Question

If $\sin A+\sin B=\surd 3(\cos B-\cos A)$, then $\sin 3A+\sin 3B$ is equal to

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1: Simplify the expression

Given that, $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$

Divide both sides by $2$ we get:

$\begin{array}{rcl}\frac{1}{2}\left(\sin A+\sin B\right)& =& \frac{\sqrt{3}}{2}(\cos B-\cos A)\\ \frac{1}{2}\sin A+\frac{1}{2}\sin B& =& \frac{\sqrt{3}}{2}\cos B-\frac{\sqrt{3}}{2}\cos A\\ \cos 60°\sin A+\sin 60°\cos A& =& \cos 30°\cos B-\sin 30°\sin B\\ \sin \left(A+60°\right)& =& \cos \left(B+30°\right)\\ \sin \left(A+60°\right)& =& \sin \left({60}^o-B\right)[\cos \theta =\sin \left({90}^o-\theta \right)]\end{array}$

Step 2: Find the value of the expression

Comparing arguments we get:

$\begin{array}{rcl}A+{60}^o& =& {60}^o-B\\ A& =& -B\end{array}$

Therefore, the required expression is:

$\begin{array}{rcl}\sin 3A+\sin 3B& =& \sin 3(-B)+\sin 3B\\ & =& -\sin 3B+\sin 3B\\ & =& 0\end{array}$

Hence, option A is correct.