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

We are given the equality

$$\sin^4\alpha + 4\cos^4\beta + 2 = 4\sqrt2\;\sin\alpha\cos\beta\,,\qquad \alpha,\,\beta\in[0,\pi].$$

Because $$\sin\alpha\ge 0$$ for every $$\alpha\in[0,\pi],$$ we set

$$\sin\alpha=\sqrt{s}\quad(s\ge 0),\qquad \cos\beta = C\quad(C\in[-1,1]).$$

Then $$s=\sin^2\alpha\in[0,1]$$ and $$C^2=\cos^2\beta\in[0,1].$$ Substituting these symbols in the given equation we obtain

$$s^2 + 4C^4 + 2 = 4\sqrt2\,\sqrt{s}\,C. \quad -(1)$$

The sign of the right-hand side is the sign of $$C$$. The left-hand side is clearly positive. Hence $$C$$ must be non-negative, i.e. $$C\ge 0\;(0\le\beta\le\pi/2).$$ Therefore we can write $$C=\sqrt{c}\,(c=C^2).$$ It is more convenient to remove the radicals completely, so we introduce two new non-negative variables

$$v=\sqrt{s}=\sin\alpha,\qquad u=\sqrt2\,C=\sqrt2\,\cos\beta.$$

Because $$0\le s\le 1$$ and $$0\le C\le 1,$$ we have $$0\le v\le 1,\;0\le u\le\sqrt2.$$ Now (1) becomes

$$v^4+u^4+2=4uv. \quad -(2)$$

To analyse (2) we separate two non-negative parts.

First write

$$v^4+u^4=(v^2-u^2)^2+2v^2u^2,$$ so that

$$(2)\;\Longrightarrow\;(v^2-u^2)^2+2v^2u^2+2-4uv=0. \quad -(3)$$

We now estimate the second bracket in (3) with the Arithmetic-Geometric Mean inequality. For any non-negative numbers $$x,y$$ we have $$\dfrac{x+1}{2}\ge\sqrt{x}.$$ Putting $$x=v^2u^2$$ yields

$$v^2u^2+1\;\ge\;2vu\quad\Longrightarrow\quad 2v^2u^2+2\;\ge\;4vu.$$

Hence

$$2v^2u^2+2-4vu\;\ge\;0.$$

Both summands in (3) are therefore non-negative, and their sum equals zero. Consequently each must vanish.

1. $$(v^2-u^2)^2=0\;\Longrightarrow\;v^2=u^2\;\Longrightarrow\;v=u.$$

2. $$2v^2u^2+2-4vu=0\;\Longrightarrow\;v^2u^2+1=2vu.$$ Substituting $$u=v$$ from step 1 we get $$v^4+1=2v^2\;\Longrightarrow\;(v^2-1)^2=0\;\Longrightarrow\;v=1.$$

Thus

$$v=1,\qquad u=1.$$ Reverting to the original trigonometric quantities, we have

$$\sin\alpha=v=1\;\Longrightarrow\;\alpha=\dfrac{\pi}{2},$$

$$\sqrt2\,\cos\beta=u=1\;\Longrightarrow\;\cos\beta=\dfrac1{\sqrt2}\; \Longrightarrow\;\beta=\dfrac{\pi}{4}.$$

Now we evaluate the required expression. First state the identity that will be used:

$$\cos(A+B)-\cos(A-B)=-2\sin A\sin B.$$

Using $$A=\alpha,\;B=\beta$$ together with $$\sin\alpha=1,\;\sin\beta=\dfrac1{\sqrt2},$$ we get

$$\cos(\alpha+\beta)-\cos(\alpha-\beta)= -2\sin\alpha\sin\beta =-2\left(1\right)\left(\dfrac1{\sqrt2}\right)=-\sqrt2.$$

Hence, the correct answer is Option B.