Question

If $\sin \theta +\sin \varphi =\sqrt{3}(\cos \varphi -\cos \theta ),$ then the value of $\sin 3\theta +\sin 3\varphi$ is
$(\text{where}\theta \in (0,{90}^{\circ }),\varphi \in (0,{90}^{\circ }\left)\right)$

Answer 1

Given: $\sin \theta +\sin \varphi =\sqrt{3}(\cos \varphi -\cos \theta )$
$\Rightarrow \frac{\sin \theta +\sin \varphi }{\cos \varphi -\cos \theta }=\sqrt{3}\Rightarrow \frac{2\sin \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)}{-2\sin \left(\frac{\theta +\varphi }{2}\right)\sin \left(\frac{\varphi -\theta }{2}\right)}=\sqrt{3}\Rightarrow \cot \left(\frac{\theta -\varphi }{2}\right)=\sqrt{3}\Rightarrow \frac{\theta -\varphi }{2}=\frac{\pi }{6}(\because 0<\theta ,\varphi <{90}^{\circ })\Rightarrow \theta -\varphi =\frac{\pi }{3}\Rightarrow 3\theta =\pi +3\varphi$
Taking $\sin$ both sides,
$\Rightarrow \sin \left(3\theta \right)=\sin (\pi +3\varphi )\Rightarrow \sin \left(3\theta \right)=-\sin \left(3\varphi \right)\therefore \sin \left(3\theta \right)+\sin \left(3\varphi \right)=0$