<p>Given, \(\cos(a - b) = 1\) and \(\cos(a + b) = \frac{1}{e}\). Find the number of ordered pairs \((a, b)\) satisfying the relation \(\cos(2a) = \frac{1}{e}\) where \(-\pi < a < \pi\).</p>
Step-by-Step Solution
Key Concept: Use the given conditions to reduce the problem to finding solutions of \(\cos(2a) = \frac{1}{e}\) in the interval \((-2\pi, 2\pi)\). Count the number of solutions by analyzing the periodicity of cosine.
<p><strong>Step 1:</strong> From \(\cos(a - b) = 1\), we get \(\cos(a - b) = \cos(0)\), so \(a = b\).</p><p><strong>Step 2:</strong> From \(\cos(a + b) = \frac{1}{e}\) and \(a = b\), we get \(\cos(2a) = \frac{1}{e}\).</p><p><strong>Step 3:</strong> Given \(-\pi < a < \pi\), we have \(-2\pi < 2a < 2\pi\).</p><p><strong>Step 4:</strong> In the interval \((-2\pi, 2\pi)\), the equation \(\cos(2a) = \frac{1}{e}\) has exactly 4 solutions.</p><p>∴ There are 4 ordered pairs.</p>
Correct Answer: 4