# Question c2afe

Feb 14, 2018

$\cos \left(\frac{- 19 \pi}{6}\right)$ as color(purple)(cos(-theta) = cos theta

$= \cos \left(\frac{19 \pi}{6}\right)$

$= \cos \left(2 \pi + \frac{5 \pi}{6}\right)$

$= \cos \left(\frac{5 \pi}{6}\right)$ as color(purple)(cos(2npi+theta)=costheta#

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

Feb 14, 2018

The result is $\frac{\sqrt{3}}{2}$.

#### Explanation:

Since the cosine function has a period of $2 \pi$ (which means it repeats itself every $2 \pi$ units), we can add or subtract any multiples of $2 \pi$ from the inside of the parentheses to make it easier to compute:

$\textcolor{w h i t e}{=} \cos \left(- \frac{19 \pi}{6}\right)$

$= \cos \left(- \frac{19 \pi}{6} + 4 \pi\right)$

$= \cos \left(- \frac{19 \pi}{6} + \frac{24 \pi}{6}\right)$

$= \cos \left(- \frac{19 \pi}{6} + \frac{24 \pi}{6}\right)$

$= \cos \left(\frac{5 \pi}{6}\right)$

Here's that rotation on our unit circle:

We know that in this triangle with the ${30}^{\circ}$ reference angle that the cosine ratio (adjacent over hypotenuse) is $\frac{\frac{\sqrt{3}}{2}}{1}$, or just $\frac{\sqrt{3}}{2}$.