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Ayres1234
@Ayres1234
June 2022
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Помогите с алгеброй , пожалуйста ((( (10 кл)
Высислите sin 2a(альфа), cos 2в(бэта) sin (а-в) и соs (а+в), если:
sin a = 4/5, cos в = -5/13, п/2<а<п, п/2<в<п
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nKrynka
Решение
1) sin2a = 2sinacosa
sinα = 4/5 = 0,8
π/2 < α < π
cosα = - √(1 - sin²a) = - √(1 - 0,8²) = √0,36 = - 0,6
sin2a = 2*0,8*(- 0,6) = - 0,96
2) cos2β = 2cos²b - 1
cosβ = - 5/13
π/2 < β < π
cos2β = 2*(-5/13)² - 1 = (2*25)/169 - 1 = 50/169 - 1 = - 119/169
sinβ = √(1 - cos²β) = √(1 - (-5/13)²) = √144/169 = 12/13
3) sin(α - β) = sinαcosβ - cosαsinβ
sin(α - β) = (4/5)*(-5/13) + (3/5)*(12/13) = - 4/13 + 36/65 = 16/65
4) cos(α + β) = cosαcosβ - sinαsinβ
cos(α + β) = (- 3/5)*(- 5/13) - (4/5)*(12/13) = 3/13 - 48/65 = - 33/65
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Answers & Comments
1) sin2a = 2sinacosa
sinα = 4/5 = 0,8
π/2 < α < π
cosα = - √(1 - sin²a) = - √(1 - 0,8²) = √0,36 = - 0,6
sin2a = 2*0,8*(- 0,6) = - 0,96
2) cos2β = 2cos²b - 1
cosβ = - 5/13
π/2 < β < π
cos2β = 2*(-5/13)² - 1 = (2*25)/169 - 1 = 50/169 - 1 = - 119/169
sinβ = √(1 - cos²β) = √(1 - (-5/13)²) = √144/169 = 12/13
3) sin(α - β) = sinαcosβ - cosαsinβ
sin(α - β) = (4/5)*(-5/13) + (3/5)*(12/13) = - 4/13 + 36/65 = 16/65
4) cos(α + β) = cosαcosβ - sinαsinβ
cos(α + β) = (- 3/5)*(- 5/13) - (4/5)*(12/13) = 3/13 - 48/65 = - 33/65