The half-range sine series is the expansion for [tex]f(t)[/tex] with the assumption that [tex]f(t)[/tex] is considered to be an odd function over its full range, [tex]-1<t<1[/tex]. So for (a), you're essentially finding the full range expansion of the function
[tex]f(t)=\begin{cases}2-t&\text{for }0\le t<1\\-2-t&\text{for }-1<t<0\end{cases}[/tex]
with period 2 so that [tex]f(t)=f(t+2n)[/tex] for [tex]|t|<1[/tex] and integers [tex]n[/tex].
Now, since [tex]f(t)[/tex] is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with
[tex]f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L[/tex]
where
[tex]b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt[/tex]
In this case, [tex]L=1[/tex], so
[tex]b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt[/tex]
[tex]b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}[/tex]
[tex]b_n=\dfrac{4-2(-1)^n}{n\pi}[/tex]
The half-range sine series expansion for [tex]f(t)[/tex] is then
[tex]f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t[/tex]
which can be further simplified by considering the even/odd cases of [tex]n[/tex], but there's no need for that here.
The half-range cosine series is computed similarly, this time assuming [tex]f(t)[/tex] is even/symmetric across its full range. In other words, you are finding the full range series expansion for
[tex]f(t)=\begin{cases}2-t&\text{for }0\le t<1\\2+t&\text{for }-1<t<0\end{cases}[/tex]
Now the sine series expansion vanishes, leaving you with
[tex]f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L[/tex]
where
[tex]a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt[/tex]
for [tex]n\ge0[/tex]. Again, [tex]L=1[/tex]. You should find that
[tex]a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3[/tex]
[tex]a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt[/tex]
[tex]a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}[/tex]
[tex]a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}[/tex]
Here, splitting into even/odd cases actually reduces this further. Notice that when [tex]n[/tex] is even, the expression above simplifies to
[tex]a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0[/tex]
while for odd [tex]n[/tex], you have
[tex]a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}[/tex]
So the half-range cosine series expansion would be
[tex]f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t[/tex]
[tex]f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t[/tex]
[tex]f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t[/tex]
Attached are plots of the first few terms of each series overlaid onto plots of [tex]f(t)[/tex]. In the half-range sine series (right), I use [tex]n=10[/tex] terms, and in the half-range cosine series (left), I use [tex]k=2[/tex] or [tex]n=2(2)-1=3[/tex] terms. (It's a bit more difficult to distinguish [tex]f(t)[/tex] from the latter because the cosine series converges so much faster.)
