Question

Evaluate: $$ \sin \frac{\pi}{12} \cdot \sin \frac{2\pi}{12} \cdot \sin \frac{3\pi}{12} \cdot \sin \frac{4\pi}{12} \cdot \sin \frac{5\pi}{12} \cdot \sin \frac{6\pi}{12} $$

• A. \( \frac{\sqrt{3}}{16\sqrt{2}} \)
• B. \( \frac{\sqrt{3}}{8\sqrt{2}} \)
• C. \( \frac{1}{32} \)
• D. \( \frac{1}{16} \)

Hint:

Use known sine values and pair angles that sum to 90° when computing multiple sine products.

Answer 1

The Correct Option is A

We convert to angles: \[ \sin \frac{\pi}{12} = \sin 15^\circ,\ \sin \frac{2\pi}{12} = \sin 30^\circ,\ \sin 45^\circ,\ \sin 60^\circ,\ \sin 75^\circ,\ \sin 90^\circ \] So product becomes: \[ \sin 15^\circ \cdot \sin 30^\circ \cdot \sin 45^\circ \cdot \sin 60^\circ \cdot \sin 75^\circ \cdot \sin 90^\circ \] Now use values: \[ \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4},\ \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},\ \sin 30^\circ = \frac{1}{2},\ \sin 60^\circ = \frac{\sqrt{3}}{2},\ \sin 45^\circ = \frac{1}{\sqrt{2}},\ \sin 90^\circ = 1 \] Now multiply: \[ \left( \frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \right) \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \left( \frac{6 - 2}{16} \right) \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{4}{16} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{8} \cdot \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{16\sqrt{2}} \]