cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)的值为多少?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 18:25:59
cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)的值为多少?

cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)的值为多少?
cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)的值为多少?

cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)的值为多少?
cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)
=cos(α-38°-22°-α)
=cos-60
=cos60°°
=1/2

运用三角函数两角和公式:cos(α-β)=cosαcosβ+sinαsinβ
cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)
=cos[α-38°-(22°+α)]
=cos(-60°)
=cos60°
=1/2

根据余弦定理公式cosαcosβ+sinαsinβ=cos(α-β);
所以cos(α-38°)cos(22°+α)+sin(α-38°)sin(22°+α)
=cos(α-38°-22°-α)
=cos(-60º)
=cos(60°)
=1/2