Inverse Trigonometric Functions

Work backwards from ratios to angles

Have you ever looked at a ladder leaning against a wall and wondered exactly what angle it makes with the ground? Or maybe you have seen a ramp and wanted to know its steepness without pulling out a protractor. You know the measurements, you can calculate the ratio, but now you need to work backwards to find the actual angle. This is exactly what inverse trigonometric functions are for.

Think about it this way: regular trigonometric functions take an angle and give you a ratio. If you know a 30-degree angle, sine tells you the ratio is 0.5. But what if you start with the ratio 0.5 and need to find the angle? That is where inverse trig functions come in. They reverse the process, taking a ratio and returning an angle.

If you have ever used the “sin$^{-1}$” button on your calculator, you have already used an inverse trig function. In this chapter, we are going to understand what that button actually does, why it works the way it does, and how to use these functions confidently.

Core Concepts

Why We Need Restricted Domains

Here is something that might puzzle you at first: the sine of 30 degrees is 0.5, but the sine of 150 degrees is also 0.5. In fact, there are infinitely many angles whose sine equals 0.5 (30 degrees, 150 degrees, 390 degrees, and so on). So when you ask “what angle has a sine of 0.5?”, which answer should you get?

This is a real problem. For a function to have an inverse, each output must come from exactly one input. But regular sine, cosine, and tangent fail this test because they repeat their values over and over again. A single ratio like 0.5 corresponds to infinitely many angles.

The solution? We restrict the domain of each trig function to a portion where it does not repeat values. We choose a piece where the function hits every possible output value exactly once. This restricted version can then have a proper inverse.

It is a bit like how a city might have multiple “Main Streets” in different neighborhoods. If someone says “meet me on Main Street,” you need more context. But if they say “meet me on Main Street in downtown,” now there is only one location. Restricting the domain removes the ambiguity.

The Three Main Inverse Trig Functions

Arcsine: sin$^{-1}(x)$ or arcsin$(x)$

The inverse sine function answers: “What angle (in the restricted range) has this sine value?”

Domain: $[-1, 1]$ (You can only input values from -1 to 1, because sine never outputs anything outside this range.)

Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ or $[-90°, 90°]$

Why this range? We need a piece of the sine function that:

  1. Includes all possible sine outputs (from -1 to 1)
  2. Never repeats a value

The interval from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ does exactly that. On this interval, sine steadily increases from -1 to 1, hitting every value exactly once.

Example: $\arcsin(0.5) = \frac{\pi}{6}$ (or 30 degrees), because sine of 30 degrees is 0.5, and 30 degrees falls within our restricted range.

Arccosine: cos$^{-1}(x)$ or arccos$(x)$

The inverse cosine function answers: “What angle (in the restricted range) has this cosine value?”

Domain: $[-1, 1]$ (Same as arcsine, because cosine also only outputs values from -1 to 1.)

Range: $[0, \pi]$ or $[0°, 180°]$

Why this range? On the interval from 0 to $\pi$, cosine steadily decreases from 1 to -1, covering every possible cosine value exactly once.

Example: $\arccos(0.5) = \frac{\pi}{3}$ (or 60 degrees), because cosine of 60 degrees is 0.5, and 60 degrees falls within our restricted range.

Arctangent: tan$^{-1}(x)$ or arctan$(x)$

The inverse tangent function answers: “What angle (in the restricted range) has this tangent value?”

Domain: All real numbers $(-\infty, \infty)$

This is different from arcsine and arccosine. Tangent can output any real number, from negative infinity to positive infinity, so arctangent can accept any real number as input.

Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ or $(-90°, 90°)$

Notice the parentheses instead of brackets. The range does not include the endpoints because tangent is undefined at $\pm\frac{\pi}{2}$ (those are the vertical asymptotes).

Example: $\arctan(1) = \frac{\pi}{4}$ (or 45 degrees), because tangent of 45 degrees is 1.

