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

How do you simplify $$\displaystyle{\sin{{\left({{\tan}^{{-{1}}}{\left({x}\right)}}\right)}}}$$?

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Paul Mitchell
Explanation:
Knowing that
$$\displaystyle{{\sin}^{{2}}{\left(\theta\right)}}+{{\cos}^{{2}}{\left(\theta\right)}}={1}$$
We divide both sides by $$\displaystyle{{\sin}^{{2}}{\left(\theta\right)}}$$ so we have
$$\displaystyle{1}+{{\cot}^{{2}}{\left(\theta\right)}}={{\csc}^{{2}}{\left(\theta\right)}}$$
Or,
$$\displaystyle{1}+{\frac{{{1}}}{{{{\tan}^{{2}}{\left(\theta\right)}}}}}={\frac{{{1}}}{{{{\sin}^{{2}}{\left(\theta\right)}}}}}$$
Taking the least common multiple we have
$$\displaystyle{\frac{{{{\tan}^{{2}}{\left(\theta\right)}}+{1}}}{{{{\tan}^{{2}}{\left(\theta\right)}}}}}={\frac{{{1}}}{{{{\sin}^{{2}}{\left(\theta\right\rbrace}}}}}$$
Inverting both sides we have
$$\displaystyle{{\sin}^{{2}}{\left(\theta\right)}}={\frac{{{{\tan}^{{2}}{\left(\theta\right)}}}}{{{{\tan}^{{2}}{\left(\theta\right)}}+{1}}}}$$
So we say that $$\displaystyle\theta={\arctan{{\left({x}\right)}}}$$
$$\displaystyle{{\sin}^{{2}}{\left({\arctan{{\left({x}\right)}}}\right)}}={\frac{{{{\tan}^{{2}}{\left({\arctan{{\left({x}\right)}}}\right)}}}}{{{{\tan}^{{2}}{\left({\arctan{{\left({x}\right)}}}\right)}}+{1}}}}$$
Knowing that $$\displaystyle{\tan{{\left({\arctan{{\left({x}\right)}}}\right)}}}={x}$$
$$\displaystyle{{\sin}^{{2}}{\left({\arctan{{\left({x}\right)}}}\right)}}={\frac{{{x}^{{2}}}}{{{x}^{{2}}+{1}}}}$$
So we take the square root of both sides
$$\displaystyle{\sin{{\left({\arctan{{\left({x}\right)}}}\right)}}}=\pm\sqrt{{{\frac{{{x}^{{2}}}}{{{x}^{{2}}+{1}}}}}}=\pm{\frac{{{\left|{x}\right|}}}{{\sqrt{{{x}^{{2}}+{1}}}}}}$$
###### Not exactly what youâ€™re looking for?
Deufemiak7

We can use the principles of "SOH-CAH-TOA".
First, let's call $$\displaystyle{\sin{{\left({{\tan}^{{-{1}}}{\left({x}\right)}}\right)}}}={\sin{{\left(\theta\right)}}}$$ where the angle $$\displaystyle\theta={{\tan}^{{-{1}}}{\left({x}\right)}}$$
More specifically, $$\displaystyle{{\tan}^{{-{1}}}{\left({x}\right)}}=\theta$$ is the angle when $$\displaystyle{\tan{{\left(\theta\right)}}}={x}$$. We know this from the definition of inverse functions.
Since $$\displaystyle{\tan{{\left(\theta\right)}}}={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{a}{d}{j}{a}{c}{e}{n}{t}}}}$$, and here $$\displaystyle{\tan{{\left(\theta\right)}}}={\frac{{x}}{{1}}}$$ we know that
$$\displaystyle{\left\lbrace\begin{array}{c} {O}{p}{p}{o}{s}{i}{t}{e}={x}\\{a}{d}{j}{a}{c}{e}{n}{t}={1}\\{h}{y}{p}{o}{t}{e}nu{s}{e}=?\end{array}\right.}$$
Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs x and 1 has hypotenuse $$\displaystyle=\sqrt{{{x}^{{2}}+{1}}}$$
Now find $$\displaystyle{\sin{{\left({{\tan}^{{-{1}}}{\left({x}\right)}}\right)}}}$$
Since $$\displaystyle{\sin{{\left(\theta\right)}}}={\frac{{{o}{p}{p}{o}{s}{i}{t}{e}}}{{{h}{y}{p}{o}{t}{e}nu{s}{e}}}}$$, we see that
$$\displaystyle{\sin{{\left({{\tan}^{{-{1}}}{\left({x}\right)}}\right)}}}={\frac{{{x}}}{{\sqrt{{{x}^{{2}}+{1}}}}}}$$

karton