Inverse Trigonometric Functions

⚡ Key Takeaways

Inverse trigonometric functions ( $\arcsin$ , $\arccos$ , $\arctan$ , and their reciprocal cousins) recover an angle from a ratio. Key result: because trig functions are periodic and not one-to-one, each inverse is defined only on a restricted domain, giving a single "principal value." They apply wherever you must solve for an angle — navigation, physics, engineering, computer graphics. Formalized through the 17th–18th centuries (Euler standardized the notation).

💡 The Core Insight

A trig function takes an angle and outputs a ratio; its inverse takes a ratio and outputs an angle. But trig functions repeat, so they are many-to-one — to invert them we first amputate the domain down to a stretch where the function rises or falls without repeating. The chosen stretch is a human convention (the principal branch), and every property of inverse trig functions flows from that choice.

DefinitionInverse Sine, Cosine, Tangent

Let $y$ be a real number. The principal inverse trigonometric functions are defined by:

1

\quad \arcsin: [-1,1]\to\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\qquad \arcsin(y)=\theta \iff \sin\theta=y

Derivation hint: restrict $\sin$ to $[-\pi/2,\pi/2]$ , where it is strictly increasing and hence invertible.

2

\quad \arccos: [-1,1]\to[0,\pi],\qquad \arccos(y)=\theta \iff \cos\theta=y

Derivation hint: restrict $\cos$ to $[0,\pi]$ , where it is strictly decreasing.

3

\quad \arctan: (-\infty,\infty)\to\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right),\qquad \arctan(y)=\theta \iff \tan\theta=y

Derivation hint: restrict $\tan$ to the open interval $(-\pi/2,\pi/2)$ , where it is strictly increasing over all reals. Here $\theta$ is the angle (in radians), $y$ is the ratio, and the codomain interval is the principal range. The notations $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ mean the same thing and must not be read as reciprocals.

DefinitionReciprocal Inverse Functions

The remaining three are defined from the first three:

4

\quad \operatorname{arcsec}(y)=\arccos\!\left(\tfrac{1}{y}\right),\quad \operatorname{arccsc}(y)=\arcsin\!\left(\tfrac{1}{y}\right),\quad \operatorname{arccot}(y)=\tfrac{\pi}{2}-\arctan(y)

Derivation hint: invert $\sec,\csc,\cot$ by relating each to its reciprocal trig function; domains require $|y|\ge 1$ for arcsec/arccsc. The principal range of $\operatorname{arcsec}$ and $\operatorname{arccsc}$ follows from that of $\arccos$ and $\arcsin$ respectively; $\operatorname{arccot}$ maps $\mathbb{R}\to(0,\pi)$ .

TheoremFundamental Inverse Identities

On the principal ranges, the following hold:

5

\quad \sin(\arcsin y)=y,\quad \cos(\arccos y)=y,\quad \tan(\arctan y)=y

Derivation hint: composing a function with its inverse on the restricted domain returns the input by definition.

6

\quad \arcsin(\sin x)=x \ \text{ only if } x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]

Derivation hint: the reverse composition recovers $x$ only when $x$ already lies in the principal range; otherwise it returns the equivalent principal angle.

7

\quad \arcsin y + \arccos y = \tfrac{\pi}{2}\quad(|y|\le 1)

Derivation hint: if $\theta=\arcsin y$ then $\cos(\tfrac{\pi}{2}-\theta)=\sin\theta=y$ , and $\tfrac{\pi}{2}-\theta\in[0,\pi]$ .

Figure

The graph of

y=\arcsin x

(gold) as the reflection of restricted

y=\sin x

across the line

y=x

.

