Q108 / 160 · TS EAMCET 2017 · Mathematics
$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)-i \sin \left(\frac{\pi}{12}\right)}\right]^{72}=$
- A0
- B-1
- Correct answerC1
- D$\frac{1}{2}$
Solution
Consider, $\left[\frac{\left(1+\cos \frac{\pi}{12}\right)+i \sin \frac{\pi}{12}}{\left(1+\cos \frac{\pi}{12}\right)-i \sin \frac{\pi}{12}}\right]^{72}$
$$
\begin{aligned}
& =\left[\frac{2 \cos ^2 \frac{\pi}{24}+i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos ^2 \frac{\pi}{24}-i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}\right]^{72} \\
& =\left[\frac{2 \cos \frac{\pi}{24}\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{2 \cos \frac{\pi}{24}\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72} \\
& =\left(\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right)^{72} \\
& =\left(\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)} \times \frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}\right)^{72} \\
& =\left(\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^2}{\cos ^2 \frac{\pi}{24}-i^2 \sin ^2 \frac{\pi}{24}}\right)^{72} \\
& =\left(\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^2}{\cos ^2 \frac{\pi}{24}+\sin ^2 \frac{\pi}{24}}\right)^{72}=\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{144} \\
&
\end{aligned}
$$
$$
=\cos \left(\frac{\pi}{24} \times 144\right)+i \sin \left(\frac{\pi}{24} \times 144\right)
$$
[By Demoivre's theorem]
$$
\begin{aligned}
& =\cos 6 \pi+i \sin 6 \pi \\
& =1 \quad\left[\begin{array}{l}
\because \sin n \pi=0 \forall n \in Z \\
\text { and } \cos n \pi=\left\{\begin{aligned}
1, & \text { if } n \text { is even } \\
-1, & \text { if } n \text { is odd }
\end{aligned}\right]
\end{array}\right. \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.