Trigonometry & Inverse Trigonometry
Trigonometric Equations and Inequalities
Grade 12
Question:
<p>Given that \(6\cos\theta + 2\sin\theta = \sqrt{10}\), where \(\tan^{-1}(3) = \alpha\) and \(\sin(\theta + \alpha) = \frac{1}{2}\), find the range of \(\theta\).</p>
<p>(a) \(\theta \in \left(0, \frac{5\pi}{6} - \tan^{-1}(3)\right)\)</p>
<p>(b) \(\theta = \frac{5\pi}{6} - \tan^{-1}(3)\)</p>
<p>(c) \(\theta > \frac{5\pi}{6} - \tan^{-1}(3)\)</p>
<p>(d) \(\theta + \alpha = \frac{5\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Convert trigonometric equations to standard form using angle addition formulas and solve for the parameter range.
<p><strong>Step 1:</strong> From \(6\cos\theta + 2\sin\theta = \sqrt{10}\), express in the form \(R\sin(\theta + \phi)\)</p><p><strong>Step 2:</strong> Calculate: \(R = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\)</p><p><strong>Step 3:</strong> We have \(\sin(\theta + \alpha) = \frac{1}{2}\) where \(\tan\alpha = 3\)</p><p><strong>Step 4:</strong> This gives \(\theta + \alpha = \frac{5\pi}{6}\) (for the relevant branch)</p><p><strong>Step 5:</strong> Therefore: \(\theta = \frac{5\pi}{6} - \tan^{-1}(3)\) and \(\theta \in \left(0, \frac{5\pi}{6} - \tan^{-1}(3)\right)\)</p><p>∴ Answer is (a, b, c, d).</p>
Correct Answer: a, b, c, d