Question

If $\cos\theta = \frac{1}{2}\left( x + \frac{1}{x} \right)$, then $\frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} \right) =$

• A. $\sin 2\theta$
• B. $\cos 2\theta$
• C. $\tan 2\theta$
• D. $\sec 2\theta$

Answer 1

Answer: (B)

$\cos 2\theta$

Answer 2

Given that $\cos\theta = \frac{1}{2}\left( x + \frac{1}{x} \right) \Rightarrow x + \frac{1}{x} = 2\cos\theta$

We know that $x^{2} + \frac{1}{x^{2}} = \left( x + \frac{1}{x} \right)^{2} - 2$

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

$\therefore\frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} \right) = \frac{1}{2} \times 2\cos 2\theta = \cos 2\theta$