MCQ
$16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)$ is equal to
- A$\sqrt{3}$
- ✓$2 \sqrt{3}$
- C$3$
- D$4 \sqrt{3}$
✓
Answer
Correct option: B.
$2 \sqrt{3}$
b
$16 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$
$16 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$
$=16 \sin 40^{\circ} \sin 20^{\circ} \sin 80^{\circ}$
$=4(4 \sin (60-20) \sin (20) \sin (60+20))$
$=4 \times \sin \left(3 \times 20^{\circ}\right)$
${[\because \sin 3 \theta=4 \sin (60-\theta) \times \sin \theta \times \sin (60+\theta)]}$
$=4 \times \sin 60^{\circ}$
$=4 \times \frac{\sqrt{3}}{2}=2 \sqrt{3}$
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