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: 43


$2 \sin x \cos x= \sin 2x $ (Verified)

Work Step by Step

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ $\sin^2 x=\dfrac{1-\cos 2x }{2}=\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....$ Now, $\dfrac{d(\sin^2 x)}{dx}=\dfrac{d(\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....)}{dx}$ or, $2 \sin x \cos x= 2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-....$ or, $2 \sin x \cos x= \sin 2x $ Hence, the result has been verified.
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