How do you factor and simplify sin^4x-cos^4x?

Aug 29, 2016

$\left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$

Explanation:

Factorizing this algebraic expression is based on this property:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Taking ${\sin}^{2} x = a$ and ${\cos}^{2} x = b$ we have :

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

Applying the above property we have:

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

Applying the same property on${\sin}^{2} x - {\cos}^{2} x$

thus,

${\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2}$
$= \left(\sin x - C o s x\right) \left(\sin x + \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x\right)$

Knowing the Pythagorean identity, ${\sin}^{2} x + {\cos}^{2} x = 1$ we simplify the expression so,

${\left({\sin}^{2} x\right)}^{2} - {\left({\cos}^{2} x\right)}^{2}$
$= \left(\sin x - C o s x\right) \left(\sin x + \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x\right)$
$= \left(\sin x - \cos x\right) \left(\sin x + \cos x\right) \left(1\right)$
$= \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$

Therefore,
${\sin}^{4} x - {\cos}^{4} x = \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)$

Apr 18, 2017

= - cos 2x

Explanation:

${\sin}^{4} x - {\cos}^{4} x = \left({\sin}^{2} x + {\cos}^{2} x\right) \left({\sin}^{2} x - {\cos}^{2} x\right)$
Reminder:
${\sin}^{2} x + {\cos}^{2} x = 1$, and
${\cos}^{2} x - {\sin}^{2} x = \cos 2 x$
Therefore:
${\sin}^{4} x - {\cos}^{4} x = - \cos 2 x$