Arccotangent Calculator

What is Arccot?

Arccot (arc cotangent) runs the cotangent backwards. From a value x, it hands back the angle θ where cot(θ) = x, and in practice you get it via arccot(x) = π/2 − arctan(x).

The answer always falls inside (0°, 180°). For instance: arccot(1) = 45°, arccot(0) = 90°, arccot(−1) = 135°.

Any real number works as input.

The arccotangent function

Arccotangent undoes the cotangent. Write acot(x) = θ when cot(θ) = x. Most references put the domain at ℝ and the range at (0, π), which gives you a continuous curve that decreases throughout, with horizontal asymptotes at 0 and π. A handful of textbooks go with [−π/2, π/2] \ {0} instead, so it pays to check which one your tool uses before you trust a value. To take a quick example, acot(1) = π/4 = 45°. Its derivative comes out as (acot x)' = −1/(1 + x²); the minus sign is what separates it from the arctangent derivative, and nothing else. And the two are tied together directly, since atan(x) + acot(x) = π/2 holds for every x > 0, making them complementary.

Applications: integration, optics and Fourier series

It gets taught less often than the arctangent at the elementary level, but the arccotangent still turns up in a few places. You'll find it in integration tables, where the antiderivative ∫ 1/(1+x²) dx can be written either as atan x + C or, just as validly, −acot x + C. It also shows up in geometric optics for some angle-of-incidence formulas, in Fourier series that come out of rational functions, and as a substitute expression in certain numerical methods.

FAQ

Why are there two conventions for the range? Picking (0, π) keeps the curve continuous, whereas [−π/2, π/2] \ {0} makes acot an odd function that matches acot(x) = atan(1/x). Modern texts and computer algebra systems usually side with continuity.

How is acot related to atan? When x > 0, you have acot(x) = atan(1/x) = π/2 − atan(x). Once x drops below zero, the formula picks up a π correction whose sign depends on the convention you've chosen.

What is acot(0)? If you're using the (0, π) convention, then acot(0) = π/2 = 90°, because cot(π/2) = 0.

Why is the function decreasing? The cotangent is itself decreasing on each of its principal branches, and an inverse keeps that monotonicity, so arccotangent decreases too.