Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 105

Answer

$ \tan\left ( \sin ^{-1}\frac{u}{\sqrt{u^2+2}} \right )=\frac{u\sqrt 2}{2} $

Work Step by Step

Let $ \theta = \sin ^{-1}\frac{u}{\sqrt{u^2+2}} $ then $ \sin\theta = \frac{u}{\sqrt{u^2+2}} $ By using trigonometric identities. $ \cos^2\theta+\sin^2\theta=1 $ Isolate $ \cos \theta $. $ \cos \theta = \sqrt{1-\sin^2 \theta} $ Substitute the value of $ \sin \theta $ . $ \cos \theta = \sqrt{1-\left (\frac{u}{\sqrt{u^2+2}} \right )^2} $ Clear parentheses. $ \cos \theta = \sqrt{1-\frac{u^2}{u^2+2} } $ Simplify. $ \cos \theta = \sqrt{\frac{u^2+2-u^2}{u^2+2} } $ $\cos \theta = \sqrt{\frac{2}{u^2+2} } $ $\cos \theta = \frac{\sqrt{2}}{\sqrt{u^2+2}} $ By using trigonometric ratios $\tan \theta = \frac{\sin \theta }{\cos \theta } $ Substitute the values of $ \sin \theta $ and $\cos \theta $. $ \tan \theta = \frac{\frac{u}{\sqrt{u^2+2}} }{\frac{\sqrt{2}}{\sqrt{u^2+2}} } $ Simplify. $\tan \theta = \frac{u}{\sqrt{2}}=\frac{u\sqrt 2}{2} $
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