If s i n 4 α + 4 c o s 4 β + 2 = 4 2 s i n α c o s β , α ,   β ∈ 0 …

We have,

${\sin }^4\alpha +4{\cos }^4\beta +2=4\sqrt{2}\sin \mathrm{ }\alpha \cos \beta$

Applying $\mathrm{AM}\ge \mathrm{GM}$, we get

$\left(\frac{{\sin }^4\alpha +4{\cos }^4\beta +1+1}{4}\right)\ge {\left(4\mathrm{s}\mathrm{i}{\mathrm{n}}^4\alpha ·\mathrm{c}\mathrm{o}{\mathrm{s}}^4\beta ·1·1\right)}^{\frac{1}{4}}$

$\Rightarrow {\sin }^4\alpha +4{\cos }^4\beta +2\ge 8\sin \alpha \cos \beta$

And, $\mathrm{AM}=\mathrm{GM}\Rightarrow$numbers are equal and positive

$\Rightarrow {\sin }^4\alpha =4{\cos }^4\beta =1$

$\therefore \mathrm{s}\mathrm{i}{\mathrm{n}}^4\alpha =1\Rightarrow \alpha =\frac{\mathrm{\pi }}{2}$ and

$\mathrm{c}\mathrm{o}\mathrm{s}\beta =\frac{1}{\sqrt{2}}\Rightarrow \beta =\frac{\mathrm{\pi }}{4}$

Hence,

$\cos \left(\alpha +\beta \right)-\cos \left(\alpha -\beta \right)$

$=-2\sin \alpha \sin \beta$

$=-2\times 1\times \frac{1}{\sqrt{2}}=-\sqrt{2}$