Why is $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$? I was studying math.. and I just realized that I only just memorized these trigonometric equations, but I don't really know the reason behind them. So um... 
Why is $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$? 
What happens if $x=3\theta$? 
Would the equations change to something like 
$\cos(6\theta)=\cos^2(3\theta)-\sin^2(3\theta)$ and 
$\sin(6\theta)=2\sin(3\theta)\cos(3\theta)$?
 A: One way to see and also remember easier the always evolving trigonometric identities is using (a representation of) complex numbers and DeMoivre's theorem (or more correctly Euler's identity) like this:
$$e^{ix} = \cos(x) + i\sin(x)$$
Then
$$e^{i2x} = e^{ix} \times e^{ix}$$
$$\cos(2x) + i\sin(2x) = (\cos(x) + i\sin(x)) \times (\cos(x) + i\sin(x))$$
by equating real and imaginary parts respectively one gets the identities and so on..
Now this is purely formal as stated, but by actually taking into account the geometrical meaning (or representation) of a complex number (as is assumed in Euler's representation) this is made way more intuitive. For example multiplication of 2 complex numbers (of modulus $1$ i.e $e^{ia}$) has a direct meaning as geometrical rotation (on the unit circle) and the rest follow from that.
Furthermore, this representation of complex numbers of modulus one as representing rotations is further used in groups like $SU(n)$ and $U(1)$ which play a central part in physics and engineering (for example Quantum Mechanics)
A: 
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ Proof without words:

$\qquad\quad$ 

$\qquad\quad$ Image credit: "Blue's Blog: The Bloog!". See also this Math.SE answer by Blue.

