# Arcsin — Formula, Graph, Domain and Range

TL;DR

Arcsin (written sin⁡−1x or arcsin⁡x) is the inverse sine function. It takes a number in 
\[-1,1\] and returns the angle in \[-\pi/2,\pi/2\] whose sine equals that number. Its graph is a smooth, strictly-increasing S-curve passing through the origin, with endpoints (−1,−π/2) and (1,π/2).

## A Function Built on a Promise

To define the _inverse_ of sine, mathematicians had to make a promise: pick one angle per sine value, and _stick to it_. Sine is periodic — infinitely many angles share each sine value. The promise was made in the 1820s by Cauchy and locked in ever since: arcsin returns the angle in \[-\pi/2,\pi/2\], the slice of sine closest to zero where the function is one-to-one.

## What Is Arcsin?

**Arcsin** is the inverse of the sine function. If sin⁡θ=x for some \( \theta \in \[-\pi/2,\pi/2\] \), then arcsin⁡x=θ. Equivalently:

\[ y = \arcsin x \iff \sin y = x \text{ and } y \in \[-\pi/2, \pi/2\] \]

Arcsin is also written sin⁡−1x — but the "−1" is _not_ a reciprocal. The reciprocal of sin⁡x is csc⁡x (cosecant). sin⁡−1x means "the inverse function evaluated at x."

## The Arcsin Formula

The defining relationship:

\[ y = \arcsin x \iff \sin y = x \text{ with } −1≤x≤1 	ext{ and } −\pi/2≤y≤\pi/2 \]

### Standard values you should know cold:

| xxx | arcsin⁡x (radians) | arcsin⁡x (degrees) |
| --- | --- | --- |
| −1 | −π/2 | −90° |
| −√3/2 | −π/3 | −60° |
| −√2/2 | −π/4 | −45° |
| −1/2 | −π/6 | −30° |
| 0 | 0 | 0° |
| 1/2 | π/6 | 30° |
| √2/2 | π/4 | 45° |
| √3/2 | π/3 | 60° |
| 1 | π/2 | 90° |

The pattern is symmetric — arcsin is an _odd function_, so arcsin⁡(−x)=−arcsin⁡(x). This is what gives the table its mirror symmetry around zero.

## Domain and Range of Arcsin

| Property      | Value                                           |
|---------------|------------------------------------------------|
| **Domain**    | \[-1,1\] — closed interval                    |
| **Range**     | \[-\pi/2,\pi/2\] — closed interval         |
| **Continuity**| Continuous on \[-1,1\], smooth on (−1,1)    |
| **Symmetry**  | Odd function — arcsin⁡(−x)=−arcsin⁡(x)      |
| **Monotonicity**| Strictly increasing                          |
| **Endpoints** | (−1,−π/2) and (1,π/2) — both reached         |

The domain is the closed interval \[-1,1\] because sine outputs values only in \[-1,1\] — the inverse can't accept anything outside that range. The range \[-\pi/2,\pi/2\] is the **principal branch** — the largest interval containing zero on which sine is one-to-one and continuous.

## The Arcsin Graph

The arcsin graph is the reflection of the restricted sine curve on \[-\pi/2,\pi/2\] across the line y=x.

Three features worth pinning down:

- **Strictly increasing.** Higher xxx always gives higher arcsin⁡x.
- **Passes through the origin.** arcsin⁡(0)=0.
- **Vertical tangents at ±1.** The derivative goes to infinity as x→±1 — the graph turns vertical at the endpoints. This is the geometric mirror of sine's horizontal tangents at ±π/2.

## Arcsin Identities

The most useful arcsin identities:

1. arcsin⁡(−x)=−arcsin⁡(x) (odd function)
2. arcsin⁡(x)+arccos⁡(x)=π/2 for all x∈\[-1,1\]
3. arcsin⁡(sin⁡θ)=θ only when θ∈\[-\pi/2,\pi/2\]
4. sin⁡(arcsin⁡x)=x for all x∈\[-1,1\]

### Arcsin derivative and integral

\[ \frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}, \quad x \in (-1, 1) \]

\[ \int \arcsin(x), dx = x \arcsin(x) + \sqrt{1-x^2} + C \]

The derivative blows up at x=±1 — matching the vertical tangents of the graph.

## Three Worked Examples — Quick, Standard, Stretch

### Quick

**Find arcsin⁡(1/2).**

We want the angle θ∈\[-π/2,π/2\] with sin⁡θ=1/2.

From the unit circle, sin⁡(π/6)=1/2, and π/6 is within the principal range.

arcsin⁡(1/2)=π/6=30°.

### A Solve You Can Trust — Standard Example

**Find arcsin⁡(sin⁡(5π/6)).**

_The wrong path._ A student writes: "Since arcsin is the inverse of sine, the two cancel: arcsin⁡(sin⁡(5π/6))=5π/6."

That's the textbook trap. The "inverse cancellation" only works _one-way_ — the side that lives in the original function's principal range.

**Sanity check.** Arcsin's range is \[-\pi/2,\pi/2\]. The angle 5π/6 is out of that range, so arcsin⁡(sin⁡(5π/6)) cannot equal 5π/6.

_The correct path._ Compute sin⁡(5π/6) first. 5π/6 is in quadrant II, with reference angle π−5π/6=π/6. Sine is positive in quadrant II:

sin⁡(5π/6)=sin⁡(π/6)=1/2.

Then:

arcsin⁡(sin⁡(5π/6))=arcsin⁡(1/2)=π/6.

### Stretch

**Find the exact value of cos⁡(arcsin⁡(3/5)).**

Let θ=arcsin⁡(3/5), so sin⁡θ=3/5 and θ∈\[-π/2,π/2\].

Since 3/5 is positive, θ is in the first quadrant, so cos⁡θ>0.

Imagine a right triangle with opposite side 3 and hypotenuse 5. By Pythagoras, the adjacent side is √(25−9)=4.

cos⁡θ=4/5.

So:

cos⁡(arcsin⁡(3/5))=4/5.

## Where Arcsin Shows Up in the Real World

Arcsin is the function that recovers an angle from a _vertical_ measurement — the height-over-hypotenuse ratio.

1. **Snell's law and refraction.** 
2. **Astronomy.** 
3. **Aviation.** 
4. **Computer vision.**

## The Mathematicians Who Shaped Arcsin

1. **Aryabhata** (476–550 CE, Indian)
2. **Daniel Bernoulli** (1700–1782, Swiss)
3. **Augustin-Louis Cauchy** (1789–1857, French)

**Why it matters:** Every modern calculator's arcsin button is executing Cauchy's convention.

## Key Takeaways

- **Arcsin** is the inverse sine — input in \[-1,1\], output in \[-\pi/2,\pi/2\].
- The graph is a strictly increasing S-curve from (−1,−π/2) to (1,π/2), passing through the origin.
- The derivative is \[ \frac{1}{\sqrt{1-x^2}} \] — diverges at the endpoints.
- arcsin⁡(sin⁡x)=x only when x is already in the principal range.
- sin⁡−1x≠1/sin⁡x — the former is the inverse function; the latter is cosecant.