- #1
Appleton
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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.