Question

Let $x+\dfrac{1}{x}=2cosα$ .For any $n∈N$,the value of $x^n+\dfrac{1}{x^n}​$ is .

• A. $cos(nα)$
• B. $2cos(nα)$
• C. $2isin(nα)$
• D. $isin(nα)$
• E. $4cos(nα)$

Answer 1

Answer: (B)

$2cos(nα)$

Answer 2

Give that

Let $x+\dfrac{1}{x}=2cosα$. For any $n∈N$

$x=cosα+isinα=e^{iα}$

So, $\dfrac{1}{x}​=x^{−1}=e^{−iα}=cosα−isinα$

$x+\dfrac{1}{x}​=2cosα$

So now , $x^n+\dfrac{1}{x^n}​=x^n+x^{−n}$

         $ =(cosα+isinα)^n+(cosα+isinα)^{−n}$

          $=cosnα+isinnα+sinnα−isinnα$  
          $=2cosnα$ (_Ans)