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: 22

Answer

$sin$(420)$^{\circ}$ = $\frac{\sqrt3}{2}$ $cos$(420)$^{\circ}$ = $\frac{1}{2}$ $tan$(420)$^{\circ}$ =$\sqrt3$ $cot$(420)$^{\circ}$ = $\frac{\sqrt3}{3}$ $sec$(420)$^{\circ}$ = $\frac{2\sqrt3}{\sqrt3}$ $csc$(420)$^{\circ}$ = 2

Work Step by Step

420$^{\circ}$ 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. 420$^{\circ}$ - 360$^{\circ}$ = 60$^{\circ}$ $sin$(60)$^{\circ}$ = $\frac{\sqrt3}{2}$ $cos$(60)$^{\circ}$ = $\frac{1}{2}$ $tan$(60)$^{\circ}$ = $frac{\sqrt3}{1}$ =$\sqrt3$ $cot$(60)$^{\circ}$ = $\frac{1}{\sqrt3}$ = $\frac{sqrt3}{3}$ $sec$(60)$^{\circ}$ = $\frac{2}{\sqrt3}$ = $\frac{2\sqrt3}{\sqrt3}$ $csc$(60)$^{\circ}$ = $\frac{2}{1}$ = 2
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.