Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 25

Answer

$sin$(570)$^{\circ}$ = $\frac{-1}{2}$ $cos$(570)$^{\circ}$ = $\frac{-\sqrt3}{2}$ $tan$(570)$^{\circ}$ = $\frac{1}{\sqrt3}$ = $\frac{\sqrt3}{3}$ $cot$(570)$^{\circ}$ = $\frac{\sqrt3}{1}$ = $\sqrt3$ $csc$(570)$^{\circ}$ = $\frac{2}{-1}$ = -2 $sec$(570)$^{\circ}$ = $\frac{2}{-\sqrt3}$ = $\frac{2\sqrt3}{3}$

Work Step by Step

570$^{\circ}$ We can solve for the functions by using the coterminal angle. We can find the coterminal angle by adding or subtracting 360$^{\circ}$ as many times as needed. 570$^{\circ}$ - 360$^{\circ}$ = 210$^{\circ}$ Next we find the reference angle: 210$^{\circ}$ - 180$^{\circ}$ = 30$^{\circ}$ -$sin$(30)$^{\circ}$ = $\frac{-1}{2}$ -$cos$(30)$^{\circ}$ = $\frac{-\sqrt3}{2}$ $tan$(30)$^{\circ}$ = $\frac{1}{\sqrt3}$ = $\frac{\sqrt3}{3}$ $cot$(30)$^{\circ}$ = $\frac{\sqrt3}{1}$ = $\sqrt3$ -$csc$(30)$^{\circ}$ = $\frac{2}{-1}$ = -2 -$sec$(30)$^{\circ}$ = $\frac{2}{-\sqrt3}$ = $\frac{2\sqrt3}{3}$
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.