Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.11 Application of Taylor Polynomials - 11.11 Exercises - Page 821: 24


$\approx 0.61566$

Work Step by Step

We are given that $f(x)=\sin x$ and $a=\dfrac{\pi}{6}$ Here, we have $38^{\circ} \times \dfrac{\pi}{180^{\circ}} \approx \dfrac{19\pi}{90}$ Check the Taylor inequality for the interval $\dfrac{\pi}{6}-\dfrac{2\pi}{45} \leq x\leq \dfrac{\pi}{6}+\dfrac{2\pi}{45}$ Therefore, $|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$ This gives: $|R_4(x)|\leq \dfrac{1}{5!}|\dfrac{19\pi}{90}-\dfrac{\pi}{6}|^{5}\approx 0.00000041$ and $f(x) \approx T_4(x)\approx |\dfrac{1}{2}+\dfrac{\sqrt 3}{2}(x-\dfrac{\pi}{6})-\dfrac{1}{4}(x-\dfrac{\pi}{6}))^2 +\dfrac{\sqrt 3}{12}-(x-\dfrac{\pi}{2})^6+\dfrac{1}{48}(x-\dfrac{\pi}{6})^4-\sin x |$ Thus, $ T_4(38^{\circ})=\dfrac{1}{2}+\dfrac{\sqrt 3}{2}(x-\dfrac{\pi}{6})-\dfrac{1}{4}(x-\dfrac{\pi}{6}))^2 +\dfrac{\sqrt 3}{12}-(x-\dfrac{\pi}{2})^6+\dfrac{1}{48}(x-\dfrac{\pi}{6})^4 =T_4(\dfrac{19\pi}{90}) \approx 0.61566$
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