Trigonometry & Inverse Trigonometry
Multiple Angle Formulas
Grade 12
Question:
<p>If \(4x^3 - 3x - p = 0\), where \(-1 \leq p \leq 1\) has a unique root in \([-1, 1]\), then the root is</p>
<p>(a) \(\cos\left(\frac{1}{3}\cos^{-1}p\right)\)</p>
<p>(b) \(\cos\left(\frac{1}{3}\cos^{-1}p\right)\)</p>
<p>(c) \(\cos(\cos^{-1}p)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that \(4x^3 - 3x\) corresponds to the triple angle formula for cosine, and use substitution \(x = \cos\theta\) to convert to an inverse trigonometric equation.
<p><strong>Step 1:</strong> Observe that \(4x^3 - 3x\) is the multiple angle formula for cosine: \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\)</p><p><strong>Step 2:</strong> Let \(x = \cos\theta\), then the equation becomes \(4\cos^3\theta - 3\cos\theta - p = 0\)</p><p><strong>Step 3:</strong> This simplifies to \(\cos 3\theta = p\)</p><p><strong>Step 4:</strong> Therefore \(3\theta = \cos^{-1}p\), which gives \(\theta = \frac{1}{3}\cos^{-1}p\)</p><p><strong>Step 5:</strong> Since \(x = \cos\theta\), the root is \(x = \cos\left(\frac{1}{3}\cos^{-1}p\right)\)</p><p>∴ Answer is (b).</p>
Correct Answer: B