Question

已知 tana=3tan(a+b),b=π/6,求sin(2a+b)的值

Answer 1

tan a=3tan(a+b),b=π/6代入得：√3tanA^2+6tanA+3√3＝0

tanA=－√3

如果是会万能公式,直接代入即可.

sin(2A+B)=sin2AcosB+cos2AsinB＝√3sinAcosA+(2cosA^2-1)/2

=√3sinAcosA+cosA^2-1/2=[(√3sinAcosA+cosA^2)/(sinA^2+cosA^2)]-1/2

=(√3tanA+1)/(tanA^2+1)-1/2

=-1

解法二：请看图片：