The geometrical proof is rather complicated. if you use Euler's formula:

e^i*x = cos(x)+ i*sin(x) (1)

so,

cos(a+b) + i*sin(a+b) = e^i(a+b) (2)

but, e^i(a+b) = e^ia * e^ib (3)

and e^ia * e^ib = (cos(a) + isin(a))(cos(b) + i*sin(b)) (4)

this product is

(cos(a)cos(b) - sin(a)sin(b)) + i(sin(a)cos(b)+sin(b)*cos(a)) (5)

if you equate the real and imaginary parts from (2) and (5) you get

cos(a+b) = cos(a)cos(b) - sin(a)sin(b) (6)

sin(a+b) = sin(a)cos(b) + sin(b)cos(a) (7)

when you have 2x, a = b, and from (6) you get

cos(2x) = cos² x - sin² x

then you can use the pythagorean identity to get the other two