Question:

<p><strong>Step 1:</strong> Note that \(\cos\left(\frac{8\pi}{7}\right) = \cos\left(\pi + \frac{\pi}{7}\right) = -\cos\left(\frac{\pi}{7}\right)\)</p><p><strong>Step 2:</strong> So the product becomes \(-\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right)\)</p><p><strong>Step 3:</strong> Using the product formula repeatedly with \(2\sin\theta\cos\theta = \sin(2\theta)\):</p><p>\(= -\frac{1}{2\sin\left(\frac{\pi}{7}\right)}\left[2\sin\left(\frac{\pi}{7}\right)\cos\left(\frac{\pi}{7}\right)\right]\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right)\)</p><p>\(= -\frac{1}{2\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{2\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right)\)</p><p><strong>Step 4:</strong> Continuing this process:</p><p>\(= -\frac{1}{4\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{4\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right) = -\frac{1}{8\sin\left(\frac{\pi}{7}\right)}\sin\left(\frac{8\pi}{7}\right)\)</p><p><strong>Step 5:</strong> Since \(\sin\left(\frac{8\pi}{7}\right) = \sin\left(\pi + \frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)\):</p><p>\(= -\frac{1}{8\sin\left(\frac{\pi}{7}\right)} \cdot \left(-\sin\left(\frac{\pi}{7}\right)\right) = \frac{1}{8}\)</p><p><strong>Correction:</strong> The answer is \(-\frac{1}{8}\)</p>