Trigonometric Ratios and Identities
Important Identities
JEE Main 2022
Grade 11

Question:

$16 \sin 20^\circ \sin 40^\circ \sin 80^\circ$ is equal to:
(1) $\sqrt{3}$
(2) $2\sqrt{3}$
(3) $4\sqrt{3}$
(4) $2$

Step-by-Step Solution

Key Concept: Recall $\sin A \sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4} \sin 3A$. Set $A = 20^\circ$, so $\sin 60^\circ = 4 \sin 20^\circ \sin 40^\circ \sin 80^\circ$. Then scale by 16.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)

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