QJEEM22Ma2dc5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma2dc5 Solution: 16 sin ⁡ ( 20 ∘ ) sin ⁡ ( 40 ∘ ) sin ⁡ ( 80 ∘ ) is equal to :

Question

Verified Solution • 2025–26 Session moderate

$16 \sin (20^{\circ}) \sin (40^{\circ}) \sin (80^{\circ})$ is equal to :

Accepted Answer

$16 \sin 20^{\circ} . \sin 40^{\circ} . \sin 80^{\circ}$

$= 4 \sin 60^{\circ}$ { $∵$ $4 \sin θ . \sin (60^{\circ} - θ) . \sin (60^{\circ} + θ) = \sin 3 θ$ }

$= 2 \sqrt{3}$

Step-by-Step Verified Solution