8. Given $\theta=\arctan \frac{5}{12}$. Then, the principal value of the argument of the complex number $z=\frac{\cos 2 \theta+\mathrm{i} \sin 2 \theta}{239+\mathrm{i}}$ is $\qquad$
$$
\begin{array}{l}
\tan \theta=\frac{5}{12} \Rightarrow \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{119}{169}, \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{120}{169} . \\
\text { Therefore, } \arg z=\arg \left[\frac{\cos 2 \theta+\mathrm{i} \sin 2 \theta}{(\sin 2 \theta+\cos 2 \theta)+\mathrm{i}(\sin 2 \theta-\cos 2 \theta)}\right] \\
=\arg \left[\frac{\cos 2 \theta+\mathrm{i} \sin 2 \theta}{\cos \left(2 \theta-\frac{\pi}{4}\right)+\mathrm{i} \sin \left(2 \theta-\frac{\pi}{4}\right)}\right]=\frac{\pi}{4} .
\end{array}
$$