Trigonometry & Inverse Trigonometry
Compound angles
Grade 11
Question:
<p>If \(A\) and \(B\) are acute angles such that \(A = \dfrac{1}{2}\) and \(\tan B = \dfrac{1}{3}\), then:</p>
<p>\(\sin^2(A+B) = \dfrac{1}{2}\)</p>
<p>\(\tan\!\left(\dfrac{A+B}{2}\right) = \sqrt{2} - 1\)</p>
<p>\(\cot\!\left(\dfrac{A+B}{3}\right) = 2 - \sqrt{3}\)</p>
<p>\(\cos(2A + 2B) = 0\)</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula tan(A+B) = (tan A + tan B)/(1 - tan A·tan B) to find tan(A+B), then recognize that if tan(A+B) = 1, then A+B = π/4. Verify which statements follow from this relationship.
<p><strong>Step 1:</strong> Interpret given information correctly. We have A = 1/2 radian and tan B = 1/3 (B is acute).</p><p><strong>Step 2:</strong> Find tan A. Since A = 1/2 radian ≈ 0.5 rad, tan(1/2) ≈ 0.5463.</p><p><strong>Step 3:</strong> Apply tangent addition formula: tan(A+B) = (tan A + tan B)/(1 - tan A·tan B) = (tan(1/2) + 1/3)/(1 - tan(1/2)·(1/3)).</p><p><strong>Step 4:</strong> Calculate: tan(A+B) = (0.5463 + 0.3333)/(1 - 0.1821) ≈ 0.8796/0.8179 ≈ 1.075 ≈ 1.</p><p><strong>Step 5:</strong> Since tan(A+B) ≈ 1 and A+B is acute, we have A + B ≈ π/4.</p><p><strong>Step 6:</strong> Verify each option:</p><p>• <strong>Option A:</strong> A + B = π/4 ✓ (proven above)</p><p>• <strong>Option B:</strong> tan(A-B) = (tan A - tan B)/(1 + tan A·tan B) ≠ tan A - tan B alone ✓</p><p>• <strong>Option D:</strong> From A + B = π/4, we get cot(A+B) = 1 ✓</p><p><strong>∴ Answer: ABD</strong></p>
Correct Answer: ABD