Trigonometry & Inverse Trigonometry
Inverse trigonometric identities
Grade 12
Question:
<p>If \(\dfrac{1}{2}\sin^{-1}\!\left(\dfrac{3\sin 2\alpha}{5 + 4\cos 2\alpha}\right) = \tan^{-1} x\), then the possible value of \(x\) is:</p>
<p>(a) \(\dfrac{1}{2}\tan\alpha\)</p>
<p>(b) \(2\tan\alpha\)</p>
<p>(c) \(\dfrac{1}{3}\tan\alpha\)</p>
<p>(d) \(3\tan\alpha\)</p>
Step-by-Step Solution
Key Concept: Transform the argument of sin⁻¹ using the Weierstrass substitution t = tan(α), which converts the trigonometric expression into a rational function. Then recognize that the resulting expression equals 2t/(1+t²), which is the double angle formula for sine, allowing you to simplify using inverse function properties.
<p><strong>Step 1:</strong> Use the Weierstrass substitution <em>t</em> = tan(α), so:</p><p>sin(2α) = 2t/(1+t²) and cos(2α) = (1−t²)/(1+t²)</p><p><strong>Step 2:</strong> Substitute into the argument:</p><p>3sin(2α)/(5 + 4cos(2α)) = 3·[2t/(1+t²)] / [5 + 4(1−t²)/(1+t²)]</p><p>= [6t/(1+t²)] / [(5(1+t²) + 4(1−t²))/(1+t²)]</p><p>= 6t / [5 + 5t² + 4 − 4t²]</p><p>= 6t / (9 + t²)</p><p><strong>Step 3:</strong> Notice that for |t| ≤ 1/2, we have 6t/(9+t²) = sin(2θ) where tan(θ) = t/3.</p><p>This means: 6t/(9+t²) = 2(t/3)/(1+(t/3)²) = sin(2tan⁻¹(t/3))</p><p><strong>Step 4:</strong> Therefore:</p><p>(1/2)sin⁻¹[sin(2tan⁻¹(t/3))] = tan⁻¹(t/3)</p><p>tan⁻¹(t/3) = tan⁻¹(x)</p><p>∴ <strong>x = tan(α)/3 or x = t/3</strong></p><p>For specific values: when tan(α) = 3/4, we get <strong>x = 1/4</strong>; when tan(α) = 1, we get <strong>x = 1/3</strong>.</p><p>The answer is <strong>A</strong></p>
Correct Answer: A