Trigonometry & Inverse Trigonometry
Inverse trigonometric identities
Grade 12
Question:
<p><strong>276.</strong> 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: Express the argument of sin⁻¹ in the form of a tangent using the Weierstrass substitution t = tan(α). The expression 3sin(2α)/(5+4cos(2α)) simplifies to a function of t that equals 2t/(1+t²) when properly manipulated.
<p><strong>Step 1:</strong> Use the identity: if sin⁻¹(y) = 2tan⁻¹(x), then y = sin(2tan⁻¹(x)).</p><p><strong>Step 2:</strong> Apply the double angle formula. Let tan⁻¹(x) = θ, so sin(2θ) = 2sin(θ)cos(θ).</p><p>With tan(θ) = x: sin(θ) = x/√(1+x²) and cos(θ) = 1/√(1+x²)</p><p>Therefore: sin(2θ) = 2x/(1+x²)</p><p><strong>Step 3:</strong> Equate expressions:</p><p>3sin(2α)/(5+4cos(2α)) = 2x/(1+x²)</p><p><strong>Step 4:</strong> Use Weierstrass substitution with t = tan(α):</p><p>sin(2α) = 2t/(1+t²) and cos(2α) = (1-t²)/(1+t²)</p><p><strong>Step 5:</strong> Substitute:</p><p>3·[2t/(1+t²)] / [5 + 4·(1-t²)/(1+t²)] = 3·[2t/(1+t²)] / [(5(1+t²) + 4(1-t²))/(1+t²)]</p><p>= 6t / (5+5t²+4-4t²) = 6t / (9+t²)</p><p><strong>Step 6:</strong> Set equal to 2x/(1+x²):</p><p>6t/(9+t²) = 2x/(1+x²)</p><p>This gives x = t/3 = tan(α)/3</p><p><strong>Step 7:</strong> For the constraint |argument| ≤ 1, we need specific values. Testing t=0 gives x=0, and examining the range, the possible value is:</p><p>∴ Answer: x = tan(α)/3 (or x = 0 for α = 0)</p>
Correct Answer: A