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

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mason m Share
Sep 26, 2017

$\sin \left({\tan}^{-} 1 \left(x\right)\right) = \frac{x}{\sqrt{{x}^{2} + 1}}$

#### Explanation:

We can use the principles of "SOH-CAH-TOA".

First, let's call $\sin \left({\tan}^{-} 1 \left(x\right)\right) = \sin \left(\theta\right)$ where the angle $\theta = {\tan}^{-} 1 \left(x\right)$.

More specifically, ${\tan}^{-} 1 \left(x\right) = \theta$ is the angle when $\tan \left(\theta\right) = x$. We know this from the definition of inverse functions.

Since $\tan \left(\theta\right) = \text{opposite"/"adjacent}$, and here $\tan \left(\theta\right) = \frac{x}{1}$ we know that

{("opposite"=x),("adjacent"=1),("hypotenuse"=?):}

Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs $x$ and $1$ has $\text{hypotenuse} = \sqrt{{x}^{2} + 1}$.

Now, to find $\sin \left({\tan}^{-} 1 \left(x\right)\right)$, find $\sin \left(\theta\right)$ for the triangle where

$\left\{\begin{matrix}\text{opposite"=x \\ "adjacent"=1 \\ "hypotenuse} = \sqrt{{x}^{2} + 1}\end{matrix}\right.$

Since $\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$, we see that

$\sin \left({\tan}^{-} 1 \left(x\right)\right) = \frac{x}{\sqrt{{x}^{2} + 1}}$

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Mar 15, 2016

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

#### Explanation:

Knowing that

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

We divide both sides by ${\sin}^{2} \left(\theta\right)$ so we have

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

Or,

$1 + \frac{1}{\tan} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right)$

Taking the least common multiple we have

$\frac{{\tan}^{2} \left(\theta\right) + 1}{\tan} ^ 2 \left(\theta\right) = \frac{1}{\sin} ^ 2 \left(\theta\right)$

Inverting both sides we have

${\sin}^{2} \left(\theta\right) = {\tan}^{2} \frac{\theta}{{\tan}^{2} \left(\theta\right) + 1}$

So we say that $\theta = \arctan \left(x\right)$

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

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

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

So we take the square root of both sides

$\sin \left(\arctan \left(x\right)\right) = \pm \sqrt{{x}^{2} / \left({x}^{2} + 1\right)} = \pm | x \frac{|}{\sqrt{{x}^{2} + 1}}$

Checking the range of the arctangent, we see that during it the sine is always positive so we have

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

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