∴
sinB |
sinA |
∴C为钝角,sinB=-3sinAcosC,
又sinB=sin(A+C)=sinAcosC+cosAsinC,
∴sinAcosC+cosAsinC=-3sinAcosC,即cosAsinC=-4sinAcosC,
∴tanC=-4tanA,
∴tanB=-tan(A+C)=-
tanA+tanC |
1-tanAtanC |
-3tanA |
1+4tan2A |
3 | ||
|
3 |
4 |
当且仅当
1 |
tanA |
1 |
2 |
则tanB的最大值为
3 |
4 |
故答案为:
3 |
4 |
sinB |
sinA |
sinB |
sinA |
tanA+tanC |
1-tanAtanC |
-3tanA |
1+4tan2A |
3 | ||
|
3 |
4 |
1 |
tanA |
1 |
2 |
3 |
4 |
3 |
4 |