How would you find the exact value of the six trigonometric function of 330 degrees?

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How would you find the exact value of the six trigonometric function of 330 degrees?

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Answer 1 · 2017-01-25T05:47:14

Use trig table of special arcs and unit circle -->
$sin 330 = sin (- 30 + 360) = sin (- 30) = - sin 30 = - \frac{1}{2}$ $sin 330 = sin (-30 + 360) = sin (-30) = - sin 30 = - 1/2$
$cos (330) = cos (- 30 + 360) = cos (- 30) = cos 30 = \frac{\sqrt{3}}{2}$ $cos (330) = cos (-30 + 360) = cos (- 30) = cos 30 = sqrt3/2$
$tan 330 = \frac{sin}{cos} = (- \frac{1}{2}) (\frac{2}{\sqrt{3}}) = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$ $tan 330 = sin/(cos) = (-1/2)(2/sqrt3) = - 1/sqrt3 = - sqrt3/3$
$cot 330 = \frac{1}{tan} = - \sqrt{3}$ $cot 330 = 1/(tan) = -sqrt3$
$sec 330 = \frac{1}{cos} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$ $sec 330 = 1/(cos) = 2/sqrt3 = (2sqrt3)/3$
$csc 330 = \frac{1}{sin} = - 2$ $csc 330 = 1/(sin) = - 2$

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How would you find the exact value of the six trigonometric function of 330 degrees?

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