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waterina
@waterina
July 2022
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Вычислить : sin ( arccos (sqrt(2)/3)+ arctg(sqrt(5))
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Sin( arccos(√2/3) + arctg√5)
Арккосинус √2/3 - это угол, α, косинус которого равен √2/3.
arccos(√2/3) = α α∈[0 ; π]
cos α = √2/3
arctg√5 = β, β∈[ - π/2 ; π/2]
tgβ = √5
sin( arccos(√2/3) + arctg√5) = sin(α + β) = sinα·cosβ + cosα·sinβ
sinα = √(1 - cos²α) = √(1 - 2/9) = √7/3
tg²β + 1 = 1/cos²β
5 + 1 = 1/cos²β
cos²β = 1/6
cosβ = 1/√6
sinβ = √(1 - cos²β) = √(1 - 1/6) = √(5/6)
sinα·cosβ + cosα·sinβ = √7/3 · 1/√6 + √2/3 · √5/√6 =
= (√7 + √10)/(3√6)
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waterina
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Verified answer
Sin( arccos(√2/3) + arctg√5)Арккосинус √2/3 - это угол, α, косинус которого равен √2/3.
arccos(√2/3) = α α∈[0 ; π]
cos α = √2/3
arctg√5 = β, β∈[ - π/2 ; π/2]
tgβ = √5
sin( arccos(√2/3) + arctg√5) = sin(α + β) = sinα·cosβ + cosα·sinβ
sinα = √(1 - cos²α) = √(1 - 2/9) = √7/3
tg²β + 1 = 1/cos²β
5 + 1 = 1/cos²β
cos²β = 1/6
cosβ = 1/√6
sinβ = √(1 - cos²β) = √(1 - 1/6) = √(5/6)
sinα·cosβ + cosα·sinβ = √7/3 · 1/√6 + √2/3 · √5/√6 =
= (√7 + √10)/(3√6)