Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 34


$e^{\pi i}+1=0$

Work Step by Step

We are given that $e^{a+ib}=e^a (\cos b+i \sin b)$ Plug $a=0; b=\pi$ $e^{0+\pi i}=e^{0} (\cos \pi+i \sin \pi)$ This gives: $e^{\pi i}=-1+0$ and $e^{\pi i}+1=0$ Hence, the result has been proved.
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