<s>[INST] 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!

 [/INST] From $\cos(A-B)=\frac{7}{8}$, we have $\sin(\frac{A-B}{2})=\sqrt{\frac{1-\cos(A-B)}{2}}=\frac{1}{4}. {\tag 1}$
This is a nice property about interior angles: $$\sin(\frac{A-B}{2})=\frac{a-b}{c}\cos(\frac{C}{2}).$$ From this property and $(1)$ with $a=5$, $b=4$, we have $\cos(\frac C2)=\frac c4.\tag 2$
By using the half-angle formula $\cos C=2\cos^2(\frac C2)-1$, we have $\cos(C)=\frac{c^2-8}{8}.\tag3$
By using the law of cosines $c^2=a^2+b^2-2ab\cos C$ for the side $c$ and $(3)$ we can easily find $c=\frac{3\sqrt3}{\sqrt2}.$ And finally, by using $(3)$ again
$$\cos C=\frac{\frac{27}{2}-8}{8}=\frac{11}{16}.$$
</s>