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canary651
@canary651
August 2022
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Помогите сделать номер 5
Совсем запуталась там :)
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Verified answer
Tgα=sinα/cosα cosα=2/√5
α∈(1,5π;2π)
sinα=-√(1-cos²α)=-√(1-4/5)=-√(1/5)=-1/√5
tgα=(-1/√5) / (2/√5)=-1/2=-0,5
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canary651
Спасибо большое)
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Answers & Comments
Verified answer
Tgα=sinα/cosα cosα=2/√5α∈(1,5π;2π)
sinα=-√(1-cos²α)=-√(1-4/5)=-√(1/5)=-1/√5
tgα=(-1/√5) / (2/√5)=-1/2=-0,5