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

Routine facts are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); results on real and imaginary parts and conjugates are read for the current complex numbers by The Earlier and the New Versions of the Definitions of the Natural Numbers, the Complex Numbers, the Conjugate and the Modulus Agree, and the Two Versions of Real Correspond (ii). Let $\exp$ be the complex exponential function.

Values at zero. Since $i0=0$, The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus (i) gives $\exp(i0)=\exp0=1$. As $1=1+0\,i$ with $1$ and $0$ real, the uniqueness of real and imaginary parts (Real and Imaginary Parts of a Complex Number (i)) gives $\operatorname{Re}1=1$ and $\operatorname{Im}1=0$. By the definitions of the cosine and the sine (The Real Sine and Cosine Functions (i), The Real Sine and Cosine Functions (ii)), therefore,

$$\cos0=\operatorname{Re}\exp(i0)=1,\qquad \sin0=\operatorname{Im}\exp(i0)=0 ,$$

which is (i).

Parity. We compute $\exp(-ix)$ in two ways. Since $i(-x)=-ix$, Euler's formula Euler's Formula: the Complex Exponential on the Imaginary Axis, Its Modulus and Its Conjugate (i), applied with $t=-x$, gives $\exp(-ix)=\cos(-x)+i\sin(-x)$. On the other hand, Euler's Formula: the Complex Exponential on the Imaginary Axis, Its Modulus and Its Conjugate (iv), applied with $t=x$, gives $\exp(-ix)=\overline{\exp(ix)}$; and since $\operatorname{Re}\exp(ix)=\cos x$ and $\operatorname{Im}\exp(ix)=\sin x$ (The Real Sine and Cosine Functions (i), The Real Sine and Cosine Functions (ii)), the definition of the conjugate (The Complex Conjugate of a Complex Number (i)) gives $\overline{\exp(ix)}=\cos x-(\sin x)\,i$. Altogether,

$$\cos(-x)+i\sin(-x)=\exp(-ix)=\overline{\exp(ix)}=\cos x-(\sin x)\,i .$$

Both ends have the form $p+qi$ with $p$ and $q$ real, so taking real and imaginary parts (Real and Imaginary Parts of a Complex Number (i)) gives $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$. This is (ii).

Double angle. The addition formulas Addition Formulas for Sine and Cosine (ii) and Addition Formulas for Sine and Cosine (i), applied with $x$ in place of both $a$ and $x$, give

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

The Pythagorean identity The Pythagorean Identity for Sine and Cosine (i) gives $(\sin x)^{2}=1-(\cos x)^{2}$, and substituting this into the formula for $\cos(2x)$ gives $\cos(2x)=2(\cos x)^{2}-1$; it also gives $(\cos x)^{2}=1-(\sin x)^{2}$, and substituting this instead gives $\cos(2x)=1-2(\sin x)^{2}$. This is (iii).

A difference of squared sines. The addition formula Addition Formulas for Sine and Cosine (ii), applied with $(a,x)=(x,y)$, gives $\sin(x+y)=\sin x\cos y+\cos x\sin y$; applied with $(a,x)=(x,-y)$, and combined with the parity (ii) just proved, it gives $\sin(x-y)=\sin x\cos y-\cos x\sin y$. Multiplying,

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

By the Pythagorean identity The Pythagorean Identity for Sine and Cosine (i), for $y$ and for $x$, $(\cos y)^{2}=1-(\sin y)^{2}$ and $(\cos x)^{2}=1-(\sin x)^{2}$, so the right side is

$$(\sin x)^{2}\bigl(1-(\sin y)^{2}\bigr)-\bigl(1-(\sin x)^{2}\bigr)(\sin y)^{2}=(\sin x)^{2}-(\sin y)^{2}.$$

This is (iv).