# How do you calculate sin(tan^-1(4x))?

Sep 22, 2015

$\frac{4 x}{\sqrt{16 {x}^{2} + 1}} = \sin \left(\arctan \left(4 x\right)\right)$

#### Explanation:

Assuming that by ${\tan}^{- 1} \left(4 x\right)$ you mean $\arctan \left(4 x\right)$ and not $\cot \left(4 x\right)$.

We know the pythagorean identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, so if we divide both sides by #sin^2(theta) we'd have:

${\sin}^{2} \frac{\theta}{\sin} ^ 2 \left(\theta\right) + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right) \rightarrow$
$1 + {\cot}^{2} \left(\theta\right) = {\csc}^{2} \left(\theta\right)$

So, if we switch in $\theta$ for $\arctan \left(4 x\right)$, we have
$1 + \frac{1}{\tan} ^ 2 \left(\arctan \left(4 x\right)\right) = \frac{1}{\sin} ^ 2 \left(\arctan \left(4 x\right)\right)$

Knowing that $\tan \left(\arctan \left(\theta\right)\right) = \theta$

$1 + \frac{1}{4 x} ^ 2 = \frac{1}{\sin} ^ 2 \left(\arctan \left(4 x\right)\right)$

Taking the least common multiple so we can sum the two fractions on the right side:

$\frac{16 {x}^{2} + 1}{16 {x}^{2}} = \frac{1}{\sin} ^ 2 \left(\arctan \left(4 x\right)\right)$

Invert both sides
$\frac{16 {x}^{2}}{16 {x}^{2} + 1} = {\sin}^{2} \left(\arctan \left(4 x\right)\right)$

Taking the square root

$\sqrt{\frac{16 {x}^{2}}{16 {x}^{2} + 1}} = \sin \left(\arctan \left(4 x\right)\right)$

Simplifying

$\frac{4 x}{\sqrt{16 {x}^{2} + 1}} = \sin \left(\arctan \left(4 x\right)\right)$

You can rationalize it if you want
$\frac{4 x \cdot \sqrt{16 {x}^{2} + 1}}{16 {x}^{2} + 1} = \sin \left(\arctan \left(4 x\right)\right)$

And last but not least, gotta list out the values $x$ can't be because of the liberities we took during our process. We can't have denominators being zero nor negatives in square roots, so we have:
${\left(4 x\right)}^{2} \ne 0$
$16 {x}^{2} + 1 > 0$
$\sin \left(\arctan \left(4 x\right)\right) \ne 0 \rightarrow \frac{4 x}{\sqrt{16 {x}^{2} + 1}} \ne 0$

From these we see that $x \ne 0$