proof
$$\begin{align} \cos{(A + B)} & = \frac{OT}{OR} \newline & = \frac{OP - TP}{OR} \newline & = \frac{OP}{OR} - \frac{TP}{OR} \newline & = \frac{OP}{OR} - \frac{SQ}{OR} \newline & = \frac{OP}{OQ} \cdot \frac{OQ}{OR} - \frac{SQ}{RQ} \cdot \frac{RQ}{OR} \newline & = \cos B \cdot \cos A - \sin B \cdot \sin A \end{align} $$
Hence,
$$ \cos {(A+B)} \equiv \cos A \cos B - \sin A \sin B $$
double angle
Set B = A, then
$$\begin{align} \cos {2A} & = \cos{(A+A)} \newline & =\cos A \cos A - \sin A \sin A \newline & = \cos^2 {A} - \sin^2 {A}\end{align}$$
Now, with the Pythagorean identity we can rearrange this identity.
From $ \sin^2 {A} + \cos^2 {A} \equiv 1 $ we get either
$$ \sin^2 {A} = 1 - \cos^2 {A} $$ or
$$ \cos^2 {A} = 1 - \sin^2 {A} $$
Substituting either we get
$$\begin{align} \cos {2A} & = \cos^2 {A} - \sin^2 {A} \newline & = \cos^2 {A} - (1 - \cos^2 {A}) \newline & = 2\cos^2{A} - 1 \end{align}$$
or
$$\begin{align} \cos {2A} & = \cos^2 {A} - \sin^2 {A} \newline & = (1 - \sin^2 {A}) - \sin^2 {A} \newline & = 1 - 2\sin^2{A} \end{align}$$
So,
$$ \cos {2A} \equiv \cos^2 {A} - \sin^2 {A} \equiv 2\cos^2{A} - 1 \equiv 1 - 2\sin^2{A} $$