# How do you verify the identity of: cos^3x sin^2x=(sin^2x-sin^4x)cosx?

Apr 22, 2015

From basic definitions and the Pythagorean Theorem
${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$
or
${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

First consider
(sin^2(x)-sin^4(x)

$= \left({\sin}^{2} \left(x\right)\right) \cdot \left(1 - {\sin}^{2} \left(x\right)\right)$

$= {\underbrace{{\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right)}}_{u s e d b e l o w}$

So
${\cos}^{3} \left(x\right) {\sin}^{2} \left(x\right)$

$= \left(\cos \left(x\right)\right) \cdot \left[{\underbrace{\left({\cos}^{2} \left(x\right) {\sin}^{2} \left(x\right)\right)}}_{a s a b o v e}\right]$

$= \left(\cos \left(x\right)\right) \cdot \left({\sin}^{2} \left(x\right) - {\sin}^{4} \left(x\right)\right)$

$= \left({\sin}^{2} \left(x\right) - {\sin}^{4} \left(x\right)\right) \cos \left(x\right)$