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

Show Answer

Answer:

Correct Answer: 4. (c)

Solution:

  1. 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} $



NCERT Chapter Video Solution

Dual Pane