The value of $(sin\frac{\pi }{8}+i\cos \frac{\pi }{8})^{8}{(sin\frac{\pi }{8}-i cos \frac{\pi }{8})^{8}}$ is -

${\left( {\sin \pi /8 + i\cos \pi /8} \right)^8}{\left( {\sin \frac{\pi }{8} - 2\cos \frac{\pi }{8}} \right)^8}$

$\Rightarrow \left( {\sin \pi + i\cos \pi } \right)\left( {\sin \pi - i\cos \pi } \right)$

$\Rightarrow - {i^2}\left( {\cos \pi - i\sin \pi } \right)\left( {\cos \pi + 2\sin \pi } \right)$

$\Rightarrow 1 \times {e^{ - \pi i}} \cdot {e^{\pi 2}}\,\,\,\,\,\left[ {\because {i^{ - 2}} = - i} \right]$

$\Rightarrow {e^0} = 1\,\,\,\,Ans.$