Navigation: Wiki Home > Skills > Inverse Trigonometric Functions: Definitions and Exact Values
| Function | Domain | Range | Definition |
|---|---|---|---|
| $\sin^{-1}(x)$ or $\arcsin(x)$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ | The angle in $[-\pi/2, \pi/2]$ whose sine is $x$ |
| $\cos^{-1}(x)$ or $\arccos(x)$ | $[-1, 1]$ | $[0, \pi]$ | The angle in $[0, \pi]$ whose cosine is $x$ |
| $\tan^{-1}(x)$ or $\arctan(x)$ | $(-\infty, \infty)$ | $(-\pi/2, \pi/2)$ | The angle in $(-\pi/2, \pi/2)$ whose tangent is $x$ |
Key Identity: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for all $x \in [-1, 1]$
You know that $\sin(\pi/6) = 1/2$. So what's "the angle whose sine is $1/2$"?
The problem: there are infinitely many such angles! Both $\pi/6$ and $5\pi/6$ have sine equal to $1/2$, and adding any multiple of $2\pi$ gives more solutions. The sine function is not one-to-one, so it doesn't have an inverse... unless we restrict its domain.
Mental Model: Think of the restricted sine as a single flight of stairs going from the basement ($y = -1$) to the first floor ($y = 1$). Every height on the staircase corresponds to exactly one step. Arcsine tells you which step you're on when you know your height.
This section shows how mathematicians restrict the trigonometric functions to make them invertible, and how to evaluate these inverse functions at exact values.
Can you answer these questions? If not, review the linked skill first.
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Week | Week 2 |
| Difficulty | Beginner |
| Time | ~20 minutes |
Since $\sin x$ fails the horizontal line test over its full domain, we restrict it to $[-\pi/2, \pi/2]$ where it IS one-to-one. On this interval, sine:
$$\sin^{-1}(x) = y \quad \Longleftrightarrow \quad \sin(y) = x \text{ and } -\frac{\pi}{2} \le y \le \frac{\pi}{2}$$
In words: $\sin^{-1}(x)$ is the angle in $[-\pi/2, \pi/2]$ whose sine equals $x$.
| Function | Notation | Domain | Range | Definition |
|---|---|---|---|---|
| Inverse Sine | $\sin^{-1}(x)$ or $\arcsin(x)$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ | $\sin^{-1}(x) = y \Leftrightarrow \sin(y) = x$ |
| Inverse Cosine | $\cos^{-1}(x)$ or $\arccos(x)$ | $[-1, 1]$ | $[0, \pi]$ | $\cos^{-1}(x) = y \Leftrightarrow \cos(y) = x$ |
| Inverse Tangent | $\tan^{-1}(x)$ or $\arctan(x)$ | $(-\infty, \infty)$ | $(-\pi/2, \pi/2)$ | $\tan^{-1}(x) = y \Leftrightarrow \tan(y) = x$ |
Arcsine: [-Ο/2, Ο/2] Arccosine: [0, Ο] Arctangent: (-Ο/2, Ο/2)
Ο/2 ββ Ο ββ Ο/2 ββ β β β
β β± ββ² β±
0 ββΌβββ Ο/2ββΌ β² 0 ββββΌβββββ
ββ± β β² β±β
-Ο/2 ββ 0 ββ β² -Ο/2 β β β ββ
-1 1 -1 1 -β βEach range is chosen so the restricted trig function is one-to-one AND covers all possible output values.
When composing a function with its inverse:
$$\sin(\sin^{-1}(x)) = x \quad \text{for } -1 \le x \le 1$$ $$\sin^{-1}(\sin(x)) = x \quad \text{for } -\frac{\pi}{2} \le x \le \frac{\pi}{2}$$
Warning: The second equation requires $x$ to be in the restricted domain!
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| $\sin^{-1}(\sin(2\pi)) = 2\pi$ | $2\pi$ is outside $[-\pi/2, \pi/2]$ | Compute $\sin(2\pi) = 0$, then $\sin^{-1}(0) = 0$ |
| $\cos^{-1}(-1/2) = -\pi/3$ | Arccos range is $[0, \pi]$, so output can't be negative | Find angle in $[0, \pi]$: answer is $2\pi/3$ |
| Confusing $\sin^{-1}(x)$ with $\frac{1}{\sin(x)}$ | Different notation conventions | $\sin^{-1}(x) = \arcsin(x)$ is the inverse; $(\sin x)^{-1} = \csc x$ is the reciprocal |
| $\arctan(\tan(3\pi/4)) = 3\pi/4$ | $3\pi/4$ is outside $(-\pi/2, \pi/2)$ | Compute $\tan(3\pi/4) = -1$, then $\arctan(-1) = -\pi/4$ |
| $x$ | $\sin^{-1}(x)$ | $\cos^{-1}(x)$ | $\tan^{-1}(x)$ |
|---|---|---|---|
| $0$ | $0$ | $\pi/2$ | $0$ |
| $1/2$ | $\pi/6$ | $\pi/3$ | β |
| $\sqrt{2}/2$ | $\pi/4$ | $\pi/4$ | β |
| $\sqrt{3}/2$ | $\pi/3$ | $\pi/6$ | β |
| $1$ | $\pi/2$ | $0$ | $\pi/4$ |
| $\sqrt{3}$ | β | β | $\pi/3$ |
$$\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2} \quad \text{and} \quad \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$$
This follows from the vertical asymptotes of tangent at $\pm\pi/2$.
Find the exact value of $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
Find the exact value of $\cos^{-1}\left(-\frac{1}{2}\right)$.
Evaluate $\arcsin\left(\sin\left(\frac{7\pi}{6}\right)\right)$.
Evaluate $\displaystyle\lim_{x \to 0^+} \arctan\left(\frac{1}{x}\right)$.
Prove that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Question: If $\sin^{-1}(a) = 0.5$ (in radians), which of the following is true?
(A) $\sin(0.5) = a$ (B) $a = 0.5$ (C) $\sin(a) = 0.5$ (D) $a = \sin(0.5)$
(A) By definition, $\sin^{-1}(a) = 0.5$ means $\sin(0.5) = a$.
Common error: Choosing (C) reverses the input and output of the inverse function.
Understanding inverse trig definitions is the foundation for several important topics:
| Topic | How This Skill Connects |
|---|---|
| Derivatives of Inverse Trig | You'll use implicit differentiation on equations like $\sin y = x$ |
| Integrals Yielding Inverse Trig | Recognizing $\frac{1}{\sqrt{1-x^2}}$ as the derivative of $\arcsin$ |
| Trig Substitution (Ch. 7) | When you see $\sqrt{1-x^2}$, substitute $x = \sin\theta$ with $\theta \in [-\pi/2, \pi/2]$ |
| Limits at Infinity | The horizontal asymptotes of $\arctan x$ are $\pm\pi/2$ |
The "arc" in arcsin, arccos, and arctan comes from the arc length on a unit circle. If you travel along the unit circle starting from $(1, 0)$, the arc length you travel equals the angle (in radians) you've swept. So "$\arcsin(x)$" literally means "the arc (angle) whose sine is $x$."
This connection to circles is why we require radian measureβdegrees would break the elegant relationship between arc length and angle.
| Previous | Up | Next |
|---|---|---|
| Exponential Growth/Decay | Skills Index | Simplifying Inverse Trig Expressions |
Last updated: 2026-01-22