Question

If $ \sqrt{x} + \frac{1}{\sqrt{x}} = 2 \, \cos \theta $, then $x^6 + x^6 = $

• A. $ 2 \, \cos 12 \theta $
• B. $ 2 \, \cos 6\theta $
• C. $ 2 \, \sin 3 \theta $
• D. $ 2 \, \cos 3 \theta $

Answer 1

The Correct Option is A

$ \sqrt{x} + \frac{1}{\sqrt{x}} = 2 \, \cos \theta $ ...(i)
Squaring (i) both sides, we get
$x + \frac{1}{x} + 2 = 4 \cos^{2} \theta $
$\Rightarrow x + \frac{1}{x} = 4 \cos ^{2} \theta - 2 = 2\left(2 \cos ^{2} \theta-1\right) $
$\Rightarrow x + \frac{1}{x} = 2\left( \cos 2\theta\right) $...(ii)
Cubing (ii) both sides, we get
$\Rightarrow \left(x +\frac{1}{x}\right)^{3} = 8 \cos^{3} 2\theta$
$ \Rightarrow x^{3} + \frac{1}{x^{3}} + 3\left(x +\frac{1}{x}\right) = 8 \cos^{3} 2 \theta $
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 8 \cos ^{3} 2\theta - 6 \cos 2\theta = 2\left(4 \cos ^{3} 2\theta - 3 \cos 2\theta\right) $
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 2\left(\cos 6\theta\right)$...(iii)
Squaring (iii) both sides we get
$\Rightarrow \left(x^{3} + \frac{1}{x^{3}}\right)^{2} = 4 \cos ^{2} 6 \theta$
$ x^{6} + \frac{1}{x^{6}} + 2 = 4 \cos ^{2} 6 \theta $
$= 4 \cos ^{2} 6 \theta - 2 = 2\left(2 \cos ^{2} 6\theta - 1\right)$
$\Rightarrow x^{6} + \frac{1}{x^{6} } = 2 \cos 12 \theta$