# Math Help - Prove sinA+sinB>sinC>|sinA-sinB|

1. ## Prove sinA+sinB>sinC>|sinA-sinB|

If A+B+C=180, then prove $\sin A+\sin B>\sin C>|\sin A-\sin B|$

I can prove $\sin A+\sin B>\sin C$ by

$2\sin \frac{A+B}{2}\cos \frac{A-B}{2}>2\sin \frac{C}{2}\cos \frac{C}{2}$

$\cos \frac{A-B}{2}>\cos \frac{A+B}{2}$

$\cos \frac{A-B}{2}-\cos \frac{A+B}{2}>0$

$2\sin A\sin B>0$

Since sin is positive from 0 to 180, $\sin A\sin B>0$ and $\sin A+\sin B>\sin C$

I can't prove the next part. Help?

2. To prove sinC> sin(A-B)

2sin(C/2)cos(C/2) > 2cos([A+B)/2]*sin[(A-B)/2]

sin[(A+B)/2] > sin[A-B)/2]

A+B and A-B are less than 180 degrees.

3. To prove sinC> sin(A-B)

2sin(C/2)cos(C/2) > 2cos([A+B)/2]*sin[(A-B)/2]

sin[(A+B)/2] > sin[A-B)/2]

A+B and A-B are less than 180 degrees.