Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 634: 70

Answer

(a) $cosθ$=$\frac{e^{iθ} + e^{-iθ}}{2}$=$coshiθ$ (b) $i.sinθ$=$\frac{e^{iθ} - e^{-iθ}}{2}$=$sinhiθ$

Work Step by Step

We know that $e^{iθ}$=$cosθ+isinθ$ and $e^{-iθ}$=$cosθ-isinθ$ Using these formulas (a) $e^{iθ}$ + $e^{-iθ}$=$(cosθ +isinθ)$ + $(cosθ -isinθ)$=$2cosθ$ $cosθ$=$\frac{e^{iθ} + e^{-iθ}}{2}$=$coshiθ$ (b) $e^{iθ}$ $-$ $e^{-iθ}$=$(cosθ +isinθ)$ $-$ $(cosθ -isinθ)$=$2i .sinθ$ $i.sinθ$=$\frac{e^{iθ} - e^{-iθ}}{2}$=$sinhiθ$
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