Inverse Trigonometric Functions

So far, you have learned how to find the sine, cosine, or tangent of an angle. Given an angle like $30°$ or $\frac{\pi}{4}$, you can calculate the ratio. But what if you know the ratio and need to find the angle? What if someone tells you “the sine of this angle is $\frac{1}{2}$” and you need to figure out what angle that is?

If this seems like working backwards, that is exactly what it is. And if the idea of “inverse” functions sounds intimidating, here is the thing: you already do this all the time. When you see the equation $x + 3 = 7$ and think “what number plus 3 gives 7?”, you are using the inverse of addition (which is subtraction). When you see $3x = 12$ and think “what number times 3 gives 12?”, you are using the inverse of multiplication (which is division). Inverse trigonometric functions work the same way: they undo what the regular trig functions do.

Core Concepts

Why We Need Inverse Trig Functions

Imagine you are building a ramp. You know the ramp rises 3 feet over a horizontal distance of 4 feet. What is the angle of the ramp? You can set up the problem: $\tan(\theta) = \frac{3}{4} = 0.75$. But now you are stuck. You need to find the angle $\theta$ whose tangent is 0.75. There is no simple way to solve this using what you already know about tangent.

This is where inverse trigonometric functions come in. Instead of asking “what is the tangent of this angle?”, they answer the question “what angle has this tangent?” For our ramp, the answer is $\theta = \tan^{-1}(0.75) \approx 36.87°$. The inverse tangent (also called arctangent) gives us the angle when we know the ratio.

The same idea applies to sine and cosine. If you know a ratio, the inverse function tells you the angle that produces that ratio.

The Problem with Inverses: Trigonometry Is Not One-to-One

Here is where things get a little tricky, and honestly, this is where a lot of people get confused. So if this feels complicated, you are not alone.

For a function to have an inverse, it must be one-to-one: each output must come from exactly one input. But sine, cosine, and tangent are not one-to-one. They repeat their values over and over.

Think about it: $\sin(30°) = \frac{1}{2}$, but so does $\sin(150°)$. And so does $\sin(390°)$ and $\sin(-210°)$ and infinitely many other angles. If you ask “what angle has sine equal to $\frac{1}{2}$?”, there are infinitely many correct answers.

This is a problem. If we want an inverse function, we need exactly one answer, not infinitely many.

The Solution: Restricting the Domain

The solution is clever: we restrict ourselves to a portion of each trigonometric function where it is one-to-one. We pick a range of angles where each ratio appears exactly once, and we agree that the inverse function will only give answers from that restricted range.

This restricted answer is called the principal value. When you ask your calculator for $\sin^{-1}(\frac{1}{2})$, it gives you $30°$ (or $\frac{\pi}{6}$), not $150°$ or any of the other possibilities. That is because $30°$ is the principal value: the answer that falls within the agreed-upon range.

Let us look at how this works for each inverse function.

Inverse Sine: $\sin^{-1}x$ or $\arcsin x$

The inverse sine function asks: “What angle has this sine?”

To make sine one-to-one, we restrict it to angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ (or $-90°$ to $90°$). On this interval, sine takes every value from $-1$ to $1$ exactly once.

Domain of $\sin^{-1}x$: $[-1, 1]$

You can only take the inverse sine of a number between $-1$ and $1$. This makes sense: sine never outputs anything outside this range, so you cannot ask “what angle has sine 2?” There is no such angle.

Range of $\sin^{-1}x$: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (or $[-90°, 90°]$)

The output is always an angle in this range. This means $\sin^{-1}x$ always gives you an angle in the right half of the unit circle: Quadrant I for positive values, Quadrant IV for negative values.

Key values to remember:

• $\sin^{-1}(0) = 0$
• $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (30°)
• $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ (45°)
• $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ (60°)
• $\sin^{-1}(1) = \frac{\pi}{2}$ (90°)
• $\sin^{-1}(-1) = -\frac{\pi}{2}$ (-90°)

Inverse Cosine: $\cos^{-1}x$ or $\arccos x$

The inverse cosine function asks: “What angle has this cosine?”

To make cosine one-to-one, we restrict it to angles from $0$ to $\pi$ (or $0°$ to $180°$). On this interval, cosine takes every value from $-1$ to $1$ exactly once (going from $1$ down to $-1$).

Domain of $\cos^{-1}x$: $[-1, 1]$

Same as inverse sine: you can only input values that cosine actually outputs.

