# Thread: Trigonometry Triangle Problem

1. ## Trigonometry Triangle Problem

Hi it's me again..

I'm having a bit of trouble visualizing this question. It's the last one in my "Holiday Homework package, and I have no idea what it means...

In any triangle ABC, prove that:

(a-b)/b= (2 sin(C/2) sin((A-B)/2))/ sinB

P.S: I didn't want to revive a dead thread, but I would like to thank Soroban for answering the question on my last thread!

2. From the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

we have $a=2R\sin A, \ b=2R\sin B$

Then $\frac{a-b}{b}=\frac{\sin A-\sin B}{\sin B}=$

$=\frac{2\sin\frac{A-B}{2}\cos\frac{A+B}{2}}{\sin B}=\frac{2\sin\frac{A-B}{2}\sin\frac{C}{2}}{\sin B}$

3. THank you very much for posting an answer. I understand the first part of the question very clearly. I just have a question about the last part:

How does cos((A+B/2) become sin (C/2)?

Thanks!

4. Originally Posted by KelvinScale
THank you very much for posting an answer. I understand the first part of the question very clearly. I just have a question about the last part:

How does cos((A+B/2) become sin (C/2)?

Thanks!
$\cos\frac{A+B}{2}=\cos\frac{\pi-C}{2}=\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)=\sin\frac{C}{2}$

because $\cos\left(\frac{\pi}{2}-x\right)=\sin x$