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Inverse Trigonometric Functions

Q: Why is the range of arccos different from arcsin?

The range of arccos is $[0, \pi]$ (Quadrants I and II) to ensure it is a function that covers all possible cosine values from -1 to 1. If we used the same range as arcsin (Quadrants I and IV), we would only get positive cosine values, and we wouldn't be able to find the arccos of a negative number.

Q: Can I find the arcsin of 2?

No, you cannot. The domain of the arcsin function is $[-1, 1]$ because the sine function only produces values within that range. Since there is no angle $\theta$ for which $\sin(\theta) = 2$, the expression $\arcsin(2)$ is undefined.

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Ever wondered how to find an angle when you only know the sides of a right triangle? Regular trig functions give you a ratio from an angle, but inverse trigonometric functions do the opposite—they find the angle from the ratio. Let's dive into how these powerful functions work.

Inverse Trigonometric Functions — an original Algebra911 reference diagram defining inverse trigonometric functions with its key formula and a worked example. — Inverse Trigonometric Functions: A Complete Guide

What Are Inverse Trigonometric Functions?

Inverse trigonometric functions are functions that “undo” the regular trigonometric functions (sine, cosine, tangent) to find an angle given its trigonometric ratio. For example, if we know that $\sin (θ) = x$ , the inverse sine function allows us to find the angle $θ$ . In essence, they answer the question, “What angle gives me this specific sine, cosine, or tangent value?”

These functions are fundamental in geometry, physics, and engineering for solving problems involving angles. There are two common notations for inverse trigonometric functions:

Arcsine, Arccosine, Arctangent: Written as $\arcsin (x)$ , $\arccos (x)$ , and $\arctan (x)$ . The 'arc' prefix refers to the arc length on a unit circle that corresponds to the given angle.
Inverse Function Notation: Written as $\sin^{- 1} (x)$ , $\cos^{- 1} (x)$ , and $\tan^{- 1} (x)$ . This is the notation often found on calculators.

It is critically important to understand that the $- 1$ in $\sin^{- 1} (x)$ is not an exponent. It denotes an inverse function, not a reciprocal.

\sin^{- 1} (x) \neq \frac{1}{\sin (x)} = \csc (x)

To avoid this confusion, many mathematicians prefer the $\arcsin$ notation. Throughout this lesson, we will use both notations interchangeably as you are likely to encounter both.

Why Do We Need to Restrict the Domain of Trig Functions?

A core concept in algebra is that for a function to have a well-defined inverse, it must be one-to-one. This means that every output value corresponds to exactly one input value. A simple way to check this is the Horizontal Line Test: if you can draw a horizontal line that crosses the function's graph more than once, the function is not one-to-one, and its inverse will not be a function.

Let's look at the graph of $y = \sin (x)$ . It's a wave that repeats forever. A horizontal line, like $y = 0.5$ , intersects the graph infinitely many times. This means there are infinite angles whose sine is $0.5$ (e.g., $30^{\circ}$ , $150^{\circ}$ , $390^{\circ}$ , etc.). If we tried to define $\arcsin (0.5)$ , which angle should it be? There's no single, unique answer.

To solve this problem, we restrict the domain of the original trigonometric function to a specific interval where it is one-to-one. By doing this, we ensure that for any given ratio, there is only one possible angle as the output of the inverse function. This output is called the principal value.

For $y = \sin (x)$ , we restrict its domain to $[- \frac{π}{2}, \frac{π}{2}]$ . In this interval, the graph passes the Horizontal Line Test, and we can define a proper inverse function, $y = \arcsin (x)$ . Similar restrictions are applied to cosine and tangent to create their inverse functions. These restricted ranges are essential and must be memorized.

A Deep Dive: The Inverse Sine Function (Arcsine)

The inverse sine function, denoted $\arcsin (x)$ or $\sin^{- 1} (x)$ , answers the question: “Which angle in the restricted range has a sine equal to $x$ ?”

The relationship is formally defined as:

y = \arcsin (x) ⟺ \sin (y) = x

Because we restricted the domain of sine to create arcsine, the domain and range of $y = \arcsin (x)$ are swapped and restricted from the original function. Here are the key properties:

Property	Value	Explanation
Domain	$[- 1, 1]$	The input $x$ must be a value that sine can produce, which is between -1 and 1, inclusive.
Range	$[- \frac{π}{2}, \frac{π}{2}]$	The output angle $y$ is restricted to Quadrant I and Quadrant IV of the unit circle.

