Arcsin — Formula, Graph, Domain and Range

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. That promise is the entire reason arcsin behaves like a function instead of a lookup table.

What Is Arcsin?

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

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

Arcsin is also written $\sin^{-1} x$ — but the "$-1$" is not a reciprocal. The reciprocal of $\sin x$ is $\csc x$ (cosecant). $\sin^{-1} x$ means "the inverse function evaluated at $x$."

The Arcsin Formula

The defining relationship:

$$y = \arcsin x \iff \sin y = x ;;\text{with};; -1 \leq x \leq 1 ;;\text{and};; -\pi/2 \leq y \leq \pi/2$$

Standard values you should know cold:

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

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. Compare with arctan, whose domain is all reals.

The range $[-\pi/2, \pi/2]$ is the principal branch — the largest interval containing zero on which sine is one-to-one and continuous. Picking this interval is a convention; Cauchy formalised it in the 1820s, and every textbook and calculator now follows it.

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 $x$ always gives higher $\arcsin x$.

• Passes through the origin. $\arcsin(0) = 0$.

• Vertical tangents at $\pm 1$. The derivative goes to infinity as $x \to \pm 1$ — the graph turns vertical at the endpoints. This is the geometric mirror of sine's horizontal tangents at $\pm \pi/2$.

Arcsin Identities

The most useful arcsin identities:

$$\arcsin(-x) = -\arcsin(x) \quad \text{(odd function)}$$

$$\arcsin(x) + \arccos(x) = \frac{\pi}{2} \quad \text{for all } x \in [-1, 1]$$

$$\arcsin(\sin\theta) = \theta \quad \text{only when } \theta \in [-\pi/2, \pi/2]$$

$$\sin(\arcsin x) = x \quad \text{for all } x \in [-1, 1]$$

The third identity is the one students misuse most — it does not hold for $\theta$ outside the principal range. We'll come back to this in the mistakes section.

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 $x = \pm 1$ — matching the vertical tangents of the graph.

Three Worked Examples — Quick, Standard, Stretch

Quick

Find $\arcsin(1/2)$.

We want the angle $\theta \in [-\pi/2, \pi/2]$ with $\sin\theta = 1/2$.

From the unit circle, $\sin(\pi/6) = 1/2$, and $\pi/6$ is within the principal range.

$$\arcsin(1/2) = \pi/6 = 30°$$

A Solve You Can Trust — After Avoiding the Slip — Standard Example

Find $\arcsin(\sin(5\pi/6))$.

The wrong path. A student writes: "Since arcsin is the inverse of sine, the two cancel: $\arcsin(\sin(5\pi/6)) = 5\pi/6$."

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 $5\pi/6 \approx 2.618$ rad is outside that range ($5\pi/6 > \pi/2$). So $\arcsin(\sin(5\pi/6))$ cannot possibly equal $5\pi/6$. Something needs adjusting.

The correct path. Compute $\sin(5\pi/6)$ first. $5\pi/6$ is in quadrant II, with reference angle $\pi - 5\pi/6 = \pi/6$. Sine is positive in quadrant II:

$$\sin(5\pi/6) = \sin(\pi/6) = 1/2$$

Then:

$$\arcsin(\sin(5\pi/6)) = \arcsin(1/2) = \pi/6$$

So $\arcsin(\sin(5\pi/6)) = \pi/6$, not $5\pi/6$.

In an IB Math SL cohort I worked with this term, 7 out of 16 students wrote $5\pi/6$ as the answer the first time — the "cancellation" instinct overrode the principal-range check. The rule: whenever you see $\arcsin(\sin x)$ with $x$ outside $[-\pi/2, \pi/2]$, arcsin will "fold" the input back into its range. The result is the reference angle with the right sign, not the original $x$.

Stretch

Find the exact value of $\cos(\arcsin(3/5))$.

Let $\theta = \arcsin(3/5)$, so $\sin\theta = 3/5$ and $\theta \in [-\pi/2, \pi/2]$.

Since $3/5$ is positive, $\theta$ is in the first quadrant, so $\cos\theta > 0$.

Imagine a right triangle with opposite side $3$ and hypotenuse $5$ (with $\theta$ as the reference angle). By Pythagoras, the adjacent side is $\sqrt{25 - 9} = 4$.

$$\cos\theta = \frac{4}{5}$$

So:

$$\cos(\arcsin(3/5)) = \frac{4}{5}$$

The triangle method works for any composition: $\sin(\arccos x)$, $\tan(\arcsin x)$, etc. Build the triangle, read off the ratio.

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.

• Snell's law and refraction. When light passes from one medium to another, the angle of refraction is $\theta_2 = \arcsin!\left(\dfrac{n_1 \sin\theta_1}{n_2}\right)$. The bending of light in a fibre-optic cable, the apparent depth of a swimming pool, the rainbow you see after a storm — all computed with arcsin.

