如题所述
你参考看看～
后面的我算了等于3
追答你算给我看看呢～
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第1个回答 2017-05-16
解∵π/6<α<π/2
∴π/2<α+π/3<2π/3
∴(α+π/3)为第二象限角
又sin(α+π/3)=2√5/5
∴cos(α+π/3)=-√[1-sin²(α+π/3)]
=-√[1-(2√5/5)²]
=-√5/5
∴tan(α+π/3)=sin(α+π/3)/cos(α+π/3)=-2
∴tanα=tan[(α+π/3)-π/3]
=[tan(α+π/3)-tan(π/3)]/[1+tan(α+π/3)×tan(π/3)]
=(-2-√3)/[1+(-2)×√3]
=2+√3)2√3+1
=(5√3+8)/11追答
∴π/2<α+π/3<2π/3
∴(α+π/3)为第二象限角
又sin(α+π/3)=2√5/5
∴cos(α+π/3)=-√[1-sin²(α+π/3)]
=-√[1-(2√5/5)²]
=-√5/5
∴tan(α+π/3)=sin(α+π/3)/cos(α+π/3)=-2
∴tanα=tan[(α+π/3)-π/3]
=[tan(α+π/3)-tan(π/3)]/[1+tan(α+π/3)×tan(π/3)]
=(-2-√3)/[1+(-2)×√3]
=2+√3)2√3+1
=(5√3+8)/11追答
=2+√3)2√3+1这一行，不用理会，是我在计算，简单写上去，却忘了删除
