Question

If $A+B+C=2S$, then prove that
$\cos (S-A)+\cos (S-B)+\cos (S-C)+\cos S=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

Answer 1

A+B+C=25 (Given)

Also, $A+B+C=\pi$ (Sum of angles of triangle)

L.H.S $\Rightarrow \cos (s-A)+\cos (s-B)+\cos (s-C)\Rightarrow \cos (90°-A)+\cos (90°-B)+\cos (90°-C)\Rightarrow \sin A+\sin B+\sin C$

$\Rightarrow 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\sin C$

$[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$

$\Rightarrow 2\cos \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+2\sin \frac{C}{2}\cos \frac{C}{2}\Rightarrow 2\cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right]$

$[\because \sin \frac{C}{2}=\cos \left(\frac{A+B}{2}\right)]$

$\Rightarrow 2\cos \frac{C}{2}\times 2\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\Rightarrow 4\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

L.H.S=R.H.S

Hence, Proved.