Trigonometry & Inverse Trigonometry
Periodicity of Trigonometric Functions
Grade 11
Question:
<p>The period of \(f(\theta) = \sin^2\theta\) is:</p>
<p>\(2\pi\)</p>
<p>\(\pi\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: sin²θ can be rewritten using the double angle formula: sin²θ = (1 - cos2θ)/2. The period is determined by the argument of the cosine function, which is 2θ, giving period π rather than 2π.
<p><strong>Step 1:</strong> Apply the double angle formula to simplify sin²θ.</p><p>sin²θ = (1 - cos2θ)/2</p><p><strong>Step 2:</strong> Identify the period from the transformed expression.</p><p>The fundamental period of cos(2θ) is 2π/2 = π</p><p><strong>Step 3:</strong> Verify by checking f(θ + π).</p><p>f(θ + π) = sin²(θ + π) = (-sinθ)² = sin²θ = f(θ) ✓</p><p><strong>Step 4:</strong> Confirm π is the smallest positive period.</p><p>For θ = π/4: f(π/4) = sin²(π/4) = 1/2, but f(π/4 + π/2) = sin²(3π/4) = 1/2, yet f(π/4 + π/4) = sin²(π/2) = 1 ≠ 1/2</p><p>∴ Period of sin²θ is <strong>π</strong></p>
Correct Answer: B