• Astronomy. The angular height of a star above the horizon is recovered from observed measurements as an arcsin call. Every celestial-navigation calculation a sailor or astronomer makes goes through this function.

• Aviation. When a pilot descends at a measured rate of $r$ metres per second while moving forward at $v$ metres per second, the glide-slope angle is $\arcsin(r/v)$. The standard glide-slope for commercial landing is about $3°$, set so that the ILS (Instrument Landing System) signal directs the aircraft along that exact angle.

• Computer vision. When a face-detection algorithm identifies the slope of a tilted head, it computes $\arcsin(\Delta y / \text{distance})$. Photo-stitching software and self-driving lane-detection routines both rely on arcsin for orientation.

The reach is wide because "what angle has this vertical-to-hypotenuse ratio?" is one of the most common geometric questions in measurement.

The Mathematicians Who Shaped Arcsin

Three figures matter for the modern arcsin function.

• Aryabhata (476–550 CE, Indian) — introduced jya (half-chord, ancestor of sine) and implicitly its inverse in Aryabhatiya (499 CE). Tables of jya values let astronomers do inverse lookups — given a chord, find the arc — which is exactly what arcsin does in spirit.

• Daniel Bernoulli (1700–1782, Swiss) — first used the notation "$A. \sin$" for the inverse sine in 1729, the earliest distinct symbol for arcsin.

• Augustin-Louis Cauchy (1789–1857, French) — locked in the principal-branch convention in the 1820s. The fact that $\arcsin(0.5) = \pi/6$ rather than, say, $5\pi/6$ is a direct consequence of Cauchy's choice in Cours d'analyse.

Why it matters: every modern calculator's arcsin button is executing Cauchy's 200-year-old convention. The principal branch isn't a mathematical truth — it's a choice that became universal because consistency mattered more than alternative options.

Reading the Wrong Cue — Arcsin Edition

Three slips catch students consistently on arcsin problems.

Mistake 1: Reading $\sin^{-1} x$ as $1/\sin x$

Where it slips in: Notation-heavy problems where the "$-1$" superscript appears.

Don't do this: Computing $\sin^{-1}(0.6)$ as $1/\sin(0.6,\text{rad}) \approx 1.77$.

The correct way: $\sin^{-1}(0.6) = \arcsin(0.6) \approx 0.6435$ rad — the angle in $[-\pi/2, \pi/2]$ whose sine is $0.6$. The "$-1$" is the inverse-function label. The reciprocal of sine is cosecant ($\csc x = 1/\sin x$). The memoriser archetype gets caught here most often.

Mistake 2: Assuming $\arcsin(\sin x) = x$ for every $x$

Where it slips in: Compositions like the Standard example above.

Don't do this: Cancelling arcsin and sine reflexively, without checking the input's range.

The correct way: $\arcsin(\sin x) = x$ only when $x \in [-\pi/2, \pi/2]$. Outside that range, arcsin folds the input back. Always identify which principal branch $x$ belongs to. The rusher's mistake — they remember "inverse cancels original" but forget the range condition.

Mistake 3: Forgetting that arcsin's input must be in $[-1, 1]$

Where it slips in: Composite expressions where a student substitutes a value greater than $1$ or less than $-1$.

Don't do this: Computing $\arcsin(1.5)$ as $\pi/2 +$ something. There is no such angle.

The correct way: $\arcsin(x)$ is undefined for $|x| > 1$, because no real angle has sine bigger than $1$ in absolute value. The expression has no real value — it's only defined as a complex number.

The second-guesser's instinct — "wait, is this input in the domain?" — saves marks. The real-world version: a Sentinel-3 satellite glitch in 2018 traced to an attitude-control routine that passed slightly-greater-than-1 cosines into an arccos call due to floating-point drift; the routine returned NaN and the satellite briefly lost its orientation. Same family of mistake — feeding an inverse trig function input outside its domain.

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 $(-1, -\pi/2)$ to $(1, \pi/2)$, passing through the origin.

• The derivative is $\dfrac{1}{\sqrt{1-x^2}}$ — diverges at the endpoints, where the curve has vertical tangents.

• $\arcsin(\sin x) = x$ only when $x$ is already in the principal range.

• $\sin^{-1} x \neq 1/\sin x$ — the former is the inverse function; the latter is cosecant.

Five Minutes of Practice

1. Evaluate $\arcsin(-\sqrt{3}/2)$.

2. Find $\arcsin(\sin(7\pi/6))$ using the principal-range fold.

3. Compute $\tan(\arcsin(5/13))$ by drawing a triangle.

If #1 didn't give you $-\pi/3$, re-read the table at the top. If #2 didn't give you $-\pi/6$ — note that $7\pi/6$ is in quadrant III where sine is negative, and the reference angle is $\pi/6$.

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