos c-\cos d=\frac{2\sin +d}{2}.\sin \frac{d-c}{2}$