Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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

share|cite|improve this answer
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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.