已知三角形ABC中,a,b,c分别为角ABC的对边,sin(2C-π/2)＝1/2,且a^2+b^2

Question

已知三角形ABC中,a,b,c分别为角ABC的对边,sin(2C-π/2)＝1/2,且a^2+b^2

Answer 1

1/2=sin(2C-π/2)=-sin(π/2-2C)=-cos2C
cos2C=-1/2,
2(cosC)^2-1=-1/2
(cosC)^2=1/4
因为a^2+b^2π/2,cosC1
(2/√3)=2√3/3≥cos[(A-B)/2]/(√3/2)>1
即得(a+b)/c的取值范围
1