In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!


Here is another way to solve for $\cos(C)$, by area calculations, done in 2 ways.
From Bob Dobb' answer, we have:
$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = \sqrt{\frac{128}{5}}$
$\displaystyle Δ = \frac{a\,b\,c}{4\,R} \qquad \qquad \quad
→ Δ^2 = \frac{125}{32}\,c^2$
Using my own triangle area formula:
$\displaystyle z = \frac{c^2 - (a-b)^2}{4} = \frac{c^2-1}{4}$
$\displaystyle Δ = \sqrt{(a\,b-z)\,z} \qquad → Δ^2 = \frac{-81+82\,c^2-c^4}{16}$
Area squared calculations, done 2 ways, must match:
$\displaystyle \frac{125}{32}\,c^2 - \frac{-81+82\,c^2-c^4}{16} = \frac{(c^2-6)(c^2-\frac{27}{2})}{16} = 0$
$c^2=6 \qquad → \cos(C) = \frac{5^2 + 4^2 - 6}{2×5×4} = \frac{7}{8}$
$c^2=\frac{27}{2} \quad\; → \cos(C) = \frac{5^2 + 4^2 - \frac{27}{2}}{2×5×4} = \frac{11}{16}$
It is interesting we have 2 solutions for $\cos(C)\;$!
If we pick one solution for $\cos(C)$, the other is $\cos(A-B)\;$!
$\displaystyle \cos(A-B) = \frac{7}{8} \quad → \cos(C) = \frac{11}{16}$
