Question

同角三角比的关系

Answer 1

解答过程如下：

证明
（1）
（sina-cosa)²
=sin²a-2sinacosa+cos²a
=sin²a+cos²-2sinacosa
=1-2sinacosa
左边=右边

命题成立
（2）
(tana+tanβ)/(cota+cotβ)
=(sina/cosa+sinβ/cosβ)/(cosa/sina+cosβ/sinβ)
=[(sinacosβ+cosasinβ)/(cosacosβ)]/[(sinacosβ+cosasinβ)/(sinasinβ)]
=(sinasinβ)/(cosacosβ)
=tana*tanβ
左边=右边
命题成立

请点击输入图片描述cosΘ=4/5 it me 

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
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