## Using y''+y=0 to prove trig identities

Knowing that sine and cosine are a fundamental set of solutions to y''+y=o, and by assuming d(cos(x))/dx=-sin(x), d(sin(x))/dx=cos(x), sin(0)=0, and cos(0)=1, show that sin(x+u)=sin(x)cos(u)+cos(x)sin(u).