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Prove that

Posted July 19th, 2009 by alpha
in
$$(a-b)\cos\frac{C}{2}= c \sin\frac{A-B}{2}$$
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answer

On July 19th, 2009 Structure says:
$$2R(\sin A-\sin B)\cos\frac{C}{2}=4R\sin\frac{A-B}{2}\cos \frac{A+B}{2}\cos\frac{C}{2}=$$
$$=4R\sin\frac{A-B}{2}\sin \frac{C}{2}\cos\frac{C}{2}=2R \sin C \sin\frac{A-B}{2}=c\sin\frac{A-B}{2}$$

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