ProofWhy a restricted domain is forced, and the derivative of arcsin

Part A — necessity of restriction. 1. $\sin$ is periodic: $\sin(\theta)=\sin(\theta+2\pi)$ , so for any value $y\in[-1,1]$ infinitely many $\theta$ satisfy $\sin\theta=y$ . 2. A function assigns exactly one output to each input. So $\arcsin$ cannot return all those angles. 3. To invert, we need a stretch on which $\sin$ is strictly monotonic (one-to-one). The interval $[-\pi/2,\pi/2]$ works: $\sin$ rises from $-1$ to $1$ , hitting each value exactly once. 4. By the inverse function theorem for monotone continuous functions, the restricted $\sin$ has a continuous strictly increasing inverse, defined as $\arcsin$ on $[-1,1]$ .

Part B — derivative. 5. Let $\theta=\arcsin x$ , so $\sin\theta=x$ with $\theta\in[-\pi/2,\pi/2]$ . 6. Differentiate implicitly: $\cos\theta\,\dfrac{d\theta}{dx}=1$ . 7. On $[-\pi/2,\pi/2]$ , $\cos\theta\ge 0$ , so $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}$ . 8. Therefore:

8

\quad \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}},\qquad |x|<1

Derivation hint: implicit differentiation of $\sin\theta=x$ plus the Pythagorean identity with the correct sign of $\cos$ . Analogously:

9

\quad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\arctan x = \frac{1}{1+x^2}

Derivation hint: same implicit method; for $\arctan$ , use $\sec^2\theta=1+\tan^2\theta=1+x^2$ .

∎

mathematician · Leonhard Euler | 1707–1783

Euler did not invent the inverse trig functions, but he systematized and popularized their analysis. In his Introductio in analysin infinitorum (1748) he treated arc-functions through infinite series and connected $\arctan$ to logarithms of complex numbers via the relation $\arctan x = \tfrac{1}{2i}\ln\frac{1+ix}{1-ix}$ . Euler's fluency with series let him compute $\pi$ from arctangent identities and embed these functions firmly in the new edifice of analysis. The familiar "arc" terminology — emphasizing that the output is an arc length on the unit circle (numerically equal to the radian angle) — also stems from this era.

🧠 Intuition

Picture the unit circle. A trig function "wraps" an angle onto the circle and reads off a coordinate; the inverse "unwraps" a coordinate back into an angle. The trouble is that the wrapping goes around forever, so unwrapping is ambiguous — like asking "what time is it?" and getting "3 o'clock" when it could be AM or PM. The principal range is the convention that fixes one answer, just as a 24-hour clock disambiguates. For $\arcsin$ and $\arctan$ we choose the answer nearest to $0$ (the right half, from $-90°$ to $+90°$ ); for $\arccos$ we choose the upper half, from $0°$ to $180°$ . Notice the asymmetry: $\arccos$ lives in $[0,\pi]$ because cosine is the horizontal coordinate, which decreases cleanly across the top of the circle.

ExampleFinding a ramp angle

Given: A wheelchair ramp rises $0.80\text{ m}$ over a horizontal run of $9.6\text{ m}$ . Find the angle of inclination.

Steps: 1. The tangent of the angle is rise/run: $\tan\theta = \dfrac{0.80}{9.6}=0.08\overline{3}$ . 2. Apply the inverse: $\theta=\arctan(0.0833)$ . 3. Evaluate (radians): $\arctan(0.0833)\approx 0.08314\text{ rad}$ . 4. Convert to degrees: $0.08314\times \dfrac{180}{\pi}\approx 4.76°$ .

\boxed{\theta \approx 4.8^{\circ}}

(Comfortably below the typical $4.76°$ ≈ $1{:}12$ accessibility maximum.)

ExampleComposite evaluation without a calculator

Given: Evaluate $\cos\!\left(\arcsin\tfrac{3}{5}\right)$ .

Steps: 1. Let $\theta=\arcsin\tfrac{3}{5}$ , so $\sin\theta=\tfrac{3}{5}$ and $\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ . 2. On that range $\cos\theta\ge 0$ , so $\cos\theta=+\sqrt{1-\sin^2\theta}$ . 3. Compute: $\cos\theta=\sqrt{1-\tfrac{9}{25}}=\sqrt{\tfrac{16}{25}}=\tfrac{4}{5}$ . 4. (Geometric check: a 3-4-5 right triangle with opposite side 3, hypotenuse 5, adjacent 4.)

