Question

sin A + sin B = $\sqrt{3}$ (cos B – cos A) $\Rightarrow$ sin 3A + sin 3B =

• A. \N
• B. 2
• C. 1
• D. -1

Answer 1

The correct option is A \N

$SinA+sinB=\sqrt{3}(cosB-cosA)\Rightarrow \sqrt{3}cosA+sinA=\sqrt{3}cosB-sinB$

$\Rightarrow cosA\left(\frac{\sqrt{3}}{2}\right)+sinA\left(\frac{1}{2}\right)=cosB\left(\frac{\sqrt{3}}{2}\right)-sinB\left(\frac{1}{2}\right)\Rightarrow cos(A-\frac{\pi }{6})=cos(B+\frac{\pi }{6})$

$\Rightarrow A-\frac{\pi }{6}=\pm (B+\frac{\pi }{6})\Rightarrow \text{A = - B or A - B =}\frac{\pi }{3}$

Clearly A = - B satisfies the given equation. $\therefore 3A=-3B\Rightarrow sin3A=sin(-3B)\Rightarrow sin3A+sin3B=0$