<p><strong>169.</strong> If \(\sin\alpha + \sin\beta + \sin\gamma = -3\), \(\alpha, \beta, \gamma \in (0, 2\pi)\), then \(\cos 2\alpha + \cos 4\beta + \cos 6\gamma\) is equal to:</p>
Step-by-Step Solution
Key Concept: Since sine is bounded by [-1, 1] and their sum equals -3, each sine must equal its minimum value of -1. This forces α = 3π/2, β = 3π/2, γ = 3π/2.
<p><strong>Step 1:</strong> Analyze the constraint sin α + sin β + sin γ = -3</p><p>Since sin x ∈ [-1, 1] for all x, and the maximum value of a sum of three sines is 3 (when each equals 1), the only way to get -3 is if each sine equals its minimum: sin α = sin β = sin γ = -1</p><p><strong>Step 2:</strong> Find α, β, γ in (0, 2π)</p><p>sin α = -1 ⟹ α = 3π/2</p><p>sin β = -1 ⟹ β = 3π/2</p><p>sin γ = -1 ⟹ γ = 3π/2</p><p><strong>Step 3:</strong> Calculate cos 2α + cos 4β + cos 6γ</p><p>2α = 3π, 4β = 6π, 6γ = 9π</p><p>cos(3π) + cos(6π) + cos(9π) = -1 + 1 + (-1) = <strong>-1</strong></p><p>∴ Answer: C</p>
Correct Answer: C