Inverse Trigonometric Functions

Learning Objectives

After this section you will be able to:

1. Explain why the sine, cosine, and tangent functions are not invertible on their full domain, and how restricting their domain fixes the problem.
2. State the principal-branch domain and range of $\arcsin x$, $\arccos x$, and $\arctan x$ — and know why each range is chosen.
3. Use the cancellation identities $\sin(\arcsin x) = x$ and $\arcsin(\sin x) = x$ (and recognize when the second one fails).
4. Reach for atan2 instead of plain arctan whenever you have separate $x$ and $y$ components, and explain why it covers all four quadrants.
5. Implement the inverse trig functions in plain Python and use the differentiable PyTorch versions for ML pipelines.
6. Apply them to real problems: angles of elevation, robotics joint angles, phase recovery, computer-graphics surface normals.

The Problem: From Ratio Back to Angle

The forward trig functions answer the question: “Given an angle, what is the ratio?” For instance, climbing a ramp at $30^{\circ}$ means you rise $\sin 30^{\circ} = 0.5$ units for every 1 unit of slope.

But in real engineering and science the question almost always runs the other way:

I measured the rise and the slope length. What angle is the ramp at?

That is what the inverse trig functions exist to do. $\arcsin$, $\arccos$, and $\arctan$ reverse the arrow: feed them the ratio, get back the angle.

Why Sin, Cos, Tan Aren't Invertible Yet

A function $f$ is invertible only when each output comes from one and only one input — the so-called horizontal-line test: every horizontal line must hit the graph at most once.

Sine fails this test catastrophically. The horizontal line $y = 0.5$ crosses $\sin x$ at infinitely many places — every $30^{\circ}$, $150^{\circ}$, $390^{\circ}$, and so on. So if someone says “sin θ = 0.5, what is θ?”, the honest answer is “infinitely many possibilities”. That is useless.

Cosine and tangent need the same trick, but with different slices:

arcsin: The Inverse of Sine

Take the sine curve and clip it to the principal slice $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Now reflect it across the diagonal line $y = x$. That mirror image IS $y = \arcsin x$.

This is the deepest single picture in the section — drag the slider below and watch the mirror move with you.

Notice three things in the picture:

1. The blue point lives on the sine curve, with coordinates $(\theta, \sin\theta)$.
2. The pink point lives on the arcsin curve, with coordinates $(\sin\theta, \theta)$ — its x and y are simply swapped. That swap is what "reflection across $y = x$" means.
3. The pink curve only exists for inputs in $[-1, 1]$, because that is the range of sine. Try to ask $\arcsin(1.5)$ and you get a domain error: no angle has a sine of 1.5.

Formal definition. For $x \in [-1, 1]$, $\arcsin x$ is the unique angle $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ such that $\sin\theta = x$.

Reading the triangle

If a right triangle has opposite side $o$ and hypotenuse $h$, then the angle opposite to side $o$ is recovered by $\theta = \arcsin(o/h)$. Try the interactive triangle below with the "arcsin" tab — slide the ratio and watch the angle change.

arccos and arctan

arccos: the "adjacent over hypotenuse" inverse

$\arccos$ takes a value in $[-1, 1]$ and returns an angle in $[0, \pi]$. Why $[0, \pi]$ and not $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$? Because cosine is even: $\cos(-\theta) = \cos\theta$, so the values it produces on $[-\tfrac{\pi}{2}, 0]$ exactly repeat what it does on $[0, \tfrac{\pi}{2}]$. We need a slice where cosine is strictly monotone, and the natural choice is $[0, \pi]$, where cosine is strictly decreasing from $+1$ to $-1$.

arctan: the "opposite over adjacent" inverse

$\arctan$ accepts any real number and returns an angle in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ — open intervals at both ends, because tangent has vertical asymptotes there. As the input grows without bound, the output approaches but never reaches $\pm\tfrac{\pi}{2}$.

This is why arctan is the only inverse trig function with a bounded range but an unbounded domain. It maps the whole real line to a finite open interval — that property makes it a popular "soft clamp" in machine learning and signal processing.

