Question

If sin⁴α + 4cos⁴β + 2 = 4√2 sinα.cosβ ; α, β ∈ [0, π], then cos(α + β) is equal to :
(A) 0 (B) –1
(C) √2 (D) –√2

Answer 1

||✪✪ QUESTION ✪✪||

If sin⁴α + 4cos⁴β + 2 = 4√2 sinα.cosβ ; α, β ∈ [0, π], then cos(α + β) is equal to ?

|| ✰✰ ANSWER ✰✰ ||

→ sin⁴α + 4cos⁴β + 2 = 4√2 sinα.cosβ

→ sin⁴α + 4cos⁴β + 1 + 1 = 4 * √2 * sinα * cosβ

Now, AM ≥ GM says That , Arithmetic mean (A.M.) is greater than geometric mean (G.M.) for same Number of variables where variables are ≥ 0...

So , Using This in LHS,

→ (sin⁴α + 4cos⁴β + 1 + 1)/4 = ⁴√(sin⁴α * 4cos⁴β * 1 * 1)

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 * [ ⁴√(sin⁴α * cos⁴β * 4) ]

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 * [ sinα * cosβ * √2) {As (√2)⁴ = 4}

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 * √2 * sinα * cosβ

So, we can say That , AM = GM.

ஃ with equality if and only a1 = a2 = a3 = ________aN.

_________________

So, we Have :-

→ sin⁴α = 4cos⁴β = 1

→ sin⁴α = 1

→ 4cos⁴β = 1

when ,

→ sin⁴α = 1

→ sin⁴α = sin90°

→ sin⁴α = sin⁴90°

→ α = 90°..

And,

→ 4cos⁴β = 1

→ cos⁴β = 1/4

→ cos⁴β = (1/√2)⁴

→ cos⁴β = (cos45°)⁴

→ cosβ = cos45°

→ β = 45°..

_______________

Now,

→ cos(α + β)

→ cos(90°+45°)

→ -sin(45°)

→ -(1/√2) (Ans.)

Hence, The value of cos(α + β) is (-1/√2)

Answer 2

CorrEcT QuEstioN:-

★If sin⁴α + 4cos⁴β + 2 = 4√2 sinα.cosβ ; α, β ∈ [0, π], then cos(α + β) is equal to :

(A) 0 (B) –1

(C) √2 (D) –√2 (E) -1√2

$\rule{200}2$

AnSwEr:-

★-1√2(E)

$\rule{200}2$

SteP By SteP ExplainaTion:-

★ sin⁴α + 4cos⁴β + 2 = 4√2 sinα.cosβ

★ sin⁴α + 4cos⁴β + 1 + 1 = 4 ×√2 ×sinα × cosβ

Arithmetic progression is always equal or greater than geometric progression;

→ (sin⁴α + 4cos⁴β + 1 + 1)/4 = ⁴√(sin⁴α ×4cos⁴β ×1 × 1)

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 × [ ⁴√(sin⁴α ×cos⁴β × 4) ]

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 ×[ sinα ×cosβ ×√2) (√2)⁴ = 4)

→ (sin⁴α + 4cos⁴β + 1 + 1) = 4 × √2 ×sinα ×cosβ

AM = GM.

a1 = a2 = a3 = .......aN.

Now,

→ sin⁴α = 4cos⁴β = 1

→ sin⁴α = 1

→ 4cos⁴β = 1

when ,

→ sin⁴α = 1

→ sin⁴α = sin90°

→ sin⁴α = sin⁴90°

→ α = 90°

Also,

→ 4cos⁴β = 1

→ cos⁴β = 1/4

→ cos⁴β = (1/√2)⁴

→ cos⁴β = (cos45°)⁴

→ cosβ = cos45°

→ β = 45°

Now,

→ cos(α + β)

→ cos(90°+45°)

→ -sin(45°)

→ -1√2

$\rule{200}2$