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  4. [(1+ cos((π/12)) + i sin((π/12))/1+ cos((π)12) - i sin((π)12)] =

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Q. $\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] = $

TS EAMCET 2017

Solution:

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