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 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 95


$sin~(arccos~u) = \sqrt{1-u^2}$

Work Step by Step

Let $~~\theta = arccos~u$ Then $~~cos~\theta = u$ $sin~\theta = \sqrt{1-cos^2~\theta}$ $sin~\theta = \sqrt{1-u^2}$ Therefore, $~~sin~(arccos~u) = \sqrt{1-u^2}$
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