## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Review - Exercises - Page 786: 45

#### Answer

$sinx=\frac{1}{2}\Sigma_{n=0}^{\infty}(-1)^{n}[\frac{1}{(2n)!} (x-\frac{\pi}{6})^{2n}+\frac{\sqrt 3}{(2n+1)!}(x-\frac{\pi}{6})^{(2n+1)}]$

#### Work Step by Step

Taylor series: $f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$ $sinx=\frac{1}{2}+\frac{\sqrt 3}{2} (x-\frac{\pi}{6})-\frac{1}{2.2!}(x-\frac{\pi}{6})^{2}-...$ $sinx=\Sigma_{n=0}^{\infty}(-1)^{n}[\frac{1}{2(2n)!} (x-\frac{\pi}{6})^{2n}+\frac{\sqrt 3}{2.(2n+1)!}(x-\frac{\pi}{6})^{(2n+1)}]$ Hence, $sinx=\frac{1}{2}\Sigma_{n=0}^{\infty}(-1)^{n}[\frac{1}{(2n)!} (x-\frac{\pi}{6})^{2n}+\frac{\sqrt 3}{(2n+1)!}(x-\frac{\pi}{6})^{(2n+1)}]$

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