1. ## Triangle Problem

Hey sorry I'm having a bit of mind blank today, can anyone give me a hint to get me started with this
If A, B and C are the angles of a triangle prove that
$
\cos \begin{pmatrix} \frac{B-C}{2} \end{pmatrix} - \sin \begin{pmatrix} \frac{A}{2} \end{pmatrix} = 2 \sin \begin{pmatrix} \frac{B}{2} \end{pmatrix} \sin \begin{pmatrix} \frac{C}{2} \end{pmatrix}
$
Cheers

Simon

2. Originally Posted by thelostchild
Hey sorry I'm having a bit of mind blank today, can anyone give me a hint to get me started with this

Cheers

Simon
Observe that $A = \pi - B - C$(Why?), so

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

Can you continue now?

3. thank you, I'm feeling pretty stupid now I kept looking for how I could do it geometrically that is much easier