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DM1824
@DM1824
July 2022
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помогите ,вообще не понимаю
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Vasily1975
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Sin(α-β)=sinα*cosβ-cosα*sinβ. Так как 3π/2<α<2π, то sinα=-√(1-cos²α)=-√9/25=-3/5. Так как π<β<3π/2, то cosβ=-√(1-sin²β)=-√16/25=-4/5. Тогда sin(α-β)=-3/5*(-4/5)-4/5*3/5=0. Ответ: 0.
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Verified answer
Sin(α-β)=sinα*cosβ-cosα*sinβ. Так как 3π/2<α<2π, то sinα=-√(1-cos²α)=-√9/25=-3/5. Так как π<β<3π/2, то cosβ=-√(1-sin²β)=-√16/25=-4/5. Тогда sin(α-β)=-3/5*(-4/5)-4/5*3/5=0. Ответ: 0.