If $x+\frac{1}{x}=2\cos \theta$, then $x^n+\frac{1}{x^n}$ is equal to

The correct option is B

$2\cos n\theta$

Explanation for the correct option:

Step 1. Find the value of $x$ and $\frac{1}{x}$.

It is given that $x+\frac{1}{x}=2\cos \theta$.

Let $x=\cos \theta +i\sin \theta$. Then the value of $\frac{1}{x}$ is given as:

$\begin{array}{rcl}\frac{1}{x}& =& \frac{1}{\cos \theta +i\sin \theta }\\ & =& \frac{\cos \theta -i\sin \theta }{\left(\cos \theta +i\sin \theta \right)\left(\cos \theta -i\sin \theta \right)}\\ & =& \frac{\cos \theta -i\sin \theta }{{\cos }^2\theta -i^2{\sin }^2\theta }\\ & =& \frac{\cos \theta -i\sin \theta }{{\cos }^2\theta +{\sin }^2\theta }\\ & =& \frac{\cos \theta -i\sin \theta }{1}\\ & =& \cos \theta -i\sin \theta \end{array}$

Now, using Euler's representation $e^{i\theta }=\cos \theta +i\sin \theta$ it can be seen that

$x=\cos \theta +i\sin \theta =e^{i\theta }$ and

$\begin{array}{rcl}\frac{1}{x}& =& \cos \theta -i\sin \theta \\ & =& \cos \left(-\theta \right)+i\sin \left(-\theta \right)\\ & =& e^{i\left(-\theta \right)}\end{array}$

Step 2. Find the value of $x^n+\frac{1}{x^n}$.

The value of $x^n+\frac{1}{x^n}$ is given as:

$\begin{array}{rcl}{x}^n+\frac{1}{x^n}& =& x^n+{\left(\frac{1}{x}\right)}^n\\ & =& {\left(e^{i\theta }\right)}^n+{\left(e^{i\left(-\theta \right)}\right)}^n\\ & =& e^{i\left(n\theta \right)}+e^{i\left(-n\theta \right)}\\ & =& \cos n\theta +i\sin n\theta +\cos (-n\theta )+i\sin (-n\theta )\left[e^{i\theta }=\cos \theta +i\sin \theta \right]\\ & =& \cos n\theta +i\sin n\theta +\cos n\theta -i\sin n\theta \\ & =& 2\cos n\theta \end{array}$

So, the value of $x^n+\frac{1}{x^n}$ is $2\cos n\theta$.

Hence, the correct option is (B) .