Trigonometry & Inverse Trigonometry
Compound Angles
Grade 11
Question:
<p>Let \(\cos(\alpha+\beta)=\dfrac{4}{5}\) and let \(\sin(\alpha-\beta)=\dfrac{5}{13}\), where \(0 \le \alpha,\ \beta \le \dfrac{\pi}{4}\), then \(\tan 2\alpha =\)</p>
<p>\(\dfrac{56}{33}\)</p>
<p>\(\dfrac{19}{12}\)</p>
<p>\(\dfrac{20}{7}\)</p>
<p>\(\dfrac{25}{16}\)</p>
Step-by-Step Solution
Key Concept: Use the sum-to-product identity: tan(2α) = tan[(α+β) + (α-β)] = [tan(α+β) + tan(α-β)] / [1 - tan(α+β)tan(α-β)]. First find tan(α+β) and tan(α-β) from the given cosine and sine values using the constraint that both angles lie in [0, π/4].
<p><strong>Step 1:</strong> From cos(α+β) = 4/5 where 0 ≤ α+β ≤ π/2, we get sin(α+β) = 3/5 (positive in first quadrant).</p><p>Therefore, tan(α+β) = 3/4</p><p><strong>Step 2:</strong> From sin(α-β) = 5/13 where -π/4 ≤ α-β ≤ π/4, cosine is always positive.</p><p>cos(α-β) = √(1 - 25/169) = √(144/169) = 12/13</p><p>Therefore, tan(α-β) = 5/12</p><p><strong>Step 3:</strong> Since 2α = (α+β) + (α-β), apply tangent addition formula:</p><p>tan(2α) = [tan(α+β) + tan(α-β)] / [1 - tan(α+β)tan(α-β)]</p><p>tan(2α) = [3/4 + 5/12] / [1 - (3/4)(5/12)]</p><p>tan(2α) = [(9+5)/12] / [1 - 15/48]</p><p>tan(2α) = [14/12] / [33/48]</p><p>tan(2α) = (14/12) × (48/33) = (14 × 4) / 33 = 56/33</p><p>∴ Answer: A (tan(2α) = 56/33)</p>
Correct Answer: A