5. Given the complex number $z=1-\sin \theta+\operatorname{i} \cos \theta\left(\frac{\pi}{2}<\theta<\pi\right)$, then the principal value of the argument of the conjugate complex number $\bar{z}$ is
5. $\frac{3}{4} \pi-\theta . \bar{z}=1-\sin \theta-\mathrm{i} \cos \theta$ Since $1-\sin \theta>0,-\cos \theta>0$ so $\bar{z}$ is in the first quadrant $\tan (\arg \bar{z})=\frac{-\cos \theta}{1-\sin \theta}=\tan \left(\theta+\frac{\pi}{4}\right)=\tan \left(\frac{3}{4} \pi-\theta\right)$