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.

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- Nov 2nd 2012, 03:18 PMEraser147Tangent 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 PMMarkFLRe: 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 PMskeeterRe: 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 PMEraser147Re: 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 PMskeeterRe: 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 PMEraser147Re: Tangent of...
Yeah my hw says its + sqrt two NOT that answer.

- Nov 2nd 2012, 06:26 PMskeeterRe: Tangent of...
- Nov 2nd 2012, 06:28 PMMarkFLRe: 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:

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

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

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