When simplifying $\sin(\arctan(x))$, why is negative $x$ not considered? Let $u = \arctan(x)$, hence $x = \tan(u)$ for $u$ belongs in $(-\frac\pi2, \frac\pi2)$. Since $u$ belongs in $(-\frac\pi2, \frac\pi2)$, we consider $\sin(u)$ where $u$ belongs in $(-\frac\pi2, \frac\pi2)$.
I used the unit circle to determine that the hypotenuse is $\sqrt{x^2 + 1}$ and got an answer $\sin(u) = \frac{x}{\sqrt{x^2 + 1}}$ when I consider than the angle $u$ lies between $(0, \frac\pi2)$.
That's what my textbook says too. However, why don't we also consider when $x$ is negative, and the angle $u$ lies between $(-\frac\pi2, 0)$ ?

To add some additional information regarding user46234's comment, we can also understand the omission of a $\pm$ sign as a simple consequence of algebra.
Here are the relevant pieces of information to conclude this:

*

*By definition, $\sqrt{x^2}=|x|$


*$\arctan$ is a strictly increasing function bounded between $(-\frac{\pi}{2},\frac{\pi}{2})$


*$\arctan(0)=0$


*$\sin(x) \geq 0$ if $x \in [0,\frac{\pi}{2})$ and $\sin(x) \lt 0$ if $x \in (-\frac{\pi}{2},0)$


*$\cos(x) \gt 0$ if $x \in (-\frac{\pi}{2},\frac{\pi}{2})$

Working forward from the above bulleted points:
Let $y=\arctan(x)$.We know that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$. By definition, $\tan(y)=\tan\circ \arctan(x)=x \implies \frac{\sin(y)}{\cos(y)}=x$
To further simplify, we use the Pythagorean identity $\cos^2(y)+\sin^2(y)=1 \implies \cos^2(y)=1-\sin^2(y)$. Taking the square root of both sides yields: $|\cos(y)|=\sqrt{1-\sin^2(y)}$.  Note that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$, which means that $\cos(y) \gt 0$...and therefore $|\cos(y)|=\cos(y)$, so that we have $\cos(y)=\sqrt{1-\sin^2(y)}$.
We then have that $x=\frac{\sin(y)}{\sqrt{1-\sin^2(y)}}$. Rearranging and then squaring gives us $\big(1-\sin^2(y)\big)\cdot x^2=\sin^2(y)$. Because $x=\tan(y)$, we can use the tangent expression of the Pythagorean identity to further simplify: $1+\frac{\sin^2(y)}{\cos^2(y)}=\frac{1}{\cos^2(y)} \implies1+\tan^2(y)=\frac{1}{1-\sin^2(y)} \implies \frac{1}{1+\tan^2(y)}=1-\sin^2(y) \implies\frac{1}{1+x^2}=1-\sin^2(y)$. Note that in order to apply this substitution, we need to make sure that $\cos(y) \neq 0$...however, given that $y \in (-\frac{\pi}{2},\frac{\pi}{2})$, we know that $\cos(y) \neq 0$. We can then apply the relevant substitution that gives us:
$$\frac{x^2}{1+x^2}=\sin^2(y)$$
Taking the square root of this expression gives us:
$$\frac{|x|}{\sqrt{1+x^2}}=|\sin(y)| \quad (\dagger_1)$$
Now, suppose $x \geq 0$. Then $y \in [0,\frac{\pi}{2})$. We then know that $\sin(y) \geq 0$, so we must have that $|x|=x$ and $|\sin(y)|=\sin(y)$. Under these circumstances, $(\dagger_1)$ is just $\frac{x}{\sqrt{1+x^2}}=\sin(y)$.
Next, suppose $x \lt 0$. Then $y \in (-\frac{\pi}{2},0)$. We then know that $\sin(y) \lt 0$, so we must have that $|x|=-x$ and $|\sin(y)|=-\sin(y)$. Under these circumstances, $(\dagger_1)$ is $\frac{-x}{\sqrt{1+x^2}}=-\sin(y)$, which can of course be rewritten as $\frac{x}{\sqrt{1+x^2}}=\sin(y)$.
Thus, across all relevant cases, we must have that $\sin(\arctan(x))=\sin(y)=\frac{x}{\sqrt{1+x^2}}$