Range of $\cos^{-1}x$: $[0, \pi]$ (or $[0°, 180°]$)

The output is always an angle in this range. This means $\cos^{-1}x$ gives you an angle in the top half of the unit circle: Quadrant I for positive values, Quadrant II for negative values.

Key values to remember:

• $\cos^{-1}(1) = 0$
• $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ (30°)
• $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ (45°)
• $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ (60°)
• $\cos^{-1}(0) = \frac{\pi}{2}$ (90°)
• $\cos^{-1}(-1) = \pi$ (180°)

Inverse Tangent: $\tan^{-1}x$ or $\arctan x$

The inverse tangent function asks: “What angle has this tangent?”

To make tangent one-to-one, we restrict it to angles from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ (not including the endpoints, where tangent is undefined). On this interval, tangent takes every real value exactly once.

Domain of $\tan^{-1}x$: $(-\infty, \infty)$

Unlike inverse sine and cosine, you can input any real number. This makes sense: tangent can output any real number (it goes to infinity in both directions).

Range of $\tan^{-1}x$: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (or $(-90°, 90°)$)

Note the parentheses instead of brackets: the endpoints are not included. The output is always an angle strictly between these values.

Key values to remember:

• $\tan^{-1}(0) = 0$
• $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$ (30°)
• $\tan^{-1}(1) = \frac{\pi}{4}$ (45°)
• $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ (60°)
• As $x \to \infty$, $\tan^{-1}(x) \to \frac{\pi}{2}$
• As $x \to -\infty$, $\tan^{-1}(x) \to -\frac{\pi}{2}$

Evaluating Inverse Trig Functions

To evaluate an inverse trig function, ask yourself: “What angle (in the restricted range) gives this ratio?”

For example, to find $\cos^{-1}\left(-\frac{1}{2}\right)$:

1. Ask: “What angle has cosine equal to $-\frac{1}{2}$?”
2. Remember the range is $[0, \pi]$, so the answer must be between $0$ and $\pi$.
3. You might recall that $\cos(60°) = \frac{1}{2}$. But we need a negative value.
4. In the range $[0, \pi]$, cosine is negative in Quadrant II.
5. The reference angle is $60°$, so the answer is $180° - 60° = 120°$, or $\frac{2\pi}{3}$.

Therefore, $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

Compositions of Trig and Inverse Trig Functions

Here is something that trips up many students. You might expect that $\sin(\sin^{-1}(x)) = x$ and $\sin^{-1}(\sin(x)) = x$. But these are not both always true!

When the function comes second: $\sin(\sin^{-1}(x)) = x$ for all $x$ in $[-1, 1]$.

This always works (as long as $x$ is in the domain). If you start with a ratio, find the angle, then take the sine of that angle, you get back your original ratio. The same is true for cosine and tangent.

When the inverse comes second: $\sin^{-1}(\sin(x)) = x$ only if $x$ is already in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

If $x$ is outside the principal value range, you will not get $x$ back. Instead, you will get the principal value: the angle in the restricted range that has the same sine as $x$.

For example: $\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$. But $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$, not $\frac{5\pi}{6}$.

So $\sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right) = \frac{\pi}{6} \neq \frac{5\pi}{6}$.

This is a common source of errors. Always pay attention to whether the angle is in the principal value range.

Using Inverse Trig Functions on Calculators

Your calculator has buttons for $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ (often labeled as second functions above sin, cos, and tan). To use them:

1. Make sure you know whether your calculator is in degree mode or radian mode. Check this before every problem!
2. Enter the ratio (the number whose inverse trig you want).
3. Press the inverse trig button.

For example, to find $\tan^{-1}(0.75)$:

• In degree mode: Enter 0.75, press $\tan^{-1}$, get approximately $36.87°$
• In radian mode: Enter 0.75, press $\tan^{-1}$, get approximately $0.6435$ radians

The calculator always gives the principal value. If you need an angle in a different quadrant, you will need to adjust the answer yourself.

Notation and Terminology

Term	Meaning	Example
$\sin^{-1}x$ or $\arcsin x$	Angle whose sine is $x$	$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
$\cos^{-1}x$ or $\arccos x$	Angle whose cosine is $x$	$\cos^{-1}(0) = \frac{\pi}{2}$
$\tan^{-1}x$ or $\arctan x$	Angle whose tangent is $x$	$\tan^{-1}(1) = \frac{\pi}{4}$
Principal value	The unique output in the restricted range	For $\sin^{-1}$, always in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
One-to-one	Each input gives exactly one output	Required for inverses to exist
Domain restriction	Limiting inputs to make a function one-to-one	We restrict sine to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

A note on notation: The notation $\sin^{-1}x$ does not mean $\frac{1}{\sin x}$. That would be written as $(\sin x)^{-1}$ or $\csc x$. The $-1$ superscript on the function name indicates an inverse function, not a reciprocal. The notation $\arcsin x$ avoids this confusion, which is why some textbooks prefer it.

