Question

If $0\le A,B,C\le \pi$ and $A+B+C=\pi$, then the minimum value of $\sin 3A+\sin 3B+\sin 3C$ is

Answer 1

Let $A\le B\le C,$

$\therefore 3A\le \pi$ $\therefore A+B+C=\pi$

$\&$ $\sin \left(3A\right)\ge 0$

So that minimum for $A=0,B=C=\frac{\pi }{2}$

Hence $\sin 3A+\sin 3B+\sin 3C$

$=\sin 0+\sin \frac{3\pi }{2}+\sin \frac{3\pi }{2}$

$=0+2\sin 3\pi 2$

$=0+2\times (-1)$

$=-2.$

Hence, the answer is $-2.$