MCQ
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=$
  • A
    $\frac{-3}{16}$
  • B
    $\frac{5}{16}$
  • $\frac{3}{16}$
  • D
    $\frac{-5}{16}$

Answer

Correct option: C.
$\frac{3}{16}$
(C)
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
$=\frac{\sqrt{3}}{2} \sin 20^{\circ} \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)$
$\sin \theta \sin \left(60^{\circ}-\theta\right) \sin \left(60^{\circ}+\theta\right)=\frac{1}{4} \sin 3 \theta$
$=\frac{\sqrt{3}}{2} \cdot \frac{1}{4} \sin 60^{\circ}$
$=\frac{\sqrt{3}}{8} \cdot \frac{\sqrt{3}}{2}$
$=\frac{3}{16}$

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