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## Homework Statement

Given a triangle ABC prove that

[itex]c sin \frac{A-B}{2} = (a-b)cos \frac{C}{2}[/itex]

## Homework Equations

## The Attempt at a Solution

It looks rather similar to a formula mentioned in my book's lead-in to this exercise:

[itex]\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}[/itex]

Which can be rearranged to

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}

[/itex]

So from this it would seem I need only prove that

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}= c[/itex]

However this doesn't seem much of a simplification and I feel i'm still at square one trying to prove another identity. I also felt my approach was a bit unorthodox. Any suggestions would be appreciated.