SATHEE: Trigonometrical Equations 1 Question 4

Trigonometrical Equations 1 Question 4

4. If $\sin^{4} α + 4 \cos^{4} β + 2 = 4 \sqrt{2} \sin α \cos β$ ;

$α, β \in [0, π]$ , then $\cos (α + β) - \cos (α - β)$ is equal to

(2019 Main, 12 Jan II)

(a) -1

(b) $\sqrt{2}$

(d) 0

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Answer:

Correct Answer: 4. (c)

Solution:

By applying $A M \geq G M$ inequality, on the numbers

$\sin^{4} α, 4 \cos^{4} β, 1$ and 1 , we get

$\begin{array}{r} \frac{\sin^{4} α + 4 \cos^{4} β + 2}{4} \geq {((\sin^{4} α) (4 \cos^{4} β) \cdot 1 \cdot 1)}^{1 / 4} \\ \Rightarrow \sin^{4} α + 4 \cos^{4} β + 2 \geq 4 \sqrt{2} \sin α \cos β \end{array}$

But, it is given that

$\sin^{4} α + 4 \cos^{4} β + 2 = 4 \sqrt{2} \sin α \cos β$

So, $\sin^{4} α = 4 \cos^{4} β = 1$

$[∵$ In $A M \geq G M$ , equality holds when all given

$\Rightarrow \sin α = 1$ and $\sin β = \frac{1}{\sqrt{2}}$ positive quantities are equal.]

Now, $\cos (α + β) - \cos (α - β) = - 2 \sin α \sin β$ $[∵ α, β \in [0, π]]$

$\begin{aligned} ∵ \cos C - \cos D = 2 \sin \frac{C + D}{2} \sin \frac{D - C}{2} \\ = - 2 \times 1 \times \frac{1}{\sqrt{2}} [from eq. (i)] \\ = - \sqrt{2} \end{aligned}$

Trigonometrical Equations 1 Question 4

4. If $\sin^{4} α + 4 \cos^{4} β + 2 = 4 \sqrt{2} \sin α \cos β$ ;

Answer:

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Trigonometrical Equations 1 Question 4

4. If sin4⁡α+4cos4⁡β+2=42sin⁡αcos⁡β;

Answer:

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4. If $\sin^{4} α + 4 \cos^{4} β + 2 = 4 \sqrt{2} \sin α \cos β$ ;