Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$ Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself)
I could find a geometric solution in Wikipedia, but is there any solution that doesnt require drawing something?
Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change.
Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?

The differential equation
$$f''=-f$$
has a unique solution for given initial conditions $f(0)=x_0$, $f'(0)=y_0$. To show uniqueness assume that $g$ is another solution for the same initial values. Then $h:=f-g$ also satisfies $h''=-h$, and $h(0)=h'(0)=0$. But then
$$\frac\partial{\partial t}(h(t)^2+h'(t)^2)=2h(t)h'(t)+2h'(t)h''(t)=2h(t)h'(t)-2h'(t)h(t)=0,
$$
and it follows that $h(t)=0$ must hold for all $x$, so $f=g$. 
This unique solution is 
$$f(t)=y_0\sin t+x_0\cos t,
$$
as one checks immediately.
Now we can apply this to the function $f(t)=\sin(t+\beta)$. It satisfies 
$f''(t)=-f(t)$, $f(0)=\sin\beta$, $f'(0)=\cos\beta$, hence we must have
$$f(t)=\cos\beta\sin t+\sin\beta\cos t
.$$

This may seem to come out of the blue at first, but if one sets $x(t)=f(t)$, $y(t)=f'(t)$, then $f''=-f$ becomes
$$\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}y\\-x\end{pmatrix},
$$
which describes a motion on a circle around the origin (the velocity is perpendicular to the vector from the origin to the current point), and of course $\sin$ and $\cos$ are all about circles. The calculation done above for $h$ shows that indeed $x(t)^2+y(t)^2$ is constant.