The Three Graphs Side by Side

Switch tabs in the viewer below to see each principal-branch graph with its domain on the x-axis, its range as a shaded band on the y-axis, and a live readout for the value you pick.

A few features deserve attention:

• arcsin: domain $[-1, 1]$, range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Passes through the origin. Odd function — symmetric about the origin.
• arccos: domain $[-1, 1]$, range $[0, \pi]$. Passes through $(0, \tfrac{\pi}{2})$. Strictly decreasing — bigger cosine means smaller angle (think: the closer the adjacent is to the hypotenuse, the more the triangle is "flattened").
• arctan: domain all of $\mathbb{R}$, range $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$. Two horizontal asymptotes. Odd function.

Cancellation Identities (and a Common Trap)

Because $\arcsin$ is the inverse of restricted sine, the cancellation works perfectly in one direction:

Same pattern for the other two:

But the other direction is delicate. Try plugging $x = 2\pi$ into $\arcsin(\sin 2\pi)$:

Concretely, if $x = 5\pi/4$ (which is $225^{\circ}$, in the third quadrant), then $\sin(5\pi/4) = -\tfrac{\sqrt{2}}{2}$, and $\arcsin(-\tfrac{\sqrt{2}}{2}) = -\tfrac{\pi}{4}$ — definitely not $5\pi/4$. Information about which "copy" of the angle we started in was destroyed by sine and cannot be recovered by arcsin alone.

Useful Pythagorean identities

Frequently the angle is just an intermediate step; what you really want is another trig value of it. These identities let you skip the angle entirely:

Both of these come from drawing a right triangle that realises the inverse-trig input as one of its sides. For $\cos(\arcsin x) = \sqrt{1-x^2}$: let $\theta = \arcsin x$, so the opposite side is $x$ and the hypotenuse is $1$; by Pythagoras the adjacent side is $\sqrt{1 - x^2}$, and $\cos\theta = \text{adj}/\text{hyp} = \sqrt{1 - x^2}$. You can derive every entry in this table that way.

Worked Example: Finding the Angle of a Roof

A construction crew is building a roof. The rise (vertical) is $3.5$ meters and the run (horizontal) is $6.0$ meters. The supervisor needs the angle of the roof in degrees so the trusses can be cut. Let's do this by hand before letting a calculator do it, so the meaning of every step is clear.

atan2: The Practical Four-Quadrant Inverse

Here is a problem you will hit in your first robotics, game, or graphics project. You know a point's $(x, y)$ in the plane, and you want the angle from the positive $x$-axis to the point. The naive answer is $\theta = \arctan(y/x)$. The naive answer is wrong half the time.

Why? $\arctan$ returns angles only in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$, which covers only the right half of the plane. And dividing $y$ by $x$ throws away the sign information you need to tell, for example, the point $(-1, 1)$ in the second quadrant apart from $(1, -1)$ in the fourth — both have $y/x = -1$.

The fix is a function that takes $y$ and $x$ separately, looks at the sign of each, and returns the correct angle anywhere in $(-\pi, \pi]$. Every language calls it the same name: atan2(y, x).

Drag the yellow point in the viewer below. The green ray shows the angle that atan2 recovers (always correct). The dashed pink ray shows what arctan(y/x) would give — watch it stay in the right half-plane even when the actual point is on the left.

The piecewise definition of atan2

Under the hood, atan2(y, x) is just arctan(y/x) with a sign correction that depends on the quadrant. The full definition (using radians) is:

The piecewise nature of the definition is exactly what makes it correct: it inspects the sign of $x$ (which arctan never sees) and pushes the answer into the correct half-plane.

Python Implementation: From Scratch

Let's build $\arctan$ ourselves using only the most basic Python — no math.atan, no NumPy — so the inner workings are exposed. The Taylor series gives us the principle; an iterative argument reduction trick makes it fast enough to use.

