ABC的内角A,B,C的对边分别为a,b,c,且2tanAtanC=tanAtanB+tanBtanC1.证明a^2,b^2,c^2成等差数列且0

问题描述:

ABC的内角A,B,C的对边分别为a,b,c,且2tanAtanC=tanAtanB+tanBtanC
1.证明a^2,b^2,c^2成等差数列且0

1.因为2tanAtanC=tanAtanB+tanBtanC,所以:tanB=2tanAtanC/(tanA+tanC)=2sinAsinC/(sinAcosC+cosAsinC)=2sinAsinC/sin(A+C)=2sinAsinC/sinB则:cosB=sin²B/(2sinAsinC)=b²/(2ac) (*) (注:应用正弦定理)...