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: 32

Answer

$\sin$(-1020) = $\frac{\sqrt3}{2}$ $\cos$(-1020) = $\frac{1}{2}$ $\tan$(-1020) = $\sqrt3$ $\cot$(-1020) = $\frac{2\sqrt3}{3}$ $\csc$(-1020) = 2 $\sec$(-1020) = $\frac{\sqrt3}{3}$

Work Step by Step

-1020$^{\circ}$ We must first fine the coterminal angle: -1020$^{\circ}$ + 360$^{\circ}$ = -660$^{\circ}$ -660$^{\circ}$ + 360$^{\circ}$ = -300$^{\circ}$ -300$^{\circ}$ + 360$^{\circ}$ = 60$^{\circ}$ -150 is in Quadrant I. Therefore all the trigonometric functions are positive. The reference angle is: $\theta$$^{1}$ = 60$^{\circ}$ $\sin$(60) = $\frac{\sqrt3}{2}$ $\cos$(60) = $\frac{1}{2}$ $\tan$(60) = $\sqrt3$ $\cot$(60) = $\frac{2\sqrt3}{3}$ $\csc$(60) = 2 $\sec$(60) = $\frac{\sqrt3}{3}$
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.