University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 61

Answer

$$\sinh^{-1}\Big(-\frac{5}{12}\Big)=\ln\frac{2}{3}$$

Work Step by Step

We have $$\sinh^{-1}x=\ln(x+\sqrt{x^2+1})$$ for $x\in(-\infty,\infty)$ Therefore, $$\sinh^{-1}\Big(-\frac{5}{12}\Big)=\ln\Big(-\frac{5}{12}+\sqrt{\frac{25}{144}+1}\Big)$$ $$=\ln\Big(-\frac{5}{12}+\sqrt{\frac{169}{144}}\Big)$$ $$=\ln\Big(-\frac{5}{12}+\frac{13}{12}\Big)$$ $$=\ln\frac{2}{3}$$
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