Answer
$n=0$: ${\ \ \ }$ ${p_0} = - 1$
$n=2$: ${\ \ \ }$ ${p_2} = - 1 + \frac{1}{2}{\left( {x - \pi } \right)^2}$
$n=4$: ${\ \ \ }$ ${p_4} = - 1 + \frac{1}{2}{\left( {x - \pi } \right)^2} - \frac{1}{{24}}{\left( {x - \pi } \right)^4}$
$n=6$: ${\ \ \ }$ ${p_6} = - 1 + \frac{1}{2}{\left( {x - \pi } \right)^2} - \frac{1}{{24}}{\left( {x - \pi } \right)^4} + \frac{1}{{720}}{\left( {x - \pi } \right)^6}$
Work Step by Step
We compute the derivatives:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}&{{f^{\left( n \right)}}\left( \pi \right)}\\
0&{\cos x}&{ - 1}\\
1&{ - \sin x}&0\\
2&{ - \cos x}&1\\
3&{\sin x}&0\\
4&{\cos x}&{ - 1}\\
5&{ - \sin x}&0\\
6&{ - \cos x}&1
\end{array}$
By Definition 9.7.3, the $n$th Taylor polynomial about $x = {x_0}$ for $f$ is
${p_n}\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \dfrac{{f{\rm{''}}\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2} + \dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{\left( {x - {x_0}} \right)^n}$
The first few Taylor polynomials about $x = {x_0} = \pi $:
$n=0$: ${\ \ \ }$ ${p_0} = f\left( \pi \right) = - 1$
$n=1$: ${\ \ \ }$ ${p_1} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) = - 1$
$n=2$: ${\ \ \ }$ ${p_2} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) + \dfrac{{f{\rm{''}}\left( \pi \right)}}{{2!}}{\left( {x - \pi } \right)^2} = - 1 + \dfrac{1}{2}{\left( {x - \pi } \right)^2}$
$n=3$: ${\ \ \ }$ ${p_3} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) + \dfrac{{f{\rm{''}}\left( \pi \right)}}{{2!}}{\left( {x - \pi } \right)^2} + \dfrac{{f{\rm{'''}}\left( \pi \right)}}{{3!}}{\left( {x - \pi } \right)^3} = - 1 + \dfrac{1}{2}{\left( {x - \pi } \right)^2}$
$n=4$: ${\ \ \ }$ ${p_4} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) + \dfrac{{f{\rm{''}}\left( \pi \right)}}{{2!}}{\left( {x - \pi } \right)^2} + \dfrac{{f{\rm{'''}}\left( \pi \right)}}{{3!}}{\left( {x - \pi } \right)^3} + \dfrac{{{f^{\left( 4 \right)}}\left( \pi \right)}}{{4!}}{\left( {x - \pi } \right)^4}$
$ = - 1 + \dfrac{1}{2}{\left( {x - \pi } \right)^2} - \dfrac{1}{{24}}{\left( {x - \pi } \right)^4}$
$n=5$: ${\ \ \ }$ ${p_5} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{\left( {x - \pi } \right)^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{\left( {x - \pi } \right)^3} + \dfrac{{{f^{\left( 4 \right)}}\left( \pi \right)}}{{4!}}{\left( {x - \pi } \right)^4} + \dfrac{{{f^{\left( 5 \right)}}\left( \pi \right)}}{{5!}}{\left( {x - \pi } \right)^5}$
$ = - 1 + \dfrac{1}{2}{\left( {x - \pi } \right)^2} - \dfrac{1}{{24}}{\left( {x - \pi } \right)^4}$
$n=6$: ${\ \ \ }$ ${p_6} = f\left( \pi \right) + f'\left( \pi \right)\left( {x - \pi } \right) + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{\left( {x - \pi } \right)^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{\left( {x - \pi } \right)^3} + \dfrac{{{f^{\left( 4 \right)}}\left( \pi \right)}}{{4!}}{\left( {x - \pi } \right)^4} + \dfrac{{{f^{\left( 5 \right)}}\left( \pi \right)}}{{5!}}{\left( {x - \pi } \right)^5} + \dfrac{{{f^{\left( 6 \right)}}\left( \pi \right)}}{{6!}}{\left( {x - \pi } \right)^6}$
$ = - 1 + \dfrac{1}{2}{\left( {x - \pi } \right)^2} - \dfrac{1}{{24}}{\left( {x - \pi } \right)^4} + \dfrac{1}{{720}}{\left( {x - \pi } \right)^6}$
The first four distinct Taylor polynomials about $x = {x_0} = \pi $ are ${p_0}$, ${p_2}$, ${p_4}$, ${p_6}$ (please see the figure attached).
