2019 · Q53
Suppose 0 < θ < 90°, then for every θ, 4sin²θ + 1 is greater than or equal to
Elementary MathematicsTrigonometry
- A2
- B4sinθ
- C4cosθ
- D4tanθ
Explanation
By AM–GM, 4 sin^2 θ + 1 ≥ 2√(4 sin^2 θ) = 4|sin θ| = 4 sin θ for 0 < θ < 90°. So 4 sin^2 θ + 1 ≥ 4 sin θ.