In the principles of Mathematical Analysis, here is the formula (60)
$f(x)=\sum_{-N}^{N} c_{n}e^{inx}$ ($x$ is real).
Let us multiply (60) by $e^{-imx}$, where $m$ is an integer; if we integrate the product, (61) shows that
$c_m = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-imx} dx$ (62)
for $|m| \leq N$. If $|m| > N$, the integral in (62) is 0.
The following observation can be read off from (60) and (62): The trigonometric polynomial $f$, given by (60), is real if and only if $c_{-n}=\overline{c_n}$ for $n=0,\cdots, N$.
How can we easily read off?
Inspired by @Phicar, I come up with the proof about the inverse without introducing the concept of orthogonal:
If $f$ is real, then $c_m = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-imx} dx = \frac{1}{2\pi}\int_{-\pi}^{\pi} \overline{f(x)}e^{-imx} dx=\frac{1}{2\pi}\int_{-\pi}^{\pi} \sum_{-N}^{N} \overline{c_{n}} e^{-i(m+n)x} dx=\frac{1}{2\pi} \sum_{-N}^{N} \overline{c_{n}}\int_{-\pi}^{\pi} e^{-i(m+n)x} dx=\overline{c_{-m}}$.
Is that correct?
Well, $f(x)$ is real iff $\overline{f(x)}=f(x)$ but by (60) $$\overline{f(x)}=\overline{\sum _{n=-N}^Nc_ne^{inx}}=\sum _{n=-N}^N\overline{c_ne^{inx}}= \sum _{n=-N}^N\overline{c_n}e^{-inx},$$ this last equality comes by the euler formula and the fact cosine is even and sine is odd. Now the $e^{inx}$ are orthogonal so it can only be $f(x)$ if $\overline{c_n}=c_{-n}.$ The converse is the same.