How to prove that the lattice potential can be separated into two components in diatomic lattice? For a 1-D diatomic lattice with overall period $a$ and atoms $A$, $B$ placed at $\pm \frac{a}{4}(1-\delta)$, I am struggling to prove that $$ U_\frac{2\pi}{a} = \sin(\frac{\pi\delta}{2})(U^A_\frac{2\pi}{a}+U^B_\frac{2\pi}{a})-i\cos(\frac{\pi\delta}{2})(U^A_\frac{2\pi}{a}-U^B_\frac{2\pi}{a})\text{, where}$$ $$ U^{A, B}_\frac{2\pi}{a} = \frac{N}{V} \int_{\text{unit cell}} e^{-i(\frac{2\pi}{a}\cdot r)}U^{A,B}(r)\space dr.$$
A huge source of confusion is that I'm not quite sure what the $U^{A,B}(r)$ are supposed to represent. If they are contributions to the potential from $A, B$ respectively, shouldn't they simply superpose?
For context, I have proven everything in the below image thus far, and am struggling wtith 4.144/4.145:
 A: We define the Fourier coefficient of the full potential $U(x)$ to be
$$
U_{G}=\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U(x)\,dx\,.
$$
Now, we note that the potential can be written as a sum
$$
U(x) = U^A(x-a(1-\delta)/4)+U^B(x+a(1-\delta)/4)\,,
$$
where we have explicitly built in the positions of the two atoms in the basis. Then, the Fourier coefficient becomes
\begin{align*}
U_{G}&=\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^A(x-a(1-\delta)/4)\,dx
+\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^B(x+a(1-\delta)/4)\,dx
\\
&=
\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG (x+a(1-\delta)/4)}U^A(x)\,dx
+\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG (x-a(1-\delta)/4)}U^B(x)\,dx
\\
&=
e^{-iG a(1-\delta)/4}\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^A(x)\,dx
+e^{iG a(1-\delta)/4}\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^B(x)\,dx\,,
\end{align*}
which for $G=2\pi/a$ becomes
\begin{align*}
U_{2\pi/a}
&=
-ie^{i\pi\delta/2}\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^A(x)\,dx
+ie^{-i\pi\delta/2}\frac{N}{V}\int_{\textrm{unit cell}}e^{-iG x}U^B(x)\,dx
\\
&=-ie^{i\pi\delta/2}U^A_{2\pi/a}+ie^{-i\pi\delta/2}U^B_{2\pi/a}
\,.
\end{align*}
This can then be written in terms of trigonometric functions as
\begin{align*}
U_{2\pi/a}
&=-i(\cos(\pi\delta/2)+i\sin(\pi\delta/2))U^A_{2\pi/a}
+i(\cos(\pi\delta/2)-i\sin(\pi\delta/2))U^B_{2\pi/a}
\\
&=\sin(\pi\delta/2)(U^A_{2\pi/a}+U^B_{2\pi/a})
+i\cos(\pi\delta/2)(U^B_{2\pi/a}-U^A_{2\pi/a})
\,.
\end{align*}
This is the same as the expression in the OP.

Now, my guess as to why you would want to do things in this way is to make it so that the dependence on the relative positions of the two atoms in the basis (parameterized by $\delta$) shows up explicitly in the expressions for the Fourier coefficients and therefore the energies, etc.  For instance, if the two atoms in the basis are the same, then $U^A(x)=U^B(x)$, in which case the expression becomes
\begin{align*}
U_{2\pi/a}
&=2\sin(\pi\delta/2)U^A_{2\pi/a}
\,,
\end{align*}
and so $\delta$ will explicitly parameterize the splitting that occurs when we take a monatomic lattice of lattice constant $a/2$ and create a diatomic lattice of lattice constant $a$ by moving every other atom by a small amount $\delta a/2$.
