Exams › JEE Advanced › Maths

Evaluate the value of sin(7.5 deg).

1. (1/2) sqrt(2 - sqrt(2 + sqrt(3)))
2. (1/2) sqrt(2 + sqrt(2 + sqrt(3)))
3. (1/2) sqrt(2 - sqrt(2 - sqrt(3)))
4. (1/2) sqrt(2 + sqrt(2 - sqrt(3)))

Correct answer: (1/2) sqrt(2 - sqrt(2 + sqrt(3)))

Solution

Using sin²(7.5 deg) = (1 - cos 15 deg)/2 and cos 15 deg = sqrt(2 + sqrt(3))/2 gives sin 7.5 deg = (1/2) sqrt(2 - sqrt(2 + sqrt(3))).

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 →