How do you evaluate $cos ((5pi)/6)$ ?

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How do you evaluate $cos ((5pi)/6)$ ?

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Answer 1 · 2016-04-14T16:29:17

$-sqrt3/2$

Explanation:

Trig unit circle, trig table, and property of supplement arcs -->
$cos ((5pi)/6) = cos (-pi/6 + (6pi)/6) = cos (-pi/6 + pi) =$
$= - cos (-pi/6) = - cos (pi/6) = - sqrt3/2$

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How do you evaluate $cos ((5pi)/6)$ ?

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How do you evaluate $cos ((5pi)/6)$ ?