Answer
$\sqrt{3}$
Work Step by Step
$\cot{(\theta+180^{\circ} k) } = \cot{\theta}$
$\therefore \cot{390^{\circ}} = \cot{(30^{\circ}+2 \times 180^{\circ})} = \cot{30^{\circ}}$
$\cot{30^{\circ}} = \boxed{\sqrt{3}}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 417: 17
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