Just use:
$$e^{ikx} = \cos(kx) + i\sin(kx)$$
Generally, if the fourier transform is real, you'll have terms like $e^{ikx}+e^{-ikx}$ that cancel out the imaginary part. In your case, divide the infinite series into three parts:
$$\sum_{k=-\infty}^{\infty} =\sum_{k=-\infty}^{-1} + \sum_{k=1}^{\infty} + \sum_{k=0}^{0}$$
Notice how you now get terms like $e^{ikx}+e^{-ikx}=2\cos(kx)$, or $e^{ikx}-e^{-ikx}=2i\sin(kx)$.
$\newcommand{\Vector}[1]{\mathbf{#1}}\newcommand{\vece}{\Vector{e}}$The linked questions provide good answers, but may be at the technical end of "intuitive". Here's a fast-and-loose conceptual motivation:
If $\bigl(V, \langle\ ,\ \rangle\bigr)$ is an $N$-dimensional real inner product space, and if $\{\vece_{n}\}_{n=1}^{N}$ is an (ordered) orthonormal basis, then an arbitrary vector $v$ in $V$ may be written as a linear combination
$$
v = \sum_{n=1}^{N} \langle v, \vece_{n}\rangle \vece_{n}.
\tag{1}
$$
Indeed, $\{\vece_{n}\}$ is a basis of $V$, so there exist real coefficients $a_{k}$ such that
$$
v = \sum_{k=1}^{N} a_{k} \vece_{k}.
\tag{2}
$$
Taking the inner product of each side with $\vece_{n}$ gives
$\langle v, \vece_{n}\rangle = a_{n}$ because the basis $\{\vece_{n}\}$ is orthonormal.
Loosely, one might expect a similar conclusion to hold if $V$ is infinite-dimensional. Getting the definitions and hypotheses right, and proving a version of (1) in this new setting, is why any "honest" answer is bound to be technical. Phrases in quotes below are not mathematically correct, and therefore require careful inspection and/or justification.
Intuitively, let $L > 0$ be real, let $V$ be "the space of real-valued functions" on $[-L, L]$, and define an "inner product" by
$$
\langle f, g\rangle = \frac{1}{L} \int_{-L}^{L} f(t) g(t)\, dt.
$$
The functions
$$
C_{n}(t) = \begin{cases}
1/\sqrt{2}, & n = 0, \\
\cos(n\pi t/L), & n > 0;
\end{cases}\qquad
S_{n}(t) = \sin(n\pi t/L),\quad n > 0;
$$
turn out (by elementary calculus and trigonometry) to form an "orthonormal basis" of $V$.
Loosely, we expect that if $f$ is a function, we can express $f$ as an infinite sum of these basis functions, and the coefficients are the inner products of $f$ with the basis elements, i.e. (for $n > 0$),
\begin{align*}
a_{0} &= \langle f, 1\rangle = \frac{1}{L} \int_{-L}^{L} f(t)\, dt, \\
a_{n} &= \langle f, C_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \cos(n\pi t/L)\, dt, \\
b_{n} &= \langle f, S_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \sin(n\pi t/L)\, dt, \\
f(t) &= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n} C_{n}(t) + b_{n} S_{n}(t), \\
&= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos(n\pi t/L) + b_{n} \sin(n\pi t/L).
\end{align*}
(The "special" factor of $1/2$ on the constant term arises because $C_{0} = 1/\sqrt{2} \neq 1$.)
Best Answer
Thanks to the formula $e^{ix} = \cos(x) + i \sin(x)$ you can transform $$\sum_{n = -\infty}^\infty c_n(l) e^{in\pi x \, /\, l}$$ into the following: $$\sum_{n = -\infty}^\infty c_n(l) \cos( n \pi x \, / \, l) + \sum_{n = -\infty}^\infty i c_n(l) \sin(n\pi x\,/\,l).$$ Then, by exploiting the identities $\cos(-x)=\cos(x)$ and $\sin(-x) = -\sin(x)$, you get $$c_0(l) + \sum_{n = 1}^\infty [c_n(l)+c_{-n}(l)] \cos( n \pi x \, / \, l) + \sum_{n = 1}^\infty i [c_n(l)-c_{-n}(l)] \sin(n\pi x\,/\,l) $$