2000

JEE Main 2000 Mathematics question on Trigonometry

Let f(θ)=sin⁡θ(sin⁡θ+sin⁡3θ)f(\theta) = \sin\theta(\sin\theta + \sin 3\theta). Then, f(θ)f(\theta)
  1. A.
    ≥0\geq 0 only when θ≥0\theta \geq 0
  2. B.
    ≤0\leq 0 for all real θ\theta
  3. C.
    ≥0\geq 0 for all real θ\theta
  4. D.
    ≤0\leq 0 only when θ≤0\theta \leq 0
Show answer and solution

Correct option: C

Solution

f(θ) = sinθ(sinθ + sin3θ). Use the sum-to-product identity: sinθ + sin3θ = 2sin2θcosθ. So f(θ) = sinθ · 2sin2θcosθ = 2sinθcosθ · sin2θ = sin2θ · sin2θ = sin²(2θ). Since sin²(2θ) ≥ 0 for all real θ, f(θ) ≥ 0 always.

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