Convert Complex to Trigonometric. How do we convert $me^{ik}+ne^{-ik}$ to a purely trigonometric expression?
I am guessing it has something to do with $e^{it}=\cos(t)+i\sin(t)$ but I can't figure out exactly how.
Edit:- I should have been clearer. In particular, I was looking for how to convert the above to $A\cos(k+x)$ where $x$ and $A$ are constants.
Edit 2:- Here is what I have achieved so far.
$me^{ik}+ne^{-ik}\\$
$=m(\cos(k)+i\sin(k))+n(\cos(-k)+i\sin(-k))\\$
$=m(\cos(k)+i\sin(k))+n(\cos(k)-i\sin(k))\\$
$=\cos(k)(m+n)+i\sin(k)(m-n)\\$
Now what?
 A: \begin{equation}
me^{ik}+ne^{-ik}\\
=m[\cos(k)+i\sin(k)]+n[\cos(k)-i\sin(k)]\\
=(m+n)\cos(k)+i(m-n)\sin(k)\\
=\cos(k)\cos(x)-\sin(k)\sin(x)\\
=\cos(k+x)
\end{equation}
where$~~\cos(x)=m+n=$ and $~\sin(x)=-i(m-n)=~$ are two constants.

  
*
  
*Trigonometric Rule: $~\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)~$

Answer for to your last edit: You can take $~~A\cos(x)=m+n=$ and $~A\sin(x)=-i(m-n)=~$ are two constants [where $~A,~x~$ are constants]. You can get 
$$(m+n)\cos(k)+i(m-n)\sin(k)\\
=A\cos(k)\cos(x)-A\sin(k)\sin(x)\\
=A\cos(k+x)$$
A: OP has cross-posted at PSE here so I will post my answer below.
We have as a solution to an ODE
$$x(t) = Ae^{i\omega t} + Be^{−i\omega t}$$
where $A$ and $B$ are some (complex) constants and $x(t)$ is real. But I want to write this as
$$x(t) = C\cos(\omega t + \phi)$$
First, since $x(t)$ is real, it must be that $B = A^*$, i.e., that $A$ and $B$ are complex conjugates.
Then, using the polar form $A = |A|e^{i\phi}$, write the first expression as
$$x(t) = |A|e^{i\phi}e^{i\omega t} + |A|e^{-i\phi}e^{-i\omega t}$$
Finally, combine the products of exponential terms and use $e^{i\theta}+e^{-i\theta}=2\cos\theta$ to write this as
$$x(t) = |A|\left( e^{i(\omega t+\phi)} + e^{-i(\omega t+\phi)}\right) = 2|A|\cos(\omega t + \phi)$$
Thus, by inspection
$$C = 2|A|$$
$$\phi = \angle A$$


Okay, the key point I seemed to be missing was that A and B are
  complex conjugates. But why exactly is that? How does x(t) being real
  imply that?

You may recall that the sum of a complex number and its conjugate is twice the real part (which is, of course, real)
$$Z + Z^* = (a + ib) + (a - ib) = 2a$$
So, if $B=A^*$ then $x(t)$ has that form since
$$\left(Ae^{i\omega t}\right)^* = A^*e^{-i\omega t}$$
and then $x(t)$ is real.
Another way to see this is to explicitly write the imaginary part of $x(t)$ and then find the condition for which it is zero
$$i2\cdot\mathfrak{Im}\{x(t)\} = \left(x(t) - x^*(t)\right)$$
$$= Ae^{i\omega t}+Be^{-i\omega t}-A^*e^{-i\omega t}-B^*e^{i\omega t} = (A - B^*)e^{i\omega t}-(A^* - B)e^{-i\omega t}$$
If you stare at that for just a bit, you'll see that this identical to zero when $B = A^*$.
