Question

$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$. Find the value of $\sin 3A+\sin 3B$

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$0$

$\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$
$\frac{1}{2}\sin A+\frac{1}{2}\sin B=\frac{\sqrt{3}}{2}\cos B-\frac{\sqrt{3}}{2}\cos A$
$\frac{1}{2}\sin A+\frac{\sqrt{3}}{2}\cos A=\frac{\sqrt{3}}{2}\cos B-\frac{1}{2}\sin B$

$\sin (A+{60}^0)=\cos (B+{30}^0)$
$\sin (A+{60}^0)=\sin ({60}^0-B)$
$\Rightarrow A+{60}^0={60}^0-B$

So, $\sin 3A+\sin 3B=sin(-3B)+\sin 3B=0$