JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
\(16 \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(80^{\circ}\right)\) is equal to
- A \(\sqrt{3}\)
- B \(2 \sqrt{3}\)
- C \(3\)
- D \(4 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(16 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}\) \(=16 \sin 40^{\circ} \sin 20^{\circ} \sin 80^{\circ}\) \(=4(4 \sin (60-20) \sin (20) \sin (60+20))\) \(=4 \times \sin \left(3 \times 20^{\circ}\right)\)…
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