Unit 6, lesson 56

The Arctangent Function

How arctangent reverses the central tangent branch, why it accepts every real number but never returns a right angle in either direction, and how its graph levels off toward two horizontal asymptotes.

  • Theory part 111
  • Practice part 112
  • Inverse Trigonometric Functions

A road sign says the grade is 10%10\%, which means the road climbs 0.10.1 meters per meter of horizontal distance. That number is the tangent of the road’s angle, and to get the angle back you need arctangent. Any slope at all can be fed in, however steep, yet the angle that comes out never quite reaches a right angle.

Key terms

Restricted tangent branch

Definition The restricted tangent branch is the tangent function on the open interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Tangent is undefined at the endpoints ±π2\pm\frac{\pi}{2}, where its graph has the vertical asymptotes x=±π2x = \pm\frac{\pi}{2}. On this interval tangent is increasing and takes every real value exactly once, so the branch is one-to-one with range (−∞,∞)(-\infty, \infty).

In plain words The middle piece of the tangent graph climbs from far below to far above, crossing each height once. That makes it possible to run it backwards.

Example The equation tan⁡x=1\tan x = 1 has the solutions π4+kπ\frac{\pi}{4} + k\pi for every integer kk, such as π4\frac{\pi}{4}, 5π4\frac{5\pi}{4} and −3π4-\frac{3\pi}{4}. Only π4\frac{\pi}{4} lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Arctangent

Definition For every real number uu, the arctangent of uu is the unique angle θ\theta in radians with

−π2<θ<π2andtan⁡θ=u.-\frac{\pi}{2} < \theta < \frac{\pi}{2} \quad \text{and} \quad \tan \theta = u.

It is written θ=arctan⁡u\theta = \arctan u or θ=tan⁡−1u\theta = \tan^{-1} u, where tan⁡−1\tan^{-1} means the inverse function and not the reciprocal 1tan⁡u=cot⁡u\frac{1}{\tan u} = \cot u.

In plain words The expression arctan⁡u\arctan u asks which angle strictly between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} has tangent uu. There is always exactly one.

Example Since tan⁡π3=3\tan \frac{\pi}{3} = \sqrt{3} and π3\frac{\pi}{3} lies in the open interval, arctan⁡3=π3\arctan \sqrt{3} = \frac{\pi}{3}. For the grade in the opening, the angle is arctan⁡0.1≈0.0997\arctan 0.1 \approx 0.0997 radians, a little under 66 degrees.

Principal angle

Definition The value arctan⁡u\arctan u is the principal angle with tangent uu. It is a signed angle, negative when u<0u < 0, zero when u=0u = 0 and positive when u>0u > 0. Every other angle with tangent uu is arctan⁡u+kπ\arctan u + k\pi for some integer kk. The inverse identities are

tan⁡(arctan⁡u)=u for every real u,arctan⁡(tan⁡θ)=θ only for θ∈(−π2,π2).\begin{aligned} &\tan(\arctan u) = u \\[0.4em] &\quad \text{ for every real } u, \\[0.4em] &\arctan(\tan \theta) = \theta \\[0.4em] &\quad \text{ only for } \theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right). \end{aligned}

In plain words Arctangent picks the angle on the right half of the unit circle, never at its top or bottom point, using a negative angle for a negative tangent.

Example Since tan⁡3π4=−1\tan \frac{3\pi}{4} = -1 and the principal angle with tangent −1-1 is −π4-\frac{\pi}{4},

arctan⁡(tan⁡3π4)=arctan⁡(−1)=−π4.\begin{aligned} \arctan\left(\tan \frac{3\pi}{4}\right) &= \arctan(-1) \\[0.4em] &= -\frac{\pi}{4}. \end{aligned}

The angle 3π4\frac{3\pi}{4} is outside the central interval, so the second identity does not apply, and the answer differs from it by π\pi.

All real domain

Definition The domain of arctangent is the set of all real numbers, (−∞,∞)(-\infty, \infty), because the restricted tangent branch takes every real value. Unlike arcsine and arccosine, arctangent needs no condition on its input. A composition arctan⁡(g(x))\arctan(g(x)) is defined exactly where gg is defined.

In plain words Any number at all is a possible tangent, so arctangent accepts every input.

Example The value arctan⁡1000\arctan 1000 exists, about 1.56981.5698, very close to π2≈1.5708\frac{\pi}{2} \approx 1.5708. The function arctan⁡1x\arctan \frac{1}{x} is defined for every x≠0x \ne 0, because 1x\frac{1}{x} is.

Open output interval

Definition The range of arctangent is the open interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). The endpoints ±π2\pm\frac{\pi}{2} are never attained, since tan⁡θ\tan \theta is undefined at θ=±π2\theta = \pm\frac{\pi}{2} and no real uu can have them as its arctangent. On a graph these heights are marked by dashed lines, never by dots.

In plain words Every output is strictly between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}. The ends of that interval are boundaries the function never reaches.

Example The equation arctan⁡x=2\arctan x = 2 has no solution, because 2>π2≈1.5712 > \frac{\pi}{2} \approx 1.571. For g(x)=2arctan⁡xg(x) = 2\arctan x every output doubles, so the range of gg is the open interval (−π,π)(-\pi, \pi), still with both ends left out.

