Question

Find the value of $2\sin 3\theta \cos \theta -\sin 4\theta -\sin 2\theta$

Answer 1

$2\sin 3\theta cos\theta -sin4\theta -\sin 2\theta$
= $2\sin 3\theta cos\theta -(sin4\theta +\sin 2\theta )$
= $2\sin 3\theta cos\theta -\left(2sin3\theta \cos theta\right)$ ($\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right))$
= $0$