Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation. 
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$
$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$
I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?

Let's modify ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} $
to ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} =K$
And ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$
to $K = {p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {K\sqrt{A^2 +B^2}\over \sqrt{A^2 + B^2}}={\sqrt{(-AK)^2 +(-BK)^2}\over \sqrt{A^2 + B^2}}={\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$
