Inverse Trigonometric Functions: Definitions and Exact Values
Quick Reference: The Three Main Inverse Trig Functions
| 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 $x \in [-1, 1]$
Why Can’t We Just “Undo” Sine?
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 are on when you know your height.
Restricting the trigonometric functions to an interval where each is one-to-one makes them invertible, and on that interval the inverse can be evaluated at exact values.
📋 Before You Start: Check Your Prerequisites
Can you answer these questions? If not, review the linked skill first.
Trig values at standard angles: What is $\sin(\pi/4)$? What is $\cos(\pi/3)$?
- If unsure: Review Trig Functions Review
Inverse functions: If $f(x) = 2x + 3$ and $f^{-1}(y) = ?$, what is $f^{-1}(y)$?
- If unsure: Review Inverse Functions Basics
Horizontal line test: Why does $f(x) = x^2$ fail the horizontal line test?
- If unsure: Review One-to-One Functions
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Inverse Functions |
| Difficulty | Beginner |
| Time | ~20 minutes |
Key Concepts
The Restriction Strategy
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:
- Goes from $-1$ to $+1$ (covers the full range)
- Is strictly increasing (passes horizontal line test)
- Has an inverse we call $\arcsin$ or $\sin^{-1}$
Definition: Arcsine (Inverse 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 $x$.
The Three Main Inverse Trig Functions
| 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 $\arctan(x)$ | $(-\infty, \infty)$ | $(-\pi/2, \pi/2)$ | $\tan^{-1}(x) = y \Leftrightarrow \tan(y) = x$ |
Why These Ranges?
Arcsine: [-π/2, π/2] Arccosine: [0, π] Arctangent: (-π/2, π/2)
π/2 ─┐ π ─┐ π/2 ── ─ ─ ─
│ ╱ │╲ ╱
0 ─┼─── π/2─┼ ╲ 0 ───┼─────
│╱ │ ╲ ╱│
-π/2 ─┘ 0 ─┘ ╲ -π/2 ─ ─ ─ ──
-1 1 -1 1 -∞ ∞Each range is chosen so the restricted trig function is one-to-one AND covers all possible output values.
Cancellation Equations
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!
- $\sin^{-1}(\sin(\pi)) = \sin^{-1}(0) = 0 \ne \pi$
Common Pitfalls
| 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 $\arctan(-1) = -\pi/4$ |
Standard Values to Memorize
| $x$ | $\sin^{-1}(x)$ | $\cos^{-1}(x)$ | $\tan^{-1}(x)$ |
|---|---|---|---|
| $0$ | $0$ | $\pi/2$ | $0$ |
| $1/2$ | $\pi/6$ | $\pi/3$ | N/A |
| $\sqrt{2}/2$ | $\pi/4$ | $\pi/4$ | N/A |
| $\sqrt{3}/2$ | $\pi/3$ | $\pi/6$ | N/A |
| $1$ | $\pi/2$ | $0$ | $\pi/4$ |
| $\sqrt{3}$ | N/A | N/A | $\pi/3$ |
The Horizontal Asymptotes of Arctangent
$$\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2} \quad \text{and} \quad \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$$
This follows from the vertical asymptotes of tangent at $\pm\pi/2$.
Practice Problems
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 $x \in [-1, 1]$.
Conceptual Check (CCI-Style)
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) $a = \sin(0.5)$
Answer
(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.
Common Misconceptions
$\sin^{-1}(x)$ means $\frac{1}{\sin(x)}$, the reciprocal of sine.
This is the trig-as-algebra-symbols error. The notation $\sin^{-1}$ (or equivalently $\arcsin$) denotes the inverse function of the restricted sine, not its reciprocal. The reciprocal of $\sin(x)$ is $\csc(x)$, written $(\sin x)^{-1}$. For $x = \frac{1}{2}$, $\sin^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{6}$, a value in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, while $\frac{1}{\sin(1/2)} \approx 2.086$; these are entirely different quantities.
Mastery Checklist
Looking Ahead
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 $\pm\pi/2$ |
📚 Historical Note: Why "Arc"?
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 measure: degrees 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