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Miscellaneous Exercise 3 (II) · Q110

Q.Prove the following: sin⁡20°sin⁡40°sin⁡80°=38\sin20°\sin40°\sin80°=\dfrac{\sqrt3}{8}

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Step 1: Note sin⁡40°=sin⁡(60°−20°)\sin40°=\sin(60°-20°) and sin⁡80°=sin⁡(60°+20°)\sin80°=\sin(60°+20°).

Step 2: Use the standard identity sin⁡θsin⁡(60°−θ)sin⁡(60°+θ)=14sin⁡3θ\sin\theta\sin(60°-\theta)\sin(60°+\theta)=\dfrac14\sin3\theta with θ=20°\theta=20°: sin⁡20°sin⁡40°sin⁡80°=14sin⁡60°\sin20°\sin40°\sin80°=\dfrac14\sin60°. …

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