This range means that when you evaluate an arcsin expression, your answer must be an angle between $- 90^{\circ}$ and $90^{\circ}$ (or $- \frac{π}{2}$ and $\frac{π}{2}$ radians).

Example 1

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

Solution:

We are looking for an angle $θ$ such that $\sin (θ) = \frac{\sqrt{3}}{2}$ and $θ$ is in the interval $[- \frac{π}{2}, \frac{π}{2}]$ .

Ask yourself: What angle has a sine of $\frac{\sqrt{3}}{2}$ ? From our knowledge of the unit circle, we know that $\sin (60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin (120^{\circ}) = \frac{\sqrt{3}}{2}$ . In radians, this is $\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ and $\sin (\frac{2 π}{3}) = \frac{\sqrt{3}}{2}$ .
Now, check which of these angles falls within the restricted range of arcsine, which is $[- \frac{π}{2}, \frac{π}{2}]$ .
The angle $\frac{π}{3}$ is within this range. The angle $\frac{2 π}{3}$ is not.
Therefore, the principal value is $\frac{π}{3}$ .

Answer: $\arcsin (\frac{\sqrt{3}}{2}) = \frac{π}{3}$ .

Exploring the Inverse Cosine Function (Arccosine)

The inverse cosine function, $\arccos (x)$ or $\cos^{- 1} (x)$ , finds the angle whose cosine is $x$ , within a specific range.

y = \arccos (x) ⟺ \cos (y) = x

To make cosine one-to-one, we restrict its domain differently than we did for sine. We choose the interval $[0, π]$ because it covers all possible output values of cosine (from $- 1$ to $1$ ) exactly once. This leads to the following properties for arccosine:

Property	Value	Explanation
Domain	$[- 1, 1]$	Like arcsine, the input $x$ must be a valid cosine value.
Range	$[0, π]$	The output angle $y$ is restricted to Quadrant I and Quadrant II of the unit circle.

This means any answer for an arccosine problem must be an angle between $0^{\circ}$ and $180^{\circ}$ (or $0$ and $π$ radians). This range is chosen to provide a unique output for every valid input, including negative values.

Example 2

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

Solution:

We need to find an angle $θ$ such that $\cos (θ) = - \frac{1}{2}$ and $θ$ is in the interval $[0, π]$ .

First, think of the reference angle. What angle has a cosine of $+ \frac{1}{2}$ ? That would be $\frac{π}{3}$ or $60^{\circ}$ .
Cosine is negative in Quadrant II and Quadrant III.
The range of arccosine is $[0, π]$ , which covers Quadrants I and II. Therefore, we must find the angle in Quadrant II that has a reference angle of $\frac{π}{3}$ .
The angle in Quadrant II is calculated as $π - reference angle$ . So, $θ = π - \frac{π}{3} = \frac{2 π}{3}$ .
The angle $\frac{2 π}{3}$ is within the required range $[0, π]$ .

Answer: $\arccos (- \frac{1}{2}) = \frac{2 π}{3}$ .

Understanding the Inverse Tangent Function (Arctangent)

The inverse tangent function, $\arctan (x)$ or $\tan^{- 1} (x)$ , gives the angle whose tangent is $x$ .

y = \arctan (x) ⟺ \tan (y) = x

The tangent function has a range of all real numbers. Consequently, the domain of the inverse tangent function is also all real numbers. To make tangent one-to-one, we restrict it to the interval between its vertical asymptotes, $(- \frac{π}{2}, \frac{π}{2})$ .

Property	Value	Explanation
Domain	$(- \infty, \infty)$	The input $x$ can be any real number, since the tangent function can produce any real number as output.
Range	$(- \frac{π}{2}, \frac{π}{2})$	The output angle $y$ is in Quadrant I or IV, but does not include the endpoints $- \frac{π}{2}$ and $\frac{π}{2}$ (where tangent is undefined).

Notice the range uses parentheses, not brackets, because tangent is undefined at $- 90^{\circ}$ and $90^{\circ}$ .

