## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 16

#### Answer

$(\dfrac{x}{2})-\dfrac{1}{2^3 \cdot 3!} x^3+\dfrac{1}{2^{5} \cdot 5!}x^5-...$

#### Work Step by Step

Consider the Maclaurin Series for $\sin x$ as follows: $\sin (x)=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ Plug $\dfrac{x}{2}$ instead of $x$ in the above series. This implies that $\sin (\dfrac{x}{2})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x}{2})^{2n+1}}{(2n+1)!}$ and $\sin (\dfrac{x}{2})=(\dfrac{x}{2})-\dfrac{1}{2^3 \cdot 3!} x^3+\dfrac{1}{2^{5} \cdot 5!}x^5-...$

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