Simplify $\sin[\arctan(4x)]$ for $ - \frac 14 \le x \le \frac 14 $ I am coming up with the wrong answer.
The problem is $\sin[\arctan(4x)]$
Here are the reciprocal functions I am using $\sin\theta= \frac {1}{\csc\theta}$ and $\cot\theta = \frac{1}{\tan\theta}$
Here is the Pythagorean identity I am using $\csc^2 \theta = 1+\cot^2 \theta$
I am trying to rewrite the problem $\sin[\arctan(4x)]$ in terms of the Pythagorean Identity using the reciprocal identities.  $$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac{1}{1+\tan^2 \theta}}$$  This seems to be a stretch, I do not think I am expressing the Pythagorean Identity right, and I am not sure how to either.  I did start to try to plug in the numbers.
$$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac {1}{(4x^2)}}$$
$$ \frac{1}{\csc \theta} = \sqrt{1+4x} $$
$$\sin\theta = \frac{1}{\sqrt{1+4x}} $$
I could rationalize the denominator and multiply the by its conjugate to get; $$\frac{\sqrt{1-4x}}{\sqrt{(1+4x)(1-4x)}} $$   But, at this point I know I am wrong.  The answer is $$ \frac{4x}{\sqrt{16x^2+1}}$$

As egreg said, the condition is redundant. On the other words, for any real $x$,$$\sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}}.$$
I like to prove it by 2 cases. Let $y=\arctan 4x$.
A. When $x\geq 0,$
$$\sin ^{2} y=\tan ^{2} y \cos ^{2} y=16 x^{2}\left(1-\sin ^{2} y\right) \Rightarrow \sin ^{2} y=\frac{16 x^{2}}{16 x^{2}+1} \Rightarrow \sin y=\frac{4 x}{\sqrt{16 x^{2}+1}} $$
Hence $$
\sin (\arctan (4 x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
B. When $x< 0,$ let $z=-x>0,$ then using the result in case A yields
$$
\sin (\arctan 4 z)=\frac{4 z}{\sqrt{16 z^{2}+1}}.
$$
Now putting $z=-x$ yields $$
\sin (\arctan 4(-x))=\frac{-4 x}{\sqrt{16 (-x)^{2}+1}}.
$$
Using the facts that $\sin x$ and $\arctan x $are odd functions, we have$$
-\sin (\arctan (4x))=-\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
$$
\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.
$$
Therefore we can conclude that for any real $x$, $$
\boxed{\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}}.
$$
In general, for any real $x$, $$\boxed{\sin (\arctan (x))=\frac{x}{\sqrt{x^{2}+1}}}$$