Example 3

Evaluate $\arctan (- 1)$ .

Solution:

We are searching for an angle $θ$ where $\tan (θ) = - 1$ and $θ$ is in the interval $(- \frac{π}{2}, \frac{π}{2})$ .

The reference angle for $\tan (θ) = 1$ is $\frac{π}{4}$ .
Tangent is negative in Quadrants II and IV.
The range of arctangent is $(- \frac{π}{2}, \frac{π}{2})$ , which covers Quadrant I (for positive inputs) and Quadrant IV (for negative inputs). We must choose the angle in Quadrant IV.
Angles in Quadrant IV are represented as negative angles in this range. So, we use the negative of the reference angle.
$θ = - \frac{π}{4}$ . This value is within the specified range.

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

Key formulas for inverse trigonometric functions by Algebra911.

How Do You Evaluate Combined Trigonometric Expressions?

Sometimes you will encounter expressions where trigonometric and inverse trigonometric functions are composed, such as $\cos (\arcsin (x))$ or $\arctan (\tan (x))$ . There are two main scenarios.

1. Inverse Function of a Function: $\arcsin (\sin (x))$

You might think that $\arcsin (\sin (x)) = x$ is always true. This is only true if $x$ is within the restricted range of arcsine, $[- \frac{π}{2}, \frac{π}{2}]$ . If $x$ is outside this range, you must first find an angle within the range that has the same sine value. For example, $\arcsin (\sin (π)) = \arcsin (0) = 0$ , not $π$ .

2. Function of an Inverse Function: $\cos (\arcsin (x))$

For expressions like this, where the inner function is an inverse, it's often best to use a right-triangle approach. Let the inner part equal an angle $θ$ , use the definition of the inverse function to set up side ratios, and then find the required trigonometric value.

Example 4

Evaluate $\tan (\arccos (\frac{2}{3}))$ .

Solution:

This expression looks intimidating, but we can break it down.

Define an angle. Let $θ = \arccos (\frac{2}{3})$ . This is the angle we are interested in. Our goal is now to find $\tan (θ)$ .
Use the definition of arccosine. The statement $θ = \arccos (\frac{2}{3})$ means that $\cos (θ) = \frac{2}{3}$ . It also tells us that $θ$ is in the interval $[0, π]$ . Since the cosine value is positive, $θ$ must be in Quadrant I.
Draw a right triangle. Since $\cos (θ) = \frac{adjacent}{hypotenuse}$ , we can draw a right triangle in Quadrant I with an adjacent side of length $2$ and a hypotenuse of length $3$ .
Find the missing side. Using the Pythagorean theorem ( $a^{2} + b^{2} = c^{2}$ ), we can find the length of the opposite side (let's call it $o$ ):
$2^{2} + o^{2} = 3^{2}$
$4 + o^{2} = 9$
$o^{2} = 5$
$o = \sqrt{5}$
Calculate the final value. Now that we know all three sides of the triangle, we can find $\tan (θ)$ .
$\tan (θ) = \frac{opposite}{adjacent} = \frac{\sqrt{5}}{2}$

Answer: $\tan (\arccos (\frac{2}{3})) = \frac{\sqrt{5}}{2}$ .

Common Mistakes to Avoid with Inverse Trig Functions

Inverse trigonometric functions can be tricky, and a few common errors trip up many students. Be on the lookout for these pitfalls:

Confusing Inverse with Reciprocal: This is the most frequent mistake. Always remember that $\sin^{- 1} (x)$ means arcsin, not $\frac{1}{\sin (x)}$ . The reciprocal of sine is cosecant ( $\csc (x)$ ). The notation is unfortunate, but it's a rule you must internalize.
Forgetting the Range Restrictions: Every inverse trig function has a strictly defined output range. You cannot get an answer of $\frac{7 π}{4}$ from an arctan calculation, even though $\tan (\frac{7 π}{4}) = - 1$ . The correct principal value is $- \frac{π}{4}$ . Always check if your answer is in the correct range.
Incorrectly Canceling Functions: Assuming that $\arccos (\cos (x)) = x$ for any $x$ is a mistake. For example, $\arccos (\cos (2 π)) = \arccos (1) = 0$ . The functions only cancel directly if $x$ is already in the defined range for the inverse function.
Calculator Mode Errors: Your calculator can give answers in degrees or radians. If a problem is given in radians (e.g., involves $π$ ), your calculator must be in radian mode. If the problem uses degrees, switch to degree mode. A wrong mode will always lead to a wrong answer.
Domain Errors: You cannot take the arcsin or arccos of a number greater than $1$ or less than $- 1$ . An expression like $\arcsin (2)$ is undefined because there is no angle whose sine is $2$ .

