Q59 / 155 · WBJEE 2016 · Mathematics
$1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$ equals
- Correct answerA$\left(2 \cos \frac{\theta}{2}\right)^{n} \cos \frac{n \theta}{2}$
- B$2 \cos ^{2} \frac{n \theta}{2}$
- C$2 \cos ^{2 n} \frac{\theta}{2}$
- D$\left(2 \cos ^{2} \frac{\theta}{2}\right)^{n}$
Solution
Given, $1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$
Which is real part of complex number. $\left({ }^{n} C_{0}+{ }^{n} C_{1} e^{i\theta}+\ldots\right)$
ie. $\operatorname{Re}\left({ }^{n} \mathrm{C}_{0}+{ }^{n} \mathrm{C}_{1} e^{i\theta}+\ldots\right)$
$=\operatorname{Re}\left(1+e^{\theta i}\right)^{n}=\operatorname{Ro}(1+\cos \theta+i \sin \theta)^{n}$
$=$ Re $\left(2 \cos ^{2} n \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}\right)$
$\left[\begin{array}{c}\because 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2} \text { and } \\ \sin \theta=2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}\end{array}\right]$
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \operatorname{Re}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)^{n}$
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \operatorname{Re}\left(\cos \frac{n \theta}{2}+i \sin \frac{\theta}{2}\right)$
[by De moivre's theorem]
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \cdot \cos \frac{n \theta}{2}$
Hence, the correct answer is option: 'A'.