Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises: 30

Answer

$\sin$(-390) = -$\frac{1}{2}$ $\cos$(-390) = $\frac{\sqrt3}{2}$ $\tan$(-390) = -$\frac{\sqrt3}{3}$ $\cot$(-390) = -$\sqrt3$ $\csc$(-390) = -2 $\sec$(-390) = $\frac{2\sqrt3}{3}$

Work Step by Step

-390$^{\circ}$ We must first fine the coterminal angle: -390$^{\circ}$ + 360$^{\circ}$ = -30$^{\circ}$ -30 is in Quadrant IV. Therefore all the trigonometric functions are negative with the exception of $\cos$ and $\sec$ $\sin$(-30) = -$\frac{1}{2}$ $\cos$(-30) = $\frac{\sqrt3}{2}$ $\tan$(-30) = -$\frac{\sqrt3}{3}$ $\cot$(-30) = -$\sqrt3$ $\csc$(-30) = -2 $\sec$(-30) = $\frac{2\sqrt3}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.