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!

 A: Here's my approach:

1.) First, we draw a line $AD$ from $A$ such that $\angle ABD=\angle DAB=\beta$, this means that $\angle DAC=\alpha-\beta$ ($\angle BAC=\alpha$). This implies that $AD=BD=x$.
Now, we know that:
$$\cos(A-B)=\frac{7}{8}$$.
We can apply the Law of Cosines in $\triangle DAC$:
$$\cos(A-B)=\frac{x^2+4^2-(5-x)^2}{8x}$$
$$\frac{7}{8}=\frac{x^2+16-25+10x-x^2}{8x}$$
$$7x=10x-9$$
Therefore, $x=3$.
Now, we can apply the Law of Cosines again in $\triangle DAC$:
$$\cos(C)=\frac{4+16-9}{16}$$
$$\cos(C)=\frac{11}{16}$$
A: 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.
A: We can use Law of Tangents
Let $t_1 = \tan\frac{A-B}{2}$
Let $t_2 = \tan\frac{A+B}{2}$
$\displaystyle \cos(A-B) = \frac{7}{8} = \frac{1-t_1^2}{1+t_1^2}
\quad → t_1^2 = \frac{8-7}{8+7} = \frac{1}{15}$
$\displaystyle \frac{\tan\frac{A-B}{2}}{\tan \frac{A+B}{2}} 
=\frac{a-b}{a+b}$
$\displaystyle \frac{t_1}{t_2} = \frac{5-4}{5+4} = \frac{1}{9}
\qquad\qquad\quad → t_2^2 = 81×t_1^2 = \frac{27}{5}$
$\displaystyle \cos(C) = -\cos(A+B) = -\frac{1-t_2^2}{1+t_2^2} 
= \frac{27-5}{27+5} = \frac{11}{16}$
A: 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 $\cos(A-B)\;$!
$\displaystyle \cos(A-B) = \frac{7}{8} \quad → \cos(C) = \frac{11}{16}$
A: A novel way to solve the problem is to change the question, with equvalent answer.
$\cos(A-B) \;=\; \sin(A)\sin(B) \;+\; \cos(A)\cos(B)$
$\cos(\quad C \quad) \;=\; \sin(A)\sin(B) \;-\; \cos(A)\cos(B)$
Only difference between the two are the sign of cosine product term.
We change the question! $\; a=5,\; b=4,\;\cos(C)=7/8,\;\text{find}\;\cos(A-B)$
$c^2 = a^2 + b^2 - 2\,a\,b\,\cos(C) = 5^2+4^2-2×5×4×\frac{7}{8} = 6$
$\displaystyle 2R = \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} 
= \sqrt{\frac{6}{1-(7/8)^2}} = \sqrt{\frac{128}{5}}$
Compare with original question, we have the same $R$
$\displaystyle \cos(A-B) + \cos(C) = 2\,\sin(A)\sin(B)$
$\displaystyle \cos(A-B) = -\cos(C) + \frac{2\,a\,b}{4R^2} 
= -\frac{7}{8} \;+\; \frac{2×5×4}{128/5}
= \frac{11}{16}$
--> Original question, $\displaystyle\;\cos(C) = \frac{11}{16}$
A: 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
$$\cos C=\frac{\frac{27}{2}-8}{8}=\frac{11}{16}.$$
