# How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle (7pi)/12?

May 1, 2017

$\sin \left(\frac{7 \pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\cos \left(\frac{7 \pi}{12}\right) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\tan \left(\frac{7 \pi}{12}\right) = - \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

#### Explanation:

Given: Use the half angle formulas and angle $\frac{7 \pi}{12} = {105}^{\circ}$

This angle is found in the second quadrant where sine is positive, cosine is negative and tangent is negative.

Let $\theta = \frac{7 \pi}{6} \text{ since } \frac{\frac{7 \pi}{6}}{2} = \frac{\frac{7 \pi}{6}}{\frac{2}{1}} = \frac{7 \pi}{6} \cdot \frac{1}{2} = \frac{7 \pi}{12}$

$\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \text{ sign based on quadrant}$

$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \text{ sign based on quadrant}$

$\tan \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \text{ sign based on quadrant}$

From a trigonometric circle or a $30 - 60 - {90}^{\circ}$ in the 3rd quadrant where $\frac{7 \pi}{6} = {210}^{\circ}$:

$\cos \left(\frac{7 \pi}{6}\right) = \frac{- \sqrt{3}}{2}$

$\sin \left(\frac{7 \pi}{12}\right) = \sqrt{\frac{1 - \frac{- \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}}$

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

$\cos \left(\frac{7 \pi}{12}\right) = - \sqrt{\frac{1 + \frac{- \sqrt{3}}{2}}{2}} = - \sqrt{\frac{2 - \sqrt{3}}{2} \cdot \frac{1}{2}}$

$= - \sqrt{\frac{2 - \sqrt{3}}{4}} = - \frac{\sqrt{2 - \sqrt{3}}}{2}$

$\tan \left(\frac{7 \pi}{12}\right) = - \sqrt{\frac{1 - \frac{- \sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} = - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}}$

$= - \sqrt{\frac{2 + \sqrt{3}}{\cancel{2}} \cdot \frac{\cancel{2}}{2 - \sqrt{3}}} = - \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$