Sets, Relations & Functions
Operations on Sets
Grade None
Question:
<p>Let \(P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta\}\) and \(Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta\}\) be two sets. Then</p>
<p>\(P \subset Q\) and \(Q - P \neq \phi\)</p>
<p>\(Q \not\subset P\)</p>
<p>\(P = Q\)</p>
<p>\(P \not\subset Q\)</p>
Step-by-Step Solution
Key Concept: Transform both equations into standard forms by isolating trigonometric terms and squaring strategically, or rewrite using the identity a·sinθ + b·cosθ = √(a²+b²)·sin(θ+φ). The key is recognizing that both sets define specific angle relationships where tangent equals a particular value.
<p><strong>Step 1:</strong> Simplify set P: sin θ - cos θ = √2 cos θ</p><p>sin θ = cos θ + √2 cos θ = (1 + √2)cos θ</p><p>tan θ = 1 + √2</p><p>So P = {θ : tan θ = 1 + √2}</p><p><strong>Step 2:</strong> Simplify set Q: sin θ + cos θ = √2 sin θ</p><p>cos θ = √2 sin θ - sin θ = (√2 - 1)sin θ</p><p>tan θ = 1/(√2 - 1) = (√2 + 1)/(2 - 1) = √2 + 1</p><p>So Q = {θ : tan θ = √2 + 1}</p><p><strong>Step 3:</strong> Observe that 1 + √2 = √2 + 1, therefore P = Q</p><p>∴ Answer: C (both sets are equal)</p>
Correct Answer: C