sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 19:11:36
![sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)](/uploads/image/z/7639480-64-0.jpg?t=sin%28%CE%B1-%CE%B2%29sin%28%CE%B2-r%29-cos%28%CE%B1-%CE%B2%29cos%28r-%CE%B2%29+1.sin%28%CE%B1-%CE%B2%29sin%28%CE%B2-r%29-cos%28%CE%B1-%CE%B2%29cos%28r-%CE%B2%292.%EF%BC%88+tan4%E5%88%86%E4%B9%8B5%CF%80%2Btan12%E5%88%86%E4%B9%8B5%CF%80%EF%BC%89%2F%EF%BC%881-tan12%E5%88%86%E4%B9%8B5%CF%80%EF%BC%893.%5B+sin%EF%BC%88%CE%B1%2B%CE%B2%EF%BC%89-2sin%CE%B1cos%CE%B2%5D%2F2sin%CE%B1sin%CE%B2%2Bcos%EF%BC%88%CE%B1%2B%CE%B2%EF%BC%89)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)
3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
=-[cos(α-β)cos(r-β)-sin(α-β)sin(β-r)]
=-[cos(α-β)cos(β-r)-sin(α-β)sin(β-r)]
=-[cos(α-β)+(β-r)]
=-cos(α-β+β-r)
=-cos(α-r)
( tan5π/4+tan5π/12)/(1-tan5π/12)
=[ tan(π+π/4)+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tanπ/4tan5π/12)
= tan(π/4+5π/12)
= tan(3π/4)
= tan(π-π/4)
=- tanπ/4
=-1
[ sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]
= [ sinαcosβ+cosαsinβ-2sinαcosβ]/[2sinαsinβ+cosαcosβ-sinαsinβ]
= [ cosαsinβ-sinαcosβ]/[cosαcosβ+sinαsinβ]
=-[ sinαcosβ-cosαsinβ]/[cosαcosβ+sinαsinβ]
=- sin(α-β)/cos(α-β)
=-tan(α-β)