Exams › JEE Advanced › Maths

Find the value of the product cos(pi/15)*cos(2pi/15)*cos(3pi/15)*cos(4pi/15)*cos(5pi/15)*cos(6pi/15)*cos(7pi/15).

1. 1/128
2. 1/64
3. 1/256
4. -1/128

Correct answer: 1/128

Solution

Using the known identity productₖ₌₁ⁿ cos(k*pi/(2n+1)) = 1/2ⁿ with n = 7, the product equals 1/2⁷ = 1/128.

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 →