Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 34



Work Step by Step

Given: $arctanx=arccos\frac{5}{13}$ Consider $arccos\frac{5}{13}=P$ $cos P=\frac{5}{13}$ Apply trigonometric identity for cosine. $cosx=\frac{adj}{hyp}=\frac{5}{13}$ Thus, $opp=\sqrt {13^{2}-5^{2}}=\sqrt {169-25}=12$ Likewise, $tanP=\frac{opp}{adj}=\frac{12}{5}$ $arctanx=P$ gives $x=tanP$ Hence, $x=\frac{12}{5}$
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