# Chapter 11 - Infinite Sequences and Series - 11.11 Application of Taylor Polynomials - 11.11 Exercises - Page 821: 23

$\approx 0.17365$

#### Work Step by Step

We are given that $f(x)=\cos x$ and $a=\dfrac{\pi}{2}$ Here, we have $80^{\circ} \times \dfrac{\pi}{180^{\circ}} \approx \dfrac{4\pi}{9}$ Check the Taylor inequality for the interval $\dfrac{4\pi}{9} \leq x\leq \dfrac{\pi}{2}$ Now, $|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$ and $|R_3(x)|\leq \dfrac{1}{5!}|\dfrac{4\pi}{9}-\dfrac{\pi}{2}|^{5}\approx 0.00000135$ and $f(x) \approx T_3(x)\approx -(x-\dfrac{\pi}{2}) +\dfrac{1}{6}-(x-\dfrac{\pi}{2})^3$ Thus, we have $f(x) \approx T_3(\dfrac{4\pi}{9})\approx -(\dfrac{4\pi}{9}-\dfrac{\pi}{2}) +\dfrac{1}{6}-(\dfrac{4\pi}{9}-\dfrac{\pi}{2})^3\approx 0.17365$

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