MCQ
If $x + \frac{1}{x} = 2\,\cos \theta ,$ then ${x^3} + \frac{1}{{{x^3}}} = $
  • A
    $\cos \,\,3\theta $
  • $2\,\cos \,3\theta $
  • C
    $\frac{1}{2}\cos \,3\theta $
  • D
    $\frac{1}{3}\cos \,3\theta $

Answer

Correct option: B.
$2\,\cos \,3\theta $
b
(b) We have $x + \frac{1}{x} = 2\cos \theta $,

Now ${x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3x\frac{1}{x}\left( {x + \frac{1}{x}} \right)$ 

$=  {(2\cos \theta )^3} - 3(2\cos \theta ) = 8{\cos ^3}\theta - 6\cos \theta $

$=  2(4{\cos ^3}\theta - 3\cos \theta ) = 2\cos 3\theta $. 

Trick : Put $x = 1$ $ \Rightarrow $ $\theta = {0^\circ }$. 

Then ${x^3} + \frac{1}{{{x^3}}} = 2 = 2\cos 3\theta $.

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