= Solution
{c}
For the <Fourier cosine series>, <integration by parts> gives
$$
a_0=\frac2\pi\int_0^\pi(1+x)\,dx=2+\pi,
$$
and, for $n\geq1$,
$$
\begin{aligned}
a_n&=\frac2\pi\int_0^\pi(1+x)\cos(nx)\,dx\\
&=\frac2\pi\left[\frac{(1+x)\sin(nx)}n+\frac{\cos(nx)}{n^2}\right]_0^\pi
=\frac{2\bigl((-1)^n-1\bigr)}{\pi n^2}.
\end{aligned}
$$
Only odd <Fourier harmonics> remain, so
$$
\boxed{1+x=1+\frac\pi2-\frac4\pi
\sum_{m=0}^\infty\frac{\cos((2m+1)x)}{(2m+1)^2}
\quad(0\leq x\leq\pi)}.
$$
For the <Fourier sine series>, another <integration by parts> gives
$$
\begin{aligned}
b_n&=\frac2\pi\int_0^\pi(1+x)\sin(nx)\,dx\\
&=\frac2\pi\left[-\frac{(1+x)\cos(nx)}n+\frac{\sin(nx)}{n^2}\right]_0^\pi
=\frac{2\bigl[1-(1+\pi)(-1)^n\bigr]}{\pi n}.
\end{aligned}
$$
Consequently
$$
\boxed{1+x=\frac2\pi\sum_{n=1}^\infty
\frac{1-(1+\pi)(-1)^n}{n}\sin(nx)
\quad(0<x<\pi)}.
$$
At $x=0$ and $x=\pi$, the <Fourier sine series> instead has value zero.
To describe the requested graphs precisely, take $t\in[-\pi,\pi)$ congruent to $x$ modulo $2\pi$. The <Fourier cosine series> converges to the continuous triangular <periodic extension>
$$
\boxed{F_c(x)=1+|t|}.
$$
Its minima occur at the even multiples of $\pi$, with value $1$, and its maxima at the odd multiples, with value $1+\pi$. The <Fourier sine series> converges to
$$
\boxed{F_s(x)=
\begin{cases}
1+t,&0<t<\pi,\\
t-1,&-\pi<t<0,\\
0,&x\in\pi\mathbb Z.
\end{cases}}
$$
The <jump discontinuities> at even multiples of $\pi$ go from $-1$ to $1$; those at odd multiples go from $1+\pi$ to $-(1+\pi)$. Each jump is assigned its midpoint, zero, by the <Fourier series>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-resit-fourier-extensions.png]
{title=Limits of the <Fourier cosine series> and <Fourier sine series> of $1+x$ on $[-3\pi,3\pi]$}
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