(a) Extend the definition of f(x) to make it an even function f*(x),
[tex]f^*(x)=\begin{cases}0&\text{if }-2\le x<-1\\1&\text{if }0\le x\le1\\0&\text{if }1<x\le2\end{cases}[/tex]
and we take f* to be periodic over an interval of length P = 4. We compute the coefficients of the cosine series:
[tex]a_n = \displaystyle\frac2P \int_{-2}^2 f^*(x)\cos\left(\frac{2n\pi}Px\right)\,\mathrm dx = \frac{2\sin\left(\frac{n\pi}2\right)}{n\pi}[/tex]
Note that a₀ = 1 (you can compute the integral again without the cosine, or just take the limit as n -> 0). For all other even integers n, the numerator vanishes, so we split off the odd case for n = 2k - 1 :
[tex]a_{2k-1} = \dfrac{2\sin\left(\frac{(2k-1)\pi}2\right)}{(2k-1)\pi}[/tex]
Then the cosine series of f(x) is
[tex]\displaystyle \frac{a_0}2 + \sum_{k=1}^\infty a_{2k-1}\cos\left(\frac{(2k-1)\pi}2x\right)[/tex]
(b) For the sine series, you instead extend f(x) to an odd function f*(x),
[tex]f^*(x)=\begin{cases}-1&\text{if }-2\le x\le-1\\x&\text{if }-1<x<1\\1&\text{if }1\le x\le2\end{cases}[/tex]
Again, P = 4, and the coefficient of the sine series are given by
[tex]b_n = \displaystyle\frac2P\int_{-2}^2f^*(x)\sin\left(\frac{2n\pi}Px\right)\,\mathrm dx[/tex]
which we can again split into the even/odd cases,
[tex]b_{2k} = -\dfrac1{k\pi}[/tex]
[tex]b_{2k-1} = \dfrac{4(-1)^{k+1}+2(2k-1)\pi}{(2k-1)^2\pi^2}[/tex]
So the sine series is
[tex]\displaystyle \sum_{k=1}^\infty\left(b_{2k}\sin\left(k\pi x\right) + b_{2k-1}\sin\left(\frac{(2k-1)\pi}2x\right)\right)[/tex]
I've attached plots of the extended versions of f(x) along with the corresponding series up to degree 4.
