Arctan and Arccot

6 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

Separate tangent inversion from reciprocal notation.

Builds on Inverse Trig: Choosing One Angle · Arcsin · Arccos

The bigger question: How does an angle become a number?

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Idea

Arcsin and arccos only accept numbers from $-1$ to $1$ . The last two inverses accept everything.

Tangent runs from $-\infty$ to $\infty$ between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ , climbing the whole way. Cotangent runs from $\infty$ down to $-\infty$ across $(0, \pi)$ . Each passes every real number exactly once, so each window inverts cleanly - and each window's ends are excluded, because the function is undefined there.

Rule

$y = \arctan x \iff \tan y = x \qquad D = (-\infty, \infty) \qquad R = \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$y = \text{arccot}\,x \iff \cot y = x \qquad D = (-\infty, \infty) \qquad R = (0, \pi)$

Both ranges are open: the ends are approached, never reached. The two functions answer in different quadrants.

$\arctan(\text{positive}) \rightarrow$ first quadrant, $\arctan(\text{negative}) \rightarrow$ fourth

$\text{arccot}(\text{positive}) \rightarrow$ first quadrant, $\text{arccot}(\text{negative}) \rightarrow$ second

How it is used

1. One is odd, the other reflects

Arctan's window straddles zero, so it follows arcsin's rule. Arccot's holds no negative angles, so it follows arccos's.

\arctan(-x) = -\arctan x

\text{arccot}(-x) = \pi - \text{arccot}\,x

The picture below shows arccot: drag the probe and the two readings always add to $\pi$ , never to zero. Do not let the "tan" in the name pull you to the wrong rule - the window decides, not the name.

Inverse trigonometric functions

Try this. Move the probe from positive x to negative x. Compare the two outputs in radians and the indicated symmetry or sum.

Probe x0.5

arccot(0.5) = 1.107 rad · arccot(-0.5) = 2.034 rad · Sum = 3.142 rad

2. The values worth knowing

$x$	$-\sqrt{3}$	$-1$	$-\dfrac{1}{\sqrt{3}}$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\arctan x$	$-60^\circ$	$-45^\circ$	$-30^\circ$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$
$\text{arccot}\,x$	$150^\circ$	$135^\circ$	$120^\circ$	$90^\circ$	$60^\circ$	$45^\circ$	$30^\circ$

Read the zero column twice. $\arctan 0 = 0$ , but $\text{arccot}\,0 = 90^\circ$ - cotangent is $\dfrac{\cos x}{\sin x}$ , which is $0$ where the cosine is. That difference catches people out.

A fourth-quadrant angle can be written two ways: $-45^\circ$ and $315^\circ$ are the same point. Arctan always reports the negative one, because that is the reading inside its range.

3. The ends are asymptotes

Arctan takes every real number yet never leaves a fixed interval: far to the right it flattens towards $\dfrac{\pi}{2}$ , far to the left towards $-\dfrac{\pi}{2}$ . Those two levels are horizontal asymptotes, and arctan is the standard example when limits at infinity are taught.

Arccot does the same with its own ends: large positive inputs push it towards $0$ , large negative ones towards $\pi$ . Neither end is reached, which is what the open ranges mean.

4. Converting one into the other

\arctan x + \text{arccot}\,x = \frac{\pi}{2}

True for every real $x$ . It is the fastest way to change one function into the other, and the safest: it gets the sign right for you.

5. What carries over from the earlier lessons

The round trips work the same way: $\tan(\arctan x) = x$ and $\cot(\text{arccot}\,x) = x$ , for every real $x$ .

The releasing pattern is unchanged too. Arctan releases a tangent, arccot a cotangent.

No special angle? Draw the triangle, as in Arcsin, section 4 - starting from $\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}}$ or $\cot\alpha = \dfrac{\text{adjacent}}{\text{opposite}}$ .

Worked example

Find the inverse of $f(x) = \arctan(x + 2) - 3$ .

Remember

$\tan$ is what releases $x$ from inside an arctan $\tan(\arctan A) = A$ , so applying it cancels the arctan finish by swapping the letters, since an inverse is written in $x$

Write $y$ for $f(x)$ , then work towards $x$ .

y = \arctan(x + 2) - 3

Move the constant, leaving the arctan alone.

y + 3 = \arctan(x + 2)

Apply tangent to both sides. It releases the whole bracket, not $x$ alone.

x + 2 = \tan(y + 3)

Subtract $2$ , then swap the letters.

f^{-1}(x) = \tan(x + 3) - 2

Note which function did the releasing. Here $x$ was trapped inside an arctan, so tangent freed it. In $f(x) = 2\tan(x - 4) + 1$ it is the other way round, and arctan does the work.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

What is

\arctan(-1)

$-1$

$-\pi/4$

$3\pi/4$

Hint 1 · Find a starting point

Arctan returns an angle strictly between −π/2 and π/2.

Hint 2 · Take the next step

Tangent is −1 at a clockwise angle of 45°.

Show the reasoning

Answer: $-\pi/4$

The principal value is $-\pi/4$ . Another angle may have the same tangent, but it is not the arctan output.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example - a ratio with no special angle

Evaluate $\sin(\arctan 2)$ .

Remember

no special angle? Name it and draw the triangle - Arcsin, section 4 $\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}}$

Name the angle. There is no special angle for $\arctan 2$ , so do not look for one.

\arctan 2 = \alpha \quad \text{so} \quad \tan\alpha = \frac{2}{1}

Tangent is opposite over adjacent, so the triangle has opposite $2$ and adjacent $1$ .
Pythagoras supplies the hypotenuse.

\sqrt{2^2 + 1^2} = \sqrt{5}

Now read the sine straight off the completed triangle.

\sin\alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}

The angle itself was never needed. The same triangle answers $\cos(\text{arccot}\,2)$ - only the side names change, because cotangent reads adjacent over opposite.

A right triangle

Try this. Compare the labeled lengths with sine, cosine and tangent. If size is adjustable, scale the triangle: lengths change together, but the ratios stay fixed.

θ = 63.435° · sin θ = 0.894 · cos θ = 0.447 · tan θ = 2

Explore

Drag the probe far to the right on the first graph. The input passes $2$ , $5$ , $8$ - the reading crawls toward $1.571$ and never gets there. The dashed line is the ceiling: $\pi$ over $2$ . Far to the left the floor is $-\pi$ over $2$ . Every real number in, and the answer never escapes that band.

On the circle below it, predict what cot does at $90$ degrees before you drag - tan breaks there. Now move the point across $90$ . The tan readout jumps to undefined, but cot passes through $0$ calmly: they trade behaviour, because each is the other upside down. Drag on to $180$ : now cot breaks. That is why arccot's window runs between the breaks, from $0$ to $180$ degrees.

Inverse trigonometric functions

Try this. Move the probe from positive x to negative x. Compare the two outputs in radians and the indicated symmetry or sum.

Probe x0.5

arctan(0.5) = 0.464 rad · arctan(-0.5) = -0.464 rad

The unit circle

Try this. Try 0°, 90°, 180° and 270°. The horizontal projection is cosine; the vertical projection is sine. Track their signs between axes.

Angle (degrees)135

θ = 135° = 0.75π rad · cos θ = -0.707 · sin θ = 0.707 · arccot window: 0 to 180 degrees: inside

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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