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!


If the Law of cosines works the Law of sines probably works. From $\frac{\sin A}{5}=\frac{\sin B}{4}=k$ and $\cos A\cos B+\sin A\sin B=\cos(A-B)$ we get the equation $$\sqrt{1-25k^2}\sqrt{1-16k^2}+20k^2=\frac{7}{8}$$
and $k=\frac{\sqrt{5}}{8\sqrt{2}}$. Hence, $\sin A=\frac{5\sqrt{5}}{8\sqrt{2}}$, $\sin B=\frac{4\sqrt{5}}{8\sqrt{2}}$, $\cos A=\frac{\sqrt{6}}{16}$, $\cos B=\frac{\sqrt{6}}{4}$ and
$$\cos C=-\cos(A+B)=\sin A\sin B-\cos A\cos B=\frac{5\sqrt{5}}{8\sqrt{2}}\frac{4\sqrt{5}}{8\sqrt{2}}-\frac{\sqrt{6}}{16}\frac{\sqrt{6}}{4}=\frac{25}{32}-\frac{3}{32}=\frac{11}{16.}$$
Note: WA output for Albert Chan's comment. Is this enough? I hope so.
