Proof of The Arctangent: Continuity, Monotonicity, Derivative, Special Values and Oddness

Routine arithmetic and routine facts about functions are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); results stated for Derivative of a Real Function on an Interval are read for Derivative of a Real Function on an Interval by The Earlier and the New Versions of the Definitions of the Derivative on an Interval and of the Riemann Integral Agree (ii) and The Earlier and the New Versions of the Definitions of the Derivative on an Interval and of the Riemann Integral Agree (iii).

Write $I=(-\pi/2,\pi/2)$ and let $\tan:I\to\mathbb{R}$ be the tangent function, with $\sin$ and $\cos$ as there. Since $\pi>0$ by Quarter-Turn Identities and Periodicity of Sine and Cosine (i), $I$ is an interval containing two distinct points.

The inverse-function lemma applies. By The Tangent: Derivative, Monotonicity, Bijectivity onto $\mathbb{R}$ and Special Values (ii), $\tan$ is continuous on $I$ and strictly increasing, that is, $\tan s<\tan t$ for all $s,t\in I$ with $s<t$. So Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative applies on the interval $I$ to the function $f=\tan$. Its image $J=\tan(I)$ is an interval containing two distinct points, by Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative (i), so derivatives of maps $J\to\mathbb{R}$ are defined. Since $\tan$ maps $I$ onto $\mathbb{R}$ by The Tangent: Derivative, Monotonicity, Bijectivity onto $\mathbb{R}$ and Special Values (iii), we have $J=\mathbb{R}$. Consequently the inverse $g:J\to I$ of Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative (iii) sends each $y\in\mathbb{R}$ to the unique $x\in I$ with $\tan x=y$; that is, $g=\arctan$, by The Arctangent Function (i).

Continuity and monotonicity. With $g=\arctan$ and $J=\mathbb{R}$, Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative (iii) says that $\arctan y<\arctan z$ for all $y,z\in\mathbb{R}$ with $y<z$, and Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative (iv) says that $\arctan$ is continuous on $\mathbb{R}$. This is (i).

Derivative. Let $y\in\mathbb{R}$ and put $x=\arctan y$, so that $x\in I$ and $\tan x=y$. By The Tangent: Derivative, Monotonicity, Bijectivity onto $\mathbb{R}$ and Special Values (i), $\tan$ is differentiable at $x$, with

$$\tan'(x)=1+\tan^{2}x=1+y^{2}>0 ;$$

in particular $\tan'(x)\neq0$. Hence, by Inverse of a Continuous Strictly Increasing Function on an Interval: Continuity and Derivative (v) at the point $x$, the inverse $\arctan$ is differentiable at $\tan x=y$, with

$$\arctan'(y)=\frac{1}{\tan'(x)}=\frac{1}{1+y^{2}} .$$

As $y\in\mathbb{R}$ was arbitrary, $\arctan$, read as a map $\mathbb{R}\to\mathbb{R}$, is differentiable at every point of $J=\mathbb{R}$, with the derivative claimed in (ii).

Special values. By The Tangent: Derivative, Monotonicity, Bijectivity onto $\mathbb{R}$ and Special Values (iv), $\tan 0=0$ and $\tan(\pi/4)=1$, where $0$ and $\pi/4$ lie in $I$. So $0$ is the element of $I$ at which $\tan$ takes the value $0$, and $\pi/4$ the element at which it takes the value $1$; by the uniqueness in The Arctangent Function (i), $\arctan 0=0$ and $\arctan 1=\pi/4$. This is (iii).

Oddness. Let $y\in\mathbb{R}$ and put $x=\arctan y$, so that $x\in I$ and $\tan x=y$. The interval $I$ is symmetric about $0$, so $-x\in I$, and by Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (ii), $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$. Therefore

$$\tan(-x)=\frac{-\sin x}{\cos x}=-\tan x=-y .$$

Thus $-x$ is an element of $I$ at which $\tan$ takes the value $-y$, and by the uniqueness in The Arctangent Function (i), $\arctan(-y)=-x=-\arctan y$. This is (iv).