Question

If $A+B+C=\pi$,

show that $\cos \left(A/2\right)+\cos \left(B/2\right)\cos \left(C/2\right)=4\cos \left\{\right(\pi -A\left)/4\right\}\cos \left\{\right(\pi -B\left)/4\right\}\cos \left\{\right(\pi -C\left)/4\right\}=4\cos \left\{\right(B+C\left)/4\right\}\cos \left\{\right(C+A\left)/4\right\}\cos \left\{\right(A+B\left)/4\right\}$.

Answer 1

$A+B+C=\pi$
$\therefore \left\{\right(AB\left)/4\right\}=\left\{\right(\pi -C\left)/4\right\}$
etc. for $2$nd form $\left(1\right)$
L.H.S. $=2\cos \frac{A+B}{4}\cos \frac{A-B}{4}+\sin (\frac{\pi }{2}-\frac{C}{2})$ (Note)
$=2\cos \frac{\pi -C}{4}\cos \frac{A-B}{4}+2\sin \frac{\pi -C}{4}\cos \frac{\pi -C}{4}$
$[\because \sin \theta =2\sin (\theta /2\left)\cos \right(\theta /2\left)\right]$
$=2\cos \frac{\pi -C}{4}[\cos \frac{A-B}{4}+\sin \frac{A+B}{4}]$, by $\left(1\right)$
$=2\cos \frac{\pi -C}{4}[\cos \frac{A-B}{4}+\cos (\frac{\pi }{2}-\frac{A+B}{4})]$
$=2\cos \frac{\pi -C}{4}\left[2\cos \frac{\pi -A}{4}\cos \frac{\pi -B}{4}\right]$
$=4\cos \frac{\pi -A}{4}\cos \frac{\pi -B}{4}\cos \frac{\pi -C}{4}$.