If $A=\sin {48}^0+\cos {48}^0,B=\sin {44}^0+\cos {44}^0$ then

$4{\tan }^2\theta =\sin \theta \cos \theta \Rightarrow \sin \theta ={\cos }^3\theta$

Given expression can be written is 

${\sin }^6\theta -3{\sin }^2\theta +3{\sin }^2\theta -1+3+{\sin }^2\theta$

${\left({\sin }^2\theta -1\right)}^3+3+{\sin }^2\theta =-{\cos }^6\theta +3+{\sin }^2\theta$