Inverse Trigonometric Functions: Quick Reference

Memorizing the domain and range for each function is essential for success. Use this table as a quick reference guide and study tool.

Function	Notation	Domain	Range (Radians)	Range (Degrees)	Quadrants
Inverse Sine	$y = \arcsin (x)$ $y = \sin^{- 1} (x)$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$	$[- 90^{\circ}, 90^{\circ}]$	I and IV
Inverse Cosine	$y = \arccos (x)$ $y = \cos^{- 1} (x)$	$[- 1, 1]$	$[0, π]$	$[0^{\circ}, 180^{\circ}]$	I and II
Inverse Tangent	$y = \arctan (x)$ $y = \tan^{- 1} (x)$	$(- \infty, \infty)$	$(- \frac{π}{2}, \frac{π}{2})$	$(- 90^{\circ}, 90^{\circ})$	I and IV

Frequently Asked Questions

What's the difference between arcsin(x) and sin⁻¹(x)?

There is no difference in their meaning; they are two different notations for the exact same function, the inverse sine. Many people prefer the 'arcsin' notation because the 'sin⁻¹(x)' notation can be easily confused with the reciprocal, (sin(x))⁻¹, which is completely different.

Why is the range of arccos different from arcsin?

The range of arccos is $[0, π]$ (Quadrants I and II) to ensure it is a function that covers all possible cosine values from -1 to 1. If we used the same range as arcsin (Quadrants I and IV), we would only get positive cosine values, and we wouldn't be able to find the arccos of a negative number.

Can I find the arcsin of 2?

No, you cannot. The domain of the arcsin function is $[- 1, 1]$ because the sine function only produces values within that range. Since there is no angle $θ$ for which $\sin (θ) = 2$ , the expression $\arcsin (2)$ is undefined.

Do my calculator answers for inverse trig functions always make sense?

Your calculator is programmed to give the correct principal value within the defined range. However, you must ensure your calculator is in the correct mode (degrees or radians) for the problem you are solving. A correct numerical answer in the wrong units is still a wrong answer.

Are there inverses for csc, sec, and cot?

Yes, the other three trigonometric functions also have inverses: arccosecant (arccsc), arcsecant (arcsec), and arccotangent (arccot). They are used less frequently but follow the same principles, including needing a restricted domain on the original function to be defined.

How are inverse trig functions used in real life?

Inverse trig functions are used extensively in fields like physics, engineering, and computer graphics. For example, they can be used to find the angle of elevation to an object, determine the launch angle of a projectile, or calculate angles in navigation and robotics.

What's the most important thing to remember about inverse trig functions?

The single most important thing to remember is the restricted range for each function: $[- \frac{π}{2}, \frac{π}{2}]$ for arcsin, $[0, π]$ for arccos, and $(- \frac{π}{2}, \frac{π}{2})$ for arctan. Nearly all common mistakes stem from forgetting these ranges.

Inverse Trigonometric Functions

What Are Inverse Trigonometric Functions?

Why Do We Need to Restrict the Domain of Trig Functions?

A Deep Dive: The Inverse Sine Function (Arcsine)

Exploring the Inverse Cosine Function (Arccosine)

Understanding the Inverse Tangent Function (Arctangent)

How Do You Evaluate Combined Trigonometric Expressions?

1. Inverse Function of a Function: arcsin⁡(sin⁡(x))

2. Function of an Inverse Function: cos⁡(arcsin⁡(x))

Common Mistakes to Avoid with Inverse Trig Functions

Inverse Trigonometric Functions: Quick Reference

Frequently Asked Questions

1. Inverse Function of a Function: $\arcsin (\sin (x))$

2. Function of an Inverse Function: $\cos (\arcsin (x))$