8. In $\triangle A B C, B C=5, A C=4$ and $\cos (A-B)=\frac{7}{8}$. Find the value of $\cos C$.在 $\triangle A B C$ 中, $B C=5 、 A C=4$ 而 $\cos (A-B)=\frac{7}{8}$ 。求 $\cos C$ 的值。
8. $\frac{11}{16}$
8. Let $D$ be a point on $B C$ such that $D A=D B$. Then $\cos \angle D A C=\cos (A-B)=\frac{7}{8}$. Let $D B=D A=x$. Then $D C=5-x$ and applying cosine law in $\triangle D A C$ gives
$$
(5-x)^{2}=x^{2}+4^{2}-2(x)(4)\left(\frac{7}{8}\right),
$$
or $x=3$, i.e. $D A=3$ and $D C=2$. Applying cosine law in $\triangle D A C$ again, we have
$$
\cos C=\frac{4^{2}+2^{2}-3^{2}}{2(4)(2)}=\frac{11}{16} .
$$