<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 sin α, sin β, sin γ ∈ [-1, 1] and their sum equals -3, each must equal -1 individually. This forces α = 3π/2, β = 3π/2, γ = 3π/2, uniquely determining the angles.
<p><strong>Step 1:</strong> Analyze the constraint. Since sin α, sin β, sin γ ∈ [-1, 1] for all real angles, and their sum is -3, we need: sin α + sin β + sin γ = -3</p><p><strong>Step 2:</strong> The only way three numbers from [-1, 1] sum to -3 is if each equals -1: sin α = -1, sin β = -1, sin γ = -1</p><p><strong>Step 3:</strong> For α, β, γ ∈ (0, 2π), sin θ = -1 only when θ = 3π/2. Therefore: α = 3π/2, β = 3π/2, γ = 3π/2</p><p><strong>Step 4:</strong> Calculate cos 2α + cos 4β + cos 6γ:</p><p>cos(2 · 3π/2) + cos(4 · 3π/2) + cos(6 · 3π/2)</p><p>= cos(3π) + cos(6π) + cos(9π)</p><p>= cos(π) + cos(0) + cos(π)</p><p>= -1 + 1 + (-1)</p><p>= -1</p><p><strong>Step 5:</strong> Note: cos(3π) = cos(π), cos(6π) = cos(0), cos(9π) = cos(π) using periodicity of cosine</p><p>∴ Answer: -1</p>
Correct Answer: C