Question #c2afe

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Question #c2afe

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Answer 1 · 2018-02-14T03:19:54

$cos (\frac{- 19 π}{6})$ $cos ((-19pi)/6)$ as $cos (- θ) = cos θ$ $color(purple)(cos(-theta) = cos theta$

$= cos (\frac{19 π}{6})$ $=cos ((19pi)/6)$

$= cos (2 π + \frac{5 π}{6})$ $=cos (2pi + (5pi)/6)$

$= cos (\frac{5 π}{6})$ $=cos ((5pi)/6)$ as $cos (2 n π + θ) = cos θ$ $color(purple)(cos(2npi+theta)=costheta$

$= - \frac{\sqrt{3}}{2}$ $=-sqrt3/2$

Answer 2 · 2018-02-14T03:52:55

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

Explanation:

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

$= cos (- \frac{19 π}{6})$ $color(white)=cos(-(19pi)/6)$

$= cos (- \frac{19 π}{6} + 4 π)$ $=cos(-(19pi)/6+4pi)$

$= cos (- \frac{19 π}{6} + \frac{24 π}{6})$ $=cos(-(19pi)/6+(24pi)/6)$

$= cos (\frac{5 π}{6})$ $=cos((5pi)/6)$

Here's that rotation on our unit circle:

![https://www.desmos.com/calculator](useruploads.socratic.org)

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

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Question #c2afe

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