Question

已知sin(α-β)=1/3,cosαsinβ=1/6,求cos(2α+2β)

Answer 1

问：已知sin(α-β)=1/3,cosαsinβ=1/6,求cos(2α+2β)

答：亲，你好！为您找寻的答案： cos(2α+2β) = 11/18 - 4sqrt(2)/9。 根据三角函数的和差公式，有： sin(α-β) = sinαcosβ - cosαsinβ 代入已知条件，得: 1/3 = sinαcosβ - cosαsinβ 1/6 = cosαsinβ 由此可得： cosαcosβ = sqrt(1 - cos^2α) * sqrt(1 - cos^2β) = sqrt(1 - (cosαsinβ)^2) = sqrt(1 - 1/36) = sqrt(35/36) = sqrt(35)/ sinαsinβ = 2cosαsinβ/2 = cos(π/2 - α)sinβ = sin(α-π/2)sinβ = -cosαcos(π/2-β) = cosαsinβ = 1/6 接下来，计算cos(2α+2β)： cos(2α+2β) = cos^2(α+β) - sin^2(α+β) = (cosαcosβ - sinαsinβ)^2 - (sinαcosβ + cosαsinβ)^2 = cos^2αcos^2β - 2cosαsinαsinβcosβ + sin^2αsin^2β - sin^2αcos^2β - 2cosαsinαsinβsinβ - cos^2αsin^2β = cos^2α(cos^2β - sin^2β) - sin^2α(sin^2β - cos^2β) - 2cosαsinαsinβ(cos^2β + sin^2β) = (cos^2α - sin^2α)(cos^2β - sin^2β) - 2cosαsinαsinβ = cos2αcos2β - sin2αsin2β - 2cosαsinαsinβ = cos2αcos2β - sin2αsin2β - cos(α-β) + cos(α-β) = cos2αcos2β - sin2αsin2β - cos(α-β) + cos(β-α) = cos2αcos2β - cos(α-β) - cos(α-β) = cos2αcos2β - 2cos(α-β) 代入已知条件，"

答：亲亲， 可得： cos(α-β) = sqrt(1 - sin^2(α-β)) = sqrt(8/9) = 2sqrt(2)/3 cos(2α+2β) = cos2αcos2β - 2cos(α-β) = (cos^2α - sin^2α)(cos^2β - sin^2β) - 2cos(α-β) = (1 - 4/36)(35/36 - 1/36) - 2 * 2sqrt(2)/3 = 11/18 - 4sqrt(2)/9 因此，cos(2α+2β) = 11/18 - 4sqrt(2)/9。