# What is cos(pi/12)?

Feb 19, 2015

The answer is: $\frac{\sqrt{6} + \sqrt{2}}{4}$

Remembering the formula:

$\cos \left(\frac{\alpha}{2}\right) = \pm \sqrt{\frac{1 + \cos \alpha}{2}}$

than, since $\frac{\pi}{12}$ is an angle of the first quadrant and its cosine is positive so the $\pm$ becomes $+$,

$\cos \left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos \left(2 \cdot \frac{\pi}{12}\right)}{2}} = \sqrt{\frac{1 + \cos \left(\frac{\pi}{6}\right)}{2}} =$

$= \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

And now, remembering the formula of the double radical:

$\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{{a}^{2} - b}}{2}} \pm \sqrt{\frac{a - \sqrt{{a}^{2} - b}}{2}}$

useful when ${a}^{2} - b$ is a square,

$\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{1}{2} \left(\sqrt{\frac{2 + \sqrt{4 - 3}}{2}} + \sqrt{\frac{2 - \sqrt{4 - 3}}{2}}\right) =$

$\frac{1}{2} \left(\sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}\right) = \frac{1}{2} \left(\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{1}{2} \left(\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}\right) =$

$\frac{\sqrt{6} + \sqrt{2}}{4}$