A new proof of angle sum identity for the sine

Several interesting proofs of angle sum identity are given on MathSE. In this note, I will provide another proof that is possible because, although traditionally presented in textbooks as a consequence of the angle sum identity $\sin{(x+y)}$, the double-angle formula for the sine, $\sin{(2x)}=2\sin{(x)}\cos{(x)}$, can be derived independently of it (see the wonderful proof without words of "Start wearing purple"). Then, as Blue has pointed out, the fact that supplementary angles have the same sine is an easy consequence of the double-angle formula for the sine.

The following proof is valid for $0\le\alpha\le\pi$, $0\le\beta\le\dfrac{\pi}{2}$, $0\le\alpha+\beta\le\pi$.

Theorem.  The sum identity for sine states that

$$\sin{(\alpha+\beta)}=\sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)}.\tag{1}$$

Proof. Suppose $\triangle{ABC}$ is a triangle with sides $\overline{BC}=a$, $\overline{AC}=b$ and $\overline{AB}=c$. Let $\angle{BAC}=\alpha$, $\angle{CBA}=\beta$ and $\angle{ACB}=\gamma$. Recalling that the area of a triangle can be expressed as $\Delta=\frac12bc\sin{(\alpha)}$ and substituting from the cosine rule on the right-hand side of $(1)$, we have  

\[\begin{aligned} \sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}\sin{(\beta)} &= \left(\frac{2\Delta}{bc}\right)\cos{(\beta)}+\left(\frac{2\Delta}{ac}\right)\cos{(\alpha)}\\&=\frac{2\Delta}{c}\left(\frac{\cos{(\beta)}}{b}+\frac{\cos{(\alpha)}}{a}\right)\\ &=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin{(\gamma)}\\&=\sin{(\pi-(\alpha+\beta))}\\&=\sin{2\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}\\&=2\color{red}{\sin{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\color{blue}{\cos{\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)}}\\&=2\color{red}{\cos{\left(\frac{\alpha+\beta}{2}\right)}}\color{blue}{\sin{\left(\frac{\alpha+\beta}{2}\right)}}\\&= \sin{\left(\alpha+\beta\right)}.\end{aligned}\]

$\square$