MCQ
જો $\sqrt x + \frac{1}{{\sqrt x }} = 2\cos \theta ,$ તો ${x^6} + {x^{ - 6}} = $
- A$2\cos 6\theta $
- ✓$2 \cos 12\theta$
- C$2\cos 3\theta $
- D$2\sin 3\theta $
On squaring both sides,
we get $x + \frac{1}{x} + 2 = 4\,{\cos ^2}\theta $
==> $x + \frac{1}{x} = 4{\cos ^2}\theta - 2$
==> $x + \frac{1}{x} = $$2(2{\cos ^2}\theta - 1)$ $ = 2\cos 2\theta $…..$(ii)$
Again squaring both sides, ${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}2\theta $
==> ${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}2\theta - 2$$ = 2(2{\cos ^2}2\theta - 1)$
==> ${x^2} + \frac{1}{{{x^2}}} = 2\cos 4\theta $…..$(iii)$
Now take cube of both sides, ${\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^3} = {(2\cos 4\theta )^3}$
==> ${x^6} + \frac{1}{{{x^6}}} + 3{x^2} \times \frac{1}{{{x^2}}}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 8{\cos ^3}4\theta $
==> ${x^6} + \frac{1}{{{x^6}}} + 3\,(2\cos 4\theta ) = 8{\cos ^3}4\theta $
$ \Rightarrow {x^6} + \frac{1}{{{x^6}}} = 8{\cos ^3}4\theta - 6\cos 4\theta $
$= 2\,(4{\cos ^3}4\theta - 3\cos 4\theta )$
$= 2\cos 3(4\theta ) = 2\cos 12\theta $.
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