\boxed{\cos\!\left(\arcsin\tfrac{3}{5}\right)=\tfrac{4}{5}=0.8}

⚠️

\sin^{-1}x

is NOT

1/\sin x

The superscript $-1$ in $\sin^{-1}x$ denotes functional inverse, not a reciprocal exponent. The reciprocal $1/\sin x$ is $\csc x$ . This clash of notation is why many sources prefer the unambiguous " $\arcsin$ ." Also beware: $\sin^2 x$ does mean $(\sin x)^2$ by convention, so the notation is genuinely inconsistent — a historical accident you must simply memorize.

CounterexampleThe reverse composition can betray you

One might expect $\arcsin(\sin x)=x$ always. Take $x=\dfrac{3\pi}{4}=135°$ .

$\sin\left(\tfrac{3\pi}{4}\right)=\tfrac{\sqrt2}{2}$ .
$\arcsin\left(\tfrac{\sqrt2}{2}\right)=\tfrac{\pi}{4}=45°$ , not $\tfrac{3\pi}{4}$ .

The function returned the principal angle with the same sine, because $\tfrac{3\pi}{4}$ lies outside $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ . Equation [6] holds only inside the principal range; outside it, $\arcsin(\sin x)$ "folds" $x$ back into that range.

connection

Inverse trig functions sit at a crossroads of several topics. They are the inverses of Trigonometric Functions, so everything about period and monotonicity reflects directly into their domains and ranges. Their derivatives [8]–[9] make them essential in Integration Techniques: integrals like $\int \frac{dx}{1+x^2}=\arctan x + C$ and $\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x + C$ are foundational. Through Euler's complex formula they link to the Complex Logarithm and thus to Complex Numbers. In two dimensions, the two-argument `atan2` generalizes $\arctan$ for full-circle angle recovery, connecting to Polar Coordinates and Vectors.

🔬 Where this appears

⏳ History

- ~500 CE: Indian astronomer-mathematician Aryabhata tabulates sine values; the conceptual inverse (finding arcs from chords) is implicit in such tables.

- ~1150: Bhāskara II and the Kerala school work with series related to arctangent ideas centuries before Europe.

- ~1400: Madhava of Sangamagrama discovers the power series for

\arctan

(the "Madhava–Leibniz series"), yielding

\pi

.

- 1671: James Gregory independently states the arctangent series in Europe.

- 1674: Gottfried Leibniz publishes

\tfrac{\pi}{4}=1-\tfrac13+\tfrac15-\cdots

, i.e.

\arctan 1

.

- 1736: Daniel Bernoulli and contemporaries use "arc" notation for inverse circular functions.

- 1748: Leonhard Euler systematizes the functions in Introductio, linking

\arctan

to complex logarithms.

- 19th c.: The notations

\sin^{-1}

(attributed to John Herschel, 1813) and

\arcsin

coexist, sparking the lasting ambiguity.

- 1960s–70s: The two-argument `atan2` is standardized in early FORTRAN, becoming the computing world's default for angle recovery.

open-problem · Branch choices, multivaluedness, and the complex extension

The "principal value" convention is not universal — different textbooks and software libraries occasionally choose $[0,\pi]$ versus $(-\pi/2,\pi/2)$ for $\operatorname{arcsec}$ and $\operatorname{arccot}$ , leading to sign disputes that are genuinely unsettled across the literature. More deeply, when extended to complex arguments the inverse trig functions become multivalued, requiring branch cuts in the complex plane (typically along $(-\infty,-1]$ and $[1,\infty)$ for $\arcsin$ ). Choosing where to place these cuts — and ensuring numerical libraries agree — remains a practical source of subtle bugs. There is no single "right" branch; it is a convention that must be declared, and reconciling competing conventions is an ongoing standardization concern rather than a solved theorem.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also