University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 44


$\cos^2 x=1-x^2+\dfrac{x^4}{3}-\dfrac{2x^6}{45}-....$

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and the Taylor series for $\cos x $ can be defined as: $\cos x=1-\dfrac{x^2}{2}+\dfrac{ x^3}{3}-....$ Now, $\cos^2 x=\cos 2x +\sin^2 x=(1-\dfrac{(2x)^2}{2}+\dfrac{ (2x)^3}{3}-....) + (\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+....)$ $=1-\dfrac{2x^2}{2!}+\dfrac{2^3 x^4}{4!}-\dfrac{2^5 x^6}{6!}-$ $=1-x^2+\dfrac{x^4}{3}-\dfrac{2x^6}{45}-....$ Hence, $\cos^2 x=1-x^2+\dfrac{x^4}{3}-\dfrac{2x^6}{45}-....$
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