Discussion Forum

$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i\sin \left(\frac{\pi}{12}\right)}{1+ \cos\left(\frac{\pi}{12}\right)-i\sin\left(\frac{\pi}{12}\right)}\right]^{72}=$

Answer:

Explanation:

Consider ,$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i\sin \left(\frac{\pi}{12}\right)}{1+ \cos\left(\frac{\pi}{12}\right)-i\sin\left(\frac{\pi}{12}\right)}\right]^{72}$

=$\left[\frac{2\cos^{2} \frac{\pi}{24}+i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}{2\cos^{2} \frac{\pi}{24}-i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}\right]^{72}$

 =$\left[\frac{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}+i\sin \frac{\pi}{24}\right)}{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}-i\sin \frac{\pi}{24}\right)}\right]^{72}$

       =$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$

=   $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\times \frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$

     =   $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\left(\cos^{2} \frac{\pi}{24}-i^{2} \sin^{2} \frac{\pi}{24}\right)}\right]^{72}$

=$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\cos^{2} \frac{\pi}{24}+ \sin^{2} \frac{\pi}{24}}\right]^{72}=\left( \cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{144}$

       = $\cos \left(\frac{\pi}{24}\times144\right)+i\sin\left( \frac{\pi}{24}\times144\right)$ 

                                                  [ By Dermoivre's theorm]

=$\cos 6\pi+i \sin 6 \pi$

=1                  $\left[ \because \sin n\pi=0 \forall n\in Z\right]$   and  

                                [ $\cos n \pi= \begin{cases}1, & if n is even\\-1, & if n is odd\end{cases}$]