Defining sine and cosine via ODE's So I read in Simmons book on Differential Equations that via the equation
y''+y=0 
One can define s(x), c(x) as their solutions with some given initial conditions, that is s(0)=0 s'(0)=1 ; c(0)=1, c'(0)=0
These are of course sine and cosine, but you don't actually need to know it beforehand.
Now, what the book says is that just with these information you can deduce all of the trigonometric identities, I've tried out the simplest ones: s'=c , c'= -s, s^2+c^2=1 and also that they are linearly independent, but I'm having trouble with the other identities such as:
s(x+2pi)= s(x) ; 
s(2x)=2s(x)c(x) ; 
s(x+a)=s(x)c(a)+c(x)s(a)
With pi defined as the point where s' crosses the x axis.
Any help will be deeply appreciated. 
 A: The addition formula should be your next goal:
$$\frac d{dx}s(x+a)=c(x+a)\cdot(1)=c(x+a)$$
$$\frac{d^2}{dx^2}s(x+a)=-s(x+a)\cdot(1)=-s(x+a)$$
So $s(x+a)$ satisfies the differential equation and we know that
$$s(x+a)=c_1c(x)+c_2s(x)$$
Applying initial conditions,
$$s(0+a)=s(a)=c_1$$
$$s^{\prime}(0+a)=c(0+a)=c(a)=c_2$$
So we have found that $s(x+a)=s(a)c(x)+c(a)s(x)$. From the pythagorean identity which you have already established,
$$(s(\pi))^2+(c(\pi))^2=(c(\pi))^2=1$$
And since $s(x)>0$ for $0<x<\pi$, we know that $c^{\prime}(x)=-s(x)<0$ over the same interval. Since $c(x)$ was decreasing from $1$ the entire time, it must be that $c(\pi)=-1$. Now we have
$$s(x+\pi)=s(\pi)c(x)+c(\pi)s(x)=-s(x)$$
And so $$s(x+2\pi)=-s(x+\pi)=s(x)$$
The part about $s(2x)=s(x+x)=s(x)c(s)+c(x)s(x)=2s(x)c(x)$ is easy. From this you can see that $c(\frac{\pi}2)=0$ and $s(\frac{\pi}2)=1$ Similarly you can then prove that $c(x+a)=c(a)c(x)-s(a)s(x)$ so $c(2x)=(c(x))^2-(s(x))^2=2(c(x))^2-1$ and then integrate $x^2+y^2=1$ over the first quadrant to get
$$\begin{align}\int_0^1y\,dx&=\int_0^1\sqrt{1-x^2}dx=\int_0^{\frac{\pi}2}\sqrt{1-(s(\theta))^2}c(\theta)d\theta\\
&=\int_0^{\frac{\pi}2}(c(\theta))^2d\theta=\int_0^{\frac{\pi}2}\frac12(1+c(2\theta))d\theta\\
&=\frac12\left[\theta+\frac12s(2\theta)\right]_0^{\frac{\pi}2}=\frac12\left[\frac{\pi}2+\frac12(0)-0-\frac12(0)\right]\\
&=\frac{\pi}4\end{align}$$
So now you can relate these abstract functions to a concrete bit of knowledge since you have established that the area of a quarter circle is $\frac{\pi}4r^2$.
