Find the area of given triangle In a triangle $ABC$, where $a=8$, $c=1$ and $\cos (A-C) ={16\over 65}$. 
Where $(a,b,c)$ are sides and $(A,B,C)$ are angles. 
How can we find the area of the triangle $ABC$?
A hint will be appreciated.
 A: Hint: The Law of Sines says
$$
\sin(A)=8\sin(C)\tag{1}
$$
Then
$$
\begin{align}
\frac{16}{65}
&=\cos(A)\cos(C)+\sin(A)\sin(C)\tag{2}\\
\left(\frac{16}{65}-8\sin^2(C)\right)^2
&=\left(1-64\sin^2(C)\right)\left(1-\sin^2(C)\right)\tag{3}\\
\frac{256}{4225}-\frac{256}{65}\sin^2(C)+64\sin^4(C)
&=1-65\sin^2(C)+64\sin^4(C)\tag{4}\\
\sin^2(C)
&=\frac1{65}\tag{5}\\
\end{align}
$$
This means that
$$
\sin^2(A)=\frac{64}{65}\quad\cos^2(A)=\frac1{65}\quad\sin^2(C)=\frac1{65}\quad\cos^2(C)=\frac{64}{65}\tag{6}
$$
and therefore,
$$
\begin{align}
\cos(A+C)
&=\cos(A)\cos(C)-\sin(A)\sin(C)\\
&=\frac8{65}-\frac8{65}\\[2pt]
&=0\tag{7}
\end{align}
$$
What does this say about $B$, the angle between $a$ and $c$?

Note: Since $A,C\le\pi$, we have $\sin(A),\sin(C)\ge0$. However, only given $(6)$, $\cos(A),\cos(C)$ may not both be positive. Therefore, we know that $\sin(A)\sin(C)=\frac8{65}$, but we only know that $|\cos(A)\cos(C)|=\frac8{65}$. However, $(2)$ tells us that $\cos(A)\cos(C)=\frac8{65}$.
A: Given 
$$ c,a,\quad \delta = C-A  $$

$$ C-A = \gamma- \alpha = \delta $$
Construct double angle $ 2 \alpha $ at vertex A.
Law of Sines on $ \Delta ABC $
$$ \frac{c}{a}= \frac{\sin (\delta+\alpha)}{\sin \alpha} \tag1$$
Trig expand and algebraically simplify
$$ \tan \alpha= \frac {\sin \delta}{( c/a -\cos  \delta )} \rightarrow \alpha_{1,2} \tag2 $$
$$ \alpha_2 =\alpha_1 + \pi \tag3$$
$$ \beta = (\pi- \delta- 2 \alpha _{1,2}) \tag4 $$
$$ \gamma = (\pi-  \alpha _{1,2} -\beta) \tag5 $$
One value suffices.
Area 
$$ \frac12 \,a \,c \sin \beta = .. $$
Also checked all the results numerically
EDIT1:
If we use supplementary angle for angles assuming it as an ambiguous case, all conditions given could be correctly realized, but this should be seen in a construction.
