# How do you simplify \sqrt { ( 1- \cos \theta ) ( 1+ \cos \theta ) }?

May 20, 2018

$\pm \sin \theta$

#### Explanation:

$\sqrt{\left(1 - \cos \theta\right) \left(1 + \cos \theta\right)}$

Recall: (1-costheta)(1+costheta) is an example of the difference of 2 squares. The general form is $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

= $\sqrt{1 - {\cos}^{2} \theta}$

Recall: ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
so, $1 - {\cos}^{2} \theta = {\sin}^{2} \theta$

= $\sqrt{{\sin}^{2} \theta}$

When you are square rooting a number, it can be positive or negative. Why?

Here's an example

${2}^{2} = 4$ and ${\left(- 2\right)}^{2} = 4$
If you square a negative or positive number, its result will always be positive unless its a complex number.

= $\pm \sin \theta$

May 20, 2018

$\sqrt{\left(1 - \cos \left(\theta\right)\right) \left(1 + \cos \left(\theta\right)\right)} = \textcolor{red}{\left\mid \sin \left(\theta\right) \right\mid}$

#### Explanation:

Based on the Pythagorean Theorem we know (or maybe you just simply remember)
$\textcolor{w h i t e}{\text{XXX}} {\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$
or
$\textcolor{w h i t e}{\text{XXX}} 1 - {\cos}^{2} \left(\theta\right) = {\sin}^{2} \left(\theta\right)$

$\left(1 - a\right) \left(1 + a\right) = 1 - {a}^{2}$
or, in this case
$\left(1 - \cos \left(\theta\right)\right) \left(1 + \cos \left(\theta\right)\right) = 1 - {\cos}^{2} \left(\theta\right)$
which we have already noted is the same as
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXX}} = {\sin}^{2} \left(\theta\right)$

So $\sqrt{\left(1 - \cos \left(\theta\right)\right) \left(1 + \cos \left(\theta\right)\right)} = \sqrt{{\sin}^{2} \left(\theta\right)}$

and since the square root symbol always implies the positive root
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXX}} = \left\mid \sin \left(\theta\right) \right\mid$

May 25, 2018

I'm only answering to resolve the conflict between the two other answers.

As written Alan is correct,

$\setminus \sqrt{\left(1 - \cos \theta\right) \left(1 + \cos \theta\right)} = \sqrt{1 - {\cos}^{2} \theta} = \sqrt{{\sin}^{2} \theta} = | \sin \theta |$

That's because the radical sign applied to a real number refers to the principal square root which is the positive square root, thus the absolute value is correct.

But Lucy's answer is really the one that arises in practice. We have sine or cosine and want the other one:

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

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

$\sin \theta = \pm \sqrt{1 - \setminus {\cos}^{2} \theta}$

It's hard to think clearly about the signs and the quadrants. This is just one more unnecessarily difficult thing about the way we do and teach trigonometry.

An angle is a relation between two rays. Rays are complicated. We could have set up trig so our essential measure was between two lines and avoided all the quadrant stuff.

Geometry is actually more properly done at the squared level. The basic measure of the spread between lines is ${\sin}^{2} \theta$ which becomes the replacement for angle. Instead of length we use squared length, basically area. The Pythagorean Theorem becomes $C = A + B .$ These observations form Norm Wildberger's Rational Trigonometry .