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

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(B) Given that $\cos 	heta = \frac{1}{2}\left( x + \frac{1}{x} ight)$.Multiplying by $2$,we get $x + \frac{1}{x} = 2 \cos 	heta$.We know the algeb

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$\cos 2	heta $

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(B) Given that $\cos 	heta = \frac{1}{2}\left( x + \frac{1}{x} ight)$.Multiplying by $2$,we get $x + \frac{1}{x} = 2 \cos 	heta$.We know the algebraic identity $x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} ight)^2 - 2$.Substituting the value of $(x + \frac{1}{x})$,we get $x^2 + \frac{1}{x^2} = (2 \cos 	heta)^2 - 2$.$x^2 + \frac{1}{x^2} = 4 \cos^2 	heta - 2 = 2(2 \cos^2 	heta - 1)$.Using the trigonometric identity $\cos 2	heta = 2 \cos^2 	heta - 1$,we have $x^2 + \frac{1}{x^2} = 2 \cos 2	heta$.Therefore,$\frac{1}{2}\left( x^2 + \frac{1}{x^2} ight) = \frac{1}{2} 	imes 2 \cos 2	heta = \cos 2	heta$.

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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