\( 2 \quad ( \cos 2 ^ { \circ } + \sin 2 ^ { \circ } ) ^ { 2 } + ( \cos 2 ^ { \circ } - \sin 2 ^ { \circ } ) ^ { 2 } + \)

To solve the expression \( 2 \quad ( \cos 2 ^ { \circ } + \sin 2 ^ { \circ } ) ^ { 2 } + ( \cos 2 ^ { \circ } - \sin 2 ^ { \circ } ) ^ { 2 }\), let's break it down step by step. First, simplify each squared term: 1. For \( ( \cos 2^{\circ} + \sin 2^{\circ} )^{2} \): \[ ( \cos 2^{\circ} + \sin 2^{\circ} )^{2} = \cos^2 2^{\circ} + 2 \cos 2^{\circ} \sin 2^{\circ} + \sin^2 2^{\circ} \] Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \): \[ = 1 + 2 \cos 2^{\circ} \sin 2^{\circ} = 1 + \sin 4^{\circ} \] (since \(2 \sin A \cos A = \sin 2A\)). 2. For \( ( \cos 2^{\circ} - \sin 2^{\circ} )^{2} \): \[ ( \cos 2^{\circ} - \sin 2^{\circ} )^{2} = \cos^2 2^{\circ} - 2 \cos 2^{\circ} \sin 2^{\circ} + \sin^2 2^{\circ} \] Again, using the Pythagorean identity: \[ = 1 - 2 \cos 2^{\circ} \sin 2^{\circ} = 1 - \sin 4^{\circ}. \] Now substitute these back into the original expression: \[ 2(1 + \sin 4^{\circ}) + (1 - \sin 4^{\circ}). \] This simplifies to: \[ 2 + 2\sin 4^{\circ} + 1 - \sin 4^{\circ} = 3 + \sin 4^{\circ}. \] Thus, the value of the expression is: \[ \boxed{3 + \sin 4^{\circ}}. \]