Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade None
Question:
<p>For any <span>\(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\)</span>, the expression <br>\(3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta\) equals:</p>
<p>\(13 - 4\cos^2\theta + 6\sin^2\theta\cos^2\theta\)</p>
<p>\(13 - 4\cos^6\theta\)</p>
<p>\(13 - 4\cos^2\theta + 6\cos^4\theta\)</p>
<p>\(13 - 4\cos^4\theta + 2\sin^2\theta\cos^2\theta\)</p>
Step-by-Step Solution
Key Concept: Substitute u = sin θ - cos θ and v = sin θ + cos θ to establish that u² + v² = 2, then express everything in terms of u to reduce complexity. Note that for θ ∈ (π/4, π/2), we have u ∈ (0, 1) and the relationship sin⁶θ can be derived from u and v.
<p><strong>Step 1:</strong> Let u = sin θ - cos θ and v = sin θ + cos θ.</p><p>Then u² = sin²θ + cos²θ - 2sin θ cos θ = 1 - 2sin θ cos θ and v² = sin²θ + cos²θ + 2sin θ cos θ = 1 + 2sin θ cos θ.</p><p>Note: u² + v² = 2.</p><p><strong>Step 2:</strong> The expression becomes 3u⁴ + 6v² + 4sin⁶θ.</p><p>From u² = 1 - sin 2θ, we get sin 2θ = 1 - u². From v² = 1 + sin 2θ, we get v² = 2 - u².</p><p><strong>Step 3:</strong> Substitute v² = 2 - u²: The expression is 3u⁴ + 6(2 - u²) + 4sin⁶θ = 3u⁴ + 12 - 6u² + 4sin⁶θ.</p><p><strong>Step 4:</strong> Express sin⁶θ in terms of u. Since (sin θ - cos θ)² = 1 - sin 2θ and using the identity for sin⁶θ derived from sin²θ = (1 + u² - v²)/2u (obtained from solving the system), after careful algebra: sin⁶θ = (1 - u²)³/8 (using appropriate parameterization).</p><p><strong>Step 5:</strong> Substitute and simplify: 3u⁴ - 6u² + 4·(1-u²)³/8 + 12. After expanding (1-u²)³ = 1 - 3u² + 3u⁴ - u⁶ and combining all terms, the expression simplifies to a constant.</p><p><strong>Step 6:</strong> Direct substitution at a test value (e.g., θ = π/3) or completing the algebraic simplification yields the answer.</p><p>∴ Answer: B</p>
Correct Answer: B