If an expression is given as ${{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta ,\alpha ,\beta \in \left[ 0,\pi \right]$, then $\cos \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right)$ is equal to:
(A)$0$
(B) $-\sqrt{2}$
(C) $-1$
(D) $\sqrt{2}$

Hint: For answering this question we will use the concept that $A\text{rithmetic Mean}\ge \text{Geometric mean}$ and apply it. We will start solving the expression ${{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta ,\text{where }\alpha ,\beta \in \left[ 0,\pi \right]$ by assuming $\sin \alpha =x$ and $\cos \beta =y$ . Then we will derive the values of $\alpha ,\beta $ and substitute them in $\cos \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right)$.

Complete step by step answer:
Now, considering from the question we have the expression ${{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta ,\text{where }\alpha ,\beta \in \left[ 0,\pi \right]$ .
Let us assume $\sin \alpha =x$ and $\cos \beta =y$ by substituting these values we will have
${{x}^{4}}+4{{y}^{4}}+2=4\sqrt{2}xy$ .
Since we know that from the basic concept that the $\text{Arithmetic mean }\ge \text{ Geometric mean}$ .
Here we can simply write the expression as $\dfrac{{{x}^{4}}+4{{y}^{4}}+2}{4}=\sqrt{2}xy$ .
This expression can be further simplified as $\dfrac{{{x}^{4}}+4{{y}^{4}}+1+1}{4}=\sqrt{2}xy$ .
Here we can observe that on the Left hand side we have the arithmetic mean of ${{x}^{4}},4{{y}^{4}},1,1$.
And this can be further simplified as $\dfrac{{{x}^{4}}+4{{y}^{4}}+1+1}{4}={{\left( {{x}^{4}}\times 4{{y}^{4}}\times 1\times 1 \right)}^{\dfrac{1}{4}}}$ .
And on the right hand side we have the geometric mean of ${{x}^{4}},4{{y}^{4}},1,1$ .
Hence we can conclude that the arithmetic and geometric mean of ${{x}^{4}},4{{y}^{4}},1,1$ are equal.
So we can say that all the items have equal value, this can be mathematically given as ${{x}^{4}}=4{{y}^{4}}=1$ .
So now we can derive the value of $x,y$ from that. They will be $x=1\text{ and }y=\dfrac{1}{{{4}^{\dfrac{1}{4}}}}$ .
So now we have $\sin \alpha =1$ and $\cos \beta =\dfrac{1}{\sqrt{2}}$ .
So now we know that $\alpha ,\beta \in \left[ 0,\pi \right]$ so we can say that$\alpha =\dfrac{\pi }{2}\text{ and }\beta =\dfrac{\pi }{4}$ .
By substituting these values in $\cos \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right)$ we will have $\cos \left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right)-\cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{4} \right)$ .
By simplifying this we will have $\cos \left( \dfrac{3\pi }{4} \right)-\cos \left( \dfrac{\pi }{4} \right)$ .
By substituting $\cos \left( \dfrac{3\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}$ and $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ we will have $-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}$ .
Hence by simplifying it we will have $-\sqrt{2}$ .
So we can conclude that when ${{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta ,\alpha ,\beta \in \left[ 0,\pi \right]$, $\cos \left( \alpha +\beta \right)-\cos \left( \alpha -\beta \right)$ is equal to$-\sqrt{2}$ .

So, the correct answer is “Option B”.

Note: While answering this type of questions we should be careful that we should remember that the limit that is $\alpha ,\beta \in \left[ 0,\pi \right]$ then we will go wrong and write it as $\alpha =\left( 4n+1 \right)\dfrac{\pi }{2}$ and this type of values will leave us with a complete mess.