Examples

Key Properties and Rules

Domain and Range Summary

Function	Domain	Range
$\sin^{-1}x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cos^{-1}x$	$[-1, 1]$	$[0, \pi]$
$\tan^{-1}x$	$(-\infty, \infty)$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Composition Identities

When the trig function comes second:

$$\sin(\sin^{-1}(x)) = x \quad \text{for } x \in [-1, 1]$$ $$\cos(\cos^{-1}(x)) = x \quad \text{for } x \in [-1, 1]$$ $$\tan(\tan^{-1}(x)) = x \quad \text{for all real } x$$

When the inverse comes second:

$$\sin^{-1}(\sin(x)) = x \quad \text{only if } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ $$\cos^{-1}(\cos(x)) = x \quad \text{only if } x \in [0, \pi]$$ $$\tan^{-1}(\tan(x)) = x \quad \text{only if } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Relationships Between Inverse Trig Functions

For any $x$ in $[-1, 1]$:

$$\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$$

This makes sense geometrically: if $\sin(\theta) = x$, then $\cos\left(\frac{\pi}{2} - \theta\right) = x$, and these two angles add up to $\frac{\pi}{2}$.

Negative Inputs

$$\sin^{-1}(-x) = -\sin^{-1}(x)$$ (sine inverse is an odd function) $$\tan^{-1}(-x) = -\tan^{-1}(x)$$ (tangent inverse is an odd function) $$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$$ (cosine inverse is not odd)

Real-World Applications

Finding Angles in Construction

A roof rises 8 feet over a horizontal span of 20 feet. What is the angle of the roof?

$$\tan(\theta) = \frac{8}{20} = 0.4$$ $$\theta = \tan^{-1}(0.4) \approx 21.8°$$

This angle, called the roof pitch, determines how well water and snow will run off.

Navigation and Finding Bearings

A ship travels 30 miles east and then 40 miles north. To return directly to the starting point, what bearing should it follow?

The ship needs to travel in the direction opposite to where it is now relative to the start. It is 30 miles east and 40 miles north, so:

$$\theta = \tan^{-1}\left(\frac{30}{40}\right) = \tan^{-1}(0.75) \approx 36.87°$$

The ship is $36.87°$ east of north from the starting point. To return, it should head $36.87°$ west of south, which in navigation terms is a bearing of approximately $180° + 36.87° = 216.87°$.

Physics: Angle of a Projectile

A ball is thrown with an initial horizontal velocity of 15 m/s and an initial vertical velocity of 20 m/s. At what angle above the horizontal was it thrown?

$$\theta = \tan^{-1}\left(\frac{20}{15}\right) = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13°$$

Computer Graphics: Rotation Angles

When programming animations or games, you often need to point one object toward another. If an object at position $(3, 2)$ needs to face an object at $(7, 5)$:

The direction vector is $(7-3, 5-2) = (4, 3)$.

The angle this makes with the horizontal is:

$$\theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87°$$

The $\arctan$ function (usually called atan or atan2 in programming languages) is essential for calculating rotation angles in 2D and 3D graphics.

Self-Test Problems

Summary

• Inverse trig functions answer the question: “What angle produces this ratio?” They undo the regular trig functions.
• Because trig functions repeat, we must restrict their domains to make them one-to-one before taking inverses. The answer in the restricted range is called the principal value.
• $\sin^{-1}x$ has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (outputs in Quadrants I and IV).
• $\cos^{-1}x$ has domain $[-1, 1]$ and range $[0, \pi]$ (outputs in Quadrants I and II).
• $\tan^{-1}x$ has domain $(-\infty, \infty)$ and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (outputs in Quadrants I and IV).
• Be careful with compositions: $\sin(\sin^{-1}(x)) = x$ always works (for valid inputs), but $\sin^{-1}(\sin(x)) = x$ only when $x$ is already in the principal value range.
• To evaluate compositions like $\cos(\sin^{-1}(x))$, draw a right triangle and use the Pythagorean theorem.
• The notation $\sin^{-1}x$ means the inverse function, not $\frac{1}{\sin x}$. The notation $\arcsin x$ is an alternative that avoids this confusion.
• On calculators, always check whether you are in degree or radian mode before using inverse trig functions.