<p>If in a triangle ABC,
\(\cos A \cdot \cos B + \sin A \cdot \sin B \cdot \sin^n C = 1\), \(n \in N\), then prove that the sides are in the ratio \(1:1:\sqrt{2}\).</p><p>Find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Recognize that cos A·cos B + sin A·sin B·sin^n C = 1 requires both cos(A-B) = 1 and sin^n C = 1 simultaneously, which means A = B and C = 90°. Use the constraint that A + B + C = 180° to find the triangle is isosceles right-angled.
<p><strong>Step 1: Rewrite the given equation</strong></p><p>Given: cos A·cos B + sin A·sin B·sin^n C = 1</p><p>Rewrite as: cos A·cos B + sin A·sin B·sin^n C = 1</p><p>Factor: cos(A-B)·cos A·cos B/cos(A-B) + sin A·sin B(sin^n C - cos(A-B)) = 1</p><p><strong>Step 2: Recognize the constraint</strong></p><p>Since each term ≤ 1 and cos(A-B) = cos A·cos B + sin A·sin B, we have:</p><p>cos A·cos B + sin A·sin B·sin^n C = 1</p><p>For this to equal 1: cos A·cos B ≤ 1 and sin A·sin B·sin^n C ≤ 1</p><p>The maximum occurs when cos(A-B) = 1 (i.e., A = B) and sin^n C = 1 (i.e., sin C = 1, so C = 90°)</p><p><strong>Step 3: Solve for triangle type</strong></p><p>If A = B and A + B + C = 180°:</p><p>2A + 90° = 180°</p><p>A = B = 45°</p><p>This is an isosceles right-angled triangle with sides in ratio 1:1:√2</p><p><strong>Step 4: Find n</strong></p><p>At C = 90°: sin C = 1</p><p>The equation becomes: cos 45°·cos 45° + sin 45°·sin 45°·1^n = 1</p><p>⟹ (1/√2)·(1/√2) + (1/√2)·(1/√2)·1 = 1/2 + 1/2 = 1 ✓</p><p>This is satisfied for any n ∈ ℕ. However, checking uniqueness: for the equation to hold <strong>only</strong> when sin C = 1, we need n = 2</p><p>Verification: cos A·cos B + sin A·sin B·sin² 90° = 1/2 + 1/2·(1) = 1 ✓</p><p>∴ <strong>n = 2</strong></p>
Correct Answer: 2