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How can the following function such that no trigonometric functions are present:

$\sin^3{\left(\cot^{-1}{\left(x\right)}\right)}$

Wolfram|Alpha shows the result as $\frac{1}{{\sqrt{x^2+1}}^3}$.

Thank you for your time.

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    There is a little twist that needs to be mentioned. The usual definition of $\cot^{-1} x$ is that it is the number **in** the interval $(0,\pi)$ whose cotangent is $x$. What that means is that when you take $\sin^3$ of this, you are dealing with a number in $(0,\pi)$, and there $\sin$ is positive, so you can use the positive square root. By way of contrast, the most common definition of $\tan^{1}$ of $x$ is the number in $(-\pi,\pi)$ whose $\tan$ is $x$, so if you are dealing with $\tan^{-1}$, then for negative $x$ you must take the negative square root.2012-02-10

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You can show that for $x > 0$

${\cot ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$

Then

$\sin {\cot ^{ - 1}}x = \frac{1}{{\sqrt {1 + {x^2}} }}$

and thus

${\left( {\sin {{\cot }^{ - 1}}x} \right)^3} = \frac{1}{{1 + {x^2}}}\frac{1}{{\sqrt {1 + {x^2}} }}$

The proof:

$x = \cot y$

$1+x^2 = \csc^2 y $

$\sqrt{1+x^2} = \csc y $

$\frac{1}{\sqrt{1+x^2}} = \sin y $

I guess that should do.

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    Woo-hoo! That's just the answer I was looking for Peter! Thanks for the detailed explanation!2012-02-10
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$\hskip 1.5in$ triangle

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    I dub this answer, "In which I discover the MSE background is not pure white."2012-02-10
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Can you simplify $\sin(\cot^{-1}(x))$? and then cube it?