# Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|

• Dec 10th 2008, 04:47 PM
SMA777
Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|
For this problem we're supposed to prove/verify that:

√((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|

Also, if someone could explain why the sinx in the right side has to be absolute value, that would be appreciated. Thank you :)
• Dec 10th 2008, 05:11 PM
skeeter
$\displaystyle \sqrt{\frac{1-\cos{x}}{1+\cos{x}}} \cdot \sqrt{\frac{1-\cos{x}}{1-\cos{x}}} =$

$\displaystyle \frac{1 - \cos{x}}{\sqrt{1 - \cos^2{x}}} =$

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

$\displaystyle \frac{1 - \cos{x}}{|\sin{x}|}$

basic algebra ... $\displaystyle \sqrt{a^2} = |a|$
• Dec 10th 2008, 06:06 PM
SMA777
Thank you, I understand the proof now!

A little clarification in this matter:
|a| = √a^2

is this because its a^2 and not (-a)^2? (So if it were (-a)^2 it'd be -|a| ?)

Thanks!
• Dec 10th 2008, 08:39 PM
Prove It
Quote:

Originally Posted by SMA777
Thank you, I understand the proof now!

A little clarification in this matter:
|a| = √a^2

is this because its a^2 and not (-a)^2? (So if it were (-a)^2 it'd be -|a| ?)

Thanks!

There are always two square roots to any number.

i.e. $\displaystyle \sqrt{a^2} = a$ or $\displaystyle \sqrt{a^2} = -a$

Notice that both answers have a "size" of $\displaystyle a$.

This "size" is denoted with an absolute value.

So $\displaystyle \sqrt{a^2} = |a|$ is correct, because it's saying the answer is something of "size" $\displaystyle a$.