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

Answer 1

Answer: (B)

$A.M. \ge G.M.$
$\frac{\sin^4 \alpha + 4 \cos^4 \beta + 1 + 1}{4} \, \ge (\sin^4\alpha.4\cos^4 \beta.1.1)^{\frac{1}{4}}$
$\sin^{4} \, + 4 \cos^2\beta + 2 \ge \, 4\sqrt{2} \sin \alpha \cos \beta$
given that $\sin^4 \alpha + 4 \cos^4 \beta + 2 = 4\sqrt{2} \, \sin\alpha \, \cos\beta$
$\Rightarrow A.M. \, = \, G.M. \Rightarrow \, \sin^4 \alpha = 1 =4 \cos^4 \beta$
$\sin \alpha = 1, \cos\beta \, = \pm; \frac{1}{\sqrt{2}}$
$\Rightarrow \sin \beta \, = \frac{1}{\sqrt{2}} as \beta \in [0, \pi]$
$\cos (\alpha + \beta) - \cos(\alpha - \beta) \, = \, -2 \, \sin \, \alpha \, \sin \, \beta $
$= -\sqrt{2}$