Halving An Angle Many Times

Let $x=7.5^\circ$ then the expression changes to:

$(\sin 2x - 2\sin^2 x)(\sin 2x + 2\cos^2 x)$

Using the identities $\sin^2 x = \dfrac{1-\cos 2x}{2}$ and $\cos^2 x=\dfrac{1+\cos 2x}{2}$ we have:

$=(\sin 2x +\cos 2x-1)(\sin 2x+\cos 2x+1)$

$=(\sin 2x + \cos 2x)^2 - 1$

$=\sin^2 2x + \cos^2 2x + 2\sin 2x\cos 2x-1$

Using $\sin^2 2x + \cos^2 2x = 1$ and $\sin 4x = 2\sin 2x\cos 2x$ we have:

$=1-1+\sin 4x$

$=\sin 4x$

Re-subsituting $x=7.5^\circ$ ,

$\sin 4x=\sin 30^\circ = \dfrac{1}{2}=\boxed{0.5}$

Feeling the absolute necessity to multiply things out, I got:

$\sin^{2} 15^{\circ} + 2 \sin 15^{\circ} \cos^{2} 7.5^{\circ} - 2 \sin^{2} 7.5^{\circ} \sin 15^{\circ} - 4 \sin^{2} 7.5^{\circ} \cos^{2} 7.5^{\circ}$

$= \sin^{2} 15^{\circ} + 2 \sin 15^{\circ} \cos^{2} 7.5^{\circ} - 2 \sin^{2} 7.5^{\circ} \sin 15^{\circ} - \left(2 \sin 7.5^{\circ} \cos 7.5^{\circ}\right)^2$

$= \sin^{2} 15^{\circ} + 2 \sin 15^{\circ} \cos^{2}7.5^{\circ} - 2 \sin^{2} 7.5^{\circ} \sin 15^{\circ} - \sin^{2} 15^{\circ}$

$= 2 \sin 15^{\circ} \left[ \cos^{2} 7.5^{\circ} - \sin^{2} 7.5^{\circ} \right]$

$= 2 \sin 15^{\circ} \cos 15^{\circ}$

$= \sin 30^{\circ } = \boxed{\dfrac{1}{2}}$