# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle pi/12?

Aug 6, 2017

$\frac{1}{2} \sqrt{2 - \sqrt{3}} , \frac{1}{2} \sqrt{2 + \sqrt{3}} , \sqrt{7 - 4 \sqrt{3}}$

#### Explanation:

$\text{since "pi/12" is in the first quadrant all of the trig. ratios}$
$\text{will be positive}$

•color(white)(x)sin(x/2)=+-sqrt((1-cosx)/2)

$\text{here } \left(\frac{x}{2}\right) = \frac{\pi}{12} \Rightarrow x = \frac{\pi}{6}$

sin(pi/12)=+sqrt((1-cos(pi/6))/2

color(white)(xxxxxx)=sqrt((1-sqrt3/2)/2

$\textcolor{w h i t e}{\times \times \times} = \sqrt{\frac{\left(2 - \sqrt{3}\right)}{4}} = \frac{1}{2} \sqrt{2 - \sqrt{3}}$
$\textcolor{b l u e}{\text{---------------------------------------------------------}}$

•color(white)(x)cos(x/2)=+-sqrt((1+cosx)/2)

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

$\textcolor{w h i t e}{\times \times \times} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{1}{2} \sqrt{2 + \sqrt{3}}$
$\textcolor{b l u e}{\text{----------------------------------------------------------}}$

•color(white)(x)tan(x/2)=+-sqrt((1-cosx)/(1+cosx))

$\tan \left(\frac{\pi}{12}\right) = + \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

$\text{multiply the numerator/denominator by } \left(2 - \sqrt{3}\right)$

$\Rightarrow \tan \left(\frac{\pi}{12}\right) = \sqrt{7 - 4 \sqrt{3}}$