15. In $\triangle A B C$, the three sides $a$, $b$, and $c$ satisfy $2 b=a+c$. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$.
15. From the condition $2 b=a+c$, we know $2 \sin B=\sin A+\sin C$. Also, $A+B+C=\pi$, so $4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}$. Since $\sin \frac{A+C}{2} \neq 0$, then $2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2}$, hence $2[1+\cos (A+C)]=\cos A+\cos C$ and $2[\cos A+\cos C]=1+\cos (A-C)$, which means $2[1+\cos (A-C)]=4[\cos A+\cos C]$, so $5(\cos A+\cos C)=2[2+\cos (A+C)+\cos (A-C)]=$ $4[\cos A \cos C+1]$. Therefore, $5 \cos A-4 \cos A \cos C+5 \cos C=4$.