三角比的關係

1. cosΘ=4/5
it means 鄰邊 = 4x, 斜邊 = 5x
so 對邊 = 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

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




請詳細列式,thz