Notation and Terminology

Symbol Read As Meaning
$\sin^{-1}(x)$ “inverse sine of x” or “arcsine of x” The angle whose sine is $x$
$\arcsin(x)$ “arcsine of x” or “arc-sine of x” Same as $\sin^{-1}(x)$
$\cos^{-1}(x)$ “inverse cosine of x” or “arccosine of x” The angle whose cosine is $x$
$\arccos(x)$ “arccosine of x” or “arc-cosine of x” Same as $\cos^{-1}(x)$
$\tan^{-1}(x)$ “inverse tangent of x” or “arctangent of x” The angle whose tangent is $x$
$\arctan(x)$ “arctangent of x” or “arc-tangent of x” Same as $\tan^{-1}(x)$

Important clarification: The notation $\sin^{-1}(x)$ does not mean $\frac{1}{\sin(x)}$. The -1 superscript here denotes an inverse function, not a reciprocal. If you want the reciprocal of sine, that is the cosecant function, written $\csc(x)$ or sometimes $(\sin(x))^{-1}$ with parentheses around the whole thing.

The “arc” prefix comes from the idea that we are finding the length of an arc on the unit circle. When mathematicians first developed these functions, they thought of them geometrically: given a certain sine value, what arc length (measured from the positive x-axis) gives that sine?

Evaluating Inverse Trig Functions

Using Special Angles

Many inverse trig problems can be solved by recalling the special angles from the unit circle.

Example 1: Finding arcsin of a Special Value

Evaluate $\arcsin\left(\frac{\sqrt{3}}{2}\right)$.

Step 1: Ask yourself: “What angle has a sine of $\frac{\sqrt{3}}{2}$?”

Step 2: Recall the unit circle. The sine of 60 degrees (or $\frac{\pi}{3}$) is $\frac{\sqrt{3}}{2}$.

Step 3: Check that this angle is in the range of arcsine, which is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $\frac{\pi}{3} \approx 1.047$ and $\frac{\pi}{2} \approx 1.571$, yes, $\frac{\pi}{3}$ is in the valid range.

Answer: $\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

Example 2: Finding arccos of a Special Value

Evaluate $\arccos\left(-\frac{1}{2}\right)$.

Step 1: Ask yourself: “What angle has a cosine of $-\frac{1}{2}$?”

Step 2: Cosine is negative in the second and third quadrants. From the unit circle, 120 degrees (or $\frac{2\pi}{3}$) has a cosine of $-\frac{1}{2}$.

Step 3: Check that this angle is in the range of arccosine, which is $[0, \pi]$.

Since $\frac{2\pi}{3} \approx 2.094$ and $\pi \approx 3.14$, yes, $\frac{2\pi}{3}$ is in the valid range.

Answer: $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

Example 3: Finding arctan of a Special Value

Evaluate $\arctan\left(-1\right)$.

Step 1: Ask yourself: “What angle has a tangent of $-1$?”

Step 2: Tangent equals -1 at angles like -45 degrees, 135 degrees, etc. But we need an angle in the range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Step 3: The angle $-\frac{\pi}{4}$ (or -45 degrees) is in this range and has tangent equal to -1.

Answer: $\arctan(-1) = -\frac{\pi}{4}$

Using a Calculator

For values that are not special angles, you will need a calculator.

Example 4: Calculator-Based Evaluation

A surveyor measures that the opposite side of a right triangle is 15 meters and the hypotenuse is 22 meters. What is the angle opposite to that side?

Step 1: The sine of the angle equals opposite over hypotenuse: $$\sin(\theta) = \frac{15}{22} \approx 0.6818$$

Step 2: Use the inverse sine to find the angle: $$\theta = \arcsin(0.6818)$$

Step 3: Enter this into a calculator (make sure it is in degree mode if you want degrees, or radian mode for radians).

Answer: $\theta \approx 43.0°$ or $\theta \approx 0.750$ radians

Compositions of Trig and Inverse Trig Functions

This is where things get interesting, and also where students often make mistakes. Let us carefully work through what happens when we combine trig functions with their inverses.

