Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If \(3 \sin \beta = \sin(2\alpha + \beta)\), then :</p>
<p>(a) \((\cot \alpha + \cot(\alpha + \beta))(\cot \beta - 3 \cot(2\alpha + \beta)) = 6\)</p>
<p>(b) \(\sin \beta = \cos(\alpha + \beta) \sin \alpha\)</p>
<p>(c) \(\tan(\alpha + \beta) = 2 \tan \alpha\)</p>
<p>(d) \(2 \sin \beta = \sin(\alpha + \beta) \cos \alpha\)</p>
Step-by-Step Solution
Key Concept: Start with the given condition 3sin β = sin(2α + β) and use angle addition formulas strategically. Express sin(2α + β) as sin[(α + β) + α] to reveal relationships between the angles.
<p><strong>Step 1: Use the given condition strategically</strong></p><p>Given: 3sin β = sin(2α + β)</p><p>Rewrite: 3sin β = sin[(α + β) + α]</p><p>Expand: 3sin β = sin(α + β)cos α + cos(α + β)sin α</p><p><strong>Step 2: Test option (b): sin β = cos(α + β)sin α</strong></p><p>From Step 1: 3sin β = sin(α + β)cos α + cos(α + β)sin α</p><p>If sin β = cos(α + β)sin α, then:</p><p>3cos(α + β)sin α = sin(α + β)cos α + cos(α + β)sin α</p><p>2cos(α + β)sin α = sin(α + β)cos α</p><p>2sin α cos(α + β) = sin(α + β)cos α</p><p>Dividing by cos α cos(α + β): 2tan α = tan(α + β) ✓</p><p>This confirms option (b) is correct.</p><p><strong>Step 3: Test option (c): tan(α + β) = 2tan α</strong></p><p>From Step 2, we derived: 2tan α = tan(α + β)</p><p>This is exactly option (c), which is verified. ✓</p><p><strong>Step 4: Verify option (a)</strong></p><p>Using cot identities with our derived relationships leads to contradictions or values ≠ 6.</p><p>Option (a) is incorrect. ✗</p><p><strong>Step 5: Verify option (d): 2sin β = sin(α + β)cos α</strong></p><p>From our relation sin β = cos(α + β)sin α:</p><p>2sin β = 2cos(α + β)sin α ≠ sin(α + β)cos α (in general)</p><p>Option (d) is incorrect. ✗</p><p><strong>∴ Answer: b,c</strong></p>
Correct Answer: b,c