Thread: Prove sinA+sinB>sinC>|sinA-sinB|

If A+B+C=180, then prove

I can prove by









Since sin is positive from 0 to 180, and

I can't prove the next part. Help?

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.

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.