Properties of sin(x) and cos(x) from definition as solution to differential equation y''=-y

I recently came across the interesting definition of the sine function as the unique solution to the Initial Value Problem
$$y” = -y$$
$$y(0) = 0, y'(0) = 1$$
(My first question would be why this solution is unique, but I found that proofs of uniqueness and existence are too complicated for me to understand)

Taking existence and uniqueness for granted still leaves a lot of questions about the properties of sine like:

How would one go about proving these properties using only the differential equation definition? I have no idea where to start. Usually these properties (at least for me) are intuitively understood through the unit circle but that doesn’t help here.

Solutions Collecting From Web of "Properties of sin(x) and cos(x) from definition as solution to differential equation y''=-y"

The first thing you are going to want is the cosine function. If $y=s(x)$ is the solution to $y^{\prime\prime}(x)+y(x)=0$, $y(0)=0$, $y^{\prime}(x)=1$, then let $c(x)=y^{\prime}(x)$. Then $$c^{\prime\prime}(x)+c(x)=s^{\prime\prime\prime}(x)+s^{\prime}(x)=\frac d{dx}(s^{\prime\prime}(x)+s(x))=0$$
So $c(x)$ is a solution to the same differential equation with $c(0)=s^{\prime}(0)=1$ and $c^{\prime}(0)=s^{\prime\prime}(0)=-s(0)=0$. It is independent of $c(x)$ because if $ac(x)+bs(x)=0$, then at $x=0$ we see that $a=0$ and since $ac^{\prime}(x)+bs^{\prime}(x)=0$, then at $x=0$, $b=0$, so there is no nontrivial linear combination of the two solutions that is identically $0$. Then we can say the general solution to the differential equation $y^{\prime\prime}(x)+y(x)=0$ is
$$y(x)=c_1c(x)+c_2s(x)$$
We can see that $c^{\prime}(x)=s^{\prime\prime}(x)=-s(x)$ and we can prove the pythagorean identity because
$$\frac d{dx}(c^2(x)+s^2(x))=-2c(x)s(x)+2s(x)c(x)=0$$
so $c^2(x)+s^2(x)=\text{constant}=c^2(0)+s^2(0)=1^2+0^2=1$. The pythagorean identity establishes an upper and lower bound for s(x). Then we can write down the differential equation for $s(x+a)$ because
$$\frac d{dx}s(x+a)=s^{\prime}(x+a)\frac d{dx}(x+a)=s^{\prime}(x+a)=c(x+a)$$
$$\frac {d^2}{dx^2}s(x+a)=c'(x+a)=-s(x+a)$$
Since it satisfies the same differential equation, we can match it to the general solution. $s(0+a)=s(a)=c_1% and s^{\prime}(0+a)=c(a)=c_2$. Thus $s(x+a)=s(a)c(x)+c(a)s(x)$ Similarly you can show that $c(x+a)=c(a)c(x)-s(a)s(x)$. The periodicity seems to be much harder. Initially $s(x)$ is zero and then it increases because its derivative is positive. $c(x)$ starts out at $1$ but it decreases because its derivative is negative. Do the two function ever cross? Well, if they do at $\kappa$, then
$$c^2(\kappa)+s^2(\kappa)=2c^2(\kappa)=1$$
So $c(\kappa)=s(\kappa)=\frac1{\sqrt2}$. Now, for $0<x<\kappa$, $\frac1{\sqrt2}<x<1$, so
$$s(\kappa)=s(0)+\int_0^{\kappa}c(x)dx=\int_0^{\kappa}c(x)dx$$
So $\frac{\kappa}{\sqrt2}<s(\kappa)<\kappa$ and so $\frac1{\sqrt2}<\kappa<1$ so they must cross somewhere in that region. Then $c(x+\kappa)=c(x)c(\kappa)-s(x)s(\kappa)=\frac1{\sqrt2}(c(x)-s(x))=0$ when $c(x)=s(x)$, and the first time this happens is when $x=\kappa$ so the first time $c(x)=0$ is when $x=2\kappa$. Since $s(x)$ has been increasing the whole time, this is also the first time that $s(x)=1$, since $s(x+\kappa)=c(\kappa)s(x)+s(\kappa)c(x)=\frac1{\sqrt2}(s(x)+c(x))=1$ when $x=\kappa$. Then $s(x+2\kappa)=s(x)c(2\kappa)+c(x)s(2\kappa)=c(x)$ and we know that the first time this is zero is when $x=2\kappa$ so the second time $s(x)=0$ is when $x=2\kappa$.

Now we need the duplication formulas $s(2x)=s(x+x)=s(x)c(x)+c(x)s(x)=2s(x)c(x)$ and similarly $c(2x)=c^2(x)-s^2(x)$. Then $c(4\kappa)=c^2(2\kappa)-s^2(2\kappa)=-1$ and we can find that $s(x+4\kappa)=s(x)c(4\kappa)+c(x)s(4\kappa)=-s(x)$ So the period is not yet $4\kappa$, but the next zero of $s(x)$ is when $s(x+4\kappa)=-s(x)=0$, and that is when $x+4\kappa=4\kappa+4\kappa=8\kappa$. So $c(8\kappa)=c^2(4\kappa)-s^2(4\kappa)=1$ then $s(x+8\kappa)=s(x)c(8\kappa)+c(x)s(8\kappa)=s(x)$ and $c(x+8\kappa)=c(x)c(8\kappa)-s(x)s(8\kappa)=c(x)$. That establishes the periodicity of $s(x)$ and $c(x)$. We kind of had to crawl around the unit circle because we wanted to make sure that we had found the first period.

Note that by the differential equation method we can show that $s(a-x)=s(a)c(x)-s(a)c(x)$ so that when $a=0$, we find that $s(-x)=-s(x)$ and similarly $c(-x)=c(x)$. Thus $s(-\frac{\pi}2=s(-2\kappa)=-s(\kappa)=-1$ and we have established that $s(x)$ actually attains the lower bound. For special values, we rely on
$$c((n+1)x)=c(nx)c(x)-s(nx)s(x)$$
$$c((n-1)x)=c(nx)c(-x)-s(nx)s(-x)=c(nx)c(x)+s(nx)s(x)$$
Adding,
$$c((n+1)x)+c((n-1)x)=2c(nx)c(x)$$
or
$$c((n+1)x)=2c(nx)c(x)-c((n-1)x)$$
So
$$c(3x)=2c(2x)c(x)-c(x)=2(c^2(x)-s^x(x))c(x)-c(x)=4c^3(x)-3c(x)$$
Then we know that $c(2\kappa)=c\left(\frac{\pi}2\right)=0$, so
$$4c^3\left(\frac{\pi}6\right)-3c\left(\frac{\pi}6\right)=0$$
With solution $$c\left(\frac{\pi}6\right)\in\left\{0,-\frac{\sqrt3}2,\frac{\sqrt3}2\right\}$$
And since $0<c(x)<1$ in this region, we get $c\left(\frac{\pi}6\right)=\frac{\sqrt3}2$.