# How do you prove the identity (sin^4)x-(cos^4)x?

##### 1 Answer
Sep 3, 2015

You probably mean this one:

${\sin}^{4} x - {\cos}^{4} x = - \cos 2 x$

#### Explanation:

${\sin}^{4} x - {\cos}^{4} x$

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

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

...using the difference of squares identity:

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

Now ${\sin}^{2} x + {\cos}^{2} x = 1$, so

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

$= {\sin}^{2} x - {\cos}^{2} x$

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

$= - \cos 2 x$

...using the double angle formula for $\cos$