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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

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

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 $\Delta=\dfrac{1}{2}bc\sin{\alpha}$, substituting from the cosine rule and keeping in mind the unit circle definition of sine (from where it follows directly that supplementary angles have the same sine), $$\begin{aligned} \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta) &= \left(\frac{2\Delta}{bc}\right)\cos(\beta)+\left(\frac{2\Delta}{ac}\right)\cos(\alpha)\\&=\frac{2\Delta}{c}\left(\frac{\cos(\beta)}{b}+\frac{\cos(\alpha)}{a}\right)\\ &=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin(\gamma)\\&=\sin{(\pi-(\alpha+\beta))}\\&= \sin{\left(\alpha+\beta\right)}.\end{aligned}$$ Several proofs are given here.

Is this proof correct? If it's correct, I suspect it won't be hard to find a proof for its companion $\cos{(\alpha+\beta)}$.

Emmanuel José García
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  • Everything would be fine if you could prove that $\sin(\pi-\alpha-\beta)=\sin(\alpha+\beta)$. Maybe use that $\sin\gamma=\cos(\frac \pi2-\gamma)…$ – WindSoul Apr 24 '23 at 19:03
  • @WindSoul: As said here, the rays corresponding to supplementary angles intersect the unit circles in points having the same y-coordinate, so the two angles have the same sine (and opposite cosines). – Emmanuel José García Apr 24 '23 at 19:26
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    I don't know whether this proof is new, but as to whether it's correct - which I think it is - I wouldn't be surprised if someone argues it's circular, whether or not they're right to do so. (You sometimes see people overestimate how often one trigonometric result is needed to prove others, which is why e.g. they're often convinced you can't have trigonometric proofs of one particular trigonometric fact, the Pythagorean theorem.) – J.G. Apr 24 '23 at 19:34
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    @EmmanuelJoséGarcía: The supplementary sine identity follows easily from the double-angle formula and the co-relations/definitions (which, I suspect, is what WindSoul was suggesting) ... $$\sin(\pi-x) = \sin2\left(\frac{\pi}2-\frac{x}2\right) = 2 \color{red}{\sin\left(\frac\pi2-\frac{x}2\right)} \color{blue}{\cos\left(\frac\pi2-\frac{x}2\right)} = 2 \color{red}{\cos\frac{x}2}\color{blue}{\sin\frac{x}2} = \sin x$$ – Blue Apr 26 '23 at 20:23
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    That is much simpler. Thank you. – Emmanuel José García Apr 27 '23 at 10:46
  • I have posted the final version of my proof for $\sin{(x+y)}$, removing all dependence on the xy-plane and only appealing to basic trigonometry and algebra. I have taken into account the wonderful PWW for $\sin{(2x)}$ given at https://math.stackexchange.com/questions/460281/reasoning-that-sin2x-2-sin-x-cos-x/460302#460302 and Blue's comment. – Emmanuel José García Apr 30 '23 at 17:54

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