Inverse graph

Definition The graph of y=arctan⁡xy = \arctan x is the set of points (b,a)(b, a) for which (a,b)(a, b) lies on the graph of the restricted tangent branch. With equal scales on both axes it is the reflection of that branch across the line y=xy = x. Swapping coordinates turns the vertical asymptotes x=±π2x = \pm\frac{\pi}{2} of the branch into the horizontal lines y=±π2y = \pm\frac{\pi}{2} of the inverse, and the open ends of the interval stay open.

In plain words Turn the tangent piece on its side. What used to rise along two walls now flattens out along a floor and a ceiling.

Example Swapping coordinates of points on the tangent branch gives these points on the arctangent graph.

xx−3-\sqrt{3}−1-100113\sqrt{3}
arctan⁡x\arctan x−π3-\frac{\pi}{3}−π4-\frac{\pi}{4}00π4\frac{\pi}{4}π3\frac{\pi}{3}

The curve rises through these points and is increasing on the whole real line.

Horizontal asymptote

Definition A horizontal line y=Ly = L is a horizontal asymptote of the graph of ff if f(x)f(x) stays as close to LL as you like once xx is large enough, or once xx is negative enough. This is written f(x)→Lf(x) \to L as x→∞x \to \infty, or as x→−∞x \to -\infty, and calculus makes it precise as a limit at infinity. The graph of y=arctan⁡xy = \arctan x has two horizontal asymptotes,

arctan⁡x→π2  as x→∞,arctan⁡x→−π2  as x→−∞.\begin{gathered} \arctan x \to \frac{\pi}{2} \ \text{ as } x \to \infty, \\[0.4em] \arctan x \to -\frac{\pi}{2} \ \text{ as } x \to -\infty. \end{gathered}

The reason is visible on the tangent branch. Because tangent is increasing there, for any small ε>0\varepsilon > 0 the angle arctan⁡x\arctan x lies above π2−ε\frac{\pi}{2} - \varepsilon exactly when x>tan⁡(π2−ε)x > \tan\left(\frac{\pi}{2} - \varepsilon\right). So every input beyond that number gives an output within ε\varepsilon of π2\frac{\pi}{2}, and oddness gives the same picture near −π2-\frac{\pi}{2} on the left.

In plain words Far to the right the curve levels off just under the height π2\frac{\pi}{2}, and far to the left just above −π2-\frac{\pi}{2}, without ever touching either line.

Example To be within 0.010.01 of π2\frac{\pi}{2} you need x>tan⁡(π2−0.01)≈99.997x > \tan\left(\frac{\pi}{2} - 0.01\right) \approx 99.997. The values creep up toward π2≈1.5708\frac{\pi}{2} \approx 1.5708.

xx101010010010001000
arctan⁡x\arctan x1.47111.47111.56081.56081.56981.5698

Odd symmetry

Definition Arctangent is odd, which means

arctan⁡(−u)=−arctan⁡ufor every real u.\begin{gathered} \arctan(-u) = -\arctan u \\[0.4em] \text{for every real } u. \end{gathered}

This holds because tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan \theta and −θ-\theta lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) whenever θ\theta does. The graph is symmetric about the origin.

In plain words Flip the sign of the input and the output angle flips sign too.

Example From arctan⁡13=π6\arctan \frac{1}{\sqrt{3}} = \frac{\pi}{6} we get arctan⁡(−13)=−π6\arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}, not 5π6\frac{5\pi}{6} or 11π6\frac{11\pi}{6}.

Check yourself

  1. Find arctan⁡(−3)\arctan(-\sqrt{3}) and arctan⁡(tan⁡2π3)\arctan\left(\tan \frac{2\pi}{3}\right).
Show the answer

Oddness and arctan⁡3=π3\arctan \sqrt{3} = \frac{\pi}{3} give arctan⁡(−3)=−π3\arctan(-\sqrt{3}) = -\frac{\pi}{3}. Since tan⁡2π3=−3\tan \frac{2\pi}{3} = -\sqrt{3}, the second value is also −π3-\frac{\pi}{3}, because 2π3\frac{2\pi}{3} lies outside the central interval.

  1. Give the domain, the range and the horizontal asymptotes of y=arctan⁡(x−3)+π2y = \arctan(x - 3) + \frac{\pi}{2}.
Show the answer

The domain is all real numbers. Shifting up by π2\frac{\pi}{2} moves the open range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) to (0,π)(0, \pi), with neither end attained. The asymptotes are y=0y = 0 to the left and y=πy = \pi to the right, and the graph passes through (3,π2)\left(3, \frac{\pi}{2}\right).

  1. Explain why the equation arctan⁡x=−π2\arctan x = -\frac{\pi}{2} has no solution.
Show the answer

The range of arctangent is the open interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), which does not contain −π2-\frac{\pi}{2}. A solution would need tan⁡(−π2)=x\tan\left(-\frac{\pi}{2}\right) = x, and that tangent is undefined.

How this lesson connects