1def my_arctan(x: float, n_terms: int = 25) -> float:
2    """Compute arctan(x) for any real x, in radians."""
3    # 1. Symmetry: arctan is odd, so handle the sign once and forget about it.
4    sign = 1.0 if x >= 0 else -1.0
5    x = abs(x)
6
7    # 2. Reduction: Taylor series converges slowly for |x| > 1.
8    #    Use the identity arctan(x) = pi/2 - arctan(1/x) to push large
9    #    inputs into the well-behaved region |x| <= 1.
10    use_complement = x > 1.0
11    if use_complement:
12        x = 1.0 / x
13
14    # 3. Maclaurin series: arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
15    total = 0.0
16    term = x          # the running x^(2k+1)
17    x_sq = x * x
18    for k in range(n_terms):
19        total += term / (2 * k + 1) * (1 if k % 2 == 0 else -1)
20        term *= x_sq  # advance from x^(2k+1) to x^(2k+3)
21
22    # 4. Undo the reduction.
23    if use_complement:
24        total = (3.141592653589793 / 2) - total
25
26    return sign * total
27
28
29# Try it
30print(my_arctan(0.0))    # 0.0
31print(my_arctan(1.0))    # ~ pi/4 = 0.7853981...
32print(my_arctan(10.0))   # ~ 1.4711...

Take a moment to appreciate what just happened. We turned the abstract object "the inverse of the tangent function" into a 20-line Python program that any reader can step through with a pencil. The series gives us correctness; the reduction gives us speed; the sign trick gives us coverage of negative inputs. That is what every production atan implementation in libm ultimately is, with more polish for floating-point edge cases.

PyTorch: Vectorised and Differentiable

In machine learning the inverse trig functions show up wherever an angle is hidden inside another quantity: angular regression, orientation prediction in 6-DoF pose estimation, normalizing-flow Jacobians, hyperbolic embeddings. PyTorch ships them as first-class differentiable ops. The key point is that they propagate gradients exactly the way calculus says they should.

1import torch
2
3# 1. Plain element-wise inverse trig on a batch of ratios.
4ratios = torch.tensor([-0.5, 0.0, 0.5, 0.9])
5angles_arcsin = torch.asin(ratios)      # rad in [-pi/2, pi/2]
6angles_arccos = torch.acos(ratios)      # rad in [0, pi]
7angles_arctan = torch.atan(ratios)      # rad in (-pi/2, pi/2)
8print("arcsin:", angles_arcsin)
9print("arccos:", angles_arccos)
10print("arctan:", angles_arctan)
11
12# 2. atan2 works on (y, x) pairs and returns the correct quadrant.
13y = torch.tensor([ 1.0,  1.0, -1.0, -1.0])
14x = torch.tensor([ 1.0, -1.0, -1.0,  1.0])
15theta = torch.atan2(y, x)               # rad in (-pi, pi]
16print("atan2 angles (deg):", torch.rad2deg(theta))
17
18# 3. Backpropagate through arcsin to confirm d/dx arcsin x = 1 / sqrt(1 - x^2).
19x_grad = torch.tensor([0.5], requires_grad=True)
20loss = torch.asin(x_grad).sum()
21loss.backward()
22print("analytic 1/sqrt(1 - 0.5^2) =", 1.0 / (1.0 - 0.5**2) ** 0.5)
23print("autograd dL/dx              =", x_grad.grad.item())

Real-World Applications

Robotics: inverse kinematics for a 2-link arm

A robotic arm with two segments of lengths $L_1$ and $L_2$ reaches a point $(x, y)$. The two joint angles are recovered with inverse trig:

Notice $\operatorname{atan2}$, not $\arctan$ — the target can be in any quadrant, including behind the robot. Using plain arctan would silently put the arm in the wrong half-plane.

Computer graphics: surface normals

The angle a surface normal makes with the camera is $\arccos(\mathbf{n} \cdot \mathbf{v})$ for unit vectors. This powers diffuse-lighting calculations in every raster and ray tracer. A pixel's brightness is roughly proportional to $\max(0, \cos\theta)$ where $\theta$ came from an arccos.

Signal processing: phase recovery

A complex number $z = a + bi$ has phase $\varphi = \operatorname{atan2}(b, a)$. FFT bins, OFDM demodulators, GPS carrier-tracking loops — they all unwrap phases using atan2, and they all subtly break if a developer reaches for arctan instead.

