2015/115) If $a+b+c=\pi$

Prove that
$\sin^3 a+\sin^3 b+ \sin^3 c=3\cos(\frac{a}{2})\cos(\frac{b}{2})\cos(\frac{c}{2})$
           $+ \cos(\frac{3a}{2})\cos(\frac{3b}{2})\cos(\frac{3c}{2})$

Solution
we have $\sin(3x) = 3\sin(x) - 4\sin^3(x)$
hence
$\sin ^3 x = (\frac{3}{4}\sin x - \frac{1}{4}\sin (3x))$
so 
$\sin^3 A+\sin^3 B + \sin^3 C = \frac{3}{4}( \sin A + \sin B + \sin C) - \frac{1}{4}( \sin 3A + \sin 3B + \sin 3C)\cdots(1)$
 we have if $A + B+ C = \pi$
$\sin \frac{A+B}{2} = sin (\frac{\pi}{2}- \frac{C}{2}) = \cos \frac{C}{2}$
and $\cos \frac{A+B}{2} = \cos (\frac{\pi}{2} - \frac{C}{2}) = \sin  \frac{C}{2}$

Now
$\sin A + \sin B + \sin C= 2 \sin \frac{A+B}{2} \cos\frac{A-B}{2} + 2 \sin \frac{C}{2} \cos \frac{C}{2}$
= $2 \cos\frac{C}{2} \cos\frac{A-B}{2} + 2 \cos\frac{A+B}{2}  \cos \frac{C}{2}$
= $2 \cos   (\cos \frac{A-B}{2} + \cos \frac{A+B}{2})$
= $4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}$
hence $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2}cos \frac{B}{2} \cos \frac{C}{2}\cdots(2)$
By similar argument if
$3A+3B+3C = 3\pi$
then
$\sin \frac{3A + 3B}{2} = \sin \frac{3\pi- 3c}{2} = \cos \frac{3C}{2}$
and
$\cos \frac{3A + 3B}{2} = \cos \frac{3\pi- 3c}{2} = -\sin \frac{3C}{2}$
using this we get
$\sin 3A + \sin 3B + \sin 3C = - 4 \cos \frac{3A}{2}cos \frac{3B}{2} \cos \frac{3C}{2}\cdots(3)$
using (1), (2),(3) we get the result