So, I have the function $f(x)$ over the interval $[-\pi,\pi]$ defined as under.
$$f(x)=\begin{cases}1+2x/\pi , -\pi\le x\le 0 \\ 1-2x/\pi , 0< x\le \pi\end{cases}$$
The thing is computing the Fourier coefficients for it directly is highly tedious whereas computing it for $f(x)-1$ gets us rid of the need to calculate $b_{n}$ as the function becomes an even function. So, in general, is there any way to arrive at the Fourier series expansion of $f(x)+c$ given the series expansion of $f(x)$?
On
Having a function in the form $F(x) = f(x) + g(x)$ The Fourier series is $$ F(x) = \sum_n A_n \sin(k_nx) + B_n\cos(k_nx) $$ And we find the constants using the inner product of $F(x)$ and sine or cosines... now it implies that $$ A_n = \int_L F(x)\sin(k_nx) = \int_L f(x)\sin(k_nx)dx + \int_Lg(x)\sin(k_nx)dx $$ Implying since every functions fourier series is 'equal' to the function you can add functions fourier series together with no problem at all, also note that the constants of $F(x)$ for each sine or cosine is equal to the sum of subfunctions of $F(x)$, you can also see this by just assuming two fourier series for each subfunction and write:
$$ f(x) = \sum_n C_n\sin(k_nx) + D_n\cos(k_nx) \\ g(x) = \sum_m E_m\sin(k'_mx) + F_m\cos(k'_mx) $$
since for any domain $[-L,L]$ we have $k_n \dot = k_n([-L,L])$ therefore for the same region $k_n +k'_m = 2\delta_{mn} k_n$ means that they are the same so we have: $$ F(x) =f(x) + g(x) = \sum_j (C_j + E_j)\sin(k_jx) +(D_j +F_j)\cos(k_j x)$$ now having a $g(x) = constant$ we have to calculate $E_m,F_m$s for the constant in the region and then add them to our function.
In the function you gave I think a good strategy is to have two different fourier series for each part of the function.
If $$f(t)-1\sim\sum c_ne^{int}$$then $$f(t)\sim\sum d_ne^{int},$$with $$d_n=\begin{cases}c_n,&(n\ne0),\\c_0+1,&(n=0).\end{cases}$$