Question

$\cos \alpha +\cos \beta +\cos \gamma +\cos (\alpha +\beta +\gamma )=b\cos \frac{\alpha +\beta }{2}\cos \frac{\beta +\gamma }{2}\cos \frac{\gamma +\alpha }{2}$. Find $b$

Answer 1

L.H.S. $=\cos \alpha +\cos \beta +\cos \gamma +\cos (\alpha +\beta +\gamma )$
$=(\cos \alpha +\cos \beta )[\cos \gamma +\cos (\alpha +\beta +\gamma )]$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)+2\cos \left(\frac{\alpha +\beta +\gamma +\gamma }{2}\right)\cos \left(\frac{\alpha +\beta +\gamma -\gamma }{2}\right)$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)+2\cos \left(\frac{\alpha +\beta +\gamma +\gamma }{2}\right)\cos \left(\frac{\alpha +\beta +\gamma -\gamma }{2}\right)$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)+2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha +\beta +2\gamma }{2}\right)$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\{\cos \left(\frac{\alpha -\beta }{2}\right)+\cos \left(\frac{\alpha +\beta +2\gamma }{2}\right)\}$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\{2\cos \left(\frac{\frac{\alpha -\beta }{2}+\frac{\alpha +\beta +2\gamma }{2}}{2}\right)+\cos \left(\frac{\frac{\alpha +\beta +2\gamma }{2}-\frac{\alpha -\beta }{2}}{2}\right)\}$
$=2\cos \left(\frac{\alpha +\beta }{2}\right)\left\{2\cos \left(\frac{\alpha +\gamma }{2}\right)\cos \left(\frac{\beta +\gamma }{2}\right)\right\}=4\cos \frac{\alpha +\beta }{2}\cos \frac{\beta +\gamma }{2}\cos \frac{\gamma +\alpha }{2}=$ R.H.S.
Ans: 4