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3.Trigonometrical Ratios, Functions and Identities
easy
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 $
Solution
(b) 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 $
Standard 11
Mathematics
