Question

Using De Moivre's theorem prove that
${\left(\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }\right)}^n=\cos n\theta +i\sin n\theta ,$ where $i=\sqrt{-1}$

Answer 1

$\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }=\frac{2{\cos }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}-i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\frac{2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})}{2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}-i\sin \frac{\theta }{2})}$
$=\frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}=\frac{\cos \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}$ [De Moivre's Theorem]
$=\cos [\frac{\theta }{2}-(-\frac{\theta }{2})]=\cos \theta$
$\therefore {\left(\frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta }\right)}^n=(\cos \theta {)}^n=\cos n\theta +i\sin n\theta$.