How do you find the exact values cos (pi/6) using the special triangles?

Jul 15, 2018

As below.

Explanation:

${\left(\frac{\pi}{6}\right)}^{c} = \left(\frac{\pi}{6}\right) \cdot \left(\frac{180}{\pi}\right) = {30}^{\circ}$

$\cos {\left(\frac{\pi}{6}\right)}^{c} = \cos {\left(30\right)}^{\circ}$

In the above special right triangles,

$\text{Angles are " 30^@, 60^@, 90^@ " and the sides are } a , a \sqrt{3} , 2 a$

$\cos \theta = \text{ Adj. side / Hypotenuse} = \frac{A B}{A C}$

$\cos 30 = \frac{\cancel{a} \sqrt{3}}{2 \cancel{a}} = \frac{\sqrt{3}}{2}$