# How do you factor the expression and use the fundamental identities to simplify 1-2cos^2x+cos^4x?

Oct 31, 2017

The answer is $= {\sin}^{4} x$

#### Explanation:

We need

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

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

Therefore,

$1 - 2 {\cos}^{2} x + {\cos}^{4} x = \left(1 - {\cos}^{2} x\right) - {\cos}^{2} x + {\cos}^{4} x$

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

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

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

$= {\sin}^{4} x$

Oct 31, 2017

$\textcolor{b l u e}{{\sin}^{4} x}$

#### Explanation:

This makes use of the Pythagorean identity:

${\sin}^{2} x + {\sin}^{2} x = 1$

$1 - 2 {\cos}^{2} x + {\cos}^{4} x$

${\cos}^{4} x = {\left(1 - {\sin}^{2} x\right)}^{2} = 1 - 2 {\sin}^{2} x + {\sin}^{4} x$

${\cos}^{2} x = 1 - {\sin}^{2} x$

$\therefore$

$1 - 2 \left(1 - {\sin}^{2} x\right) + 1 - 2 {\sin}^{2} x + {\sin}^{4} x$

$1 - 2 + 2 {\sin}^{2} x + 1 - 2 {\sin}^{2} x + {\sin}^{4} x$

$1 - 2 + 2 {\sin}^{2} x + 1 - 2 {\sin}^{2} x + {\sin}^{4} x = \textcolor{b l u e}{{\sin}^{4} x}$