Question

If $x=\sin \alpha ,y=\sin \beta ,z=\sin (\alpha +\beta )$, then $\cos (\alpha +\beta )$ is equal to

• A. $\frac{x^2+y^2+z^2}{2xy}$
• B. $\frac{x^2+y^2-z^2}{xy}$
• C. $\frac{z^2-x^2+y^2}{2xy}$
• D. $\frac{z^2-x^2-y^2}{2xy}$

Answer 1

The correct option is D $\frac{z^2-x^2-y^2}{2xy}$

$x=\sin \alpha ,y=\sin \beta ,z=\sin (\alpha +\beta )$
Now,
$z^2-x^2-y^2={\sin }^2(\alpha +\beta )-{\sin }^2\alpha -{\sin }^2\beta =\sin (\alpha +\beta +\alpha )\sin (\alpha +\beta -\alpha )-{\sin }^2\beta =\sin (2\alpha +\beta )\sin \beta -{\sin }^2\beta =\sin \beta (\sin (2\alpha +\beta )-\sin \beta )=\sin \beta \left(2\cos \left(\frac{2\alpha +\beta +\beta }{2}\right)\sin \left(\frac{2\alpha +\beta -\beta }{2}\right)\right)=2\sin \alpha \sin \beta \cos (\alpha +\beta )=2xy\cos (\alpha +\beta )$
Therefore,
$\cos (\alpha +\beta )=\frac{z^2-x^2-y^2}{2xy}$