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Exercise 3.4 · Q70

Q.Prove the following: sin⁡20°sin⁡40°sin⁡60°sin⁡80°=316\sin20°\sin40°\sin60°\sin80°=\dfrac{3}{16}

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Step 1: With θ=20°\theta=20°: sin⁡20°sin⁡(60°−20°)sin⁡(60°+20°)=14sin⁡60°\sin20°\sin(60°-20°)\sin(60°+20°)=\dfrac14\sin60°, i.e. sin⁡20°sin⁡40°sin⁡80°=14⋅32=38\sin20°\sin40°\sin80°=\dfrac14\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{8}. …

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