Question

If A + B + C = $\pi$, then prove that,
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}=1+4sin\left[\frac{\pi -A}{4}\right]sin\left[\frac{\pi -B}{4}\right]sin\left[\frac{\pi -C}{4}\right]$

Answer 1

$\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}$

$=2\sin \frac{A+B}{4}\cos \frac{A-B}{4}+\sin \frac{c}{2}$

$=2\sin \frac{A+B}{4}\cos \frac{A-B}{4}+1-2{\sin }^2\frac{A+B}{4}$

$[\because A+B+C=\Pi ]$

$[\frac{C}{2}=\frac{\Pi }{2}-(\frac{A}{2}+\frac{B}{2})]$

$\therefore$ $1+2\sin (\frac{\Pi }{4}-\frac{C}{4})[\cos \left(\frac{A-B}{4}\right)-\cos (\frac{\Pi }{2}-\left(\frac{A+B}{4}\right))]$

$\Rightarrow 1+2\sin (\frac{\Pi }{4}-\frac{C}{4})2\sin (\frac{\Pi }{4}-\frac{A}{4})\sin (\frac{\Pi }{4}-\frac{B}{4})$

$\Rightarrow 1+4\sin \frac{\Pi -C}{4}\sin \frac{\Pi -A}{4}\sin \frac{\Pi -B}{4}$

(proved)