# Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 53

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#### Work Step by Step

Consider the Maclaurin Series for $\sin x$ as follows: $\sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ As we are given that $x=\pi$ Then, we have $\sin (x)=\sin \pi =0$

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