Answer
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{\left( {2n + 1} \right)!}}{\left( {x - \pi } \right)^{2n + 1}}$
Work Step by Step
We compute the derivatives of $f\left( x \right) = \sin x$ and list them in the table below:
$\begin{array}{l}
\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}&{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( \pi \right)}}{{n!}}}\\
0&{\sin \left( x \right)}&{\sin \left( x \right)}&0\\
1&{\cos \left( x \right)}&{\cos \left( x \right)}&{ - 1}\\
2&{ - \sin \left( x \right)}&{ - \dfrac{{\sin \left( x \right)}}{2}}&0\\
3&{ - \cos \left( x \right)}&{ - \dfrac{{\cos \left( x \right)}}{6}}&{\dfrac{1}{6}}
\end{array}\\
\begin{array}{*{20}{c}}
4&{\sin \left( x \right)}&{\dfrac{{\sin \left( x \right)}}{{24}}}&0\\
5&{\cos \left( x \right)}&{\dfrac{{\cos \left( x \right)}}{{120}}}&{ - \dfrac{1}{{120}}}\\
6&{ - \sin \left( x \right)}&{ - \dfrac{{\sin \left( x \right)}}{{720}}}&0\\
7&{ - \cos \left( x \right)}&{ - \dfrac{{\cos \left( x \right)}}{{5040}}}&{\dfrac{1}{{5040}}}
\end{array}
\end{array}$
By Theorem 1 of Section 11.7, the Taylor series centered at $c = \pi $ is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( \pi \right)}}{{n!}}{\left( {x - \pi } \right)^n}$
Using the table above, we get
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( \pi \right)}}{{n!}}{\left( {x - \pi } \right)^n} = - \left( {x - \pi } \right) + \dfrac{1}{6}{\left( {x - \pi } \right)^3} - \dfrac{1}{{120}}{\left( {x - \pi } \right)^5} + \dfrac{1}{{5040}}{\left( {x - \pi } \right)^7} - \cdot\cdot\cdot$
Thus, the Taylor series of $f\left( x \right) = \sin x$ centered at $c = \pi $ is
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}}}{{\left( {2n + 1} \right)!}}{\left( {x - \pi } \right)^{2n + 1}}$
