TS EAMCET · Maths · Trigonometric Ratios & Identities
\(\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 60^{\circ} \cdot \sin 80^{\circ}\) is equal to
- A \(\frac{-3}{16}\)
- B \(\frac{5}{16}\)
- C \(\frac{3}{16}\)
- D \(\frac{-5}{16}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{16}\)
Step-by-step Solution
Detailed explanation
\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…
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