Triangle Proofs of the Sine and Cosine Addition Formulas

Introduction
The addition formulas for sine and cosine admit many proofs in the mathematical literature. Standard textbook treatments use coordinates on the unit circle and the distance formula; see Abramson. Other presentations use geometric arrangements of triangles and area decompositions (Bogomolny), or Ptolemy's theorem for cyclic quadrilaterals (Bogomolny). This note presents two short triangle proofs based on the area formula and the cosine rule. The hypotheses needed for a nondegenerate triangle are stated explicitly, and the boundary cases are verified separately.

Notation and hypotheses
Throughout the two proofs, assume
$$\alpha>0,\qquad \beta>0,\qquad \alpha+\beta<\pi.\tag{1}$$
Choose a triangle $ABC$ with
$$a=BC,\qquad b=CA,\qquad c=AB,\qquad \angle BAC=\alpha,\qquad \angle CBA=\beta,\qquad \angle ACB=\gamma.$$
Then $a,b,c>0$ and
$$\gamma=\pi-(\alpha+\beta)>0.$$
Write $\Delta$ for its area. We use
$$2\Delta=bc\sin\alpha=ac\sin\beta=ab\sin\gamma\tag{2}$$
and the cosine rule
$$\cos\alpha=\frac{b^2+c^2-a^2}{2bc},\qquad \cos\beta=\frac{a^2+c^2-b^2}{2ac},\qquad \cos\gamma=\frac{a^2+b^2-c^2}{2ab}.\tag{3}$$
These facts can be established using altitudes and the Pythagorean theorem, independently of the addition formulas. The unit-circle definition also gives
$$\sin(\pi-t)=\sin t,\qquad \cos(\pi-t)=-\cos t,\qquad \sin^2 t+\cos^2 t=1.$$

The sine addition formula
Theorem 1. Under the hypotheses (1),
$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.$$

Proof. Using (2) and then (3), we obtain
$$\begin{aligned}\sin\alpha\cos\beta+\cos\alpha\sin\beta&=\frac{2\Delta}{bc}\cos\beta+\frac{2\Delta}{ac}\cos\alpha\\&=\frac{2\Delta}{c}\left(\frac{\cos\beta}{b}+\frac{\cos\alpha}{a}\right)\\&=\frac{2\Delta}{c}\left(\frac{a^2+c^2-b^2}{2abc}+\frac{b^2+c^2-a^2}{2abc}\right)\\&=\frac{2\Delta}{ab}\\&=\sin\gamma\\&=\sin\bigl(\pi-(\alpha+\beta)\bigr)\\&=\sin(\alpha+\beta).\end{aligned}$$

The cosine addition formula
Theorem 2. Under the hypotheses (1),
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Proof. The first two identities in (3) give
$$c-b\cos\alpha=c-\frac{b^2+c^2-a^2}{2c}=\frac{a^2+c^2-b^2}{2c}=a\cos\beta.$$
Together with (2), this yields
$$\cos\beta=\frac{c-b\cos\alpha}{a},\qquad \sin\beta=\frac{b\sin\alpha}{a}.$$
Consequently,
$$\begin{aligned}\cos\alpha\cos\beta-\sin\alpha\sin\beta&=\frac{c\cos\alpha-b\cos^2\alpha-b\sin^2\alpha}{a}\\&=\frac{c\cos\alpha-b}{a}\\&=\frac{c^2-a^2-b^2}{2ab}\\&=-\cos\gamma\\&=\cos(\alpha+\beta).\end{aligned}$$
The last equality uses
$$\gamma=\pi-(\alpha+\beta)$$
and the unit-circle identity for the cosine of a supplementary angle.

Boundary cases
Corollary. Both addition formulas hold for
$$\alpha\ge0,\qquad \beta\ge0,\qquad \alpha+\beta\le\pi.$$

Proof. The strict inequalities were covered above. If $\alpha=0$ or $\beta=0$, both formulas follow immediately from $\sin 0=0$ and $\cos 0=1$.
If $\alpha+\beta=\pi$, then
$$\beta=\pi-\alpha,$$
so
$$\sin\beta=\sin\alpha\qquad\text{and}\qquad \cos\beta=-\cos\alpha.$$
Hence
$$\begin{aligned}\sin\alpha\cos\beta+\cos\alpha\sin\beta&=-\sin\alpha\cos\alpha+\cos\alpha\sin\alpha=0=\sin\pi,\\cos\alpha\cos\beta-\sin\alpha\sin\beta&=-\cos^2\alpha-\sin^2\alpha=-1=\cos\pi.\end{aligned}$$
This covers every boundary case without using a degenerate triangle.