$2 {\sin }^2\theta + 4 \cos (\theta + \alpha ) \sin \alpha \sin \theta + \cos 2(\theta + \alpha ) =$

$2 {\sin }^2 \theta + 2 \cos (\theta + \alpha ) · 2 \sin \alpha \sin \theta +2 {\cos }^2 (\theta + a) - 1$

$= 2 {\sin }^2 \theta + 2 \cos (\theta + \alpha ) [\cos (\theta - a) - \cos (\theta + \alpha \left)\right] + 2 {\cos }^2 (\theta + \alpha ) - 1$

$= 2 {\sin }^2 \theta + 2 ({\cos }^2\theta - {\sin }^2 \alpha ) -1$

$= 2 - 1 - 2 {\sin }^2 \alpha \Rightarrow 1 - 2 {\sin }^2 \alpha = \cos 2 \alpha$.