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