$\begin{align} & \text{If }A,B,C\text{ are angles of a triangle show that } \\ & 3\cos \tfrac{A}{2}\cos \tfrac{B}{2}\cos \tfrac{C}{2}-\cos \tfrac{3A}{2}\cos \tfrac{3B}{2}\cos \tfrac{3C}{2}=\left( {{\sin }^{3}}A+{{\sin }^{3}}A+{{\sin }^{3}}C \right) \\\end{align}$
$\begin{align} & \text{ }3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2} \\ & \text{Note the following points} \\ & \text{sin3}\theta =3\sin \theta -4{{\sin }^{3}}\theta ----(a) \\ & \sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}--(b) \\ & \sin 3A+\sin 3B+\sin 3C=4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{\cos 3C}{2}--(c) \\ & 3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}=\frac{3}{4}(\sin A+\sin B+\sin C)-\frac{1}{4}(\sin 3A+\sin 3B+\sin 3C) \\ & 3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}=\frac{1}{4}(3\sin A+3\sin B+3\sin C-\sin 3A-\sin 3B-\sin 3C) \\ & 3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}=\frac{1}{4}\left[ (3\sin A-\sin 3A)+(3\sin B-\sin 3A)+(3\sin C-\sin 3C \right] \\ & \text{Note from (}a)\text{ that }4{{\sin }^{3}}\theta =3\sin \theta -\sin 3\theta ) \\ & 3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}=\frac{1}{4}\left[ 4{{\sin }^{3}}A+4{{\sin }^{3}}B+4{{\sin }^{3}}C \right] \\ & 3\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}={{\sin }^{3}}A+{{\sin }^{3}}B+{{\sin }^{3}}C \\\end{align}$