Question

lf $\cos \theta =\frac{1}{2}(x+\frac{1}{x})$, then $\frac{1}{2}(x^2+\frac{1}{x^2})$ is equal to

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

Answer 1

The correct option is B $\cos 2\theta$

Given $\cos \theta =\frac{1}{2}(x+\frac{1}{x})$
$2\cos \theta =x+\frac{1}{x}$
$4{\cos }^2\theta =x^2+\frac{1}{x^2}+2$
$4{\cos }^2\theta -2=x^2+\frac{1}{x^2}$
$2(1+\cos 2\theta )-2=x^2+\frac{1}{x^2}[\cos 2\theta =2{\cos }^2\theta -1]$
$2\cos 2\theta =x^2+\frac{1}{x^2}$
$\Rightarrow \cos 2\theta =\frac{1}{2}(x^2+\frac{1}{x^2})$