Discussion
\(sin15^{\circ}sin45^{\circ}sin75^{\circ} =\)
Answer: B
The product is \(\frac{1}{4\sqrt{2}}\).
Since \(\sin 75^{\circ} = \cos 15^{\circ}\), the outer factors form a double angle.
\[ \begin{aligned} \sin 15^{\circ}\cos 15^{\circ} &= \tfrac{1}{2}\sin 30^{\circ} = \tfrac{1}{4} \\ \tfrac{1}{4}\sin 45^{\circ} &= \tfrac{1}{4} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{4\sqrt{2}} \end{aligned} \]
\(\frac{1}{2\sqrt{2}}\) drops the factor \(\frac{1}{2}\) from the double-angle formula. The options with \(\sqrt{3}\) would need a 30 or 60 degree factor that is not in the product.
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