# What's wrong with my proof?

• Aug 6th 2010, 02:32 PM
chengbin
What's wrong with my proof?
Prove $\cos A + \cos B + \cos C -1=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ when $A+B+C=180^\circ$

$\cos A + \cos B + \cos C -1= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-2\sin^2\frac{C}{2}$

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

$2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\sin \frac {C}{2} )$

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

$2\sin \frac{C}{2}(-2\sin\frac {A}{2}\sin \frac {B}{2})$

$-4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

I had the same steps up towards $2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$, but then the negative sign just disappeared from the next step. Isn't the formula for $\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$?
• Aug 6th 2010, 06:25 PM
eumyang
From here:
$= 2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$

to here:
$= 2\sin \frac{C}{2}(-2\sin\frac {A}{2}\sin \frac {B}{2})$

is the problem, I think. That final angle of B/2 should be negative:
$= 2\sin \frac{C}{2}\left(-2\sin\frac {A}{2}\sin \left( -\frac {B}{2} \right) \right)$

And of course, sine being an odd function, you bring the negative out:
$= 2\sin \frac{C}{2}\left(2\sin\frac {A}{2}\sin \frac {B}{2}\right)$

$= 4\sin \frac{A}{2} \sin\frac {B}{2} \sin \frac {C}{2}$