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

Alan Smith 2021-12-27 Answered
How do you simplify \(\displaystyle{\sin{{\left({{\tan}^{{-{1}}}{\left({x}\right)}}\right)}}}\)?

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Expert Answer

Paul Mitchell
Answered 2021-12-28 Author has 225 answers
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
\(\displaystyle{\sin{{\left({\arctan{{\left({x}\right)}}}\right)}}}=\pm\sqrt{{{\frac{{{x}^{{2}}}}{{{x}^{{2}}+{1}}}}}}=\pm{\frac{{{\left|{x}\right|}}}{{\sqrt{{{x}^{{2}}+{1}}}}}}\)
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Deufemiak7
Answered 2021-12-29 Author has 4845 answers

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

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karton
Answered 2022-01-08 Author has 8659 answers
Finaly find answer. Thank you.
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