= Solution
Use the product identity $\sin z\cos z=\tfrac12\sin2z$ to decompose the initial shape:
$$
y(x,0)=A\sin\frac{\pi x}l+\frac A2\sin\frac{2\pi x}l.
$$
The <Fourier sine series> is already explicit. Zero initial velocity gives cosine time factors, so
$$
\boxed{A_1=A,\qquad A_2=A/2,\qquad A_n=0\ (n\geq3),}
$$
and $y(x,t)=A\sin(\pi x/l)\cos(\pi ct/l)+(A/2)\sin(2\pi x/l)\cos(2\pi ct/l)$. Orthogonality of the sine modes makes this decomposition unique in the usual square-integrable fixed-end setting.
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