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!


We can use Law of Tangents
Let $t_1 = \tan\frac{A-B}{2}$
Let $t_2 = \tan\frac{A+B}{2}$
$\displaystyle \cos(A-B) = \frac{7}{8} = \frac{1-t_1^2}{1+t_1^2}
\quad → t_1^2 = \frac{8-7}{8+7} = \frac{1}{15}$
$\displaystyle \frac{\tan\frac{A-B}{2}}{\tan \frac{A+B}{2}} 
=\frac{a-b}{a+b}$
$\displaystyle \frac{t_1}{t_2} = \frac{5-4}{5+4} = \frac{1}{9}
\qquad\qquad\quad → t_2^2 = 81×t_1^2 = \frac{27}{5}$
$\displaystyle \cos(C) = -\cos(A+B) = -\frac{1-t_2^2}{1+t_2^2} 
= \frac{27-5}{27+5} = \frac{11}{16}$
