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

Question

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

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

Thank you for your time.

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.

Peter Tamaroff · Accepted Answer · 2012-02-10 02:39:29Z

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.

Woo-hoo! That's just the answer I was looking for Peter! Thanks for the detailed explanation!

anon · Answer 2 · 2012-02-10 02:48:24Z

up vote 5 down vote

$\hskip 1.5in$ triangle

answered Feb 10 '12 at 2:48

anon
40.3k246112

1

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

Gerry Myerson · Answer 3 · 2012-02-10 02:20:38Z

up vote 1 down vote

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

answered Feb 10 '12 at 2:20

Gerry Myerson
86k462142

3 Answers