If cos theta = frac{1}{2}left( {x + frac{1}{x | vedclass

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Question

(b) Given that $\cos 	heta = \frac{1}{2}\,\left( {x + \frac{1}{x}} ight)\,\, $

$\Rightarrow \,x + \frac{1}{x} = 2\,\cos 	heta $

We know that

Vedclass · Accepted Answer

$\cos \,2	heta $

Vedclass · Answer

(b) Given that $\cos 	heta = \frac{1}{2}\,\left( {x + \frac{1}{x}} ight)\,\, $

$\Rightarrow \,x + \frac{1}{x} = 2\,\cos 	heta $

We know that ${x^2} + \frac{1}{{{x^2}}} = {\left( {x + \frac{1}{x}} ight)^2} - 2$

$ = {(2\cos 	heta )^2} - 2 = 4\,{\cos ^2}	heta - 2 = 2\,\cos \,\,2	heta $

$	herefore \,\,\frac{1}{2}\,\left( {{x^2} + \frac{1}{{{x^2}}}} ight) = \frac{1}{2} 	imes 2\,\cos \,2	heta = \cos \,2	heta $

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

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