Arctangent Function

Approach: We need an angle $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ with $\tan \theta = 1$.

Steps:

1. Among the special angles, $\tan \frac{\pi}{4} = 1$;
2. $\frac{\pi}{4} \approx 0.785$, which lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$;
3. By uniqueness on the principal interval, $\arctan 1 = \frac{\pi}{4}$.

Answer: $\arctan 1 = \frac{\pi}{4}$.

Approach:

The arctangent is the inverse of the tangent function restricted to its principal interval $(-\frac{\pi}{2}, \frac{\pi}{2})$: the domain comes from the range of $\tan x$, and the range is the principal interval itself.

Steps:

1. The tangent function $y = \tan x$ takes every real value on $(-\frac{\pi}{2}, \frac{\pi}{2})$, so its range is $\mathbb{R}$ and the domain of $y = \arctan x$ is $\mathbb{R}$.
2. On that interval the tangent function increases from $-\infty$ to $+\infty$, so choosing $(-\frac{\pi}{2}, \frac{\pi}{2})$ as the principal interval makes the inverse exist and stay single-valued.
3. The inverse maps the angles of the principal interval back onto $\mathbb{R}$, so the range of $y = \arctan x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Answer:

Domain: $\mathbb{R}$; Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$.