sqrt{2 + sqrt{2 + 2cos 4theta}} = | vedclass

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(A) Given expression: $\sqrt{2 + \sqrt{2 + 2\cos 4	heta}}$Using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 4	heta = 2(1 + \cos 4	het

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$2\cos 	heta$

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(A) Given expression: $\sqrt{2 + \sqrt{2 + 2\cos 4	heta}}$Using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 4	heta = 2(1 + \cos 4	heta) = 2(2\cos^2 2	heta) = 4\cos^2 2	heta$.Substituting this into the expression:$= \sqrt{2 + \sqrt{4\cos^2 2	heta}}$$= \sqrt{2 + 2\cos 2	heta}$Again,using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 2	heta = 2(1 + \cos 2	heta) = 2(2\cos^2 	heta) = 4\cos^2 	heta$.$= \sqrt{4\cos^2 	heta}$$= 2\cos 	heta$.

$\sqrt{2 + \sqrt{2 + 2\cos 4\theta}} = $

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