# How do you express sin^4theta - sin^3theta *cos^2theta in terms of non-exponential trigonometric functions?

Apr 30, 2016

${\sin}^{4} \theta - {\sin}^{3} \theta \cdot {\cos}^{2} \theta$

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

now
$\sin 3 \theta = 3 \sin \theta - 4 {\sin}^{3} \theta \implies {\sin}^{3} \theta = \frac{3}{4} \sin \theta - \frac{1}{4} \sin 3 \theta$
and ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$

Putting these in the main expression it becomes
$= {\sin}^{3} \theta \left(\sin \theta - {\cos}^{2} \theta\right)$
$= \left(\frac{3}{4} \sin \theta - \frac{1}{4} \sin 3 \theta\right) \left(\sin \theta - \frac{1}{2} \left(1 + \cos 2 \theta\right)\right)$
$= \left(\frac{3}{4} {\sin}^{2} \theta - \frac{1}{4} \sin 3 \theta \sin \theta - \frac{3}{8} \sin \theta - \frac{3}{8} \sin \theta \cos 2 \theta + \frac{1}{8} \sin 3 \theta - \frac{1}{4} \sin 3 \theta \cos 2 \theta\right)$

$= \left(\frac{3}{8} \left(1 - \cos 2 \theta\right) - \frac{1}{4} \sin 3 \theta \sin \theta - \frac{3}{8} \sin \theta - \frac{3}{8} \sin \theta \cos 2 \theta + \frac{1}{8} \sin 3 \theta - \frac{1}{4} \sin 3 \theta \cos 2 \theta\right)$
$= \left(\frac{3}{8} \left(1 - \cos 2 \theta\right) - \frac{1}{8} \left(\cos 2 \theta - \cos 4 \theta\right) - \frac{3}{8} \sin \theta - \frac{3}{16} \left(\sin 3 \theta - \sin \theta\right) + \frac{1}{8} \sin 3 \theta - \frac{1}{8} \left(\sin 5 \theta - \sin \theta\right)\right)$

Pl proceed for simlification