University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 45


$$\dfrac{\sqrt 3}{2}$$

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....; |x| \lt \infty $ We have $\\=\dfrac{\pi}{3} - \dfrac{\pi^3}{3^3 \cdot 3!} + \dfrac{\pi^5}{3^5 \cdot 5!}-....\\=\dfrac{\pi}{3}-( \dfrac{1}{3!}) (\dfrac{\pi}{3})^3+( \dfrac{1}{5!}) (\dfrac{\pi}{3})^5-....\\=\sin (\dfrac{\pi}{3})\\=\dfrac{\sqrt 3}{2}$
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