Question

Is $\cos (A+B+C)=\cos A\cos B\cos C$, then
$8\sin (B+C)\sin (C+A)\sin (A+B)+\sin 2A\sin 2B\sin 2C=0$

Answer 1

Given that
$\cos A\cos (B+C)-\sin A\sin (B+C)=\cos A\cos B\cos C$
$\therefore \cos A\left[\cos \right(B+C)-\cos B\cos C]=\sin A\sin (B+C)$
or $\cos A(-\sin B\sin C)=\sin A\sin (B+C)$
$\therefore \sin (B+C)=\frac{-\cos A\sin B\sin C}{\sin A}$
write similar values of $\sin (C+A),\sin (A+B)$ and multiply and put $\sin A\cos A=\frac{1}{2}\sin 2A$