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.
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.
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.
$\hskip 1.5in$
Can you simplify $\sin(\cot^{-1}(x))$? and then cube it?