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Prove the following

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:

  1. 0

  2. $-\sqrt{2}$

  3. $-1$

  4. $\sqrt{2}$


Correct Option: , 2

Solution:

A.M. $\geq$ G.M.

$\frac{\sin ^{4} \alpha+4 \cos ^{4} \beta+1+1}{4} \geq\left(\sin ^{4} \alpha \cdot 4 \cos ^{4} \beta .1 .1\right)^{\frac{1}{4}}$

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

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 $\beta \in[0, \pi]$

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

$=-\sqrt{2}$

 

 

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