Question

${\left[\frac{1+\cos \left(\frac{\theta }{2}\right)-i\sin \left(\frac{\theta }{2}\right)}{1+\cos \left(\frac{\theta }{2}\right)+i\sin \left(\frac{\theta }{2}\right)}\right]}^{4n}$ is equal to

• A. $\cos n\theta -i\sin n\theta$
• B. $\cos n\theta +i\sin n\theta$
• C. $\cos 2n\theta -i\sin 2n\theta$
• D. $\cos 2n\theta +i\sin 2n\theta$

Answer 1

The correct option is C $\cos 2n\theta -i\sin 2n\theta$

${\left[\frac{1+\cos \left(\frac{\theta }{2}\right)-i\sin \left(\frac{\theta }{2}\right)}{1+\cos \left(\frac{\theta }{2}\right)+i\sin \left(\frac{\theta }{2}\right)}\right]}^{4n}$
$={\left[\frac{2{\cos }^2\frac{\theta }{4}-2i\sin \frac{\theta }{4}\cos \frac{\theta }{4}}{2{\cos }^2\frac{\theta }{4}-2i\sin \frac{\theta }{4}\cos \frac{\theta }{4}}\right]}^{4n}$
$={\left[\frac{e^{\frac{-i\theta }{4}}}{e^{\frac{i\theta }{4}}}\right]}^{4n}={\left[e^{\frac{-i\theta }{2}}\right]}^{4n}$
$=e^{-2in\theta }=\cos 2n\theta -i\sin 2n\theta$