Answer
The Maclaurin polynomials:
$n=0$, ${\ \ \ }$ ${p_0}\left( x \right) = 0$
$n=1$, ${\ \ \ }$ ${p_1}\left( x \right) = \pi x$
$n=2$, ${\ \ \ }$ ${p_2}\left( x \right) = \pi x$
$n=3$, ${\ \ \ }$ ${p_3}\left( x \right) = \pi x - \dfrac{{{\pi ^3}}}{{3!}}{x^3}$
$n=4$, ${\ \ \ }$ ${p_4}\left( x \right) = \pi x - \dfrac{{{\pi ^3}}}{{3!}}{x^3}$
The $n$th Maclaurin polynomials for $\sin \pi x$:
$\mathop \sum \limits_{k = 0}^n {\left( { - 1} \right)^k}\dfrac{{{{\left( {\pi x} \right)}^{2k + 1}}}}{{\left( {2k + 1} \right)!}}$
Work Step by Step
Let $f\left( x \right) = \sin \pi x$. Thus, the derivatives:
$f'\left( x \right) = \pi \cos \pi x$, ${\ \ \ \ \ }$ $f'\left( 0 \right) = \pi $
$f{\rm{''}}\left( x \right) = - {\pi ^2}\sin \pi x$, ${\ \ \ \ \ }$ $f{\rm{''}}\left( 0 \right) = 0$
$f{\rm{'''}}\left( x \right) = - {\pi ^3}\cos \pi x$, ${\ \ \ \ \ }$ $f{\rm{'''}}\left( 0 \right) = - {\pi ^3}$
$f{\rm{''''}}\left( x \right) = {\pi ^4}\sin \pi x$, ${\ \ \ \ \ }$ $f{\rm{''''}}\left( 0 \right) = 0$
The Maclaurin polynomials:
$n=0$, ${\ \ \ }$ ${p_0}\left( x \right) = f\left( 0 \right) = 0$
$n=1$, ${\ \ \ }$ ${p_1}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x = \pi x$
$n=2$, ${\ \ \ }$ ${p_2}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} = \pi x$
$n=3$, ${\ \ \ }$ ${p_3}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{x^3} = \pi x - \dfrac{{{\pi ^3}}}{{3!}}{x^3}$
$n=4$, ${\ \ \ }$ ${p_4}\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{x^3} + \dfrac{{f{\rm{''''}}\left( 0 \right)}}{{4!}}{x^4} = \pi x - \dfrac{{{\pi ^3}}}{{3!}}{x^3}$
We obtain ${p_1}\left( x \right) = {p_2}\left( x \right)$, ${p_3}\left( x \right) = {p_4}\left( x \right)$. Thus,
${p_{2k + 1}}\left( x \right) = {p_{2k + 2}}\left( x \right) = \pi x - \dfrac{{{\pi ^3}}}{{3!}}{x^3} + \cdot\cdot\cdot + {\left( { - 1} \right)^k}\dfrac{{{{\left( {\pi x} \right)}^{2k + 1}}}}{{\left( {2k + 1} \right)!}}$, ${\ \ \ }$ for $k = 0,1,2,\cdot\cdot\cdot$
The $n$th Maclaurin polynomials for $\sin \pi x$ in sigma notation:
$\mathop \sum \limits_{k = 0}^n {\left( { - 1} \right)^k}\dfrac{{{{\left( {\pi x} \right)}^{2k + 1}}}}{{\left( {2k + 1} \right)!}}$
