Let x = sin (2tan ⁻¹α ) and y = sin ( 1/2tan ⁻¹4/3 ). If S = { a ∈ R:y² = 1 - x}, then Σ_(α ∈ S) 16α ³ is equal to _______________.

Solution

Maths sir · 4 steps

Turn both inverse-trig expressions into plain numbers first; then it is just a quadratic in α.

1. Simplify x. Put α = tan θ. Then x = sin 2θ = (2 tan θ)/(1+tan² θ) = 2α/(1+α²), and this holds for every real α.
2. Simplify y. Let φ = tan⁻¹ 4/3, so φ is acute and cos φ = 3/5. By the half-angle formula, y² = sin² φ/2 = (1-cos φ)/2 = 1/5.
3. Set up the equation. The condition y² = 1 - x gives x = 4/5. So 2α/(1+α²) = 4/5, which simplifies to the quadratic below.

   2α² - 5α + 2 = 0 ⇒ (2α - 1)(α - 2) = 0

4. So. S = {2, 1/2}, and Σ 16α³ = 16(8 + 1/8) = 128 + 2 = 130.

The common slip: 128. Keeping only the root α = 2 and dropping α = 1/2 gives 128. Both roots belong to S.

The idea to keep. Whenever you see sin(2 tan⁻¹ α), write it as 2α/(1+α²). For a half angle, use sin² φ/2 = (1-cos φ)/2, not a fresh triangle.

Answer from the NTA final answer key. Solution by PrepWiser.