Q60 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

$\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ}$ is equal to

  • A
    $\frac{-3}{16}$
  • B
    $\frac{5}{16}$
  • Correct answer
    C
    $\frac{3}{16}$
  • D
    $\frac{-5}{16}$

Solution

$\begin{aligned} & \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ} \\ & =\sin 60^{\circ} \cdot\left(\frac{1}{4} \sin \left(3 \times 20^{\circ}\right)\right) \\ & =\frac{1}{4}\left(\sin 60^{\circ}\right)^2=\frac{1}{4} \times \frac{3}{4}=\frac{3}{16} \\ & \qquad\left(\because \sin A \sin \left(60^{\circ}-A\right) \sin \left(60^{\circ}+A\right)=\frac{1}{4} \sin 3 A\right)\end{aligned}$

Hence, the correct answer is option: 'C'.

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