# Question #32396

May 18, 2016

The equation as given is invalid.
See below for a proof that assumes the exponents were omitted and the equation should be:
$\sin \left(x\right) - \sin \left(x\right) {\cos}^{2} \left(x\right) = {\sin}^{3} \left(x\right)$

#### Explanation:

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

factoring:
$\textcolor{w h i t e}{\text{XXX}} = \sin \left(x\right) \cdot \left(1 - {\cos}^{2} \left(x\right)\right)$

Since ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
$\textcolor{w h i t e}{\text{XXX}}$we know this as a standard identity
$\textcolor{w h i t e}{\text{XXX}}$or we could develop it easily from Pythagorean theorem and definition of $\sin$ and $\cos$

$\sin \left(x\right) \cdot \left(\textcolor{red}{1 - {\cos}^{2} \left(x\right)}\right)$
$\textcolor{w h i t e}{\text{XXX}} = \sin \left(x\right) \cdot \textcolor{red}{{\sin}^{2} \left(x\right)}$

$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{3} \left(x\right)$