Question

Express $\sin 3A+\sin 3B+\sin 3C$ as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is ${60}^o$

Answer 1

$A+B+C=\pi ,\therefore 3A+3B+3C=3\pi$
$\frac{1}{2}(3A+3B)=\frac{3}{2}\pi -\frac{3}{2}C$
$\therefore \sin \frac{1}{2}(3A+3B)=-\cos \frac{3}{2}C$
and $\cos \frac{1}{2}(3A+3B)=-\sin \frac{3}{2}C$.
L.H.S. $=2\sin \frac{3A+3B}{2}\cos \frac{3A-3B}{2}+2\sin \frac{3C}{2}\cos \frac{3C}{2}$
$=-2\cos \frac{3C}{2}\cos \frac{3A-3B}{2}+2\sin \frac{3C}{2}\cos \frac{3C}{2}$
$=-2\cos \frac{3C}{2}[\cos \frac{3A-3B}{2}-\sin \frac{3C}{2}]$
$=-2\cos \frac{3C}{2}[\cos \frac{3A-3B}{2}+\cos \frac{3A+3B}{2}]$
$=-4\cos \frac{3A}{2}\cos \frac{3B}{2}\cos \frac{3C}{2}$
If it be zero then at least one factor is zero, hence $3A/2={90}^o$ or $A={60}^o$ etc.