When the Inverse is on the Inside: sin(arcsin(x)), cos(arccos(x)), tan(arctan(x))

These always simplify nicely:

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

Why? This is just saying “the sine of the angle whose sine is $x$” equals $x$. It is like asking “what is the name of the person whose name is Bob?” The answer is Bob.

Example 5: Simplifying sin(arcsin(x))

Simplify $\sin\left(\arcsin\left(\frac{3}{5}\right)\right)$.

Reasoning: $\arcsin\left(\frac{3}{5}\right)$ gives us the angle whose sine is $\frac{3}{5}$. Then we take the sine of that angle. We get right back to $\frac{3}{5}$.

Answer: $\sin\left(\arcsin\left(\frac{3}{5}\right)\right) = \frac{3}{5}$

When the Inverse is on the Outside: arcsin(sin(x)), arccos(cos(x)), arctan(tan(x))

This is trickier. These only simplify to $x$ when $x$ is already in the range of the inverse function.

$$\arcsin(\sin(x)) = x \quad \text{only when } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ $$\arccos(\cos(x)) = x \quad \text{only when } x \in [0, \pi]$$ $$\arctan(\tan(x)) = x \quad \text{only when } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

If $x$ is outside these ranges, you need to find the equivalent angle that is inside the range.

Example 6: When the Angle is Outside the Range

Evaluate $\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right)$.

Step 1: First, find $\sin\left(\frac{5\pi}{6}\right)$.

The angle $\frac{5\pi}{6}$ is in the second quadrant. Its reference angle is $\frac{\pi}{6}$. Sine is positive in the second quadrant, so: $$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

Step 2: Now find $\arcsin\left(\frac{1}{2}\right)$.

This asks: “What angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has a sine of $\frac{1}{2}$?”

The answer is $\frac{\pi}{6}$.

Answer: $\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right) = \frac{\pi}{6}$

Notice that we did NOT get $\frac{5\pi}{6}$ back. That is because $\frac{5\pi}{6}$ is outside the range of arcsine. The arcsine function “resets” the angle to its equivalent value within the restricted range.

Example 7: Evaluating a Mixed Composition

Evaluate $\cos\left(\arctan\left(\frac{3}{4}\right)\right)$.

Step 1: Let $\theta = \arctan\left(\frac{3}{4}\right)$. This means $\tan(\theta) = \frac{3}{4}$ and $\theta$ is in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Step 2: Since $\frac{3}{4}$ is positive, $\theta$ is in the first quadrant (between 0 and $\frac{\pi}{2}$).

Step 3: Visualize a right triangle where the opposite side is 3 and the adjacent side is 4. By the Pythagorean theorem: $$\text{hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 4: Now find the cosine: $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$$

Answer: $\cos\left(\arctan\left(\frac{3}{4}\right)\right) = \frac{4}{5}$

Graphs of Inverse Trig Functions

Understanding the graphs helps you visualize what these functions do and remember their domains and ranges.

Graph of y = arcsin(x)

The graph of $y = \arcsin(x)$ is the reflection of a restricted portion of $y = \sin(x)$ across the line $y = x$.

Key features:

  • Domain: $[-1, 1]$ (the x-values go from -1 to 1)
  • Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (the y-values go from about -1.57 to 1.57)
  • Passes through the origin: $\arcsin(0) = 0$
  • Left endpoint: $\arcsin(-1) = -\frac{\pi}{2}$
  • Right endpoint: $\arcsin(1) = \frac{\pi}{2}$
  • Increasing throughout its domain
  • S-shaped curve, steeper in the middle, flattening near the endpoints

Graph of y = arccos(x)

Key features:

  • Domain: $[-1, 1]$
  • Range: $[0, \pi]$ (the y-values go from 0 to about 3.14)
  • Passes through $(0, \frac{\pi}{2})$: $\arccos(0) = \frac{\pi}{2}$
  • Left endpoint: $\arccos(-1) = \pi$
  • Right endpoint: $\arccos(1) = 0$
  • Decreasing throughout its domain
  • Also an S-shaped curve, but going downward

Graph of y = arctan(x)

Key features:

  • Domain: All real numbers $(-\infty, \infty)$
  • Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (the y-values approach but never reach $\pm\frac{\pi}{2}$)
  • Passes through the origin: $\arctan(0) = 0$
  • Horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$
  • Increasing throughout its domain
  • As $x \to \infty$, $y \to \frac{\pi}{2}$
  • As $x \to -\infty$, $y \to -\frac{\pi}{2}$

The arctangent graph is particularly important because it shows how even extremely large tangent values correspond to angles that are still less than 90 degrees. A tangent of 1000 gives an angle very close to, but still less than, $\frac{\pi}{2}$.

