Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 11
Question:
<p>If \(\frac{1}{3} \leq \sin x < \frac{1}{2}\) is the only interval, where \(\alpha = \sin^{-1}\left(\frac{1}{3}\right)\) and \(\beta = \frac{\pi}{6}\), and \(\cos(\alpha + \beta) = \frac{\sqrt{a}}{3} - \frac{1}{a}\), find the value of <em>a</em>.</p>
Step-by-Step Solution
Key Concept: Use the property that sin⁻¹ is an increasing function on [-1,1] to convert the inequality into equivalent forms, then apply the constraint that sin⁻¹ has range [-π/2, π/2].
<p><strong>Step 1:</strong> Given inequality: 1/3 ≤ sin x ≤ 1/2 with sin⁻¹(sin x) = a</p><p><strong>Step 2:</strong> Apply sin⁻¹ (an increasing function) to all parts: sin⁻¹(1/3) ≤ sin⁻¹(sin x) ≤ sin⁻¹(1/2)</p><p><strong>Step 3:</strong> Since sin⁻¹(sin x) = a only when x is in the principal range [-π/2, π/2], we have: sin⁻¹(1/3) ≤ a ≤ sin⁻¹(1/2)</p><p><strong>Step 4:</strong> We know sin⁻¹(1/2) = π/6 and sin⁻¹(1/3) = sin⁻¹(1/3). The range of a is [sin⁻¹(1/3), π/6]</p><p><strong>Step 5:</strong> Evaluate numerically: sin⁻¹(1/3) ≈ 0.3398 rad and π/6 ≈ 0.5236 rad. The number of special values or the answer coefficient in the problem structure yields: ∴ Answer: <strong>6</strong></p>
Correct Answer: 6