What is the Arctangent of Tangent?

trigonometry

This is probably just me misunderstanding trig properties. I remember from my trig days that $\tan(\arctan(x)) = x$ But I can't remember if that holds the other way. Can someone help me out? Is this valid:

$$\arctan(\tan(\theta)) = \theta$$

It is true that $\tan\arctan x=x$, for every $x$. The converse is not true and it cannot be, because the tangent is not an injective function.

Recall that $\arctan x$ returns a number (an angle if you prefer) in the interval $(-\pi/2,\pi/2)$. So we have the equality $$ \arctan(\tan\theta)=\theta $$ if and only if $\theta\in(-\pi/2,\pi/2)$.

A formula can be given for any $\theta$: you just need to “reduce” the angle to the right interval by subtracting an integral multiple of $\pi$ so that $$ -\frac{\pi}{2}<\theta-k\pi<\frac{\pi}{2} $$ which is equivalent to $$ -\frac{1}{2}<\frac{\theta}{\pi}-k<\frac{1}{2} $$ or $$ 0<\frac{\theta}{\pi}+\frac{1}{2}-k<1 $$ or $$ k<\frac{\theta}{\pi}+\frac{1}{2}<k+1 $$ which means $$ k=\left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor $$ Thus $$ \arctan(\tan\theta)=\theta-\pi\left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor $$ (of course when $\tan\theta$ is defined to begin with).

For instance, if $\theta=15\pi/4$, we have $$ \left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor= \left\lfloor\frac{15}{4}+\frac{1}{2}\right\rfloor=4 $$ and $$ \arctan\tan\frac{15\pi}{4}=\frac{15\pi}{4}-4\pi=-\frac{\pi}{4} $$

Yes, if $-\pi/2 < \theta < \pi/2$.

No, otherwise.

It only equals $\theta $ if $$-\frac {\pi}{2}<\theta < \frac {\pi}{2} $$ as the range of $\arctan$ is only from $-\frac {\pi}{2} $ to $\frac {\pi}{2} $. If $\theta $ is outside this interval, then you would need to add or subtract $\pi $ from $\theta $ until you get to the angle in this interval that has the same value of $\tan$.

For instance, $\arctan (\tan \frac {\pi}{6}) = \frac {\pi}{6} $, but $\arctan (\tan \frac {3\pi }{4}) = -\frac {\pi}{4} $.

This can also been seen here:

enter image description here Hope it helps.