# Tangent of...

• Nov 2nd 2012, 03:18 PM
Eraser147
Tangent of...
Attachment 25527 Check the image. I know how to get the answer, but when I use the pythagorean theorem for Cosine 22.5 Why is the answer 2+Sqrt2? Why is it a PLUS instead of a minus? Please show me the steps.
• Nov 2nd 2012, 03:35 PM
MarkFL
Re: Tangent of...
What you have posted seems incomplete. You have given the value of sin(22.5°) and speak of the cosine of this angle, and your topic title is "Tangent of...". What is it you are trying to find or show?
• Nov 2nd 2012, 03:41 PM
skeeter
Re: Tangent of...
$\displaystyle \sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos{x}}{2}$

$\displaystyle \cos(45^\circ) = \frac{\sqrt{2}}{2}$

POSITIVE because $\displaystyle 22.5^\circ$ is a quadrant I angle
• Nov 2nd 2012, 04:11 PM
Eraser147
Re: Tangent of...
So when I do Pythagorean: 1 - 2$\displaystyle \sqrt{2}$ over 2 equals 2-$\displaystyle \sqrt{2}$ over $\displaystyle \sqrt{4}$, but you're saying its in quadrant one so its automatically 2+$\displaystyle \sqrt{2}$ over 4???
• Nov 2nd 2012, 05:54 PM
skeeter
Re: Tangent of...
$\displaystyle \sin^2(22.5) = \frac{1 - \frac{\sqrt{2}}{2}}{2}$

$\displaystyle \sin^2(22.5) = \frac{2 - \sqrt{2}}{4}$

$\displaystyle \sin(22.5) = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$
• Nov 2nd 2012, 06:05 PM
Eraser147
Re: Tangent of...
Yeah my hw says its + sqrt two NOT that answer.
• Nov 2nd 2012, 06:26 PM
skeeter
Re: Tangent of...
Quote:

Originally Posted by Eraser147
Yeah my hw says its + sqrt two NOT that answer.

so check it with a calculator ...
• Nov 2nd 2012, 06:28 PM
MarkFL
Re: Tangent of...
The image you posted gives the same value for sin(22.5°) that skeeter has given. If you are trying to find $\displaystyle \cos(22.5^{\circ})$ then use:

$\displaystyle \cos^2\left(\frac{45^{\circ}}{2} \right)=\frac{1+\cos(45^{\circ})}{2}$

$\displaystyle \cos^2(22.5^{\circ})=\frac{1+\frac{1}{\sqrt{2}}}{2 }=\frac{2+\sqrt{2}}{4}$

$\displaystyle \cos(22.5^{\circ})=\frac{\sqrt{2+\sqrt{2}}}{2}$