Trigonometry & Inverse Trigonometry
Compound Angles
Grade 11
Question:
<p>Given that \(\cos(\alpha + \beta) = \dfrac{4}{5}\) and \(\sin(\alpha - \beta) = \dfrac{5}{13}\), where \(\alpha + \beta \in \left[0, \dfrac{\pi}{2}\right]\) and \(\alpha - \beta \in \left[0, \dfrac{\pi}{4}\right]\), then \(\tan 2\alpha\) is equal to:</p>
<p>(A) \(\dfrac{25}{16}\)</p>
<p>(B) \(\dfrac{56}{33}\)</p>
<p>(C) \(\dfrac{19}{12}\)</p>
<p>(D) \(\dfrac{20}{7}\)</p>
Step-by-Step Solution
Key Concept: Use the product-to-sum identity: sin(A)cos(B) = ½[sin(A+B) + sin(A-B)]. Find sin(α+β) and cos(α-β) from the given constraints, then compute sin(2α) = sin[(α+β)+(α-β)] and cos(2α) = cos[(α+β)+(α-β)] to get tan(2α).
<p><strong>Step 1: Find missing trigonometric values using intervals.</strong></p><p>Since α + β ∈ [0, π/2]: cos(α+β) = 4/5 ⟹ sin(α+β) = √(1 - 16/25) = <strong>3/5</strong></p><p>Since α - β ∈ [0, π/4]: sin(α-β) = 5/13 ⟹ cos(α-β) = √(1 - 25/169) = <strong>12/13</strong></p><p><strong>Step 2: Express 2α as sum of two angles.</strong></p><p>2α = (α+β) + (α-β)</p><p><strong>Step 3: Apply sine and cosine addition formulas.</strong></p><p>sin(2α) = sin(α+β)cos(α-β) + cos(α+β)sin(α-β)</p><p>sin(2α) = (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = <strong>56/65</strong></p><p>cos(2α) = cos(α+β)cos(α-β) - sin(α+β)sin(α-β)</p><p>cos(2α) = (4/5)(12/13) - (3/5)(5/13) = 48/65 - 15/65 = <strong>33/65</strong></p><p><strong>Step 4: Calculate tan(2α).</strong></p><p>tan(2α) = sin(2α)/cos(2α) = (56/65)/(33/65) = <strong>56/33</strong></p><p>∴ Answer: B</p>
Correct Answer: B