# Question 8f7b6

##### 1 Answer
Oct 24, 2016

Given
$A + B + C = \pi$

$\implies \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$

$\implies \frac{C}{2} = \frac{\pi}{2} - \frac{A + B}{2}$

$L H S = \sin \left(\frac{A}{2}\right) + \sin \left(\frac{B}{2}\right) + \sin \left(\frac{C}{2}\right)$

$= \sin \left(\frac{A}{2}\right) + \sin \left(\frac{B}{2}\right) + \sin \left(\frac{\pi}{2} - \frac{A + B}{2}\right)$

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

$= \sin \left(\frac{A}{2}\right) + \sin \left(\frac{B}{2}\right) + 1 - {\sin}^{2} \left(\frac{A + B}{4}\right)$

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

=1+2sin((A+B)/4)(cos((A-B)/4)-sin((A+B)/4)

=1+2sin((A+B)/4)(cos((A-B)/4)-sin((A+B)/4)#

$= 1 + 2 \sin \left(\frac{\pi}{4} - \frac{C}{4}\right) \left(\cos \left(\frac{A - B}{4}\right) - \cos \left(\frac{\pi}{2} - \frac{A + B}{4}\right)\right)$

$= 1 + 4 \sin \left(\frac{\pi}{4} - \frac{C}{4}\right) \sin \left(\frac{\pi}{4} - \frac{B}{4}\right) \sin \left(\frac{\pi}{4} - \frac{A}{4}\right)$

$= 1 + 4 \sin \left(\frac{\pi}{4} - \frac{A}{4}\right) \sin \left(\frac{\pi}{4} - \frac{B}{4}\right) \sin \left(\frac{\pi}{4} - \frac{C}{4}\right)$

$= R H S$

Proved