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?