Hello arze Originally Posted by
arze Given triangle ABC prove that $\displaystyle c\sin\frac{A-B}{2}=(a-b)\cos\frac{C}{2}$
I know that $\displaystyle \cos\frac{C}{2}=\sin\frac{A+B}{2}$
but after this i'm lost.
Thanks
The relationship you've quoted is correct, but in fact you need to use a similar one; one that begins:
$\displaystyle \sin\tfrac12C = ...$ ...(1)
I'll start the proof for you. See whether you can finish it.$\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\displaystyle \Rightarrow a -b = \frac{c}{\sin C}(\sin A - \sin B)$$\displaystyle =\frac{2c\cos\tfrac12(A+B)\sin\tfrac12(A-B)}{2\sin\tfrac12C\cos\tfrac12C}$
$\displaystyle = ... $?
Can you complete the relationship I started in (1), and so complete the proof?
Grandad