Question

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

• A. $\sin 2\theta$
• B. $\cos 2\theta$
• C. $\tan 2\theta$
• D. $sec2\theta$

Answer 1

The correct option is B

$\cos 2\theta$

Find the value of $\left(\frac{1}{2}\right)\left(x^2+\frac{1}{x^2}\right)$:

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

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

Step 2. By squaring on both sides, we get

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

$\Rightarrow$ $4{\cos }^2\theta =x^2+\frac{1}{x^2}+2$

$\Rightarrow$ $4{\cos }^2\theta -2=x^2+\frac{1}{x^2}$

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

$\Rightarrow$ $2\left(\cos 2\theta \right)=x^2+\frac{1}{x^2}$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\right]$

$\mathbf{\therefore }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }$

Hence, Option ‘B’ is Correct.