Question

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 :

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $-\sqrt{2}$

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

Applying $A.M\ge G.M.$ on LHS
$\frac{{\sin }^4\alpha +4{\cos }^4\beta +1+1}{4}\ge \sqrt[4]{4{\sin }^4\alpha \times 4{\cos }^4\beta \times 1\times 1}$
$\Rightarrow {\sin }^4\alpha +4{\cos }^4\alpha +2\ge 4\sqrt{2}\sin \alpha \cos \beta$

But given that ${\sin }^4\alpha +4{\cos }^4\beta +2=4\sqrt{2}\sin \alpha \cos \beta$
This is possible only if
${\sin }^4\alpha =4{\cos }^4\beta =1$
$\therefore \sin \alpha =\pm 1$ and $\cos \beta =\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \sin \alpha =1\text{and}\cos \beta =\pm \frac{1}{\sqrt{2}}(\because \alpha ,\beta \in [0,\pi \left]\right)$
$\therefore \sin \beta =\frac{1}{\sqrt{2}}$

$\cos (\alpha +\beta )-\cos (\alpha -\beta )$
$=-2\sin \alpha \sin \beta$
$=-\sqrt{2}$