Question:
$\sin 15^{\circ} \sin 45^{\circ} \sin 75^{\circ} =$
$\sin 15^{\circ} \sin 45^{\circ} \sin 75^{\circ} =$
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Using co-function identities like $\sin(90-x) = \cos x$ often simplifies trigonometric products.
Updated On: Apr 28, 2026
- $\frac{1}{2\sqrt{2}}$
- $\frac{1}{4\sqrt{2}}$
- $\frac{1}{3\sqrt{2}}$
- $\frac{1}{4\sqrt{3}}$
- $\frac{1}{\sqrt{3}}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Note that $\sin 75^{\circ} = \cos 15^{\circ}$.
Step 2: Analysis
The expression is $(\sin 15^{\circ} \cos 15^{\circ}) \sin 45^{\circ}$. Recall $\sin 2\theta = 2\sin \theta \cos \theta \implies \sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2}\sin 30^{\circ}$.
Step 3: Calculation
Value $= (\frac{1}{2} \cdot \frac{1}{2}) \cdot \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}}$. Final Answer: (B)
Note that $\sin 75^{\circ} = \cos 15^{\circ}$.
Step 2: Analysis
The expression is $(\sin 15^{\circ} \cos 15^{\circ}) \sin 45^{\circ}$. Recall $\sin 2\theta = 2\sin \theta \cos \theta \implies \sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2}\sin 30^{\circ}$.
Step 3: Calculation
Value $= (\frac{1}{2} \cdot \frac{1}{2}) \cdot \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}}$. Final Answer: (B)
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