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# How do you simplify (1+cos x) (1-cos x)?

Apr 12, 2016

${\sin}^{2} x$

#### Explanation:

Expand the brackets using FOIL , or the method you use.

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

using the identity $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} x + {\cos}^{2} x = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

Apr 24, 2018

${\sin}^{2} x$

#### Explanation:

The expression we have fits the difference of squares pattern $\left(a + b\right) \left(a - b\right)$, where

$a = 1$ and $b = \cos x$

We know that a difference of squares pattern is equal to ${a}^{2} - {b}^{2}$, so our expression is equal to

$\textcolor{b l u e}{1 - {\cos}^{2} x}$

This expression should look familiar. It is derived from the Pythagorean Identity

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

where we can subtract ${\cos}^{2} x$ from both sides to get what we have in blue above:

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

Thus, this expression is equal to

${\sin}^{2} x$

All we did was use the difference of squares property to our advantage, recognize that the expression we had is derived from the Pythagorean Identity, use it, and simplify.

Hope this helps!

$\setminus {\sin}^{2} x$

#### Explanation:

$\left(1 + \cos x\right) \left(1 - \setminus \cos x\right)$

$= {1}^{2} - {\left(\cos x\right)}^{2} \setminus \quad \left(\setminus \because \setminus \left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}\right)$

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

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

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

$= \setminus {\sin}^{2} x$

Jul 21, 2018

${\sin}^{2} x$

#### Explanation:

$\left(1 - \cos x\right) \left(1 + \cos x\right)$

FOIL:
$1 - {\cos}^{2} x =$

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

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

${\sin}^{2} x$