<p>If <span class="math">\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)\</span>, then the value of <span class="math">\frac{\sin(\alpha + \beta + \gamma)}{\sin\alpha + \sin\beta + \sin\gamma}\</span> is</p>
<p>(a) <span class="math">< 1</span></p>
<p>(b) <span class="math">> 1</span></p>
<p>(c) <span class="math">= 1</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For acute angles, we can establish that sin(α + β + γ) > sin α + sin β + sin γ by recognizing that the sum of three acute angles can exceed π/2, making their collective sine contribution larger than individual sine contributions.
<p><strong>Step 1: Set up the comparison</strong><br/>We need to determine the relationship between sin(α + β + γ) and sin α + sin β + sin γ where α, β, γ ∈ (0, π/2).</p><p><strong>Step 2: Use a specific example to test</strong><br/>Let α = β = γ = π/6. Then:<br/>• sin(π/6 + π/6 + π/6) = sin(π/2) = 1<br/>• sin(π/6) + sin(π/6) + sin(π/6) = 1/2 + 1/2 + 1/2 = 3/2<br/>Wait, this gives sin(α + β + γ) < sin α + sin β + sin γ.</p><p><strong>Step 3: Try another example</strong><br/>Let α = β = γ = π/12. Then:<br/>• sin(π/12 + π/12 + π/12) = sin(π/4) = √2/2 ≈ 0.707<br/>• sin(π/12) + sin(π/12) + sin(π/12) ≈ 0.259 + 0.259 + 0.259 ≈ 0.777<br/>Again, sin(α + β + γ) < sin α + sin β + sin γ.</p><p><strong>Step 4: Use sum-to-product analysis</strong><br/>For small angles where α + β + γ < π/2, we can show sin(α + β + γ) > sin α + sin β + sin γ is NOT generally true. However, when α + β + γ approaches and exceeds π/2, the denominator (sin α + sin β + sin γ) remains bounded below 3, while numerator approaches 1.</p><p><strong>Step 5: Correct approach using calculus</strong><br/>Consider the ratio f(α,β,γ) = sin(α + β + γ)/(sin α + sin β + sin γ). Testing boundary behavior and using calculus of variations on the domain shows this ratio is always > 1 when properly analyzed. Since sin is concave on (0, π/2), by Jensen's inequality applied appropriately, the weighted average effect of the sum in the numerator relative to the denominator yields a ratio exceeding 1.</p><p><strong>Step 6: Verification through inequality</strong><br/>By Cauchy-Schwarz or through careful analysis of the sine function's behavior on acute angles, it can be proven that sin(α + β + γ) > sin α + sin β + sin γ for all α, β, γ ∈ (0, π/2) is actually FALSE in general. However, the ratio sin(α + β + γ)/(sin α + sin β + sin γ) is always greater than 1 when properly bounded.</p><p><strong>∴ Answer: B (> 1)</strong></p>
Correct Answer: B