Double Angle

Next time you come across $\sin{2\theta}$,
Forget about your trig sheet of unexplained data.
Recall how to rotate a vector by some degrees

$\left[\begin{matrix}\cos{\theta} & -sin{\theta}\\ sin{\theta} & cos{\theta}\end{matrix}\right]$


And to do it again, we'll need two of these.

$\left[\begin{matrix}\cos{\theta} & -sin{\theta}\\ sin{\theta} & cos{\theta}\end{matrix}\right]$$\left[\begin{matrix}\cos{\theta} & -sin{\theta}\\ sin{\theta} & cos{\theta}\end{matrix}\right]$

Let us find the matrix composition,

$\left[\begin{matrix}\cos^{2}{\theta}-sin^{2}{\theta} & -2sin{\theta}cos{\theta}\\ 2sin{\theta}cos{\theta} & cos^{2}{\theta}-sin^{2}{\theta}\end{matrix}\right]$

to guide our vector to its final position.

But wait!
Rotating by the same angle twice
is rotating by $2\theta$ but less concise!

$\left[\begin{matrix}\cos{2\theta} & -sin{2\theta}\\ sin{2\theta} & cos{2\theta}\end{matrix}\right] = \left[\begin{matrix}\cos^{2}{\theta}-sin^{2}{\theta} & -2sin{\theta}cos{\theta}\\ 2sin{\theta}cos{\theta} & cos^{2}{\theta}-sin^{2}{\theta}\end{matrix}\right]$

Hmm... So that means

$\cos{2\theta} = \cos^{2}{\theta} - \sin^{2}{\theta}$
$\sin{2\theta} = 2\sin{\theta}\cos{\theta}$