y = \operatorname{arccot} x

The Inverse Cotangent $y = \operatorname{arccot} x$

$\operatorname{arccot} x$ , the inverse cotangent, inverts $\cot x = \dfrac{\cos x}{\sin x}$ ¹. It is the one inverse trigonometric function whose principal value is genuinely contested. The convention used here takes the range to be $(0, \pi)$ , which gives the formula below.

\operatorname{arccot} x = \frac{\pi}{2} - \arctan x

Domain and range

The domain is all real numbers. Since $\cot$ maps the interval $(0, \pi)$ monotonically onto the whole real line, its inverse accepts every real number and returns values in $(0, \pi)$ . As the range of $\arctan$ is $\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$ , subtracting it from $\dfrac{\pi}{2}$ lands exactly in $(0, \pi)$ .

Monotonicity and asymptotes

The derivative differs from that of $\arctan$ only in sign.

\frac{d}{dx}\operatorname{arccot} x = -\frac{1}{1+x^{2}}

Being always negative, the function decreases over the whole line with no extrema.

$x$	$\operatorname{arccot} x$
$\to -\infty$	$\to \pi$
$-1$	$\dfrac{3\pi}{4}$
$0$	$\dfrac{\pi}{2}$
$1$	$\dfrac{\pi}{4}$
$\to +\infty$	$\to 0$

The lines $y = \pi$ and $y = 0$ are horizontal asymptotes.

Symmetry and concavity

The graph has rotational symmetry about $\left( 0, \dfrac{\pi}{2} \right)$ , inherited from the symmetry of $\arctan$ about the origin after flipping it over and lifting it by $\dfrac{\pi}{2}$ .

The second derivative is $\dfrac{2x}{(1+x^{2})^{2}}$ , which carries the sign of $x$ , so the curve is concave down for $x < 0$ and concave up for $x > 0$ , and the center of symmetry is its only inflection point.

The two conventions

Convention	Formula	Range	Continuity
Principal value in $(0, \pi)$	$\dfrac{\pi}{2} - \arctan x$	$(0, \pi)$	continuous on the whole line
Through the reciprocal	$\arctan\dfrac{1}{x}$	$\left( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right)$ without $0$	jumps by $\pi$ at the origin

The two definitions agree for $x > 0$ and differ by exactly $\pi$ for $x < 0$ . Many computer algebra systems and numerical libraries adopt the latter, and so does the formula engine of this site: entering acot(x) in the formula field draws the discontinuous curve rather than the one shown here.

Which to choose

Where a continuous decreasing function is wanted, for instance to write an antiderivative in the following form, the range $(0, \pi)$ is convenient.

\int \frac{dx}{1+x^{2}} = -\operatorname{arccot} x + C

Where one prefers to define it through a reciprocal, as with the other inverse trigonometric functions, or to compute it mechanically from $\arctan$ , the form $\arctan\dfrac{1}{x}$ is easier. Textbooks and software divide on the question, so whenever $\operatorname{arccot}$ appears it is worth checking which range is intended.

Completing the family

With this all six inverse trigonometric functions are in place. Three pairs each add up to $\dfrac{\pi}{2}$ , a neatly symmetric arrangement.

Pair	Relation
$\arcsin$ and $\arccos$	$\arcsin x + \arccos x = \dfrac{\pi}{2}$
$\arctan$ and $\operatorname{arccot}$	$\arctan x + \operatorname{arccot} x = \dfrac{\pi}{2}$
$\operatorname{arcsec}$ and $\operatorname{arccsc}$	$\operatorname{arcsec} x + \operatorname{arccsc} x = \dfrac{\pi}{2}$

Inverse trigonometric functions, Wikipedia

The Inverse Cotangent y=arccot⁡xy = \operatorname{arccot} xy=arccotx