Proof of Product-to-Sum Formulas for Sine and Cosine

Routine facts are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity).

The four expansions. The addition formulas Addition Formulas for Sine and Cosine (i) and Addition Formulas for Sine and Cosine (ii), applied with $a=u$ and $x=v$, give

$$\cos(u+v)=\cos u\cos v-\sin u\sin v,\qquad \sin(u+v)=\sin u\cos v+\cos u\sin v .$$

Applied with $a=u$ and $x=-v$, they give $\cos(u-v)=\cos u\cos(-v)-\sin u\sin(-v)$ and $\sin(u-v)=\sin u\cos(-v)+\cos u\sin(-v)$. Since the cosine is even and the sine is odd, $\cos(-v)=\cos v$ and $\sin(-v)=-\sin v$ (Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (ii)), and these become

$$\cos(u-v)=\cos u\cos v+\sin u\sin v,\qquad \sin(u-v)=\sin u\cos v-\cos u\sin v .$$

Sums and differences. Adding the two expansions of the cosine, the terms $\pm\sin u\sin v$ cancel:

$$\cos(u-v)+\cos(u+v)=2\cos u\cos v ,$$

which is (i). Subtracting $\cos(u+v)$ from $\cos(u-v)$, the terms $\cos u\cos v$ cancel instead:

$$\cos(u-v)-\cos(u+v)=2\sin u\sin v ,$$

which is (ii). Finally, adding the two expansions of the sine, the terms $\pm\cos u\sin v$ cancel:

$$\sin(u+v)+\sin(u-v)=2\sin u\cos v ,$$

which is (iii).