Question

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

• 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 $2\cos 12$ $\theta$

$\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta$

Squaring, we get $x+\frac{1}{x}=4{\cos }^2\theta -2=2\left(\cos 2\theta \right)$

Squaring, we get $x^2+\frac{1}{x^2}=4{\cos }^{2}2\theta -2=2\left(\cos 4\theta \right)$

Multiplying the two, we get $x^3+\frac{1}{x^3}+x+\frac{1}{x}=4\left(\cos 2\theta \right)\left(\cos 4\theta \right)$

$x^3+\frac{1}{x^3}=4\left(\cos 2\theta \right)\cos 4\theta -2\left(\cos 2\theta \right)=2\cos 6\theta +2\cos 2\theta -2\cos 2\theta =2\cos 6\theta$

Squaring, we get $x^6+\frac{1}{x^6}=4{\cos }^{2}6\theta -2=2\cos 12\theta$