Why is $\arctan(\tan(25\pi / 4)) = \pi/4$?

Question

Why is $\arctan(\tan(25\pi /4)) = \pi/4$, and how can I get from the expression on the left to the one on the right?

Answer 1

The ${\tan(x)}$ function is periodic, meaning it will not be injective, thus will not have an inverse. Unless you restrict the domain.

For example, in your case

$${\tan\left(\frac{25\pi}{4}\right)=\tan\left(\frac{\pi}{4}\right)}$$

So if you apply the inverse function - what value do we take? Both ${\frac{25\pi}{4}}$ and ${\frac{\pi}{4}}$ give the same answer under the tangent function. So we are the ones who must decide which value we take. In exactly the same way that we decide ${\sqrt{4}=2}$, and not ${-2}$ (even though both ${(-2)^2=4}$ as well). The decision we make is called the "principle values". For example we say the principle square root of four is two.

We take the principle values of ${\arctan(x)}$ to be from the interval ${\left(-\frac{\pi}{2},\frac{\pi}{2}\right)}$, and since ${\tan\left(\frac{25\pi}{4}\right)=\tan\left(\frac{\pi}{4}\right)}$ and ${-\frac{\pi}{2}\leq \frac{\pi}{4}\leq \frac{\pi}{2}}$, that's the answer you will get to the expression:

$${\arctan\left(\tan\left(\frac{25\pi}{4}\right)\right)=\frac{\pi}{4}}$$

Answer 2

Since the tangent function is periodic with period $\pi$, $\tan\left(\frac{25\pi}4\right)=\tan\left(\frac\pi4\right)$. And since $\arctan$ is the inverse of the restriction of $\tan$ to $\left(-\frac\pi2,\frac\pi2\right)$ and since $\frac\pi4\in\left(-\frac\pi2,\frac\pi2\right)$,$$\arctan\left(\tan\left(\frac\pi4\right)\right)=\frac\pi4.$$

Answer 3

We can apply the identity $$\arctan{(\tan{(x)})}=x$$ only for $x\in(-\pi/2,\pi/2)$. So we just need to use the fact that $$\tan{(25\pi/4)}=\tan{(\pi/4)}$$ by periodicity of the tangent.