Question

In a triangle prove that ${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\cos \left(A/2\right)\cos \left(B/2\right)\cos \left(C/2\right)+\cos \left(3A/2\right)\cos \left(3B/2\right)\cos \left(3C/2\right)$.

Answer 1

$\sin 3A=3\sin A-4{\sin }^{3}A$.
$\therefore {\sin }^{3}A=\frac{1}{2}(3\sin A-\sin 3A)$.
Hence we have to find the value of
$\frac{3}{4}(\sin A+\sin B+\sin C)-\frac{1}{4}(\sin 3A+\sin 3B+\sin 3C)$
$=\frac{3}{4}\cdot 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}+\frac{1}{4}\cdot 4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}$.