Exams › JEE Advanced › Maths

Show that cos(6*alpha) can be written as a polynomial in cos(alpha). Which of the following is the correct expansion?

1. cos(6*alpha) = 32*cos⁶(alpha) - 48*cos⁴(alpha) + 18*cos²(alpha) - 1
2. cos(6*alpha) = 32*cos⁶(alpha) - 48*cos⁴(alpha) + 18*cos²(alpha) + 1
3. cos(6*alpha) = 16*cos⁶(alpha) - 24*cos⁴(alpha) + 9*cos²(alpha) - 1
4. cos(6*alpha) = 32*cos⁶(alpha) - 32*cos⁴(alpha) + 6*cos²(alpha) - 1

Correct answer: cos(6*alpha) = 32*cos⁶(alpha) - 48*cos⁴(alpha) + 18*cos²(alpha) - 1

Solution

Writing cos(6a) via the double-angle of 3a and substituting the triple-angle expansion of cos(3a) gives the degree-6 Chebyshev polynomial in cos(a).

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 →