Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding: 17

Answer

$\sqrt3$

Work Step by Step

$\cot \alpha =\cot \left( \alpha -360^o\right) \\\Rightarrow \cot 390^o=\cot \left( 390^o-360^o\right) \\=\cot 30^o \\=\dfrac {\cos 30^o}{\sin 30^o} \\=\dfrac {\cos \dfrac {\pi }{6}}{\sin \dfrac {\pi }{6}} \\=\dfrac {\dfrac {\sqrt {3}}{2}}{\dfrac {1}{2}} \\=\sqrt {3}$
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