Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>If \(\cos 2\theta = \sin\alpha\) then the most general relation between \(\theta\) and \(\alpha\) is (where \(n \in \mathbb{Z}\))</p>
<p>(a) \(2\theta = a + \dfrac{\pi}{2}\)</p>
<p>(b) \(\theta = n\pi \pm \left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right)\)</p>
<p>(c) \(\theta = 2n\pi \pm \left(\dfrac{\pi}{2} - \alpha\right)\)</p>
<p>(d) \(\dfrac{n\pi + (-1)^n \alpha}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the complementary angle identity sin(α) = cos(π/2 - α) to convert the equation into a form where cosines can be directly equated, then apply the general solution for cos A = cos B.
<p><strong>Step 1:</strong> Start with the given equation: cos 2θ = sin α</p><p><strong>Step 2:</strong> Convert sine to cosine using the complementary angle identity: sin α = cos(π/2 - α)</p><p><strong>Step 3:</strong> Substitute into the equation: cos 2θ = cos(π/2 - α)</p><p><strong>Step 4:</strong> Apply the general solution for cos A = cos B: A = ±B + 2πn, where n ∈ ℤ</p><p><strong>Step 5:</strong> Therefore: 2θ = ±(π/2 - α) + 2πn</p><p><strong>Step 6:</strong> This gives two cases:</p><p>• Case 1: 2θ = π/2 - α + 2πn ⟹ θ = π/4 - α/2 + πn</p><p>• Case 2: 2θ = -(π/2 - α) + 2πn = -π/2 + α + 2πn ⟹ θ = -π/4 + α/2 + πn</p><p><strong>Step 7:</strong> Combining both cases into one general relation: <strong>2θ + α = π/2 + πn or θ = π/4 - α/2 + πn/2</strong> (where n ∈ ℤ)</p><p>∴ Answer: D</p>
Correct Answer: D