Key Properties

Summary of Domains and Ranges

Function Domain Range
$y = \arcsin(x)$ $[-1, 1]$ $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$y = \arccos(x)$ $[-1, 1]$ $[0, \pi]$
$y = \arctan(x)$ $(-\infty, \infty)$ $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Identities Involving Inverse Trig Functions

Complementary relationship: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$

This beautiful identity says that arcsine and arccosine of the same value always add up to 90 degrees. If you think about a right triangle, this makes sense: the two acute angles must sum to 90 degrees, and if one angle has sine equal to $x$, the other has cosine equal to $x$.

Negative input relationships: $$\arcsin(-x) = -\arcsin(x)$$ (arcsine is an odd function) $$\arccos(-x) = \pi - \arccos(x)$$ $$\arctan(-x) = -\arctan(x)$$ (arctangent is an odd function)

These help you handle negative inputs by relating them to positive inputs.

Real-World Applications

When surveyors need to determine the angle of elevation to a point on a hill, they measure distances and use inverse trig functions. If they know the height and horizontal distance, arctangent gives them the angle:

$$\text{angle of elevation} = \arctan\left(\frac{\text{height}}{\text{horizontal distance}}\right)$$

Computer Graphics and Game Development

Inverse trig functions are essential for making objects “look at” targets. If a character in a video game needs to face toward an enemy, the game calculates the direction using arctangent:

$$\text{direction angle} = \arctan\left(\frac{y_{\text{target}} - y_{\text{character}}}{x_{\text{target}} - x_{\text{character}}}\right)$$

Many programming languages provide an “atan2” function that handles this calculation more robustly.

Physics: Projectile Motion

When analyzing projectile motion, if you know the initial horizontal and vertical velocities, you can find the launch angle:

$$\theta = \arctan\left(\frac{v_y}{v_x}\right)$$

Engineering: Inclined Planes and Ramps

Building codes specify maximum ramp angles for accessibility. Engineers use inverse trig to verify their designs:

$$\text{ramp angle} = \arcsin\left(\frac{\text{rise}}{\text{ramp length}}\right)$$

or equivalently:

$$\text{ramp angle} = \arctan\left(\frac{\text{rise}}{\text{run}}\right)$$

Astronomy

Astronomers calculate the angle of celestial objects above the horizon using inverse trig functions based on their observations and the geometry of Earth.

Self-Test Problems

Problem 1: Evaluate $\arcsin\left(\frac{1}{2}\right)$. Give your answer in radians.

Show Answer

We need an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $\frac{1}{2}$.

From the unit circle, $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Since $\frac{\pi}{6}$ is in the valid range, $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$.

Problem 2: Evaluate $\arccos(0)$. Give your answer in radians.

Show Answer

We need an angle in $[0, \pi]$ whose cosine is 0.

From the unit circle, $\cos\left(\frac{\pi}{2}\right) = 0$.

Since $\frac{\pi}{2}$ is in the valid range, $\arccos(0) = \frac{\pi}{2}$.

Problem 3: Evaluate $\arctan(\sqrt{3})$. Give your answer in radians.

Show Answer

We need an angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is $\sqrt{3}$.

From the unit circle, $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

Since $\frac{\pi}{3}$ is in the valid range, $\arctan(\sqrt{3}) = \frac{\pi}{3}$.

Problem 4: Simplify $\arccos\left(\cos\left(\frac{4\pi}{3}\right)\right)$.

