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


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. – André Nicolas Feb 10 '12 at 7:06
up vote 1 down vote accepted

You can show that for $x > 0$

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


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

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! – Oliver Spryn Feb 10 '12 at 2:40

$\hskip 1.5in$ triangle

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I dub this answer, "In which I discover the MSE background is not pure white." – anon Feb 10 '12 at 2:51

Can you simplify $\sin(\cot^{-1}(x))$? and then cube it?

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