Q. If $x + \frac{1}{x} = 2\cos\theta$, then $x^{n} + \frac{1}{x^{n}}$ is equal to
A) $2\sin{n\theta}$
B) $2\cos{n\theta}$
C) $\sin{(2n\theta)}$
D) $\cos{(2n\theta)}$
Explanation
(b)
$x^{2} + \frac{1}{x^{2}} = \left( x + \frac{1}{x} \right)^{2} - 2 = 4\cos^{2}{\theta - 2}$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 2\cos{2\theta}$
Again $x^{3} + \frac{1}{x^{3}} + 3\left( x + \frac{1}{x} \right) = 8\cos^{3}\theta$
$x^{2} + \frac{1}{x^{3\ \ }} = 8\cos^{3}{\theta - 6\cos{\theta = \cos{3\theta}}}$
Similarly, $x^{n} + \frac{1}{x^{n}} = 2\cos{n\theta}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.