The value of (2 12^ + 108^ )^2 + (2 12^ - 18^ )^2 is

Let the given expression be $E = (2 \cos 12^\circ + \cos 108^\circ)^2 + (2 \sin 12^\circ - \cos 18^\circ)^2$.

Expanding both squares, we get:

$E = (4 \cos^2 12^\circ + \cos^2 108^\circ + 4 \cos 12^\circ \cos 108^\circ) + (4 \sin^2 12^\circ + \cos^2 18^\circ - 4 \sin 12^\circ \cos 18^\circ)$

Grouping the terms:

$E = 4(\cos^2 12^\circ + \sin^2 12^\circ) + (\cos^2 108^\circ + \cos^2 18^\circ) + 4(\cos 12^\circ \cos 108^\circ - \sin 12^\circ \cos 18^\circ)$

We know that $\cos 108^\circ = \cos(90^\circ + 18^\circ) = -\sin 18^\circ$.

Therefore, $\cos^2 108^\circ = \sin^2 18^\circ$, and the second term becomes:

$\sin^2 18^\circ + \cos^2 18^\circ = 1$

The first term is simply $4(1) = 4$.

Now, simplify the cross terms:

$4(\cos 12^\circ (-\sin 18^\circ) - \sin 12^\circ \cos 18^\circ)$

$= -4(\sin 18^\circ \cos 12^\circ + \cos 18^\circ \sin 12^\circ)$

Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:

$= -4 \sin(18^\circ + 12^\circ) = -4 \sin 30^\circ = -4 \left(\dfrac{1}{2}\right) = -2$

Adding all the parts together:

$E = 4 + 1 - 2 = 3$

Alternatively, evaluating the terms inside the brackets directly:

$2 \cos 12^\circ + \cos 108^\circ = \cos 12^\circ + (\cos 12^\circ + \cos 108^\circ) = \cos 12^\circ + 2 \cos 60^\circ \cos 48^\circ = \cos 12^\circ + \cos 48^\circ = 2 \cos 30^\circ \cos 18^\circ = \sqrt{3} \cos 18^\circ$

$2 \sin 12^\circ - \cos 18^\circ = 2 \sin 12^\circ - \sin 72^\circ = \sin 12^\circ + (\sin 12^\circ - \sin 72^\circ) = \sin 12^\circ - 2 \sin 30^\circ \cos 42^\circ = \sin 12^\circ - \cos 42^\circ = \sin 12^\circ - \sin 48^\circ = -2 \sin 18^\circ \cos 30^\circ = -\sqrt{3} \sin 18^\circ$

$E = (\sqrt{3} \cos 18^\circ)^2 + (-\sqrt{3} \sin 18^\circ)^2 = 3(\cos^2 18^\circ + \sin^2 18^\circ) = 3$.

Answer: $3$