若sinθ=4/5,则cos2θ=

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 07:18:13
若sinθ=4/5,则cos2θ=
x){ѽ83[}S3󋍀<"}rv6u%47iۡi H57U34@,8#[K}# 34qA$”euAH& 2M<;l uԔ - AiqMK5|a x04!b@\f4yqF6jj;CX`7)((8-H$jB&j>71L'5@C

若sinθ=4/5,则cos2θ=
若sinθ=4/5,则cos2θ=

若sinθ=4/5,则cos2θ=
sin(π/2+θ)=cosa=3/5 cosa=3/5 (cosa)^2=9/25 (sina)^2=16/25 cos2a=(cosa)^2-(sina)^2=-7/25 sin(π/2+θ)=3/5, cosθ

cos2θ=1-2sin²θ
=1-2*16/25
=1-32/25
=-7/25

cos2a=1-2sin²a=1-2×(4/5)²=-7/25

-7/25

由Sin^2(α)+Cos^2(α)=1
Cos2a=Cos^2(a)-Sin^2(a)得
Cos2a=1-2Sin^2(a)
=18/25