Question

Prove $\frac{\sin 3\theta +\sin \theta +\sin 2\theta }{\cos 3\theta +\cos \theta +\cos 3\theta }=\tan 2\theta$

Answer 1

$\begin{array}{c}\frac{\sin 3\theta +\sin \theta +\sin 2\theta }{\cos 3\theta +\cos \theta +\cos 2\theta }=\tan 2\theta \left[\begin{array}{c}\sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{z-y}{2}\\ \cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x+y}{2}\end{array}\right]\\ LHS=\frac{2\sin \frac{3\theta +}{2}\cos \frac{3\theta }{2}+\sin 2\theta }{2\cos \frac{3\theta +\theta }{2}\cos \frac{3-\theta }{2}+\cos 2\theta }\\ LHS=\frac{2\sin 2\theta \cos \theta +\sin 2\theta }{2\cos 2\theta \cos \theta +\cos 2\theta }\\ LHS=\frac{\sin 2\theta \left(2\cos \theta +1\right)}{\cos 2\theta \left(2\cos \theta +1\right)}=\tan 2\theta =RHS\end{array}$