Trigonometrical Equations 1 Question 4
4. 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
(2019 Main, 12 Jan II)
(a) -1
(b) $\sqrt{2}$
(c) $-\sqrt{2}$
(d) 0
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Answer:
Correct Answer: 4. (c)
Solution:
- By applying $AM \geq GM$ inequality, on the numbers
$\sin ^{4} \alpha, 4 \cos ^{4} \beta, 1$ and 1 , we get
$ \begin{aligned} \quad \frac{\sin ^{4} \alpha+4 \cos ^{4} \beta+2}{4} \geq\left(\left(\sin ^{4} \alpha\right)\left(4 \cos ^{4} \beta\right) \cdot 1 \cdot 1\right)^{1 / 4} \\ \Rightarrow \sin ^{4} \alpha+4 \cos ^{4} \beta+2 \geq 4 \sqrt{2} \sin \alpha \cos \beta \end{aligned} $
But, it is given that
$ \sin ^{4} \alpha+4 \cos ^{4} \beta+2=4 \sqrt{2} \sin \alpha \cos \beta $
So, $\sin ^{4} \alpha=4 \cos ^{4} \beta=1$
$[\because$ In $AM \geq GM$, equality holds when all given
$\Rightarrow \sin \alpha=1$ and $\sin \beta=\frac{1}{\sqrt{2}}$ positive quantities are equal.]
Now, $\cos (\alpha+\beta)-\cos (\alpha-\beta)=-2 \sin \alpha \sin \beta$ $[\because \alpha, \beta \in[0, \pi]]$
$ \begin{aligned} & \because \cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2} \\ & =-2 \times 1 \times \frac{1}{\sqrt{2}} \quad \text {[from eq. (i)]} \\ & =-\sqrt{2} \end{aligned} $
