Show that one of the values of ${\left(\frac{1+\sin {\displaystyle \frac{\pi }{8}}+i \cos {\displaystyle \frac{\pi }{8}}}{1+\sin \frac{\pi }{8}-i \cos \frac{\pi }{8}}\right)}^{8/3} is -1$

Consider L.H.S=${\left(\frac{1+\sin {\displaystyle \frac{\pi }{8}}+i \cos {\displaystyle \frac{\pi }{8}}}{1+\sin \frac{\pi }{8}-i \cos \frac{\pi }{8}}\right)}^{8/3}$

$$\begin{gathered} =\left(\frac{1+\cos \left({\displaystyle \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)+i \sin \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-{\displaystyle \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right) }{1+\cos \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)-i \sin \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)}\right) \\ =\left[\frac{2{\cos }^2{\displaystyle \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16$}\right.}+2i \sin {\displaystyle \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16 $}\right.}\cos {\displaystyle \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}{2{\cos }^2\raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16$}\right.-2i \sin \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16 $}\right.\cos \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$16$}\right.}\right] \\ =\left(\frac{\cos {\displaystyle \frac{3\pi }{16}}+i \sin {\displaystyle \frac{{\displaystyle 3\pi }}{16}}}{\cos {\displaystyle \frac{3\pi }{16}}-i \sin {\displaystyle \frac{3\pi }{16}}}\right)=\frac{\cos {\displaystyle \frac{\pi }{2}}+i \sin {\displaystyle \frac{\pi }{2}}}{\cos \frac{\pi }{2}-i \sin \frac{\pi }{2}} \\ =\frac{e^{{\displaystyle \raisebox{1ex}{$i\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{e^{\raisebox{1ex}{$-i\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=e^{i\pi }=-1 \end{gathered}$$