Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Let \(\theta = \sin^{-1}\left(\dfrac{3\sin 2\alpha}{5 + 4\cos 2\alpha}\right)\). Then \(\tan^{-1} x = \dfrac{\theta}{2}\) where \(x\) equals:</p>
<p>(a) \(\dfrac{1}{2}\tan\alpha\)</p>
<p>(b) \(\dfrac{1}{4}\tan\alpha\)</p>
<p>(c) \(\dfrac{1}{3}\tan\alpha\)</p>
<p>(d) \(\dfrac{2}{3}\tan\alpha\)</p>
Step-by-Step Solution
Key Concept: Recognize that the expression inside sin⁻¹ can be rewritten using the tangent half-angle substitution t = tan α, converting it to a form that reveals θ/2 is the half-angle whose tangent is x.
<p><strong>Step 1:</strong> Use the substitution t = tan α. Then sin 2α = 2t/(1+t²) and cos 2α = (1-t²)/(1+t²).</p><p><strong>Step 2:</strong> Substitute into the expression:</p><p>sin θ = (3 · 2t/(1+t²))/(5 + 4(1-t²)/(1+t²)) = (6t/(1+t²))/((5(1+t²) + 4(1-t²))/(1+t²))</p><p><strong>Step 3:</strong> Simplify the denominator: 5(1+t²) + 4(1-t²) = 5 + 5t² + 4 - 4t² = 9 + t²</p><p><strong>Step 4:</strong> Therefore sin θ = 6t/(9+t²)</p><p><strong>Step 5:</strong> Recognize this as sin θ = 2·(3t)/(9+t²). If we set tan(θ/2) = x, then sin θ = 2x/(1+x²) (double angle formula for sine).</p><p><strong>Step 6:</strong> Comparing 2x/(1+x²) = 6t/(9+t²), we get x = 3t = <strong>3tan α</strong></p><p>∴ Answer: <strong>x = 3tan α</strong></p>
Correct Answer: C