# How do you rewrite (sin x - cos x)(sin x + cos x)? Thank you.

Apr 6, 2018

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

#### Explanation:

You're probably used to dealing with this only in quadratics, but the expression is in the difference of squares pattern

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

where $\textcolor{red}{a = \sin x}$ and $\textcolor{b l u e}{b = \cos x}$

We can just plug in our values for $a$ and $b$ into our difference of squares expression, and we get:

${\textcolor{red}{\left(\sin x\right)}}^{2} - {\textcolor{b l u e}{\left(\cos x\right)}}^{2}$

Which can be rewritten as

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

If you don't believe me, we can FOIL this expression to make sure:

With FOIL, we multiply the first, outside, inside and last terms and add the result. Thus, we have:

• First terms: $\sin x \cdot \sin x = \textcolor{red}{\sin {x}^{2}}$
• Outside terms: $\sin x \cdot \cos x = \sin x \cos x$
• Inside terms: $\sin x \cdot - \cos x = - \sin x \cos x$
• Last terms: $- \cos x \cdot \cos x = - \textcolor{b l u e}{\cos {x}^{2}}$

Now we have

${\textcolor{red}{\left(\sin x\right)}}^{2} + \cancel{\sin x \cos x - \sin x \cos x} - {\textcolor{b l u e}{\left(\cos x\right)}}^{2}$

The middle terms obviously cancel out, and we can rewrite this as

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

The key realization is that the original expression in question was in a difference of squares pattern.

If you have a $\left(a - b\right) \left(a + b\right)$ pattern, you can always use the difference of squares identity to quickly simplify it.

Hope this helps!

Apr 6, 2018

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