## Calculus 8th Edition

$sin3\theta+sin\theta=2sin2\theta$ $cos\theta$
Need to prove the identity $sin3\theta+sin\theta=2sin2\theta$ $cos\theta$ Since $sin3\theta$ can be written as: $sin3\theta=sin(2\theta+\theta)$ Use sum identity for sine. $sin3\theta=sin(2\theta+\theta)=sin2\theta cos\theta+cos2\theta sin\theta$ Thus, $sin3\theta+sin\theta=sin2\theta cos\theta+cos2\theta sin\theta+sin\theta$ $=sin2\theta cos\theta+(2cos^{2}\theta-1) sin\theta+sin\theta$ $=sin2\theta cos\theta+2cos^{2}\theta sin\theta$ $=sin2\theta cos\theta+(2 sin\theta cos\theta) cos \theta$ $=sin2\theta cos\theta+ sin2\theta cos \theta$ Hence, $sin3\theta+sin\theta=2sin2\theta$ $cos\theta$