Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If <span class="math">\sin\theta = 3\sin(\theta + 2\alpha)\</span>, then the value of <span class="math">\tan(\theta + \alpha) + 2\tan\alpha\</span> is</p>
<p>(a) 3</p>
<p>(b) 2</p>
<p>(c) -1</p>
<p>(d) 0</p>
Step-by-Step Solution
Key Concept: Expand the given trigonometric equation using angle addition formulas and rearrange to establish a relationship between the angles. Use the tangent addition formula to express the required quantity in terms of this relationship.
<p><strong>Step 1:</strong> Start with the given equation: sin θ = 3sin(θ + 2α)</p><p><strong>Step 2:</strong> Expand sin(θ + 2α) using the addition formula:</p><p>sin θ = 3[sin θ cos 2α + cos θ sin 2α]</p><p>sin θ = 3sin θ cos 2α + 3cos θ sin 2α</p><p><strong>Step 3:</strong> Rearrange the equation:</p><p>sin θ - 3sin θ cos 2α = 3cos θ sin 2α</p><p>sin θ(1 - 3cos 2α) = 3cos θ sin 2α</p><p><strong>Step 4:</strong> Divide both sides by cos θ cos 2α (assuming they're non-zero):</p><p>tan θ(1 - 3cos 2α) = 3sin 2α</p><p><strong>Step 5:</strong> Note that we need to find tan(θ + α) + 2tan α. Rewrite using angle relationships. Let's use: sin(θ + 2α) can be written as sin[(θ + α) + α]</p><p><strong>Step 6:</strong> From the original equation: sin θ = 3sin(θ + 2α), we can write:</p><p>sin θ - 3sin(θ + 2α) = 0</p><p>sin θ - 3sin[(θ + α) + α] = 0</p><p><strong>Step 7:</strong> This means: sin θ = 3[sin(θ + α)cos α + cos(θ + α)sin α]</p><p>Also, sin θ = sin[(θ + α) - α] = sin(θ + α)cos α - cos(θ + α)sin α</p><p><strong>Step 8:</strong> Equating both expressions:</p><p>sin(θ + α)cos α - cos(θ + α)sin α = 3sin(θ + α)cos α + 3cos(θ + α)sin α</p><p>-2sin(θ + α)cos α = 4cos(θ + α)sin α</p><p><strong>Step 9:</strong> Divide both sides by 2cos(θ + α)cos α:</p><p>-tan(θ + α) = 2tan α</p><p>tan(θ + α) + 2tan α = 0</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D