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Question

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

A

$2\mathrm{sin}n\theta$

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B

$2\mathrm{cos}n\theta$

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C

$\mathrm{sin}\left({2}^{n}\theta \right)$

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D

$\mathrm{cos}\left({2}^{n}\theta \right)$

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Solution

## The correct option is B $2\mathrm{cos}n\theta$Explanation for the correct option:Step 1. Find the value of $x$ and $\frac{1}{x}$.It is given that $x+\frac{1}{x}=2\mathrm{cos}\theta$.Let $x=\mathrm{cos}\theta +i\mathrm{sin}\theta$. Then the value of $\frac{1}{x}$ is given as:$\begin{array}{rcl}\frac{1}{x}& =& \frac{1}{\mathrm{cos}\theta +i\mathrm{sin}\theta }\\ & =& \frac{\mathrm{cos}\theta -i\mathrm{sin}\theta }{\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta -i\mathrm{sin}\theta \right)}\\ & =& \frac{\mathrm{cos}\theta -i\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta -{i}^{2}{\mathrm{sin}}^{2}\theta }\\ & =& \frac{\mathrm{cos}\theta -i\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }\\ & =& \frac{\mathrm{cos}\theta -i\mathrm{sin}\theta }{1}\\ & =& \mathrm{cos}\theta -i\mathrm{sin}\theta \end{array}$Now, using Euler's representation ${e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta$ it can be seen that$x=\mathrm{cos}\theta +i\mathrm{sin}\theta ={e}^{i\theta }$ and$\begin{array}{rcl}\frac{1}{x}& =& \mathrm{cos}\theta -i\mathrm{sin}\theta \\ & =& \mathrm{cos}\left(-\theta \right)+i\mathrm{sin}\left(-\theta \right)\\ & =& {e}^{i\left(-\theta \right)}\end{array}$Step 2. Find the value of ${x}^{n}+\frac{1}{{x}^{n}}$.The value of ${x}^{n}+\frac{1}{{x}^{n}}$ is given as:$\begin{array}{rcl}{x}^{n}+\frac{1}{{x}^{n}}& =& {x}^{n}+{\left(\frac{1}{x}\right)}^{n}\\ & =& {\left({e}^{i\theta }\right)}^{n}+{\left({e}^{i\left(-\theta \right)}\right)}^{n}\\ & =& {e}^{i\left(n\theta \right)}+{e}^{i\left(-n\theta \right)}\\ & =& \mathrm{cos}n\theta +i\mathrm{sin}n\theta +\mathrm{cos}\left(-n\theta \right)+i\mathrm{sin}\left(-n\theta \right)\left[{e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta \right]\\ & =& \mathrm{cos}n\theta +i\mathrm{sin}n\theta +\mathrm{cos}n\theta -i\mathrm{sin}n\theta \\ & =& 2\mathrm{cos}n\theta \end{array}$So, the value of ${x}^{n}+\frac{1}{{x}^{n}}$ is $2\mathrm{cos}n\theta$.Hence, the correct option is (B) .

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