Proof of Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine

Each result cited is universally quantified over the data in its own statement. Every step below specialises one of the three product-to-sum formulas, whose claims 1, 2 and 3 assert, for all real $u,v$,

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

and uses the identity $(\cos w)^{2}+(\sin w)^{2}=1$ of The Pythagorean Identity for Sine and Cosine (i). The number $2$ is positive, hence nonzero and invertible, by claim 8 of Elementary Order Arithmetic in an Ordered Field; cancelling a factor $2$ below means multiplying by its inverse.

Claim 1. Take $u=v=x$ in Product-to-Sum Formulas for Sine and Cosine (i) and Product-to-Sum Formulas for Sine and Cosine (ii). Since $x-x=0$ and $x+x=2x$, they read

$$\cos 0+\cos(2x)=2(\cos x)^{2}, \qquad \cos 0-\cos(2x)=2(\sin x)^{2}.$$

Adding the two and using the Pythagorean identity gives $2\cos 0=2\bigl((\cos x)^{2}+(\sin x)^{2}\bigr)=2$, so $\cos 0=1$ after cancelling the factor $2$.

Applying the Pythagorean identity at $w=0$ now gives $1+(\sin 0)^{2}=1$, hence $(\sin 0)^{2}=0$, hence $\sin 0=0$ by claim 3 of Zero Products and Elementary Identities in a Field, a field having no zero divisors.

Claim 2. Take $u=0$ and $v=x$, so that $u-v=-x$ and $u+v=x$. Product-to-Sum Formulas for Sine and Cosine (i) gives

$$\cos(-x)+\cos x=2\cos 0\,\cos x=2\cos x,$$

using $\cos 0=1$ from claim 1; subtracting $\cos x$ gives $\cos(-x)=\cos x$. Claim 3 of that lemma gives

$$\sin x+\sin(-x)=2\sin 0\,\cos x=0,$$

using $\sin 0=0$ from claim 1 and claim 1 of Zero Products and Elementary Identities in a Field; hence $\sin(-x)=-\sin x$.

Claim 3. Apply Product-to-Sum Formulas for Sine and Cosine (iii) twice, first with $u=x$ and $v=y$, then with $u=y$ and $v=x$:

$$\sin(x+y)+\sin(x-y)=2\sin x\,\cos y, \qquad \sin(y+x)+\sin(y-x)=2\sin y\,\cos x .$$

Addition in $\mathbb{R}$ is commutative, so $\sin(y+x)=\sin(x+y)$; and $y-x=-(x-y)$, so $\sin(y-x)=-\sin(x-y)$ by claim 2. Adding the two displayed equations therefore cancels the terms $\sin(x-y)$ and $\sin(y-x)$ and leaves

$$2\sin(x+y)=2\sin x\,\cos y+2\cos x\,\sin y ,$$

which gives the first identity after cancelling the factor $2$.

For the second, subtract Product-to-Sum Formulas for Sine and Cosine (ii) from Product-to-Sum Formulas for Sine and Cosine (i), both taken with $u=x$ and $v=y$: the terms $\cos(x-y)$ cancel and

$$2\cos(x+y)=2\cos x\,\cos y-2\sin x\,\sin y ,$$

which gives the second identity after cancelling the factor $2$.

Claim 4. The two displays in the proof of claim 1, together with $\cos 0=1$ proved there, give

$$\cos(2x)=2(\cos x)^{2}-1, \qquad \cos(2x)=1-2(\sin x)^{2}.$$

Taking $u=v=x$ in Product-to-Sum Formulas for Sine and Cosine (iii) gives $\sin(2x)+\sin 0=2\sin x\,\cos x$, and $\sin 0=0$ by claim 1, so $\sin(2x)=2\sin x\,\cos x$.

Claim 5. Take $u=x+y$ and $v=x-y$ in Product-to-Sum Formulas for Sine and Cosine (ii). Then $u-v=2y$ and $u+v=2x$, so

$$\cos(2y)-\cos(2x)=2\sin(x+y)\,\sin(x-y).$$

By claim 4, $\cos(2y)=1-2(\sin y)^{2}$ and $\cos(2x)=1-2(\sin x)^{2}$, so the left-hand side equals $2(\sin x)^{2}-2(\sin y)^{2}$. Cancelling the factor $2$ gives the assertion.