Prove

**arze** · February 22nd 2010, 11:07 PM

Given triangle ABC prove that $c\sin\frac{A-B}{2}=(a-b)\cos\frac{C}{2}$
I know that $\cos\frac{C}{2}=\sin\frac{A+B}{2}$
but after this i'm lost.
Thanks

**Grandad** · February 23rd 2010, 02:47 AM

Hello arze

Originally Posted by arze

Given triangle ABC prove that $c\sin\frac{A-B}{2}=(a-b)\cos\frac{C}{2}$
I know that $\cos\frac{C}{2}=\sin\frac{A+B}{2}$
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:

$\sin\tfrac12C = ...$ ...(1)

I'll start the proof for you. See whether you can finish it.

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow a -b = \frac{c}{\sin C}(\sin A - \sin B)$

$=\frac{2c\cos\tfrac12(A+B)\sin\tfrac12(A-B)}{2\sin\tfrac12C\cos\tfrac12C}$

$= ...$ ?

Can you complete the relationship I started in (1), and so complete the proof?

Grandad

Thread: Prove

LinkBack

Thread Tools

Display

Prove

Similar Math Help Forum Discussions

Prove

Prove(10)

prove that if sequence (an) converges, use def of limit of sequence to prove...

How do I prove b^x * b^y = b ^ x + y on R?

Prove:

Search Tags