If ${\sin }^4\alpha +4{\cos }^4\beta +2=4\sqrt{2}$ $\sin \alpha \cos \beta ;\alpha ,\beta \in [0,\pi ]$ then  $\cos (\alpha +\beta )-\cos (\alpha -\beta )$ is equal to 

 

Since,$\text{ A.M. }\ge G.M$

$\frac{{\sin }^4\alpha +4{\cos }^4\beta +1+1}{4}\ge {\left({\sin }^4\alpha \cdot 4{\cos }^4\beta \cdot 1\cdot 1\right)}^{1/4}$

$\Rightarrow {\sin }^4\alpha +4{\cos }^4\beta +2\ge 4\sqrt{2}\sin \alpha \cos \beta$

But  according to question, we have

$\Rightarrow \text{ A.M. }=\text{ G.M. }\Rightarrow \sin \alpha =1=4{\cos }^4\beta$

$\Rightarrow \sin \alpha =1,\cos \beta =\pm 1/\sqrt{2}$

$\therefore \sin \alpha =1\text{ and }\sin \beta =1/\sqrt{2}\text{ as }\alpha ,\beta \in [0,\pi ]$

Now, $\cos (\alpha +\beta )-\cos (\alpha -\beta )=-2\sin \alpha \sin \beta$

$=-2\cdot 1\cdot 1/\sqrt{2}=-\sqrt{2}$