Question:

<p><strong>Step 1: Apply the half-angle formula</strong></p><p>Recall that $1 + \cos\theta = 2\cos^2(\theta/2)$.</p><p>For the first term: $1 + \cos\frac{\pi}{8} = 2\cos^2\frac{\pi}{16}$</p><p>For the second term: $1 + \cos\frac{3\pi}{8} = 2\cos^2\frac{3\pi}{16}$</p><p><strong>Step 2: Take square roots</strong></p><p>$\sqrt{1 + \cos\frac{\pi}{8}} = \sqrt{2}\cos\frac{\pi}{16}$ (positive since $\frac{\pi}{16} \in (0, \frac{\pi}{2})$)</p><p>$\sqrt{1 + \cos\frac{3\pi}{8}} = \sqrt{2}\cos\frac{3\pi}{16}$ (positive since $\frac{3\pi}{16} \in (0, \frac{\pi}{2})$)</p><p><strong>Step 3: Multiply the results</strong></p><p>$\sqrt{1 + \cos\frac{\pi}{8}} \cdot \sqrt{1 + \cos\frac{3\pi}{8}} = (\sqrt{2}\cos\frac{\pi}{16})(\sqrt{2}\cos\frac{3\pi}{16})$</p><p>$= 2\cos\frac{\pi}{16}\cos\frac{3\pi}{16}$</p><p><strong>Step 4: Use the complementary angle relationship</strong></p><p>Note that $\frac{\pi}{16} + \frac{3\pi}{16} = \frac{4\pi}{16} = \frac{\pi}{4}$</p><p>Therefore: $\cos\frac{3\pi}{16} = \cos(\frac{\pi}{4} - \frac{\pi}{16}) = \sin\frac{\pi}{16}$</p><p><strong>Step 5: Apply the product-to-sum formula</strong></p><p>$2\cos\frac{\pi}{16}\sin\frac{\pi}{16} = \sin\frac{\pi}{8}$ (using $2\sin A\cos A = \sin 2A$)</p><p><strong>Step 6: Evaluate $\sin\frac{\pi}{8}$</strong></p><p>Using $\sin\frac{\pi}{8} = \sqrt{\frac{1-\cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2-\sqrt{2}}{4}} = \frac{\sqrt{2-\sqrt{2}}}{2}$</p><p>Alternatively, note that $\sin\frac{\pi}{8} = \sin 22.5° = \frac{1}{\sqrt{2}} \cdot \sqrt{1 - \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}$</p><p><strong>∴ Answer: B</strong></p>