One way to derive it is to use Euler's formula!

The reason I was asking was because the derivation of Euler's formula that I saw used it at a part...

Yeah. This is how I proved Euler's Formula in my differential equations class a few months ago.

In general, there is no such thing as "the definition" of a mathematical object like sine. Rather, there is a web of equivalent statements, any one of which can be used as "the definition". So to prove it in an entirely satisfactory way, I would need to know the definition you are starting with.

Another way to solve the equation y'' = -y is to search for a power series solution. Suppose we want the solution which matches y(0) = 0, y'(0) = 1. Then

y = 0 + x + a2 x^2 + a3 x^3 + ...

y'' = 2 a2 + 2*3 a3 x + 3*4 a4 x^2 + ...

Since we want y'' = -y, we match up coefficients and solve. You'll arrive at the Taylor series for sin(x).

Aaaaaaaaand I don't understand it. I guess I'll wait until I know more to think about it.

Let me try one more. I think the definition of sine you have in mind is geometric. Here's a derivation that is based on the unit circle.

It is easier for me to think of if the independent variable is t for time. In fact, let's rephrase the problem as looking for the function x(t) for which x'' = - x.

A common technique is to remove the second derivative by introducing a new function, y(t) = x'(t). Then our differential equation just says y'(t) = -x(t). In other words, we now have two equations with two unknown functions (but no second derivatives!)

x' = y

y' = -x

At each time t, let P(t) be the point in the plane with x coordinate x(t) and y coordinate y(t). So P(t) = ( x(t), y(t) ). As time proceeds, in which direction will this point move? For that we need to see what the rate of change of P is. That's the derivative dP/dt, which in this case is P'(t) = ( x'(t), y'(t) ) = (y,-x) from our pair of differential equations.

Now, the direction from the origin to (x,y) is perpendicular with to the direction from the origin to (-y,x), So P(t) moves around the origin in a circle with a constant speed.

Now suppose our initial conditions are x(0) = 1, x'(0) = 0. Well, y(0) = x'(0) = 0, so the point starts at (1,0). The speed is given by v^2 = (x')^2 + (y')^2 = (y)^2 + (-x)^2 = y^2 + x^2. At t=0 this gives v(0)=1, and since the speed is constant we have v(t) = 1. Thus the distance traveled in time t is just v*t = 1*t = t, and so t is equal to the angle swept out as measured in radians. The x-coordinate is therefore x(t) = cos(t).