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 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 43


$2 \sin x \cos x= \sin 2x$ ; Verified result

Work Step by Step

The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-...........$ We need to prove the result. We have: $\sin^2 x=\dfrac{1-\cos 2x }{2}$ or, $=\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....$ $\implies \dfrac{d}{dx} (\sin^2 x)=\dfrac{d}{dx}[(\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....)]$ $\implies 2 \sin x \cos x= 2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-....$ $\implies 2 \sin x \cos x= \sin 2x$ So, the result has been proven.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.