= Solution
The <Fourier sine series> has coefficients
$$
b_n=\frac2L\int_0^Lx\sin\frac{n\pi x}{L}\,dx=\frac{2L}{n\pi}(-1)^{n+1},
$$
by <integration by parts>. Thus
$$
\boxed{x\sim\frac{2L}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}{L}}.
$$
Its odd extension is $x$ on $(-L,L)$, continued with period $2L$. The usual Fourier convergence result for a piecewise smooth periodic function gives the value at a continuous point and the mean of one-sided limits at a jump. At $x=3L/2$, subtracting one period gives $-L/2$, a continuous point of this extension. Therefore the sine series converges to \b[$-L/2$] there.
For the <Fourier cosine series>, the even extension is $|x|$ on $[-L,L]$, continued with period $2L$. Its <triangular wave> is continuous even at the joining points, so the cosine series converges to that extension everywhere. On the requested interval its graph is
$$
g(x)=\begin{cases}2L+x,&-2L\leq x\leq-L,\\-x,&-L\leq x\leq0,\\x,&0\leq x\leq L,\\2L-x,&L\leq x\leq2L.\end{cases}
$$
The vertices are $(-2L,0),(-L,L),(0,0),(L,L),(2L,0)$. In particular, its value at $3L/2$ is \b[$L/2$].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-1-cosine-extension.png]
{title=The even periodic extension represented by the cosine series}
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