Question

$\tan^{-1}\tan x$ where $x \in [\frac{2027\pi}{2},\frac{2029\pi}{2}]$

Answer 1

x-1014\pi

Answer 2

We know that

$$\tan^{-1}(\tan x)=x-n\pi,\; \text{where } n \text{ is chosen so that } x-n\pi\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Given $x\in \left[\frac{2027\pi}{2},\,\frac{2029\pi}{2}\right]$, note that:

$$\frac{2027\pi}{2}=1013.5\pi \quad \text{and} \quad \frac{2029\pi}{2}=1014.5\pi.$$

Choosing $n=1014$ gives:

$$x-1014\pi\in\left[1013.5\pi-1014\pi,\; 1014.5\pi-1014\pi\right] = \left[-\frac{\pi}{2},\frac{\pi}{2}\right].$$

Thus, for every $x$ in the interval, we have:

$$\tan^{-1}(\tan x)=x-1014\pi.$$

Since tan is periodic with period $\pi$, select $n=1014$ such that $x-1014\pi \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Therefore,

$$\tan^{-1}(\tan x)=x-1014\pi.$$