Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>If \(\alpha\), \(\beta\) are two values of \(\theta\) obtained from the equation \(a\cos\theta + b\sin\theta = c\) then the value of \(\tan\frac{\alpha+\beta}{2}\) is</p>
<p>(a) \(\frac{a}{b}\)</p>
<p>(b) \(\frac{b}{a}\)</p>
<p>(c) \(\frac{c}{a}\)</p>
<p>(d) \(\frac{c}{b}\)</p>
Step-by-Step Solution
Key Concept: Use the sum-to-product identity by converting the linear combination a·cos θ + b·sin θ into a single trigonometric function R·sin(θ + φ), then apply the condition that α and β are two solutions to find their sum using Vieta's formulas on the resulting quadratic.
<p><strong>Step 1:</strong> Rewrite a·cos θ + b·sin θ = c in standard form.</p><p>Let R·sin(θ + φ) = c, where R = √(a² + b²), tan φ = a/b</p><p>So: √(a² + b²)·sin(θ + φ) = c, giving sin(θ + φ) = c/√(a² + b²)</p><p><strong>Step 2:</strong> If α and β are two solutions, then sin(α + φ) = sin(β + φ) = c/√(a² + b²)</p><p>This means: α + φ = π - (β + φ) + 2πk, so α + β = π - 2φ + 2πk</p><p><strong>Step 3:</strong> Therefore: (α + β)/2 = π/2 - φ + πk</p><p>Thus: tan((α + β)/2) = tan(π/2 - φ) = cot φ = b/a</p><p><strong>Step 4:</strong> Alternatively, using t = tan(θ/2) substitution on the original equation yields a quadratic where the sum of roots gives tan((α + β)/2) = b/a</p><p>∴ Answer: <strong>b/a</strong></p>
Correct Answer: B