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 $\sin^2\alpha+\cos^2\alpha=1$?
Why sin(?+?) = sin(?)cos(?) + sin(?)cos(?)
trigonometry
Best Answer
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}