Geolocation: bearing between two GPS points

Given two latitude/longitude pairs, the compass bearing from one to the other is computed with atan2 of the cross-track and along-track displacements on the sphere. Every navigation app you have ever opened uses this formula thousands of times per second.

Machine learning: angular losses

For tasks where the output is an angle (head pose, wind direction, magnetic declination), naive MSE on the raw value fails because of the wraparound: predicting $359^{\circ}$ when the truth is $1^{\circ}$ is a 358° error in raw MSE, but only 2° geometrically. A clean fix: predict the unit-vector $(\cos\theta, \sin\theta)$ and recover the angle with $\operatorname{atan2}$.

Summary

Key takeaways

1. Inverse trig functions exist to recover an angle from a ratio. They are the everyday-use direction.
2. Sin, cos, tan are not invertible globally. Each one is restricted to a principal branch where it is strictly monotone — that branch determines the range of the inverse.
3. $\sin(\arcsin x) = x$ always, but $\arcsin(\sin x) = x$ only inside the principal branch. Outside, the composition folds the input back into the branch.
4. $\arctan(y/x)$ is wrong half the time. Always prefer atan2(y, x) when both components are available.
5. PyTorch's asin, acos, atan, and atan2 are fully differentiable. Their gradients are the textbook derivatives $\pm 1/\sqrt{1-x^2}$ and $1/(1+x^2)$.
6. The Pythagorean identities like $\cos(\arcsin x) = \sqrt{1 - x^2}$ let you skip computing the angle altogether when only another trig ratio is needed downstream.

Exercises

Conceptual

1. Explain in your own words why $\arccos$ uses the range $[0, \pi]$ instead of $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ like $\arcsin$. Hint: think about what symmetry cosine has that sine doesn't.
2. Without computing, decide whether each statement is true or false. Justify briefly.
   • (a) $\arcsin(\sin(2)) = 2$
   • (b) $\sin(\arcsin(0.7)) = 0.7$
   • (c) $\arctan(\tan(10)) = 10$
3. Sketch $\arctan x$ on the entire real line. Mark the two horizontal asymptotes and the point of inflection.

Computational

1. A ladder of length 5 m leans against a wall. Its base is 1.5 m from the wall. What angle does the ladder make with the ground? Use $\arccos$ and show the steps.
2. Simplify, leaving no inverse trig in the answer: $\sin(\arccos(x))$ for $x \in [-1, 1]$.
3. Evaluate exactly (no calculator):
   • (a) $\arcsin\!\left(-\tfrac{\sqrt{3}}{2}\right)$
   • (b) $\arctan(1)$
   • (c) $\arccos\!\left(\tfrac{1}{2}\right)$
4. Compute by hand the bearing from the origin to the point $(-3, 4)$ using both $\arctan(y/x)$ and atan2(y, x). Which one gives the geometrically correct angle? By how much do they differ?

Programming

1. Extend my_arctan from this section to handle x = ±inf and x = NaN gracefully (returning $\pm\pi/2$ and NaN respectively). Test with the edge cases.
2. Implement my_atan2(y, x) in plain Python by composing your my_arctan with the piecewise sign-correction rules. Verify on the four points (1, 1), (-1, 1), (-1, -1), (1, -1).
3. Train a tiny PyTorch model that predicts the angle of a 2D unit vector by outputting $(\cos\hat\theta, \sin\hat\theta)$ and recovering $\hat\theta = \operatorname{atan2}$. Compare its accuracy to a model that predicts $\hat\theta$ directly with MSE. Explain the gap.

Exploration

1. The Leibniz series for $\pi/4$ is $\arctan(1) = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \dots$. How many terms do you need to compute $\pi$ to 6 decimal places? Now use Machin's formula $\pi/4 = 4\arctan(1/5) - \arctan(1/239)$ — how many fewer terms are needed?
2. Investigate the unwrap operation in numpy.unwrap. Why is it needed when you compute the phase of a time-varying complex signal with atan2?

Next we'll meet hyperbolic functions — close cousins to sine and cosine, defined by exponentials instead of circles. They will give us the same kind of story all over again: forward, inverse, identities, derivatives — but with a hyperbola in the picture instead of a unit circle.