If $Z=\sin \frac {6\pi}5+i(1+\cos \frac {6\pi }5)$ then

Given that

$\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)$

Notice, the point $\sin \left(\frac{6 \pi}{5}\right)+i\left(1+\cos 6 \frac{\pi}{5}\right)$ or

$-(\sin (\pi / 5)+1(1-\cos \pi / 5)) \text { lies in the second }$

quadrant of complex plane hence its argument is given as

$\begin{aligned} \arg (z)= & \pi-\tan ^{-1}|y / x|~~~~~~~~~~~~~[z = x + iy ] \\ & ( \forall x<0, y \geqslant 0) \\ = & \pi-\tan ^{-1}\left|\frac{1-\cos (\pi / 5)}{\sin \pi / 5}\right| \\ = & \pi-\tan ^{-1}\left|\frac{2 \sin ^{2} \pi / 10}{2 \sin \pi / 10 \cos \pi / 10}\right|=\pi-\tan ^{-1}\left|\frac{\sin \pi / 10}{\cos \pi / 10}\right| \\ = & \pi-\tan ^{-1}|\tan \pi / 10| \quad\left(\because \tan ^{-1} \pi / 10>1\right) \\ = & \pi-\pi / 10 \quad\left(\because-\pi / 2 \leq \tan ^{-1} x \leq \pi / 2\right) \\ = & \frac{9 \pi}{10} \end{aligned}$