1,α+β=π/4,tan(α+β)=(tanα+tanβ)/(1-tanα*tanβ)
2,cosα=cos(-α)=cos(π/6-α-π/6)=cos(π/6-α)*cos(π/6)+sin(π/6-α)*sin(π/6)=(15/17)*(√3 /2)+(-(8/17))*(1/2)=(15√3-8)/34
其中因为α∈(π/6,π/2),所以-(π/2)
1,α+β=π/4,tan(α+β)=(tanα+tanβ)/(1-tanα*tanβ)
2,cosα=cos(-α)=cos(π/6-α-π/6)=cos(π/6-α)*cos(π/6)+sin(π/6-α)*sin(π/6)=(15/17)*(√3 /2)+(-(8/17))*(1/2)=(15√3-8)/34
其中因为α∈(π/6,π/2),所以-(π/2)