Show Answer

Step 1: Find $\cos\left(\frac{4\pi}{3}\right)$.

The angle $\frac{4\pi}{3}$ is in the third quadrant with reference angle $\frac{\pi}{3}$. Cosine is negative in the third quadrant: $$\cos\left(\frac{4\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$

Step 2: Find $\arccos\left(-\frac{1}{2}\right)$.

We need an angle in $[0, \pi]$ whose cosine is $-\frac{1}{2}$. That angle is $\frac{2\pi}{3}$.

Answer: $\arccos\left(\cos\left(\frac{4\pi}{3}\right)\right) = \frac{2\pi}{3}$

Note: We did not get $\frac{4\pi}{3}$ back because it is outside the range of arccosine.

Problem 5: Evaluate $\sin\left(\arccos\left(\frac{5}{13}\right)\right)$.

Show Answer

Step 1: Let $\theta = \arccos\left(\frac{5}{13}\right)$. This means $\cos(\theta) = \frac{5}{13}$ and $\theta$ is in $[0, \pi]$.

Step 2: Since $\frac{5}{13}$ is positive, $\theta$ is in the first quadrant.

Step 3: Use a right triangle where the adjacent side is 5 and the hypotenuse is 13. Find the opposite side: $$\text{opposite} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$$

Step 4: Calculate sine: $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}$$

Answer: $\sin\left(\arccos\left(\frac{5}{13}\right)\right) = \frac{12}{13}$

Problem 6: A 20-foot ladder leans against a wall, with its base 8 feet from the wall. What angle does the ladder make with the ground?

Show Answer

Step 1: Draw the situation. The ladder is the hypotenuse (20 ft), the distance from the wall is the adjacent side (8 ft).

Step 2: Use cosine: $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{20} = \frac{2}{5} = 0.4$

Step 3: Apply arccosine: $$\theta = \arccos(0.4) \approx 66.4°$$

Answer: The ladder makes an angle of approximately 66.4 degrees (or about 1.16 radians) with the ground.

Problem 7: Verify that $\arcsin\left(\frac{3}{5}\right) + \arccos\left(\frac{3}{5}\right) = \frac{\pi}{2}$.

Show Answer

Using a calculator:

  • $\arcsin\left(\frac{3}{5}\right) = \arcsin(0.6) \approx 0.6435$ radians
  • $\arccos\left(\frac{3}{5}\right) = \arccos(0.6) \approx 0.9273$ radians

Adding these: $0.6435 + 0.9273 = 1.5708$

And $\frac{\pi}{2} \approx 1.5708$.

The identity holds. This is true for any value $x$ in $[-1, 1]$: the arcsine and arccosine of the same value always sum to $\frac{\pi}{2}$.

Summary

  • Inverse trig functions reverse the process of regular trig functions: they take a ratio and return an angle.

  • Restricted domains are necessary because trig functions repeat their values. By restricting to a portion where each value appears exactly once, we can define proper inverses.

  • Arcsine $(\arcsin x)$: Domain $[-1, 1]$, Range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Returns angles in the first or fourth quadrant.

  • Arccosine $(\arccos x)$: Domain $[-1, 1]$, Range $[0, \pi]$. Returns angles in the first or second quadrant.

  • Arctangent $(\arctan x)$: Domain all reals, Range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Returns angles in the first or fourth quadrant.

  • Composition rules:

    • $\sin(\arcsin(x)) = x$ for $x \in [-1, 1]$ (and similarly for cos and tan)
    • $\arcsin(\sin(x)) = x$ only when $x$ is already in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; otherwise, find the equivalent angle in that range
  • The notation $\sin^{-1}(x)$ means inverse sine, not $\frac{1}{\sin(x)}$.

  • Useful identity: $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$

  • Applications include navigation, computer graphics, physics, engineering, and astronomy, anywhere you need to find an angle from known measurements.

The key to mastering inverse trig functions is remembering that they always return an angle in a specific range. When working with compositions, always check whether your angle falls within that range. With practice, these functions will become as natural to you as the regular trig functions you already know.