If $\phi(x)$ is any function on $(0, l)$, derive the expansion $\displaystyle\phi(x) = \sum c_n \sin\left(\left(n + \frac{1}{2}\right) \frac{\pi x}{l}\right)$ for $0 < x < l$ by the following method.
Extend $\phi(x)$ to the function $\tilde{\phi}(x)$ defined by $$\tilde{\phi}(x) = \begin{cases}\phi(x) & 0 \leq x \leq l\\ \phi(2l − x) & l \leq x \leq 2l\end{cases}.$$
(This means that you are extending it evenly across $x = l$.)
Write the Fourier sine series for $\tilde{\phi}(x)$ on the interval $(0, 2l)$ and write the formula for the coefficients.
Show that every second coefficient vanishes.
I'm lost on this problem from Strauss. How does one extend beyond $0 < x < l$ ? Do I have the addition of two summations now?
The extension is already done for you. You have a function $\tilde \phi$ defined on the interval $[0,2l]$ by the formula given above. Its Fourier sine coefficients are $$ A_n = \frac{2}{2l} \int_0^{2l} \tilde \phi(x)\,\sin \frac{\pi n}{2l}x\,dx \tag{1} $$ Next step is to compute this integral somehow.
Since $\tilde \phi$ is case-defined, the integral in (1) needs to be split accordingly: $$ \int_0^{2l} \tilde \phi(x)\,\sin \frac{\pi n}{2l}x\,dx = \int_0^{l} \phi(x)\,\sin \frac{\pi n}{2l}x\,dx + \int_l^{2l} \phi(2l-x)\,\sin \frac{\pi n}{2l}x\,dx $$ The last integral here invites a change of variable $y=2l-x$ $$ \begin{split} \int_l^{2l} \phi(2l-x)\,\sin \frac{\pi n}{2l}x\,dx &= -\int_l^{0} \phi(y)\,\sin \frac{\pi n}{2l}(2l-y)\,dy \\ &= \int_0^{l} \phi(y)\, \sin \left(\pi n-\frac{\pi n}{2l}y\right)\,dy \end{split} $$ One more hint: $\sin (\pi n-x)=-\sin x$ when $n$ is even.