Given Arg (- 1 + 2I) = α, find sin (2 α + 2 π / 3)

Since Arg (- 1 + 2I) = α, sin α = 2 / √ 5, cos α = - 1 / √ 5. So we know sin2 α = 2Sin α · cos α = - 4 / 5. Cos2 α = 2cos α · cos α - 1 = - 3 / 5. Sin (2 α + 2 π / 3) = sin2 α · Cos2 π / 3 + Cos2 α · sin2 π / 3 = (- 4 / 5) × √ 3 / 2 + (- 3 / 5) × (- 1 / 2) = (3-4 √ 3) / 10

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