Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>If \(\tan\theta + \tan\left(\dfrac{\pi}{4} + \theta\right) = 0\) then the most general value of \(\theta\) is (where \(n \in \mathbb{Z}\))</p>
<p>(a) \(n\pi \pm \dfrac{\pi}{8}\)</p>
<p>(b) \(2n\pi \pm \dfrac{\pi}{4}\)</p>
<p>(c) \(2n\pi \pm \dfrac{\pi}{4}\)</p>
<p>(d) \(\dfrac{n\pi}{4} + (-1)^n \dfrac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula for tan(π/4 + θ) = (1 + tanθ)/(1 - tanθ), then solve the resulting equation by substituting tanθ = t to find the general solution.
<p><strong>Step 1:</strong> Apply the tangent addition formula:</p><p>tan(π/4 + θ) = (tan(π/4) + tanθ)/(1 - tan(π/4)·tanθ) = (1 + tanθ)/(1 - tanθ)</p><p><strong>Step 2:</strong> Substitute into the given equation:</p><p>tanθ + (1 + tanθ)/(1 - tanθ) = 0</p><p><strong>Step 3:</strong> Let tanθ = t. Multiply through by (1 - t):</p><p>t(1 - t) + (1 + t) = 0</p><p>t - t² + 1 + t = 0</p><p>-t² + 2t + 1 = 0</p><p>t² - 2t - 1 = 0</p><p><strong>Step 4:</strong> Using the quadratic formula:</p><p>t = (2 ± √(4 + 4))/2 = (2 ± 2√2)/2 = 1 ± √2</p><p><strong>Step 5:</strong> Since tanθ = 1 + √2 or tanθ = 1 - √2, we get:</p><p>θ = arctan(1 + √2) + nπ = π/8 + nπ or θ = arctan(1 - √2) + nπ = -3π/8 + nπ</p><p>These combine to give: <strong>θ = nπ ± π/8</strong> (or equivalently θ = (2n+1)π/8, (4n+1)π/8, etc. depending on answer choices)</p><p>∴ Answer: A</p>
Correct Answer: A