Срочно решите, ПОЖАЛуЙсТа 4 и 5 задания. Created by Sashulykut. algebra-ru

Срочно решите, ПОЖАЛуЙсТа 4 и 5 задания...

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Срочно решите, ПОЖАЛуЙсТа 4 и 5 задания

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nKrynka Решение
sinα = 24/25; sinβ= 4/5, π/2 < α < π, 0 < β < π/2
4.
1) sin(α + β) = sinαcosβ + cosαsinβ
cosα = - √(1 - sin²α) = - √(1 - (24/25)²) = - √(1 - 576/625) =
= - √49/625 = - 7/25
cosβ = √(1 - sin²β) = √(1 - (4/5)²) = √(1 - 16/25) = √9/25 = 3/5
sin(α + β) = sinαcosβ + cosαsinβ = (24/25)*(3/5) + (-7/25)*(4/5) = 44/125
2) cos(α - β) = cosαcosβ + sinαsinβ = (-7/25)*(3/5) + (24/25)*(4/5) =
= 75/125 = 3/5
5.
1) sin(π - α) = sinπ*cosα- cosπ*sinα = sinα
2) cos(3π/2 - α) = cos(3π/2)*cosα + sin(3π/2)*sinα = - sinα

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