(9) (20 points) The angles of triangle $ABC$ satisfy: $\frac{A}{B}=\frac{B}{C}=\frac{1}{3}$; Find the value of $T=\cos A+\cos B+\cos C$.
9) Let $A=\theta, B=3 \theta, C=9 \theta$, from $\theta+3 \theta+9 \theta=\pi$, we get $\theta=\frac{\pi}{13}$.
$$
\begin{aligned}
T & =\cos \theta+\cos 3 \theta+\cos 9 \theta \\
& =\cos \theta+\cos 3 \theta-\cos 4 \theta \\
& =2 \cos \theta \cos 2 \theta-2 \cos ^{2} 2 \theta+1 \\
& >2 \cos ^{2} 2 \theta-2 \cos ^{2} 2 \theta+1=1, \\
T^{2} & =(\cos \theta+\cos 3 \theta+\cos 9 \theta)^{2} \\
& =\cos ^{2} \theta+\cos ^{2} 3 \theta+\cos ^{2} 9 \theta+2 \cos \theta \cos 3 \theta+2 \cos \theta \cos 9 \theta+
\end{aligned}
$$
$2 \cos 3 \theta \cos 9 \theta$
$$
\begin{aligned}
= & \frac{1+\cos 2 \theta}{2}+\frac{1+\cos 6 \theta}{2}+\frac{1+\cos 8 \theta}{2}+(\cos 2 \theta+\cos 4 \theta)+ \\
& (\cos 8 \theta+\cos 10 \theta)+(\cos 6 \theta+\cos 12 \theta),
\end{aligned}
$$

and
$$
T=\cos \theta+\cos 3 \theta+\cos 9 \theta=-\cos 12 \theta-\cos 10 \theta-\cos 4 \theta,
$$

so
$$
\begin{aligned}
2 T^{2}-T= & 3+3(\cos 2 \theta+\cos 4 \theta+\cos 6 \theta+\cos 8 \theta+ \\
& \cos 10 \theta+\cos 12 \theta) .
\end{aligned}
$$

Let
$$
P=\cos 2 \theta+\cos 4 \theta+\cos 6 \theta+\cos 8 \theta+\cos 10 \theta+\cos 12 \theta,
$$

then
$$
\begin{array}{l}
P \cdot 2 \sin \theta=(\sin 3 \theta-\sin \theta)+(\sin 5 \theta-\sin 3 \theta)+(\sin 7 \theta-\sin 5 \theta)+ \\
(\sin 9 \theta-\sin 7 \theta)+(\sin 11 \theta-\sin 9 \theta)+(\sin 13 \theta-\sin 11 \theta)=-\sin \theta,
\end{array}
$$

so $P=-\frac{1}{2}$. Thus $2 T^{2}-T=3-\frac{3}{2}=\frac{3}{2}$, which means $4 T^{2}-2 T-3=0$. Since $T>1$, solving this equation gives $T=\frac{1+\sqrt{13}}{4}$.