Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Since $A+B+C=\pi$, all three of $\sin 3A$, $\sin 3B$, $\sin 3C$ cannot be negative (that would need each angle in $\left(\frac{\pi }{3},\frac{2\pi }{3}\right)$, making the sum exceed $\pi$ ). Each is at least -1, so the sum is at least -2 . Take $A=B=\frac{\pi }{2},C=0:\sin 3A=-1,\sin 3B=-1,\sin 3C=0$. So the minimum value is -2.