University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 180: 5

Answer

a) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$ b) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(-\frac{1}{\sqrt 2})=(\frac{3\pi}{4})$ c) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{6}$

Work Step by Step

a) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$ b) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(-\frac{1}{\sqrt 2})=(\frac{3\pi}{4})$ c) Need to caculate the value of inverse trigonometric funtion $cos$. $cos^{-1}(\frac{\sqrt 3}{2})=\frac{\pi}{6}$
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