$${Expand} \; f(x)= \begin{cases} 2{A\over L}x & 0\leq x\leq {L\over 2} \\ \\ 2{A\over L}\left(L-x\right) & {L\over 2}\leq x\leq L \end{cases} $$
I have determined $A_0$ (but omitted) to be $A_0=A$. For $\ A_n \ $ and $\ B_n$, I'm rather confused where to start. For $A_n$: $$A_n={2\over L}\left[2{A\over L}\int_{0}^{L\over 2}x\cos\left({2n\pi x}\over L\right)dx+2{A\over L}\left(L\int_{L\over 2}^L\cos\left({2n\pi x}\over L\right)dx-\int_{L\over 2}^Lx\cos\left({2n\pi x}\over L\right)dx\right)\right]$$ And a similar case to $ B_n$, I'm not quite sure if I'm on the right path.
That looks correct for $A_n$, you'll just need to evaluate some of the integrals by parts from here. For $B_n$, note that the function is even, and so $B_n = 0$ for all $n$, since they are coefficients of $\sin$ which is odd.