We have equation $y_{xx}=y_{tt}$ with conditions
$y(0,t)=y(2,t)=0$
and $y(x,0)=f(x)$ and $y_t(x,0)=g(x)=0$ for $0 < x < 2$
where $ f(x)= \begin{cases} 0.1x&\text{if}\, 0 \le x \le1\\ 0.1(2-x)&\text{if}\, 1 \le x \le 2\\ \end{cases} $
I have the formula $$ f(x) =\sum_{n=1}^\infty A_n \sin(\frac{n\pi}{L})x$$ $$A_n=\frac{2}{L} \int_{0}^{L}f(x)\sin(\frac{n\pi x}{L}) dx$$ $$g(x) =\sum_{n=1}^\infty B_n \sin(\frac{n\pi}{L})x $$
$$B_n=\frac{2}{L} \int_{0}^{L}g(x)\sin(\frac{n\pi x}{L}) dx$$
Now the sine series of $$f(x)=\sum_{n=1}^\infty \frac{0.8}{n^2\pi^2}sin(\frac{n\pi x}{2})sin(\frac{n\pi x}{2})$$
but I have no idea how to arrive at this. They didn't add the piecewise sections together otherwise a product wouldn't show up in the answer. What did they use for $f(x)$?
You have $$ f(x) = \sum_{n = 1}^\infty A_n \sin\left(\frac{\pi n}{2}x\right), $$ where $$ A_n = 0.1 \int_0^1 x \sin\left(\frac{\pi n}{2}x\right) dx + 0.1 \int_1^2 (2 - x) \sin\left(\frac{\pi n}{2}x\right) dx = 0.1 \frac{8}{\pi^2 n^2}\sin\left(\frac{\pi n}{2}\right), $$ therefore $$ f(x) = \sum_{n = 1}^\infty \frac{0.8}{\pi^2 n^2} \sin\left(\frac{\pi n}{2}\right)\sin\left(\frac{\pi n}{2}x\right). $$
I think you/someone else just made a typo in the expression of $\sin\left(\frac{\pi n}{2}\right)$ adding $x$ in the argument.