Trigonometry & Inverse Trigonometry
Compound Angles
Grade 11
Question:
<p>If <span class="math">\cos(\theta - \alpha) = a\</span> and <span class="math">\cos(\theta - \beta) = b\</span>, then <span class="math">\sin^2(\alpha - \beta) + 2ab\cos(\alpha - \beta)\</span> is equal to</p>
<p>(a) <span class="math">a^2 + b^2</span></p>
<p>(b) <span class="math">a^2 - b^2</span></p>
<p>(c) <span class="math">b^2 - a^2</span></p>
<p>(d) <span class="math">-a^2 - b^2</span></p>
Step-by-Step Solution
Key Concept: Use the cosine difference formula to expand cos(θ - α) and cos(θ - β), then manipulate the given expression using algebraic identities and the relationship between the given conditions.
<p><strong>Step 1:</strong> Write the given conditions using the cosine difference formula:<br/>cos(θ - α) = cos θ cos α + sin θ sin α = a<br/>cos(θ - β) = cos θ cos β + sin θ sin β = b</p><p><strong>Step 2:</strong> Use the cosine difference formula for (α - β):<br/>cos(α - β) = cos[(θ - β) - (θ - α)] = cos(θ - β)cos(θ - α) + sin(θ - β)sin(θ - α)<br/>Therefore: cos(α - β) = ab + sin(θ - β)sin(θ - α)</p><p><strong>Step 3:</strong> Find sin²(α - β) using the Pythagorean identity:<br/>sin²(α - β) = 1 - cos²(α - β)</p><p><strong>Step 4:</strong> From the given conditions:<br/>sin²(θ - α) = 1 - a² and sin²(θ - β) = 1 - b²<br/>So: sin(θ - α) = ±√(1 - a²) and sin(θ - β) = ±√(1 - b²)</p><p><strong>Step 5:</strong> Substitute into cos(α - β):<br/>cos(α - β) = ab ± √(1 - a²)√(1 - b²)</p><p><strong>Step 6:</strong> Calculate the required expression:<br/>sin²(α - β) + 2ab cos(α - β)<br/>= 1 - cos²(α - β) + 2ab cos(α - β)<br/>= 1 - [ab ± √(1 - a²)√(1 - b²)]² + 2ab[ab ± √(1 - a²)√(1 - b²)]</p><p><strong>Step 7:</strong> Expanding [ab ± √(1 - a²)√(1 - b²)]²:<br/>= a²b² ± 2ab√(1 - a²)√(1 - b²) + (1 - a²)(1 - b²)<br/>= a²b² + (1 - a² - b² + a²b²) ± 2ab√(1 - a²)√(1 - b²)<br/>= 1 - a² - b² + 2a²b² ± 2ab√(1 - a²)√(1 - b²)</p><p><strong>Step 8:</strong> Therefore:<br/>sin²(α - β) + 2ab cos(α - β)<br/>= 1 - [1 - a² - b² + 2a²b² ± 2ab√(1 - a²)√(1 - b²)] + 2ab[ab ± √(1 - a²)√(1 - b²)]<br/>= 1 - 1 + a² + b² - 2a²b² ∓ 2ab√(1 - a²)√(1 - b²) + 2a²b ± 2ab√(1 - a²)√(1 - b²)<br/>= a² + b² - 2a²b² + 2a²b<br/>= a² + b²</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A