Prove that \[ \sqrt{2+\sqrt{2+2\cos4x}}=2\cos x \] for \[ 0

Proof: \[ LHS=\sqrt{2+\sqrt{2+2\cos4x}} \] Using the identity \[ 1+\cos4x=2\cos^22x \] we get \[ 2+2\cos4x=2(1+\cos4x) \] \[ =2(2\cos^22x) \] \[ =4\cos^22x \] Therefore, \[ \sqrt{2+2\cos4x}=\sqrt{4\cos^22x} \] \[ =2\cos2x \] since \[ 00 \] Now, \[ LHS=\sqrt{2+2\cos2x} \] Again using \[ 1+\cos2x=2\cos^2x \] we obtain \[ 2+2\cos2x=2(1+\cos2x) \] \[ =2(2\cos^2x) \] \[ =4\cos^2x \] Thus, \[ LHS=\sqrt{4\cos^2x} \] \[ =2\cos x \] since \[ 00 \] Hence proved, \[ \boxed{\sqrt{2+\sqrt{2+2\cos4x}}=2\cos x} \]

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