Q95 / 180 · COMEDK 2013 · Mathematics
If $x+\frac{1}{x}=2 \cos \alpha$, then $x^{n}+\frac{1}{x^{n}}$ is equal to
- A$2^{n} \cos \alpha$
- B$2^{n} \cos n \alpha$
- C$2 i \sin n \alpha$
- Correct answerD$2 \cos n \alpha$
Solution
We have,
$x+\frac{1}{x}=2 \cos \alpha$
$\Rightarrow \quad x^{2}+1=2 x \cos \alpha$
$\Rightarrow \quad x^{2}-(2 \cos \alpha) x+1=0$
$\Rightarrow \quad x=\frac{2 \cos \alpha \pm \sqrt{4 \cos ^{2} \alpha-4}}{2}$
$=\frac{2 \cos \alpha \pm \sqrt{-4 \sin ^{2} \alpha}}{2}$
$=\frac{2 \cos \alpha \pm 2(\sin \alpha) i}{2}=\cos \alpha \pm(\sin \alpha) i$
Now, $\quad x^{n}=(\cos \alpha \pm(\sin \alpha) i)^{n}=\cos n \alpha \pm i \sin n \alpha$
and $\frac{1}{x^{n}}=(\cos \alpha \pm(\sin \alpha) i)^{-n}=\cos n \alpha \mp i \sin n \alpha$
$\therefore x^{n}+\frac{1}{x^{n}}=(\cos n \alpha \pm i \sin n \alpha)+(\cos n \alpha \mp i \sin n \alpha)$
$=2 \cos n \alpha$
Hence, the correct answer is option: 'D'.