Exams › JEE Advanced › Maths

If cos⁻¹(x) - cos⁻¹(y/2) = α, find the value of 4x² - 4xy*cos α + y².

1. 2 sin 2α
2. 4
3. 4 sin² α
4. -4 sin² α

Correct answer: 4 sin² α

Solution

With A = x, B = y/2: cos α = x*(y/2) + sqrt(1-x²)*sqrt(1 - y²/4). So cos α - xy/2 = sqrt(1-x²)*sqrt(1-y²/4). Square: cos² α - xy*cos α + x²y²/4 = (1-x²)(1 - y²/4) = 1 - y²/4 - x² + x²y²/4. Cancel x²y²/4: cos² α - xy*cos α = 1 - y²/4 - x². Multiply by 4 and rearrange: 4x² - 4xy*cos α + y² = 4 - 4cos² α = 4 sin² α.

Related JEE Advanced Maths questions

• Identify the periodic function among the following:
• Given that α, β, γ, and θ are the smallest positive angles in increasing order for which their sine values equal a positive constant k, what is the result of 4 sin(α/2) + 3 sin(β/2) + 2 sin(γ/2) + sin(θ/2)?
• Consider the function fn(θ) = tan(θ/2)(1 + sec²θ)(1 + sec⁴θ)... (1 + sec²ⁿθ). Which of the following is true?
• Given that (a−b)sin(θ+ϕ) = (a+b)sin(θ−ϕ) and a tan(θ/2) − b tan(ϕ/2) = c, which of the following holds true?
• In a right-angled triangle ABC where ∠B = 90° and the sum of the two legs a and b equals 4, the triangle's area is maximized when ∠C is:
• If the altitudes p₁, p₂, and p₃ of a triangle enclose a circle with a diameter of 4/3 units, what is the minimum possible value of p₁ + p₂ + p₃?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →