Question

$\left[\frac{1+\cos \left(\frac{\pi }{12}\right)+i\sin \left(\frac{\pi }{12}\right)}{1+\cos \left(\frac{\pi }{12}\right)-i\sin \left(\frac{\pi }{12}\right)}\right]=$

• A. 0
• B. -2
• C. 1
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Consider, ${\left[\frac{\left(1+\cos \frac{\pi }{12}\right)+i\sin \frac{\pi }{12}}{\left(1+\cos \frac{\pi }{12}\right)-i\sin \frac{\pi }{12}}\right]}^{72}$
$={\left[\frac{2{\cos }^2\frac{\pi }{24}+i2\sin \frac{\pi }{24}\cos \frac{\pi }{24}}{2{\cos }^2\frac{\pi }{24}-i2\sin \frac{\pi }{24}\cos \frac{\pi }{24}}\right]}^{72}$
$={\left[\frac{2\cos \frac{\pi }{24}\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}{2\cos \frac{\pi }{24}\left(\cos \frac{\pi }{24}-i\sin \frac{\pi }{24}\right)}\right]}^{72}$
$={\left(\frac{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}{\left(\cos \frac{\pi }{24}-i\sin \frac{\pi }{24}\right)}\right)}^{72}$
$={\left(\frac{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}{\left(\cos \frac{\pi }{24}-i\sin \frac{\pi }{24}\right)}\times \frac{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}\right)}^{72}$
$={\left(\frac{{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}^2}{{\cos }^2\frac{\pi }{24}-i^2{\sin }^2\frac{\pi }{24}}\right)}^{72}$
$={\left(\frac{{\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}^2}{{\cos }^2\frac{\pi }{24}+{\sin }^2\frac{\pi }{24}}\right)}^{72}={\left(\cos \frac{\pi }{24}+i\sin \frac{\pi }{24}\right)}^{144}$
$=\cos \left(\frac{\pi }{24}\times 144\right)+i\sin \left(\frac{\pi }{24}\times 144\right)$
[By Demoivre's theorem]
$=\cos 6\pi +i\sin 6\pi$