You CAN define cosine and sine by
1. y(x)= sin(x) is the function satisfying the differential equation d<sup>2</sup>y/dx<sup>2</sup>= -y and y(0)= 0, y'(0)= 1.
2. y(x)= cos(x) is the function satisfying the differential equation d<sup>2</sup>y/dx<sup>2</sup> = -y and y(0)= 1, y'(0)= 0.
It's easy to show that sin(x) and cos(x) are independent solutions so any solution to that equation can be written as C1cos(x)+ C2sin(x). In fact, if y satisfies y"= y, y(0)= A, y'(0)= B, then y(x)= Acos(x)+ B sin(x).
Let y= (cos(x))' (the derivative of cosine). Since cosine satisfies a second order equation, it is twice differentiable and y'= (cos(x))"= -cos(x). That means that y is twice differentiable and y"= -(cos(x))'= -y. y(0)= 0 since the derivative of cosine at 0 is 0) and y'(0)= -cos(0)= -1. Thus, y= 0cos(x)+(-1)sin(x)= -sin(x). Similarly, one can prove that (sin(x))'= cos(x).
Now, let y= sin(x+a). Then y'= cos(x+a) and y"= -sin(x+a)= -y. That is, y also satisfies y"= -y. y(0)= sin(a), y'(0)= cos(a) so y(x)= sin(x+a)= sin(a)cos(x)+ cos(a)sin(x). Let x= b and we have sin(a+b)= sin(a)cos(b)+ cos(a)sin(b).
Let y= cos(x+a). Then y'= -sin(x+a) and y"= -cos(x+a)= -y. This y also satisfies y"= -y. y(0)= cos(a), y"(0)= -sin(a) so y(x)= cos(a)cos(x)- sin(a)sin(x). Let x= b and we have cos(a+b)= cos(a)cos(b)- sin(a)sin(b).
Those are the simplest proofs I know.