# How do you simplify Cotangent(Arcsin)?

Jun 15, 2015

$\cot \left(\arcsin x\right) = \frac{\cos \left(\arcsin x\right)}{\sin \left(\arcsin x\right)}$

$= \frac{\cos \left(\arcsin x\right)}{x}$

Normally, here would be enough for a high school Pre-Calculus problem. But, let's try something creative.

$\frac{d}{\mathrm{dx}} \left[\cos \left(\arcsin x\right)\right] = - \sin \left(\arcsin x\right) \cdot \frac{1}{\sqrt{1 - {x}^{2}}} = - \frac{x}{\sqrt{1 - {x}^{2}}}$

Let:
$u = 1 - {x}^{2}$
$\mathrm{du} = - 2 x \mathrm{dx}$

$\int \frac{- x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \frac{1}{2} \int \frac{- 2 x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$= \frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{du}$

$= \frac{1}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$= \frac{1}{2} \cdot \left[2 \sqrt{u}\right] = \sqrt{u} = \sqrt{1 - {x}^{2}}$

There, now we have a representation for $\cos \left(\arcsin x\right)$.

$\int \left\{\frac{d}{\mathrm{dx}} \left[\cos \left(\arcsin x\right)\right]\right\} \mathrm{dx} = \cos \left(\arcsin x\right)$

We haven't changed the domain, since the domain of $\sin x$ and $\cos x$ are larger than that of $\arcsin x$. Thus, $\arcsin x$ decides the domain restrictions on the left and right boundaries. Remember the $x$ in the denominator adds a vertical asymptote at $x = 0$.

$\implies \frac{\cos \left(\arcsin x\right)}{x} = \frac{\sqrt{1 - {x}^{2}}}{x}$

$\forall x \in \left[- 1 , 0\right) \cup \left(0 , 1\right]$
(For all $x$ in the domain of $- 1 \le x < 0$ and $0 < x \le 1$)