Question

Simplify : $\frac{{(\sin \pi /8+i\cos \pi /8)}^8}{{(\sin \pi /8-i\cos \pi /8)}^8}$

Answer 1

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

$=(\frac{\sin \pi /8+i\cos \pi /8}{\sin \pi /8-i\cos \pi /8}\times \frac{\sin \pi /8+i\cos \pi /8}{\sin \pi /8+i\cos \pi /8})$

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

$={\left(\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}\right)}^{16}$

$={\left[\cos (\frac{\pi }{2}-\frac{\pi }{8})+i\sin (\frac{\pi }{2}-\frac{\pi }{8})\right]}^{16}$

$={\left(\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8}\right)}^{16}$

$=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}$

$=0-i$

$=-i$