# Math Help - Tangent of...

1. ## Tangent of...

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.

2. ## 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?

3. ## Re: Tangent of...

$\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos{x}}{2}$

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

POSITIVE because $22.5^\circ$ is a quadrant I angle

4. ## Re: Tangent of...

So when I do Pythagorean: 1 - 2 $\sqrt{2}$ over 2 equals 2- $\sqrt{2}$ over $\sqrt{4}$, but you're saying its in quadrant one so its automatically 2+ $\sqrt{2}$ over 4???

5. ## Re: Tangent of...

$\sin^2(22.5) = \frac{1 - \frac{\sqrt{2}}{2}}{2}$

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

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

6. ## Re: Tangent of...

Yeah my hw says its + sqrt two NOT that answer.

7. ## Re: Tangent of...

Originally Posted by Eraser147
Yeah my hw says its + sqrt two NOT that answer.
so check it with a calculator ...

8. ## 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 $\cos(22.5^{\circ})$ then use:

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

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

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