Question

$\sin \theta {\cos }^3\theta -\cos \theta {\sin }^3\theta$ is equal to ?

• A. $\frac{-1}{4}\sin \theta$
• B. $\frac{\sin 4\theta }{4}$
• C. $\frac{\cos 4\theta }{4}$
• D. $\frac{\cos 4\theta }{3}$

Answer 1

Answer: (B)

$\frac{\sin 4\theta }{4}$

$\sin \theta {\cos }^3\theta -\cos \theta {\sin }^3\theta$
$=\sin \theta \cos \theta ({\cos }^2\theta -{\sin }^2\theta )$
$=\frac{\sin 2\theta }{2}\left(\cos 2\theta \right)$
$=\frac{1}{2}\sin 2\theta \cos 2\theta$
$=\frac{1}{2}\cdot \frac{\sin 4\theta }{2}$
$=\frac{\sin 4\theta }{4}$