# What's wrong with my proof?

#### chengbin

Prove $$\displaystyle \cos A + \cos B + \cos C -1=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$ when $$\displaystyle A+B+C=180^\circ$$

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

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

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

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

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

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

I had the same steps up towards $$\displaystyle 2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$$, but then the negative sign just disappeared from the next step. Isn't the formula for $$\displaystyle \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$$?

#### eumyang

From here:
$$\displaystyle = 2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$$

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

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

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

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

chengbin