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 $
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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