# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle pi/8?

Oct 13, 2017

See the explanation below

#### Explanation:

We need

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

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

$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Here,

$x = \frac{\pi}{8}$

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

${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{1 + \frac{\sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4}$

$\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$

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

${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}$

$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$

$\tan \left(\frac{\pi}{8}\right) = \sin \frac{\frac{\pi}{8}}{\cos} \left(\frac{\pi}{8}\right) = \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}}$

$= \sqrt{\frac{\left(2 - \sqrt{2}\right) \left(2 - \sqrt{2}\right)}{\left(2 + \sqrt{2}\right) \left(2 - \sqrt{2}\right)}}$

$= \sqrt{\frac{4 + 2 - 4 \sqrt{2}}{4 - 2}}$

$\tan \left(\frac{\pi}{8}\right) = \sqrt{3 - 2 \sqrt{2}}$