How do you simplify Cotangent(Arcsin)?

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How do you simplify Cotangent(Arcsin)?

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Answer 1 · 2015-06-15T17:10:16

$cot (arcsin x) = \frac{cos (arcsin x)}{sin (arcsin x)}$ $cot(arcsinx) = (cos(arcsinx))/(sin(arcsinx))$

$= \frac{cos (arcsin x)}{x}$ $= (cos(arcsinx))/x$

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

$\frac{d}{d x} [cos (arcsin x)] = - sin (arcsin x) \cdot \frac{1}{\sqrt{1 - x^{2}}} = - \frac{x}{\sqrt{1 - x^{2}}}$ $d/(dx)[cos(arcsinx)] = -sin(arcsinx) * 1/(sqrt(1-x^2)) = -x/sqrt(1-x^2)$

Let:
$u = 1 - x^{2}$ $u = 1-x^2$
$d u = - 2 x d x$ $du = -2xdx$

$\int \frac{- x}{\sqrt{1 - x^{2}}} d x = \frac{1}{2} \int \frac{- 2 x}{\sqrt{1 - x^{2}}} d x$ $int (-x)/(sqrt(1-x^2))dx = 1/2int (-2x)/(sqrt(1-x^2))dx$

$= \frac{1}{2} \int \frac{1}{\sqrt{u}} d u$ $= 1/2int 1/(sqrt(u))du$

$= \frac{1}{2} \int u^{- \frac{1}{2}} d u$ $= 1/2int u^(-1/2)du$

$= \frac{1}{2} \cdot [2 \sqrt{u}] = \sqrt{u} = \sqrt{1 - x^{2}}$ $= 1/2*[2sqrtu] = sqrtu = sqrt(1-x^2)$

There, now we have a representation for $cos (arcsin x)$ $cos(arcsinx)$ .

$\int {\frac{d}{d x} [cos (arcsin x)]} d x = cos (arcsin x)$ $int{d/(dx)[cos(arcsinx)]}dx = cos(arcsinx)$

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

$\Rightarrow \frac{cos (arcsin x)}{x} = \frac{\sqrt{1 - x^{2}}}{x}$ $=> (cos(arcsinx))/x = sqrt(1-x^2)/x$

$\forall x \in [- 1, 0) \cup (0, 1]$ $AA x in [-1, 0) uu (0, 1]$
(For all $x$ $x$ in the domain of $- 1 \leq x < 0$ $-1 <= x < 0$ and $0 < x \leq 1$ $0 < x <= 1$ )

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How do you simplify Cotangent(Arcsin)?

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