Proving Sine Formula in Triangle ABC

[whkoh]

By using the Sine formula in triangle ABC, show that:

\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}.

I've tried:
\frac{2 sin C}{c} = \frac{sin A}{a} + \frac{sin B}{b}
\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}
\frac{a+b}{c} = \frac{b sin A + a sin B}{2 sin C}
\frac{a+b}{c} = \frac{b sin A + a sin B}{4sin\frac{c}{2}cos\frac{c}{2}}

Am I on the right track? Don't really know how to continue.

 

[VietDao29]

By using the Sine formula in triangle ABC, show that:

\frac{a+b}{c} = \frac{cos\frac{A-B}{2}}{sin\frac{c}{2}}.

I've tried:
\frac{2 sin C}{c} = \frac{sin A}{a} + \frac{sin B}{b}
\frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}
\frac{a+b}{c} = \frac{b sin A + a sin B}{2 sin C}
\frac{a+b}{c} = \frac{b sin A + a sin B}{4sin\frac{c}{2}cos\frac{c}{2}}

Am I on the right track? Don't really know how to continue.

No, you are not.
There's an error when you go from line #1 to line #2. Line #2 should read:
\frac{2 sin C}{c} = \frac{b sin A + a sin B}{ab} not \frac{2 sin C}{c} = \frac{b sin A + a sin B}{a+b}.
You may want to try this way:
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{a + b}{\sin A + \sin B} = \frac{c}{\sin C}.
Can you go from here?