三角比的关系
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1. cosΘ=4/5 it me 邻边 = 4x
斜边 = 5x so 对边 = 3x (by pyth. thm) so sinΘ = 3/5 ( sinΘ+2)/(cosΘ-1) = (3/5 + 2)(4/5 - 1) = (13/5)(-1/5) = -13/25 2. cos(4次方)Θ-sin(4次方)Θ+2sin(2次方)Θ = (cos(2次方)Θ-sin(2次方)Θ)(cos(2次方)Θ+sin(2次方)Θ) + 2sin(2次方)Θ = (cos(2次方)Θ-sin(2次方)Θ)(1) + 2sin(2次方)Θ = cos(2次方)Θ-sin(2次方)Θ+ 2sin(2次方)Θ = cos(2次方)Θ+sin(2次方)Θ = 1
参考: myself
1. 已知cosΘ=4/5 求( sinΘ+2)/(cosΘ-1) 2. cos(4次方)Θ-sin(4次方)Θ+2sin(2次方)Θ 请详细列式
thz
斜边 = 5x so 对边 = 3x (by pyth. thm) so sinΘ = 3/5 ( sinΘ+2)/(cosΘ-1) = (3/5 + 2)(4/5 - 1) = (13/5)(-1/5) = -13/25 2. cos(4次方)Θ-sin(4次方)Θ+2sin(2次方)Θ = (cos(2次方)Θ-sin(2次方)Θ)(cos(2次方)Θ+sin(2次方)Θ) + 2sin(2次方)Θ = (cos(2次方)Θ-sin(2次方)Θ)(1) + 2sin(2次方)Θ = cos(2次方)Θ-sin(2次方)Θ+ 2sin(2次方)Θ = cos(2次方)Θ+sin(2次方)Θ = 1
参考: myself
1. 已知cosΘ=4/5 求( sinΘ+2)/(cosΘ-1) 2. cos(4次方)Θ-sin(4次方)Θ+2sin(2次方)Θ 请详细列式
thz
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