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 $\sin^2\alpha+\cos^2\alpha=1$?

Using the scalar product : let $A$ and $B$ be the points on the unit circle at arc length $a$ and $b$. Then $A(\cos(a);\sin(a))$ and $B(\cos(b);\sin(b))$, and $\widehat{BOA}=a-b$. Therefore :

\begin{aligned} \overrightarrow{OB}\cdot\overrightarrow{OA} &= OA\times OB\times \cos(\widehat{\overrightarrow{OB};\overrightarrow{OA}})\\ &= 1\times 1\times \cos(\widehat{BOA})\\ &= \cos(a-b) \end{aligned}

and as $\overrightarrow{OA}\binom{\cos(a)}{\sin(a)}$, $\overrightarrow{OB}\binom{\cos(b)}{\sin(b)}$ :

$$ \overrightarrow{OB}\cdot\overrightarrow{OA} = \cos(a)\cos(b)+\sin(a)\sin(b) $$

Therefore $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

Then \begin{aligned} \sin(a-b)&=\cos\left(\frac{\pi}{2}+b-a\right)\\ &=\cos\left(\frac{\pi}{2}+b\right)\cos(a)+\sin\left(\frac{\pi}{2}+b\right)\sin(a)\\ &=-\sin(b)\cos(a)+\cos(b)\sin(a)\\ &= \sin(a)\cos(b)-\sin(b)\cos(a) \end{aligned}

I usually like to reduce these to problems involving complex exponents. Note that

$$\sin\alpha\cos\beta + \sin\beta\cos\alpha = \frac{e^{i\alpha} - e^{-i\alpha}}{2i}\frac{e^{i\beta} + e^{-i\beta}}{2} + \frac{e^{i\beta} - e^{-i\beta}}{2i}\frac{e^{i\alpha} + e^{-i\alpha}}{2}$$ and simplify this expression.

To find the RHS (assuming that we may forget the result) we write:

$$\sin(\alpha+\beta)=\Im\frac{1}{2}(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)})\\=\Im\frac{1}{2}((\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)-(\cos\alpha-i\sin\alpha)(\cos\beta-i\sin\beta))$$ and we develop and we take the imaginary part we find the desired formula.