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To prove:
$2 \sin 11^\circ 15' = \sqrt{2 - \sqrt{2 + \sqrt{2}}}$

Proof:
Let $\theta = 11^\circ 15' = \left(11\frac{1}{4}\right)^\circ = \left(\frac{45}{4}\right)^\circ$

Then, $2\theta = 22^\circ 30' = \left(\frac{45}{2}\right)^\circ$ and $4\theta = 45^\circ$

We know that:
$\cos 45^\circ = \frac{1}{\sqrt{2}}$

Using the half-angle formula:
$2\cos^2 2\theta = 1 + \cos 4\theta = 1 + \cos 45^\circ = 1 + \frac{1}{\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$
$\Rightarrow 4\cos^2 2\theta = 2 + \sqrt{2}$
$\Rightarrow 2\cos 2\theta = \sqrt{2 + \sqrt{2}}$ (since $22.5^\circ$ is in the first quadrant, $\cos 2\theta > 0$)

Now, for $\sin \theta$:
$2\sin^2 \theta = 1 - \cos 2\theta$
Multiplying both sides by 2:
$4\sin^2 \theta = 2 - 2\cos 2\theta$

Substituting the value of $2\cos 2\theta$:
$4\sin^2 \theta = 2 - \sqrt{2 + \sqrt{2}}$

Taking the square root on both sides (since $\sin 11^\circ 15' > 0$):
$2\sin \theta = \sqrt{2 - \sqrt{2 + \sqrt{2}}}$

$\therefore 2\sin 11^\circ 15' = \sqrt{2 - \sqrt{2 + \sqrt{2}}}$ (Proved)