# Math Help - Derivation of Law of Cosines (excerpt)

1. ## Derivation of Law of Cosines (excerpt)

Prove that $a=b \cos{C}+c \cos{B}$

I have that
$\cos{C}=\frac{x}{b} \implies x=b \cos{C}$

$\cos{B}=\frac{a+x}{c} \implies x=c \cos{B}-a$

Hence,

$a=c \cos{B}-b\cos{C}$

A difference instead of a sum??

2. You are using $\cos (\angle C)$ in two different ways.
Note that $\cos (\angle BCA) = - \cos (\angle DCA)$.

3. Oooh, thanks...

You're using $cos(180-\theta) = -cos(\theta)$ right?

4. You can use the sines law.
We have $b=2R\sin B, c=2R\sin C$.
We start from the right member:
$b\cos C+c\cos B=2R\sin B\cos C+2R\sin C\cos B=$
$=2R(\sin B\cos C+\sin C\cos B)=$
$=2R\sin(B+C)=$
$=2R\sin(180-A)=$
$=2R\sin A=$
$=a$